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Vector Spaces
Linear Algebra. Session 6
Dr. Marco A Roque Sol
10/02/2018
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let
S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and
S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S
be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets
of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space
V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and
S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S
is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent,
then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is
S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0
⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and
S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0
is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent,
then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is
S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If
S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S
is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent
in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and
V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V
is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace
of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,
then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S
is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent
in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set
is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set
containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0
is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Let S0 and S be subsets of a vector space V
More facts on linear independence
If S0 ⊂ S and S is linearly independent, then so is S0.
If S0 ⊂ S and S0 is linearly dependent, then so is S.
If S is linearly independent in V and V is a subspace of W,then S is linearly independent in W.
The empty set is linearly independent.
Any set containing 0 is linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors
v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2
are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent
if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if
either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them
is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple
the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0
is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and
v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0
then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then,
S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}
is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent
if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if
v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors
v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn
are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent
wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n
(i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates
is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than
the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
Two nonzero vectors v1 and v2 are linearly dependent if andonly if either of them is a scalar multiple the other.
If S0 is linearly independent and v0 ∈ V − S0 then, S0 ∪ {v0}is linearly independent if and only if v0 /∈ Span(S0).
Theorem
Vectors v1, v2, · · · , vm ∈ Rn are linearly dependent wheneverm > n (i.e., the number of coordinates is less than the number ofvectors).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m
Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then
the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality
t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the systema11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0
is equivalent to the systema11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent
to the systema11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note
that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm
are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns
of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij).
The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number
of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries
in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform
is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n.
If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n
then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are
free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, therefore
the zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution
is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Vector Spaces
proof.
Let vj = (a1j , a2j , · · · , anj) for j = 1, 2, · · · ,m Then the vectorequality t1v1 + t2v2 + · · ·+ tmvm = 0 is equivalent to the system
a11t1 + a12t2 + · · ·+ a1mtn = 0a21t1 + a22t2 + · · ·+ a2ntn = 0
...an1t1 + an2t2 + · · ·+ anmtn = 0
Note that vectors v1, v2, · · · , vm are columns of the coefficientmatrix (aij). The number of leading entries in the row echelonform is at most n. If m > n then there are free variables, thereforethe zero solution is not unique. �
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1
Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given
an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix
the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions
areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A
are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent
(as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0
is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution
of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation
Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form
of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A
has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry
in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2
Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given
a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix
the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions
areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A
are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent
(as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A
are linearly independent (as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent
(as vectors in Rn )
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
General results on linear independence in Rn
Theorem 1 Given an m × n matrix the following conditions areequivalent:
i) columns of A are linearly independent (as vectors in Rm )
ii) x = 0 is the only solution of the matrix equation Ax = 0
iii) The row echelon form of A has a leading entry in each column.
Theorem 2 Given a square matrix the following conditions areequivalent:
i) det(A) = 0
ii) Columns of A are linearly independent (as vectors in Rn )
iii) Rows of A are linearly independent (as vectors in Rn )Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1),
v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0),
v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), and
v4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4)
in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors
are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent
if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if
they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.
Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and
v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2
are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors
v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1,
v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and
v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3
are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent
if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if
thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix
A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3)
is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
−∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ =
2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Example 6.1
Consider vectors v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), andv4 = (1, 2, 4) in R3
Two vectors are linearly dependent if and only if they are parallel.Hence v1 and v2 are linearly independent.
Vectors v1, v2, and v3 are linearly independent if and only if thematrix A = (v1, v2, v3) is invertible.
det(A) =
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ = −∣∣∣∣ −1 1
1 1
∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore
v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3
are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors
in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3
are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent.
Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus
v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3
and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4
are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2
Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine
whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices
A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3
are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)
A2 =
(0 −11 −1
)A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)
A3 =
(1 00 1
)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Therefore v1, v2, v3 are linearly independent.
Four vectors in R3 are always linearly dependent. Thus v1, v2, v3and v4 are linearly dependent.
Example 6.2Let
A =
(−1 1−1 0
).
Determine whether matrices A,A2,A3 are linearly independent
Solution
We have
A =
(−1 1−1 0
)A2 =
(0 −11 −1
)A3 =
(1 00 1
)Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need
to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine
if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist
r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R
not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero
suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat
r1A + r2A2 + r3A
3 = 0. This matrix equation is equivalent toa system
−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 =
0. This matrix equation is equivalent toa system
−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0.
This matrix equation is equivalent toa system
−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation
is equivalent toa system
−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent
toa system
−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system
−r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒
−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
We need to determine if there exist r1, r2, r3 ∈ R not all zero suchthat r1A + r2A
2 + r3A3 = 0. This matrix equation is equivalent to
a system −r1 + 0r2 + r3 = 0r1 − r2 + 0r3 = 0−r1 + r2 + 0r3 = 00r1 − r2 + r3 = 0
The augmented matrix is
−1 0 1 01 −1 0 0−1 1 0 00 −1 1 0
⇒−1 0 1 00 1 −1 00 0 0 00 0 0 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form
of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix
shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is
afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable.
Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence
the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system
has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero
solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices
are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent
(one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is
A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show
that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions
ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are
linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent in
C∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that
aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0
for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all
x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R
where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where
a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, c
areconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants.
We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have
to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that
a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0.
Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate
thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity
twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independences
The row echelon form of the augmented matrix shows there is afree variable. Hence the system has a nonzero solution so that thematrices are linearly dependent (one relation is A + A2 + A3 = 0)
Example 6.3
Show that functions ex , e2x , and e3x are linearly independent inC∞(R)
Solution
Suppose that aex + be2x + ce3x = 0 for all x ∈ R where a, b, careconstants. We have to show that a = b = c = 0. Differentiate thisidentity twice:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0
aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0
aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows
that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0,
where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
,
v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
=
ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
aex + be2x + ce3x = 0aex + 2be2x + 3ce3x = 0aex + 4be2x + 9ce3x = 0
It follows that A(x)v = 0, where
A(x) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
, v =
abc
det(A(x)) =
ex e2x e3x
ex 2e2x 3e3x
ex 4e2x 9e3x
= ex
∣∣∣∣∣∣1 e2x e3x
1 2e2x 3e3x
1 4e2x 9e3x
∣∣∣∣∣∣ =
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x =
e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x =
e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ =
e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ =
e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x
∣∣∣∣ 1 13 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ =
2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since
the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x)
is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible
we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtain
A(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒
v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒
a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence,
the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set
of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions
{ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is
linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
exe2x = e3x
∣∣∣∣∣∣1 1 e3x
1 2 3e3x
1 4 9e3x
∣∣∣∣∣∣ =
e3xe3x = e6x
∣∣∣∣∣∣1 1 11 2 31 4 9
∣∣∣∣∣∣ = e6x
∣∣∣∣∣∣1 1 10 1 20 3 8
∣∣∣∣∣∣ = e6x∣∣∣∣ 1 1
3 8
∣∣∣∣ = 2e6x 6= 0
Since the matrix A(x) is invertible we obtainA(x)v = 0⇒ v = 0⇒ a = b = c = 0Hence, the set of functions {ex , e2x , e3x} is linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let
f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn
be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions
on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval
[a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b].
TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian
W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn]
is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function
on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b]
defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣
Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If
W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0
for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b]
then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then
the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functions
f1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are
linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent
in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Linear Dependence and Independence
Wronskian
Let f1, f2, ..., fn be smooth functions on an interval [a, b]. TheWronskian W[f1, f2, ..., fn] is a function on [a, b] defined by
W[f1, f2, ..., fn](x) =
∣∣∣∣∣∣∣∣∣f1(x) f2(x) · · · fn(x)f ′1(x) f ′2(x) · · · f ′n(x)
...
f(n−1)1 (x) f
(n−1)2 (x) · · · f
(n−1)n (x)
∣∣∣∣∣∣∣∣∣Theorem
If W[f1, f2, ..., fn](x0) 6= 0 for some x0 ∈ [a, b] then the functionsf1, f2, ..., fn are linearly independent in C [a, b].
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let
V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V
be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space.
Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent
spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set
Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V,
is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis.
That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means,
that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector
v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V
can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented
as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where
v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are
distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors
from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from
S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S and
r1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk
∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis and Dimension
Definition.
Let V be a vector space. Any linearly independent spanning set Sfor V, is called a basis. That means, that any vector v ∈ V can berepresented as a linear combination
v = r1v1 + r2v2 + · · ·+ rkvk
where v1, v2, · · · , vk are distinct vectors from S andr1, r2, · · · , rk ∈ R
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”,
in the above definition, implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition,
implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that
theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that theabove representation
is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
OBS
“Linearly independent”, in the above definition, implies that theabove representation is unique:
v = r1v1 + r2v2 + · · ·+ rkvk
v = r ′1v1 + r ′2v2 + · · ·+ r ′kvk ⇒
0 = (r1 − r ′1)v1 + (r2 − r ′2)v2 + · · ·+ (rk − r ′k)vk ⇒
(r1 − r ′1) = 0, (r2 − r ′2) = 0 · · · (rk − r ′k) = 0⇒
r1 = r ′1, r2 = r ′2, · · · rk = r ′k
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0),
e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0),
e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0),
· · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · ,
en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 +
x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 +
... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... +
xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.4Standard Basis for Rn
e1 = (1, 0, 0, ..., 0), e2 = (0, 1, ..., 0), · · · , en = (0, 0, ..., 1),
Solution
Indeed,
(x1, x2, ..., xn) = x1e1 + x2e2 + ... + xnen
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),
E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),
E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),
E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)=
aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 +
bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 +
cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 +
dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.5Standard Basis for M2,2
E11 =
(1 00 0
),E12 =
(0 10 0
),E21 =
(0 01 0
),E22 =
(0 00 1
)
Solution
Indeed,
A =
(a bc d
)= aE12 + bE12 + cE21 + dE22
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0,
a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x ,
a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2,
..., an−1xn−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ...,
an−1xn−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0,
a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x ,
a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2,
..., anxn, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ...,
anxn, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
Solution
POC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.6Standard Basis for Polinomial Pn
a0, a1x , a2x2, ..., an−1x
n−1
Solution
POC ... !!!!
Example 6.7Standard Basis for Polinomials P
a0, a1x , a2x2, ..., anx
n, ...
SolutionPOC ... !!!!
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let
v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and
r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R.
The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent
to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation
Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v
where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk)
x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is,
A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the
n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix
such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that
vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors
v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk
areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive
columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors
v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk
span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn
if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row
echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form
of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A
hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.
Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors
v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are
linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent
if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the row
echelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form
of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A
has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry
in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column
(no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Let v1, v2, · · · vk ∈ Rn and r1, r2, · · · rn ∈ R. The vector
v = r1v1 + r2v2 + · · ·+ rkvk
is equivalent to the matrix equation Ax = v where
A = (v1, v2, · · · vk) x =
r1r2...rk
That is, A is the n × k matrix such that vectors v1, v2, · · · vk areconsecutive columns of A.
Vectors v1, v2, · · · vk span Rn if the row echelon form of A hasno zero rows.Vectors v1, v2, · · · vk are linearly independent if the rowechelon form of A has a leading entry in each column (no freevariables).
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
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Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
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Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
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Spanning/ No Lin. Indep. No spanning/ No Lin. Indep.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
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Spanning/Lin. Indep. No spanning/Lin. Indep.
� ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗
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� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗� � ∗ ∗ ∗
∗ ∗ ∗� �
Spanning/ No Lin. Indep. No spanning/ No Lin. Indep.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗� ∗ ∗ ∗
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Spanning/Lin. Indep. No spanning/Lin. Indep.
� ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗
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∗ ∗ ∗� �
Spanning/ No Lin. Indep. No spanning/ No Lin. Indep.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
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Spanning/Lin. Indep. No spanning/Lin. Indep.
� ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗
� � ∗ ∗� ∗
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� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗� � ∗ ∗ ∗
∗ ∗ ∗� �
Spanning/ No Lin. Indep. No spanning/ No Lin. Indep.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗� ∗ ∗ ∗
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� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
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Spanning/Lin. Indep. No spanning/Lin. Indep.
� ∗ ∗ ∗ ∗ ∗ ∗� � � ∗ ∗ ∗
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� ∗ ∗ ∗ ∗ ∗ ∗� ∗ ∗ ∗ ∗ ∗
� ∗ ∗ ∗ ∗� � ∗ ∗ ∗
∗ ∗ ∗� �
Spanning/ No Lin. Indep. No spanning/ No Lin. Indep.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If
k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n
then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk
do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span
Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If
k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n
then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors
v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are
linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Let v1, v2, · · · vk ∈ Rn
Theorem 1
If k < n then the vectors v1, v2, · · · vk do not span Rn.
Theorem 2
If k > n then the vectors v1, v2, · · · vk are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If
k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then
the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions
are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn}
is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis
for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn}
is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set
for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}
is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem 3
If k = n then the following conditions are equivalent:
{v1, v2, · · · vn} is a basis for Rn
{v1, v2, · · · vn} is a spanning set for Rn
{v1, v2, · · · vn}is a linearly independent set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors
v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2
are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent
(as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel),
but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not
span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors
v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are
linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since
∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ =
2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2
6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore,
{v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3}
is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis
for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set
{v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4}
span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3
( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3}
already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ),
but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.8
v1 = (1,−1, 1), v2 = (1, 0, 0), v3 = (1, 1, 1), v4 = (1, 2, 4),
Vectors v1 and v2 are linearly independent (as they are notparallel), but they do not span R3
Vectors v1, v2 and v3 are linearly independent since∣∣∣∣∣∣1 1 1−1 0 11 0 1
∣∣∣∣∣∣ =
∣∣∣∣ −1 11 1
∣∣∣∣ = 2 6= 0
Therefore, {v1, v2, v3} is a basis for R3
The set {v1, v2, v3, v4} span R3 ( because {v1, v2, v3} already spanR3 ), but they are linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space
has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If
a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space
V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V
has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis,
then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases
for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand
have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number
of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension
of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space
V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V,
denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V),
is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber
of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements
in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any
of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Dimension
Theorem 1
Any vector space has a basis.
Theorem 2
If a vector space V has a finite basis, then all bases for V are finiteand have the same number of elements.
Definition
The dimension of a vector space V, denoted dim(V), is thenumber of elements in any of its bases.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space.
dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) =
n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices.
dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) =
4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices.
dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) =
mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.
dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) =
n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials.
dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =
∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space .
dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) =
0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.9
Rn : n-dimensional space. dim(Rn) = n
M2×2 : the space of 2× 2 matrices. dim(M2×2) = 4
Mm×n : the space of m × n matrices. dim(Mm×n) = mn
Pn : the space of polynomials of degree less than n.dim(Pn) = n
P : the space of all polynomials. dim(Pn) =∞
{0} : the trivial space . dim({0}) = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10
Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find
the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension
of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane
x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in
R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution
of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation
isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation is
x = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is,
(x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) =
(−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) =
t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) +
s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence,
the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane
is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span
of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors
v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) and
v2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1).
These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors
are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent
as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they are
not parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus,
{v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2}
is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis
so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that
the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension
of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is
2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.10Find the dimension of the plane x + 2z = 0 in R3
Solution
The general solution of the equation isx = 2sy = tz = s
(t, s,∈ R)
That is, (x , y , z) = (−2s, t, s) = t(0, 0, 1) + s(−2, 0, 1)
Hence, the plane is the span of vectors v1 = (0, 0, 1) andv2 = (−2, 0, 1). These vectors are linearly independent as they arenot parallel. Thus, {v1, v2} is a basis so that the dimension of theplane is 2.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let
S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S
be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset
of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space
V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V.
Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then,
the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions
are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S
is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis,
i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e.,
is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent
spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set
for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S
is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal
spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set
for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S
is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal
linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent
subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of
V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’
means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means:
remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element
from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and
it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer
a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’
means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means:
add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element
ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV
to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and
it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become
linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Basis
Theorem
Let S be a subset of a vector space V. Then, the followingconditions are equivalent:
i) S is a basis, i.e., is a linearly independent spanning set for V.
ii) S is a minimal spanning set for V.
iii) S is a maximal linearly independent subset of V.
’Minimal spanning set’ means: remove any element from this set,and it is no longer a spanning set.
’ Maximal linearly independent subset’ means: add any element ofV to this set, and it will become linearly dependent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let
V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V
be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space.
Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any
spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set
for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V
can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced
to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal
spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any
linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset
of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V
can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended
to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal
linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any
spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set
contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains
a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis
while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any
linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independent
set is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained
in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space
is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional
if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if
it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned
by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Theorem
Let V be a vector space. Then
i) Any spanning set for V can be reduced to a minimal spanningset.
ii) any linearly independent subset of V can be extended to amaximal linearly independent set.
Corollary 1Any spanning set contains a basis while any linearly independentset is contained in a basis.
Corollary 2
A vector space is finite-dimensional if and only if it is spanned by afinite set.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get
a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set
for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space,
then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set
to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis
dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector
at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let
v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk
be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set
for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space
V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V .
If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If
v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0
is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then
v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk
is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set
for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find a
basis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis
for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space
V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V
spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),
w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),
w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and
w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
How to find a basis?
Approach 1.
Get a spanning set for the vector space, then reduce this set to abasis dropping one vector at a time.
Theorem
Let v0, v1, ..., vk be a spanning set for a vector space V . If v0 is alinear combination of vectors v1, v2, ..., vk then v1, v2, ..., vk is alsoa spanning set for V
Example 6.11
Find abasis for the vector space V spanned by vectorsw1 = (1, 1, 0),w2 = (0, 1, 1),w3 = (2, 3, 1) and w4 = (1, 1, 1)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since
we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have
four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector
in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3,
then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are
linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and
satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy
an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation
of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where
ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R
are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all
equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero.
Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve
this system of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system
of linear equations for r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for
r1, r2, r3, r4 we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4
we apply rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply
rowreduction
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Solution
Since we have four vector in R3, then they are linearly dependent,and satisfy an equation of the form
r1w1 + r2w2 + r3w3 + r4v4 = 0
where ri ∈ R are not all equal to zero. Equivalently,
1 0 2 11 1 3 10 1 1 1
r1r2r3r4
=
0000
to solve this system of linear equations for r1, r2, r3, r4 we apply rowreduction Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
1 0 2 11 1 3 10 1 1 1
→ 1 0 2 1
0 1 1 00 1 1 1
→ 1 0 2 1
0 1 1 00 0 0 1
→ 1 0 2 0
0 1 1 00 0 0 1
→
r1 + 2r3r2 + r3r4 = 0
→
r1 = −2r3r2 = −r3r4 = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
1 0 2 11 1 3 10 1 1 1
→
1 0 2 10 1 1 00 1 1 1
→ 1 0 2 1
0 1 1 00 0 0 1
→ 1 0 2 0
0 1 1 00 0 0 1
→
r1 + 2r3r2 + r3r4 = 0
→
r1 = −2r3r2 = −r3r4 = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
1 0 2 11 1 3 10 1 1 1
→ 1 0 2 1
0 1 1 00 1 1 1
→
1 0 2 10 1 1 00 0 0 1
→ 1 0 2 0
0 1 1 00 0 0 1
→
r1 + 2r3r2 + r3r4 = 0
→
r1 = −2r3r2 = −r3r4 = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
1 0 2 11 1 3 10 1 1 1
→ 1 0 2 1
0 1 1 00 1 1 1
→ 1 0 2 1
0 1 1 00 0 0 1
→
1 0 2 00 1 1 00 0 0 1
→
r1 + 2r3r2 + r3r4 = 0
→
r1 = −2r3r2 = −r3r4 = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
1 0 2 11 1 3 10 1 1 1
→ 1 0 2 1
0 1 1 00 1 1 1
→ 1 0 2 1
0 1 1 00 0 0 1
→ 1 0 2 0
0 1 1 00 0 0 1
→
r1 + 2r3r2 + r3r4 = 0
→
r1 = −2r3r2 = −r3r4 = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
1 0 2 11 1 3 10 1 1 1
→ 1 0 2 1
0 1 1 00 1 1 1
→ 1 0 2 1
0 1 1 00 0 0 1
→ 1 0 2 0
0 1 1 00 0 0 1
→
r1 + 2r3r2 + r3r4 = 0
→
r1 = −2r3r2 = −r3r4 = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
1 0 2 11 1 3 10 1 1 1
→ 1 0 2 1
0 1 1 00 1 1 1
→ 1 0 2 1
0 1 1 00 0 0 1
→ 1 0 2 0
0 1 1 00 0 0 1
→
r1 + 2r3r2 + r3r4 = 0
→
r1 = −2r3r2 = −r3r4 = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus,
the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solution
is (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0),
t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand
a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution
is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0).
We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained
that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒
2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3.
Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence,
we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and
take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take
V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check
whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors
w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are
linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ =
1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1
6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!.
Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence
{w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4}
is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis
for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V .
Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides,
itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that
V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Thus, the general solutionis (r1, r2, r3, r4) = (−2t,−t, t, 0), t ∈ Rand a particular solution is (2, 1,−1, 0). We have obtained that2w1 + w2 −w3 = 0⇒ 2w1 + w2 = w3. Hence, we can dropp w3
and take V = Span(w1,w2,w4)
Let us check whether vectors w1,w3,w4 are linearly independent
∣∣∣∣∣∣1 0 11 1 10 1 1
∣∣∣∣∣∣ =
∣∣∣∣∣∣1 0 10 1 00 1 1
∣∣∣∣∣∣ =
∣∣∣∣ 1 01 1
∣∣∣∣ = 1 6= 0
They are!!!. Hence {w1,w3,w4} is a basis for V . Besides, itfollows that V = R3
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build
a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal
linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set
adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector
at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If
the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space
V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V
is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial,
it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has
the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If
V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0,
pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector
v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0,
if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1
spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V,
it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.
Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise
pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector
v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V
that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not
in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of
v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1.
Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. If
v1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2
span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V
they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis.
Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise
pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vector
v3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V
that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not
in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2.
And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of
the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set,
we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start
with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any
linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set
(if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one).
If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If
we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given
a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset
S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S,
it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough
to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick
new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors
only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Approach 2.
Build a maximal linearly independent set adding one vector at atime.
If the vector space V is trivial, it has the empty basis.
If V 6= 0, pick any vector v1 6= 0, if v1 spans V, it is a basis.Otherwise pick any vector v2 ∈ V that is not in the span of v1. Ifv1, v2 span V they constitute a basis. Otherwise pick any vectorv3 ∈ V that is not in the span of v1, v2. And so on.
Modifications.
Instead of the empty set, we can start with any linearlyindependent set (if we are given one). If we are given a spanningset S, it is enough to pick new vectors only in S
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark.
This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure
works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for
finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensional
vector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces.
There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is
an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure
forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces
(transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors
v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are
linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.
Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set
{v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2}
to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis
for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task
is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find
a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector
v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3
that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not
a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination
ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2.
Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now,
the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors
v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2,
span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane
x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and
thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1)
does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie
in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane.
Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence,
it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not
alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination
of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2.
Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3}
is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Remark. This inductive procedure works for finite-dimensionalvector spaces. There is an analogous procedure forinfinite-dimensional spaces (transfinite induction ).
Example 6.12
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Our task is to find a vector v3 that is not a linear combination ofv1, v2. Now, the vectors v1, v2, span the plane x + 2z = 0 and thevetor v3 = (1, 1, 1) does not lie in that plane. Hence, it is not alinear combination of v1, v2. Thus, {v1, v2, v3} is a basis for R3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors
v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and
v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1)
are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.
Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend
the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2}
to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis
for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors
e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1)
form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form a
spanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set
for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3,
at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them
can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen
as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check
that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and
{v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2}
form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis
for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ =
1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ =
2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Basis and Dimension
Example 6.13
Vectors v1 = (0, 1, 0) and v2 = (−2, 0, 1) are linearly independent.Extend the set {v1, v2} to a basis for R3
Solution
Since vectors e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) form aspanning set for R3, at least one of them can be chosen as v3
Let us check that {v1, v2, e3} and {v1, v2, e2} form a basis for R3
∣∣∣∣∣∣0 −2 11 0 00 1 0
∣∣∣∣∣∣ = 1 6= 0
∣∣∣∣∣∣0 −2 11 0 00 1 1
∣∣∣∣∣∣ = 2 6= 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space
of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n
matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A
is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace
of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension
of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space
is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called
the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank
of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank
of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A
is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal
number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number
of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent
rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A
is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in
row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form,
then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows
of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of A
are linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Row space of a matrix
The row space of an m × n matrix A is the subspace of Rn
spanned by the rows of A
The dimension of the row space is called the rank of the matrix A.
Theorem 1
The rank of a matrix A is the maximal number of linearlyindependent rows in A.
Theorem 2
If a matrix A is in row echelon form, then the nonzero rows of Aare linearly independent.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank
of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix
is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal
to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number
of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows
in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow
echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary
row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations
do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change
the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space
of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose
that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B
are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices
such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B
is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtained
from A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A
by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary
row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation.
Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am
be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rows
of A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and
b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm,
be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B.
We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have
to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show that
Span(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) =
Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Corollary
The rank of a matrix is equal to the number of nonzero rows in itsrow echelon form.
Theorem 3
Elementary row operations do not change the row space of amatrix.
proof
Suppose that A and B are m × n matrices such that B is obtainedfrom A by an elementary row operation. Let a1, ..., am be the rowsof A and b1, ...,bm, be the rows of B. We have to show thatSpan(a1, ..., am) = Span(b1, ...,bm)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However,
we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe
that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row
bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi
of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B
belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am).
Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either
bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m
(interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or
bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai
for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar
r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0
(multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), or
bi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj
for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some
i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and
r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R
( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that
Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂
Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
However, we have to observe that any row bi of B belongs toSpan(a1, ..., am). Indeed, either bi = aj for some 1 ≤ j ≤ m (interchange of two rows ), or bi = rai for some scalar r 6= 0 (multiplying any row by a scalar different from zero ), orbi = ai + raj for some i 6= j and r ∈ R ( adding to a row, amultiple of another row )
It follows that Span(b1, ...,bm) ⊂ Span(a1, ..., am)
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now,
the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A
can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also
be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained
from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementary
row operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation.
By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂
Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find
the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank
of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now, the matrix A can also be obtained from B by an elementaryrow operation. By the above,
Span(a1, ..., am) ⊂ Span(b1, ...,bm) �
Example 6.14
Find the rank of the matrix
1 1 00 1 12 3 11 1 1
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary
row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations
do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change
the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space.
Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let us
convert A to row echelon form:1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A
to row echelon form:1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒
Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒
Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus,
the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
Elementary row operations do not change the row space. Let usconvert A to row echelon form:
1 1 00 1 12 3 11 1 1
⇒
1 1 00 1 10 1 11 1 1
⇒
1 1 00 1 10 1 10 0 1
⇒
1 1 00 1 10 0 00 0 1
⇒
1 1 00 1 10 0 10 0 0
⇒Thus, the rank of A is 3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space
of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n
matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A
is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace
of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,
spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A
is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm
spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space
of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A
coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with
the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space
ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe
transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix
AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations
do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change
linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations
betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns
of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Column space of a matrix
Definition.The column space of an m × n matrix A is the subspace of Rm,spanned by columns of A is the subspace of Rm spanned bycolumns of A
Theorem 1
The column space of a matrix A coincides with the row space ofthe transpose matrix AT
Theorem 2
Elementary row operations do not change linear relations betweencolumns of a matrix.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations
do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change
the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension
of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space
of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix
(however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however
they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change
the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix
is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,
then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns
with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries
form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis
for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix,
the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and
the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space
have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Theorem 5
Elementary row operations do not change the dimension of thecolumn space of a matrix (however they can change the columnspace).
Theorem 4
If a matrix is in row echelon form,then the columns with leadingentries form a basis for the column space.
Corollary
For any matrix, the row space and the column space have thesame dimension.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find
a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis
for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space
of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space
of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A
coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides
with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space
of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT .
Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis,
we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT
to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.15
Find a basis for the column space of the matrix
A =
1 1 00 1 12 3 11 1 1
Solution
The column space of A coincides with the row space of AT . Tofind a basis, we convert AT to row echelon form:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒
1 0 2 10 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒
1 0 2 00 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒
Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒
Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors
(1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1)
form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis
for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space
of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij)
be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace
of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A,
denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A)
is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of all
n− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional
column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors
x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x
such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that
Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
AT =
1 0 2 11 1 3 10 1 1 1
⇒ 1 0 2 1
0 1 1 00 1 1 1
⇒ 1 0 2 0
0 1 1 00 0 0 1
⇒Vectors (1, 0, 2, 1), (0, 1, 1, 0), and (0, 0, 0, 1) form a basis for thecolumn space of A.
Nullspace of a matrix
Let A = (aij) be an m × n
Definition
The nullspace of the matrix A, denoted by N(A) is the set of alln− dimensional column vectors x such that Ax = 0
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A)
is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set
of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linear
homogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations
(with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A)
is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace
of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space
of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension
of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A)
is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity
of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
a11 a12 · · · a1na21 a22 · · · a2n
...am1 am2 · · · amn
x1x2...xn
=
00...0
OBS The nullspace N(A) is the solution set of a system of linearhomogeneous equations (with A as the coefficient matrix)
Theorem
N(A) is a subspace of the vector space of Rn
Definition
The dimension of the nullspace N(A) is called the nullity of thematrix A
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank of a matrix A plus the nullity of A equals the number ofcolumns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank of a matrix A plus the nullity of A equals the number ofcolumns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank
of a matrix A plus the nullity of A equals the number ofcolumns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank of a matrix A
plus the nullity of A equals the number ofcolumns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank of a matrix A plus
the nullity of A equals the number ofcolumns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank of a matrix A plus the nullity of A
equals the number ofcolumns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank of a matrix A plus the nullity of A equals
the number ofcolumns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank of a matrix A plus the nullity of A equals the number of
columns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank of a matrix A plus the nullity of A equals the number ofcolumns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Rank + Nullity
TheoremThe rank of a matrix A plus the nullity of A equals the number ofcolumns in A.
rank(A) + N(A) = n
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and the nullity of the matrix B.
Find a basis for the row space of B, then extend this basis toa basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and the nullity of the matrix B.
Find a basis for the row space of B, then extend this basis toa basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and the nullity of the matrix B.
Find a basis for the row space of B, then extend this basis toa basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and the nullity of the matrix B.
Find a basis for the row space of B, then extend this basis toa basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and
the nullity of the matrix B.
Find a basis for the row space of B, then extend this basis toa basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and the nullity of the matrix B.
Find a basis for the row space of B, then extend this basis toa basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and the nullity of the matrix B.
Find a basis for the row space of B,
then extend this basis toa basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and the nullity of the matrix B.
Find a basis for the row space of B, then extend this basis toa basis
for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and the nullity of the matrix B.
Find a basis for the row space of B, then extend this basis toa basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Example 6.16
Let B given by
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
Find the rank and the nullity of the matrix B.
Find a basis for the row space of B, then extend this basis toa basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank
(= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and
the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity
(=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace)
of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are
preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations.
We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations
toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B
into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Solution
The rank (= dimension of the row space) and the nullity (=dimension of the nullspace) of a matrix are preserved underelementary row operations. We apply such operations toconvert the matrix B into its row echelon form
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒Interchange the 1st row with the 2nd row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1
⇒Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row
1 1 2 −10 −1 4 10 3 5 −32 −1 0 1
⇒
1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1
⇒
Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row
1 1 2 −10 −1 4 10 3 5 −32 −1 0 1
⇒
1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1
⇒
Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row
1 1 2 −10 −1 4 10 3 5 −32 −1 0 1
⇒
1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1
⇒Add 3 times the 1st row to the 3rd row,
then subtract 2 times the1st row from the 4th row
1 1 2 −10 −1 4 10 3 5 −32 −1 0 1
⇒
1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1
⇒Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row
1 1 2 −10 −1 4 10 3 5 −32 −1 0 1
⇒
1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1
⇒Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row
1 1 2 −10 −1 4 10 3 5 −32 −1 0 1
⇒
1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1
⇒Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row
1 1 2 −10 −1 4 10 3 5 −32 −1 0 1
⇒
1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 4 1−3 0 −1 02 −1 0 1
⇒Add 3 times the 1st row to the 3rd row, then subtract 2 times the1st row from the 4th row
1 1 2 −10 −1 4 10 3 5 −32 −1 0 1
⇒
1 1 2 −10 −1 4 10 3 5 −30 −3 −4 3
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−11 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3
⇒Add the 4th row to the 3rd row:
1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3
⇒Add 3 times the 2nd row to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−1
1 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3
⇒Add the 4th row to the 3rd row:
1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3
⇒Add 3 times the 2nd row to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−11 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3
⇒
Add the 4th row to the 3rd row:1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3
⇒Add 3 times the 2nd row to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−11 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3
⇒
Add the 4th row to the 3rd row:1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3
⇒Add 3 times the 2nd row to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−11 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3
⇒Add the 4th row
to the 3rd row:1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3
⇒Add 3 times the 2nd row to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−11 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3
⇒Add the 4th row to the 3rd row:
1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3
⇒Add 3 times the 2nd row to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−11 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3
⇒Add the 4th row to the 3rd row:
1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3
⇒
Add 3 times the 2nd row to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−11 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3
⇒Add the 4th row to the 3rd row:
1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3
⇒
Add 3 times the 2nd row to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−11 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3
⇒Add the 4th row to the 3rd row:
1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3
⇒Add 3 times the 2nd row
to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Multiply the 2nd row by−11 1 2 −10 1 −4 −10 3 5 −30 −3 −4 3
⇒Add the 4th row to the 3rd row:
1 1 2 −10 1 −4 −10 0 1 00 −3 −4 3
⇒Add 3 times the 2nd row to the 4th row:
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0
⇒Add 16 times the 3rd row to the 4th row:
1 1 2 −10 −1 −4 −10 0 1 00 0 0 0
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0
⇒
Add 16 times the 3rd row to the 4th row:1 1 2 −10 −1 −4 −10 0 1 00 0 0 0
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0
⇒
Add 16 times the 3rd row to the 4th row:1 1 2 −10 −1 −4 −10 0 1 00 0 0 0
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0
⇒Add 16 times the 3rd row
to the 4th row:1 1 2 −10 −1 −4 −10 0 1 00 0 0 0
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0
⇒Add 16 times the 3rd row to the 4th row:
1 1 2 −10 −1 −4 −10 0 1 00 0 0 0
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0
⇒Add 16 times the 3rd row to the 4th row:
1 1 2 −10 −1 −4 −10 0 1 00 0 0 0
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
1 1 2 −10 −1 −4 −10 0 1 00 0 −16 0
⇒Add 16 times the 3rd row to the 4th row:
1 1 2 −10 −1 −4 −10 0 1 00 0 0 0
⇒
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix
is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form,
its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals
thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows,
which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3.
Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) +
(nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) =
(the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =
4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that
the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B
equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
Now that the matrix is in row echelon form, its rank equals thenumber of nonzero rows, which is 3. Since
(rank of B ) + (nullity of B ) = (the number of columns of B ) =4,
it follows that the nullity of B equals 1.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix
is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations.
Therefore the row space of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space
of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B
is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as
the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of
its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
The row space of a matrix is invariant under elementary rowoperations. Therefore the row space of the matrix B is thesame as the row space of its row echelon form:
B =
0 −1 4 11 1 2 −1−3 0 −1 02 −1 0 1
⇒
1 1 2 −10 1 −4 −10 0 1 00 0 0 0
The nonzero rows of the latter matrix are linearly independentso that they form a basis for its row space:
v1 = (1, 1, 2,−1), v2 = (0, 1,−4,−1), v3 = (0, 0, 1, 0)
To extend the basis v1, v2, v3, to a basis for R4, we need av4 ∈ R4 that is not a linear combination of v1, v2, v3.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
It is known that at least one of the vectors
e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1), can be chosen as v4
In particular, the vectors v1, v2, v3, e4 form a basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
It is known that at least one of the vectors
e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1), can be chosen as v4
In particular, the vectors v1, v2, v3, e4 form a basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
It is known that at least one of the vectors
e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1), can be chosen as v4
In particular, the vectors v1, v2, v3, e4 form a basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
It is known that at least one of the vectors
e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1),
can be chosen as v4
In particular, the vectors v1, v2, v3, e4 form a basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
It is known that at least one of the vectors
e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1), can be chosen as v4
In particular, the vectors v1, v2, v3, e4 form a basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
It is known that at least one of the vectors
e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1), can be chosen as v4
In particular, the vectors
v1, v2, v3, e4 form a basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
It is known that at least one of the vectors
e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1), can be chosen as v4
In particular, the vectors v1, v2, v3, e4
form a basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6
Vector SpacesLinear Dependence and IndependenceBasis and DimensionRow space of a matrix
Row space of a matrix
It is known that at least one of the vectors
e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), ande4 = (0, 0, 0, 1), can be chosen as v4
In particular, the vectors v1, v2, v3, e4 form a basis for R4.
Dr. Marco A Roque Sol Linear Algebra. Session 6