MATH1231 Algebra, 2017 Chapter 6 - University of New ...danielch/fyalg15b/Chapter...2 0v = 0. 3 ( 1)v = v. Here 1 is a scalar and v is the additive inverse of v. 4 If v = 0, then either

MATH1231 Algebra, 2017Chapter 6

A/Prof. Daniel Chan

School of Mathematics and StatisticsUniversity of New South Wales

[email protected]

Daniel Chan (UNSW) MATH1231 Algebra 1 / 77

Things to keep in mind before we start

1 Go to Moodle 1231 page often.

2 Do the revision questions in the “yellow” notes.

3 Lecture notes are uploaded chapter by chapter.

4 Print the lecture notes and bring them to lectures.

5 Lecture recordings will be available.

6 In this course, there are lots of new concepts and definitions. Thereare fewer computations but success requires knowing whichcomputations to perform and arguments to express.

Daniel Chan (UNSW) MATH1231 Algebra 2 / 77

Review of parametric form for lines

Recall the parametric form for a line in space passing through a indirection v:

x = a + λv, λ ∈ R.

Similarly, given a, v ∈ Rn, the parametric form above defines a line in Rn.E.g. Let’s solve the DE dydx = 2x .

Question

Is the set of solutions

y = x2 + λ = x2 + λ1, λ ∈ R

a “line” of polynomials?

Answer YES, as long as you properly define an environment in which youcan talk about lines & planes in an abstract general setting!The goal of this chapter is to define precisely such an environment.

Daniel Chan (UNSW) 6.1 Vector Spaces 2 / 77

Informal “definition” of vector spaces

A vector space is an environment in which you can talk about linearconcepts such as lines.Q What does this require?At a minimum you need the following 4 ingredients:V — a set of objects called vectors e.g. Rn or arrows,F — a set of scalars (numbers), for us either R or C,+ — an operation for adding two (possibly equal) elements u, v in V toobtain another u + v ∈ V , and∗ — an operation for multiplying an element v ∈ V by a scalar λ ∈ F, toobtain the vector λ ∗ v ∈ V which is usually abbreviated to λv.Given a “vector system” (V ,+, ∗,F), we also want the addn & scalarmultn to behave sensibly e.g. 2(2v) = 4v.Q What’s sensible mean here?The system has to satisfy the 8 axioms on the next few slides.


Definition of a Vector Space

Definition (Vector Space).

A vector space V over the set of scalars F is a vector system (V ,+, ∗,F)which obeys the following 8 axioms.

1 Associative law of addition.If u, v,w ∈ V then (u + v) + w = u + (v + w).

2 Commutative law of addition.If u, v ∈ V then u + v = v + u.

3 Existence of zero. There is a special element 0 in V called the zerovector which has the property that v + 0 = v for all v ∈ V .

4 Existence of Negative. For each v ∈ V there exists an elementw ∈ V (the negative of v, i.e. w = −v) such that v + w = 0.

and


Vector space axioms cont’d

5 Associative law of multiplication by a scalar. If λ, µ ∈ F andv ∈ V then λ(µv) = (λµ)v.

6 If v ∈ V then 1v = v.7 Scalar distributive law. If λ, µ ∈ F and v ∈ V then

(λ+ µ)v = λv + µv.

8 Vector distributive law. If λ ∈ F and u, v ∈ V thenλ(u + v) = λu + λv.

Remark

We usually abbreviate (V ,+, ∗,F) to V which is fine if +, ∗,F areunderstood.


What are axioms anyway?

Question

What’s a mammal?

A It’s a type of animal which satisfies the following axiom(s)

Warning I didn’t define what an animal was! But I also didn’t define whata set was!!Axioms are important because they

define classes of objects, and

give the starting assumptions for proofs.


The vector space Rn

The co-ordinatewise addition and scalar multiplication of Rn defined in1131 are called the usual addition and the usual scalar multiplication.

For any a =

a1...an

, b =b1...

bn

∈ Rn and λ ∈ R,

a + b =

a1 + b1...an + bn

, and λa =λa1...λan

.Theorem

The V = Rn with the usual addn & scalar multn is a vector space over R.

Hopefully, you saw all the axioms verified in MATH1131.


Droids

Vector systems can get pretty crazy. Consider following vector system(V ,+, ∗,R).V = the set of droids {C-3PO, R2-D2, BB-8}.+ — the sum of any two droids = BB-8∗ — the scalar multiple of any droid = R2-D2Q What’s 5 R2-D2 + C-3PO?

Example

Prove that V satisfies the commutative law of addition!

In fact, V also satisfies the associative laws of addn & scalar multn!


Proving vector systems are not vector spaces

Example

Prove that the vector system of droids is not a vector space.

Proof.

To show a vector system is not a vector space, you just need to show itfails one of the axioms.Think of the axioms as like a checklist — pass all & you are in goodshape, fail one & your out.


Review polynomials

Recall that a polynomial over F = R or C of degree k is a functionp : F→ F such that

p(x) = a0 + a1x + · · ·+ akxk , where a0, a1, . . . , ak ∈ F and ak 6= 0.

The zero polynomial defined by p(x) = 0 has degree −∞ by defn. LetP(F) = set of polynomials over F. Let p, q ∈ P & λ ∈ F. Definepolynomials p + q and λp by

(p + q)(x) = p(x) + q(x) and (λp)(x) = λp(x),

for all x ∈ F.NB These operations correspond to summing corresponding co-efficientsand scalar multiplying them.Eg


The vector space P

Oft abbreviate P(F) to P if F understood.

Theorem

The set P with addn & scalar multn defined above is a vector space /F.

Proof. Just check all 8 axioms.

Remark This means we can talk about linear concepts in P e.g.x2 + λ, λ ∈ R is a line of polynomials.

On the other hand, we can’t speak of lines & planes of droids. Why?


Axioms allow usual vector arithmetic

The vector space axioms allow you to perform the usual algebraicmanipulations of vector arithmetic.

Example

Let V be a vector space /R & u, v ∈ V . Simplify 3(u + v) + v stating allaxioms you use.


Other important examples of vector spaces

Examples

The following are also examples of vector spaces.

The vector space of all 2-dim geometric vectors over Rwith usual vector addn & scalar multn

The vector space of all 3-dim geometric vectors over Rwith usual vector addn & scalar multn

Cn over C.Addn & scalar multn are co-ordinatewise.

The vector space of all m × n matrices, Mmn(F) over F.with matrix addn & scalar multn.

The set of all real-valued functions with domain X , R[X ] over R.Addn & scalar multn are pointwise.


Vector subtraction

Proposition 1.

In any vector space V , the following properties hold.

1 Uniqueness of Zero. There is one and only one zero vector.

2 Cancellation Property. If u, v,w ∈ V satisfyu + v = u + w, then v = w.

3 Uniqueness of Negatives. For all v ∈ V , there exists only onew ∈ V such that v + w = 0. We write −v = w

Remark Negatives allow us to define vector subtraction

u− v = u + (−v).

Daniel Chan (UNSW) 6.2 Vector Arithmetic 14 / 77

Vector arithmetic

The axioms imply all sorts of other rules of vector arithmetic hold.

Proposition 2.

Suppose that v is a vector in a vector space V ; λ is a scalar; 0 is the zeroscalar; 0 is the zero vector in V . Then the following properties hold.

1 λ0 = 0.

2 0v = 0.

3 (−1)v = −v. Here −1 is a scalar and −v is the additive inverse of v.4 If λv = 0, then either λ = 0 or v = 0.

5 If λv = µv and v 6= 0 then λ = µ.


Sample proof

Example

Prove Proposition 2.3 assuming Proposition 2.2.

Proof.


Closure under addition

Let V be a vector space /F.

Definition

We say a subset S ⊆ V is closed under addition if for every u, v ∈ S wehave u + v ∈ S .

E.g. Let V be the vector space of 2-dim geometric vectors. Let S be thesubset of vectors parallel to the x or y axis. DRAW PICTURE.

so S is NOT closed under addition.

Daniel Chan (UNSW) 6.3 Subspaces 17 / 77

Closure under scalar multiplication

Definition

We say a subset S ⊆ V is closed under scalar multiplication if for everyv ∈ S & scalar λ we have λv ∈ S .

E.g. Let V be the vector space of 2-dim geometric vectors. Let S be thesubset of vectors parallel to the x or y axis.

so S is closed under scalar multiplication.


Subspaces

An easy way to come up with more examples of vector spaces is as follows.

Theorem-Definition

A subset S of a vector space V /F is a subspace of V ifi) the zero vector of V belongs to S ;

ii) S is closed under addn; and

iii) S is closed under scalar multn.

In this case, S is itself a vector space /F with addn & scalar multn the“same” as that in V .

Why? The axioms are all inherited from V !

In the “yellow” notes, the above result is referred to as the SubspaceTheorem.Remark i), ii), iii) are sometimes called closure axioms for a subspace.E.g. The subset 0 = {0} is a subspace called the zero subspace.


Examples of subsets which are not subspaces

To check if a subset is a subspace, just go down the “checklist” of closureaxioms.

Example

Is S ={

x ∈ R3 : x1 − 2x2 + 3x3 = 1}

is a subspace of R3?

Solution


Example of a subspace

Example

Is S = {x ∈ R3 : x1 − 2x2 + 3x3 = 0} a subspace of R3.

Solution


Yet another Rn example

Example

Is S ={

x ∈ R3 : x3 = x21 + x22}

is a subspace of R3.

Solution


The subspace Pn

Let Pn = set of polynomials of degree n or less.

Proposition

Pn is a subspace of P.

Proof.

Consider 2 polynomials in Pn which must have form

p(x) = p0 + p1x + . . .+ pnxn & q(x) = q0 + q1x + . . .+ qnx

n


An Example of a subspace of Pn.

Example

Show that S = {p ∈ P2(R) : xp′(x)− 2p(x) = 0} is a subspace of P2.

Solution


Solution (Continued)


A subspace of a subspace is a subspace

Proposition

If U is a subspace of V and V a subspace of W , then U is a subspace ofW .

E.g. S = {p ∈ P2(R) : xp′(x)− 2p(x) = 0} is a subspace of P2 & hencealso a subspace of P.


Linear Combinations

Definition (Linear Combination).

Let V = vector space /F & S = {v1, . . . , vn} ⊂ V . Then a linearcombination of S is a vector or expression of the form

λ1v1 + · · ·+ λnvn for some λ1, . . . , λn ∈ F.

Proposition

Every linear combination of S is also a vector in V .

Why?

Daniel Chan (UNSW) 6.4 Linear Combinations and Spans 27 / 77

Examples of linear combinations

Suppose that S =

13

1

, 2−2−1

,24

1

⊂ R3. Here are someexamples of linear combinations of S .

2

131

+ (−1)24

1

= 213

1

−24

1

=02

1

2

131

+ 2−2−1

−24

1

=20

0

00

0

= 013

1

+ 0 2−2−1

+ 024

1


Span

Definition (Span)

Let S = {v1, . . . , vn} ⊂ vector space V . The span of S is the set of alllinear combinations of S . It’s denoted by span(S) or span(v1, . . . , vn).

If span(S) = V , we say S is spanning set of V , or S spans V . Bydefault, define span(∅) = 0.

Example

Prove that {1, x , x2} ⊂ P2 is a spanning set for P2.

Proof.

We know span(1, x , x2) ⊆ P2 so suffice show P2 ⊆ span(S). Any p ∈ P2has form

p = a + bx + cx2 = a(1) + b(x) + c(x2) for some scalars a, b, c .∴ span({1, x , x2}) = P2 & {1, x , x2} is a spanning set for P2.


Span & parametric forms for lines & planes

Eg. 1 Span(

(11

)) = set of all λ

(11

), λ ∈ R.

Eg. 2 Span

111

,14

1

=


Determining if a vector lies in a span?

Example

Let u =

101

, v =14

1

∈ R3, and the set S =13

4

,12

3

. Isu ∈ span(S)? If so write u as a linear combination of S . How about v?

Solution


“Computing” spans i.e. Cartesian form for spans

Example (Continue from the previous example.)

Is S a spanning set for R3? If not, find conditions on b ∈ R3 to be in thespan(S). Give a geometric interpretation of span(S).

Solution


An example in Pn

Example

Is the set

S ={

1 + 2x + 3x2, 2 + 4x + x2, 1 + 2x + 8x2, 1− x + 4x2}

a spanning set for P2?

Solution



Rem A subset spans ⇐⇒ associated linear eqns always has a soln.Existence of solns ←→ spanning.


Properties of a span

Theorem

Let S = finite subset of vector space V . Then span(S) is a subspace of V .

Proof.

Span(∅) = 0 is a subspace so we may assume S = {v1, . . . , vn}.

Since 0 = 0v1 + · · ·+ 0vn, the zero vector of V is a linearcombination of S i.e. 0 ∈ Span(S).

Check Span(S) closed under addition.

Any u, v ∈ Span(S), have form

u = λ1v1 + · · ·+ λnvn, λ1, . . . , λn are scalarsv = µ1v1 + · · ·+ µnvn, µ1, . . . , µn are scalars.



Hence, by the commutative law, associative law of addition, and thescalar distributive law

u + v = (λ1v1 + · · ·+ λnvn) + (µ1v1 + · · ·+ µnvn)= (λ1 + µ1)v1 + · · ·+ (λn + µn)vn

∴ u + v is a linear combination of S and u + v ∈ Span(S).

Span(S) is closed under addition.

Check Span(S) closed under scalar multn.

For any scalar λ & u ∈ Span(S) as above we have ,λu = λ(λ1v1 + · · ·+ λnvn)

= λ(λ1v1) + · · ·+ λ(λnvn) [Vector distributive law]= (λλ1)v1 + · · ·+ (λλn)vn [Scalar assocative law]


Proof (Continued).

Hence λu is a linear combination of S and λu ∈ Span(S).

Span(S) is closed under scalar multiplication.

By the Subspace Thm-Defn, span(S) is a subspace of V .

Remark

Any subspace of V containing a finite set of vectors S contains span(S).Hence, span(S) is the smallest subspace containing S .


Subspaces of R3

Proposition

The only subspaces of R3 are {0}, lines and planes through the origin andR3 itself.

Why? Let W ⊆ R3 be a subspace of R3. If W 6= 0 then pick non-zerou ∈W .Then W contains the line span(u).

Either W = the line span(u) or we can pick v in W − span(u), so Wcontains the plane span(u, v).


Relationship between matrices and spans in Rm

Let S = {v1, . . . , vn} ⊂ Rm and b ∈ Rm. Let A = matrix (v1| · · · |vn).Then b ∈ span(S)⇐⇒ there exist scalars x1, . . . , xn such that

b = x1v1 + · · ·+ xnvn = (v1| · · · |vn)

x1...xn

.Upshot: b is a linear combination of S ⇐⇒ b = Ax for some x ∈ Rn.

In other words, b ∈ span(S)⇐⇒ Ax = b has a soln x ∈ Rn. [AlgebraNotes: Proposition 3 in 6.4]

This motivates the next slide.


Column Space

Definition

The subspace of Rm spanned by the columns of an m × n-matrix A iscalled the column space of A and is denoted by col(A).

Example

Is

111

in the column space of A =1 22 4

3 1

.Solution


Linear Independence

Definition

Let S = {v1, . . . , vn} ⊂ vector space V . We say S is linearlyindependent if

(∗) 0 = λ1v1 + · · ·+ λnvn

implies λ1 = · · · = λn = 0.Otherwise say S is linearly dependent i.e. (*) above has a soln for scalarsλ1, . . . , λn not all 0.

Eg Show S = {(

11

),

(22

)} is linearly dependent.

Next slide see linear dependence generalises of the notion of parallelvectors.

Daniel Chan (UNSW) 6.5 Linear Independence 46 / 77

Linear dependence as a generalisation of parallel vectors

Theorem

Let S = {v1, . . . , vn} ⊂ vector space V . The following statements areequivalent.

1 S is linearly independent.

2 If λ1v1 + · · ·+ λnvn = µ1v1 + · · ·+ µnvn then λ1 = µ1, . . . , λn = µn.3 None of the vectors in S is a linear combn of the other vectors in S .

Example

Two geometric vectors u, v or vectors in Rn are linearly dependent iff theyare parallel.

Why?


Checking Linear Independence

Example

Show that S =

11

3

,12

3

, 10−1

is a linearly independent set.Thm then =⇒ there is only one way to write b ∈ span(S) as a linearcombination of S .

Solution

We solve

0 = λ1

113

+ λ212

3

+ λ3 10−1


How does Span(S) change with S

Eg Find Span(

(11

)), Span(

(11

),

(01

)), Span(

(11

),

(01

),

(12

)),

Theorem

Let S = finite subset of a vector space V . For any v ∈ V , we havespan(S ∪ {v}) = span(S) if and only if v ∈ span(S).


What you can do with a linear dependence relation

Let S = {v1, v2, . . . , vn} ⊂ vector space V /F. If λ1, λ2, . . . , λn ∈ F notall zero satisfy

(∗) λ1v1 + λ2v2 + · · ·+ λnvn = 0,

then S is linearly dependent set & (*) is a called a linear dependencerelation. Can find λi 6= 0.Hence,

vi = −λ1λi

v1 − · · ·+λi−1λi

vi−1 −λi+1λi

vi+1 · · · −λnλi

vn.

∴ vi is a linear combn of v1, . . . , vi−1, vi+1, . . . , vn.Thm last slide =⇒Upshot

In this case,

span(v1, . . . , vn) = span(v1, . . . , vi−1, vi+1, . . . , vn).


Example

Suppose that

v1 =

110

, v2 =22

1

, v3 =44

1

.a) Show that the set S = {v1, v2, v3} is linearly dependent.b) Find a proper subset of S which has the same span as S .

Solution


Important theorems regarding linear independence & span

Theorem1 The span of every proper subset of S is a proper subspace of span(S)

if and only if S is linearly independent.

2 If S is linearly independent and v ∈ V but not in span(S), thenS ∪ {v} is linearly independent.

E.g. Some examples with 3-dim geometric vectors.


Verifying linear (in)dependence in Pn

Example

Is S = {1 + 2x − x2, −3− x − 2x2, 2 + 3x + x2} a linearly independentsubset of P2.

Proof.


Proof (Continued).


Why do we care about linearly independent spanning sets?

Question

Let V = vector space of 2-dim or 3-dim geometric vectors. How do youconstruct a co-ordinate system on V .

A You need co-ordinate axes & scale on axes & +ve direction for axesEquivalently, just give set S of unit positive position vectors for each axis.But when is such a choice sensible?

Upshot S gives a sensible co-ordinate system iff S is a linearly independentspanning set. In this case, no. of elements of S = no. co-ordinate axes.

Daniel Chan (UNSW) 6.6 Basis and Dimension 57 / 77

Basis and Dimension

Definition (Basis).

Let V = vector space. A subset B ⊂ V is a basis for V if a) B is linearlyindependent & b) V = span(B).

E.g. Let V = space of 2-dim geometric vectors and S = {u, v} be any 2non-parallel vectors (so are linearly independent).

Hence S is a basis for V . It is not unique!Note special case The basis for the zero vector space 0 is ∅.


Standard basis for Rn

Let ei be the vector in Rn with the i-th entry 1 and all other entries 0. Theset {e1, e2, . . . , en} is a linearly independent spanning set of Rn. Why?

(x1 x2 · · · xn)T = x1e1 + x2e2 + · · ·+ xnen.Hence, it is a basis which is called the standard basis of Rn.

Example

In R2, the standard basis is{(

10

),

(01

)}.

In R3, the standard basis is

10

0

,01

0

,00

1

.This basis is furthermore orthonormal in the sense that all basis vectorshave length 1 i.e. |ei | = 1 and vectors are mutually orthogonal i.e.ei · ej = 0 for i 6= j .


Standard basis for Pn

Example

The set {1, x , . . . , xn} is a basis for Pn called the standard basis for Pn.

Why? Every polynomial in Pn can be written uniquely in the form

p(x) = a0 + a1x + . . .+ anxn.


Verifying a Subset is a Basis

Example

Is B =

12

1

,11

0

, 1−1−2

a basis for R3?Solution

Note that verifying spanning set and linear independence involve the sameGaussian elimination so only do it once!


Example of a basis for Pn

Example

Is

S = {p1(x) = 1 + 2x + x2, p2(x) = 1 + 3x + 2x2, p3(x) = −1 + 2x + 5x2}

a basis for P2? What about P3?

Solution


Spanning sets are “bigger” than linearly independent sets

Theorem

Suppose a vector space V is spanned by a set S of s vectors. Then anylinearly independent set in V has 6 s vectors.

Proof. is hard and in notes. Result is believable e.g. V = the space of2-dim geometric vectors.


Dimension

Theorem

Any two bases of a vector space V have the same number of vectors i.e. ifB1 = {v1, v2, . . . , vn} and B2 = {u1,u2, . . . ,um} are bases for V thenm = n.

Proof.

Let V = vector space. We say V is finite dimensional if it has a (finite)basis B. In this case, the number of basis vectors is called the dimensionof V and is denoted by dim V .

dim V = no. elements in B


Dimensions of some common vector spaces

Example

dim(Rn) = n.

since the standard basis for Rn consists of n vectors.

Example

dim(Pn) = n + 1.

Since


Reducing spanning sets to bases

Let S = {v1, v2, . . . , vn} ⊂ Rm, A = (v1|v2| · · · |vn), x = (x1 . . . xn)T .Consider eqn

Ax = x1v1 + x2v2 + · · ·+ xnvn = 0.

S is linearly independent iff x = 0 is the unique soln.If x has soln with xi 6= 0, deleting vi from S does not affect span.Let’s row reduce A to a row-echelon form U. Then

S is linearly independent iff all columns of U are leading;

S is linearly dependent iff at least one of the columns of U isnon-leading;

the vectors in S corresponding to the leading columns of U form abasis for span(S).


Example of reducing spanning sets to a basis

Example

Let S =

10

3

,12

1

,11

2

.Find a basis for span(S). What’s dim span(S)?


Example of finding a basis cont’d

More gen

Theorem

If S is a finite spanning set for a vector space V , then S contains a subsetB which is a basis for V .


Extending linearly independent sets to bases

Theorem

Let W = subspace of finite dimensional vector space V . Every linearlyindependent subset of W can be extended to a basis for W . In particular,W is also finite dimensional.

In theory, we keep adding a vector to our linearly independent set S , whichdoes not belong to span(S), until we get a basis.

In practice,

Example

Let S =

11

1

,22

2

,21

1

. Find a basis for R3 containing as manyof the vectors in S as possible .


Example cont’d



Summary of algorithms for finding bases

Let S = {v1, . . . , vs} ⊂ Rm.

Theorem

Let A = (v1| . . . |vs). If U is a row-echelon form for A, the columns of Acorresponding to leading columns in U form a basis for span(S).

Theorem

Let A = (v1| . . . |vs |e1| . . . |em). If U is a row-echelon form for A, thecolumns of A corresponding to leading columns in U form a basis for Rmand this basis contains as many vectors in S as possible.In particular, if S is independent, the basis so formed will contain S as asubset.


Things you know if you know dimV

Theorem

For any finite dim vector space V with dim V = d,

1 Any spanning set S for V has > d vectors,2 Any linearly independent set S in V has 6 d vectors,3 if a spanning set S for V has only d vectors, then S is linearly

independent and so forms a basis for V ,

4 if a linearly independent subset S of V has d vectors, then S alsospans V and thus form a basis for V .

For proofs see Algebra Notes 6.6 – Theorem 3.


Examples on facts using dimension

Example

Can 5 vectors in R4 be linearly independent?

Solution

dim(R4) = 4.Thus, any linearly independent set in R4 has at most 4 vectors.No, 5 vectors in R4 cannot be linearly independent.

Example

Are there any spanning set of P3 contains only 3 polynomials?

Solution

dim(P3) = 4.Thus, any spanning set of P3 has at least 4 vectors.No, a set of 3 polynomials in P3 cannot be a spanning set of P3.


Using the dimension to check a subset is a basis

Example

Let B = {v1, v2, v3} be an orthonormal set in R3. Show that B is a basis.

Solution

We begin by showing that B is linearly independent.

Now dimR3 = 3 so by our thm, the linearly independent set B which has3 vectors must also be a basis.


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MATH1231 Algebra, 2017 Chapter 6 - University of New ...danielch/fyalg15b/Chapter...2 0v = 0. 3 ( 1)v = v. Here 1 is a scalar and v is the additive inverse of v. 4 If v = 0, then either