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A matrix is a rectangular array of numbers.
A vector is a matrix with only one column.
The components of a vector are the entries in it.
A matrix is in RREF (Reduced Row-Echelon Form) if it satisfies
all of the following:
If a row has nonzero entries, then the first nonzero entry is 1
(the leading 1 or pivot)
If a column contains a leading 1, then all other entries in that
column are zero.
If a row contains a leading 1, then the each row above it
contains a leading 1 further to the
leftSystems of linear equations can be solved by the following
algorithm:
Write the augmented matrix (also coefficient matrix) of the
system. Place a cursor in the top entry
of the first nonzero column of this matrix.
If the cursor is zero, swap the row with some row below it to
make the cursor entry nonzero.1.
Divide the cursor row by the cursor entry. (after this, the
cursor entry will be 1, which is
exactly what we want)
2.
Eliminate all other entries in the cursor column, by
substracting suitable multiples of the
cursor row from the other rows.
3.
Move the cursor one row down and one column to the right. If the
new cursor entry and all
entries below are zero, move the cursor to the right in the same
row. Repeat this step as
necessary. If there are no more rows or columns, you're
done.
4.
Return to step 1.5.
Now you have an RREF of the original augmented matrix.
Solve the resulting equations. If there is one of the form 0 =
1, then the system is inconsistent.
There are no solutions.
A system is inconsistent if and only if the RREF of its matrix
contains a row of the form [0 0 ... 0 |
1], representing the equation 0 = 1. In this case the system has
no solutions.
The rank of a matrix is defined by the number of leading 1's in
its RREF.
If the rank of the coefficient matrix of the system is smaller
than the number of variables in the
equations, then the system has either infinitely many solutions,
or none at all.
If the rank of the coefficient matrix of the system is equal to
the number of variables in theequations, then the system has at
most one solution.
The Identity matrix (I) is an N x N matrix with 1's on the
diagonal and 0's everywhere else
A linear system of equations has one solution if and only if the
RREF of its coefficient matrix is an
Identity Matrix.
The norm of a vector is the square root of the sum of the
squares of the components (Pythagoras'
theorem). The norm of some vector x can be notated as ||x||, or
as the square root of the dot product
of x and x.
The dot product of two vectors is the sum of the product of one
component of the first vector with
the component with the same index in the second vector
The sum of two matrices of the same size is defined entry by
entry (so corresponding entries aresummed to make 1 new entry in
the resulting matrix)
The product of a matrix and a scalar is defined entry by entry
(so each entry multiplied with the
scalar results in 1 new entry in the resulting matrix)
The product of a matrix and a vector is a new vector consisting
of the sum of the vectors obtained
by multiplying each component of the vector with a column in the
matrix
A linear transformation is the multiplication of one vector with
a matrix to create a new vector
A scaling is defined by a matrix of the form [k 0 ; 0 k] (Matlab
notation)
A projection is defined by a matrix of the form [u1^2 u2*u1 ;
u2*u1 u2^2], where u is a vector
on the line L we're projecting to, and u = [u1 ; u2].
A reflection is defined by a matrix of the form [a b ; b -a],
where a^2 + b^2 = 1.Furthermore, ref[L](x) = 2proj[L](x) - x (where
x is the vector being reflected, and L the
line in which it is inverted).
A rotation done counterclockwise in R2 through angle t is
[cos(t) -sin(t) ; sin(t) cos(t)].
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A horizontal shear is of the form [1 k ; 0 1] - a vertical shear
is of the form [1 0 ; k 1].
A function T from X to Y is invertible if the equation T(x) = y
has a unique solution x in X for each
y in Y.
A matrix A is invertible if and only if A is a square matrix (n
x n) and rref(A) = In
If A (an n x n matrix) is invertible, and b is a vector in Rn,
the linear system Ax = b has exactly one
solution - otherwise there are infinitely many or none at
all.
Computing the rref([A | In]) of an invertible matrix A will get
you [In | B], where B is the
inverse of A.
The product of a matrix and its inverse is an identity
matrix.
The determinant of a 2x2 matrix [a b; c d] is ad - bc.
Given matrices B and A:(BA)^-1 == A^-1 * B^-1
So order matters!
Given matrices B and A, AB does not necessarily equal BA. If it
does, A and B commute
Given matrices A, B and C, (AB)C == A(BC)
Given matrices A, B, C and D, A(C + D) == AC + AD and (A + B)C
== AC + BC (matrix
distribution)
The span of vectors v1 ... v[n] is the set of all linear
combinations of c1 * v1 + c2 * v2 + ... + c[n] *
v[n].
A subset of Rn is called a subspace of Rn iff:
it contains the zero vector.
it is closed under addition - if v1 and v2 are both in im(T),
then so is v1 + v2.
it is closed under multiplication - if v1 is in im(T), then for
any scalar k, k*v1 is in im(T) as
well.
The image of a matrix T is the span of the column vectors of
T.
im(A) is a subspace of A.
The only images for R2 are R2 itself, the zero vector, and any
of the lines through the origin.
The kernel of a matrix consists of all zeros of the
transformation, so the solutions of T(x) = Ax = 0vector.
The kernel of a matrix A is obtained by solving the linear
system Ax = 0 (so rref[A | 0], etc.).
ker(A) is a subspace of A.
A set of vectors is linearly independent if none of them can be
described as a linear combination
of the other vectors. If there is a vector that can be described
in such a way, that vector is
redundant.
If a set of vectors is linearly independent, they span V and are
in V, then they are a basis of the
subspace V.
To construct the basis of an image of matrix A, list all the
column vectors of A and omit the
dedundant ones.A relation between vectors (v1 ... v[n]) exists
if c1*v1 + ... + c[n]*v[n].
The dimension of a subspace V is the number of vectors in a
basis of V.
dim(ker A) + dim(im A) = the number of columns in A
(Rank-Nullity theorem).
Two vectors are orthogonal if their dotproduct is 0.
A vector is a unit vector if its length is 1, so the dotproduct
of the vector with itself is 1.
A set of vectors is called orthonormal if they are all unit
vectors and orthogonal to one another.
Orthonormal vectors u[1] ... u[n] in Rn form an orthonormal
basis of Rn
The orthogonal projection of some vector x on a subspace V with
orthonormal basis u1,...,u[m]
can be described as: projV(x) = (u1 . x)u1 + ... + (u[m] .
x)u[m].
The orthogonal complement of some subspace V is the kernel of
the orthogonal projection onto V,and consists of the set of vectors
x in Rn that are orthogonal to all vectors in V. The intersection
of
V and its complement consists of the zero vector alone. dim(V) +
dim(V[orthogonal]) = n.
(V[orthogonal])[orthogonal] = V.
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The orthogonal projection of a vector x is equal to x itself if
and only if x is in V.
If x and y are vectors in Rn, then the absolute value of their
dot product is smaller than or equal to
the product of their norms. (Cauchy-Schwarz inequality)
The angle between two vectors is defined as the cos((x . y) /
||x|| * ||y||).
Given a basis v1,...,v[m] of a subspace V of Rn. For j =
2,...,m, we resolve the vector v[j] into its
components parallel and perpendicular to the span of the
preceding vectors. Then we convert our
results into unit vectors by dividing each vector by its norm.
The resulting set of vectors is an
orthonormal basis of V (Gram-Schmidt proces).
Given vector v and a unit vector u, the component of vector v
perpendicular to u can be obtained as
follows: v - v[parallel] = v - projV(v) = (u1 . v)u1
For any n x n matrix A, the following are either all true or all
false:
A is invertible.
The linear system Ax = b has a unique solution x for all b in
Rn
rref[A] = In
rank(A) = n
im[A] = Rn
ker[A] = 0 vector
The column vectors of A form a basis of Rn
The column vectors of A span Rn
The column vectors of A are linearly independent
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