Linear Final

det(A^TB) =

det(B^TA)

det(A+B) DOES NOT =

det(A) + det(B)

every orthonormal basis is

an orthogonal basis

If B is the RREF of a matrix A and det(B) = 0

then det(A) = 0

Row opperation DO change

the determinant of a matrix

det(2A) =

det(A) x 2^m

• if m is the number of rows

det(AB) = 

det(A)det(B)

if det(A) NOT= 0 we know that

A is invertible

• thus a solution exists and is unique

for a set of vectors to span P_n

It must have n+1 vectors that are not scalars of each other (linearly independent)

for a matrix/set of vectors to be linearly independent

det(A) CANNOT = 0

a set of vectors is a basis for P_n if

they span P_n and are linearly independent 

a set of vectors is an orthogonal set if

all of the vetcor dot products = 0

they are linearly independent

if x^→ is NOT in the subspace W, then

x^→ - prod_wx^→ is NOT = 0

if Av=λv for a nonzero vector v,

then v is an eigenvetcor with eigenvalue λ

Algebraic multiplicity

multiplicity of eignenvalue in characteristic polynomial

geometic multiplicity

dimension of eigenspace (number of linearly independent eigenvectors)

When is a matrix diagonalizable?

when geometric multiplicity = algebraic multiplicity for all eigenvalues:

n linearly independent eigenvectors for nxn matrix

Every real symmetric matrix

is orthogonally diagonizable

to check if a matrix is diagonizable,

check if A = A^T (symmetric)

Stepsfor orthogonal diagonization:

1. find eigenvalues

2. find eigenvectors

3. normalize eigenvectors

4. for P with normalized eigenvetors, D with eigenvalues

Singular value decomp (SVD)

A=UΣVT where U and V are orthogonal, Σ is diagonal with singular values.\

• every mxn matrix has one

finding rank from SVD

rank = number of nonzero singular values (square root of eigenvalues of A^TA, always nonnegative)

Properties of U and V in SVD A:

• UU^T = I and U^TU = I

• VV^T = I and V^TV = I

• Columns are orthogonal

two linear systems are the same if

they have the same solution set

consistent system

either has one solution or infinitely many solutions

inconsistent system

has no solution

rows represent

equations (m)

columns represent

variables (n)

free variables indicate that

there are many solutions, not a unique solution

If b is in the span of the columns of A, you can also say that

b is in the column space of A

can a 3×2 matrix span R³?

No, because this matrix is 2D and cannot span a 3D subspace;

• A is unable to have a pivot in every row

can a 2×3 matrix span R²?

yes because it can have a pivot in every row

rank(A) is equal to

• the number of nonzero rows in its echelon form B

and/or

• the number of pivot positions (pivot columns) in B

Rank-Nullity theorum

rank(A) + nullity(A) = number of columns of A

the basis for Col(A) is determined by

the pivot columns in the RREF that contain a leading 1

the basis for Nul(A) is determined by

solving for RREFx = 0; just write RREF in parametric vector form

(NulA = the set of all vectors x such that Ax = 0) 

ex: the x for a 3×4 matrix will have four variables (x1, x2, x3, x4)

null space, NulA is

the set of all vectors x such that Ax = 0

For Ax=b to have a unique solution

• A is invertible

• A is row equivalent to I

• There is a pivot in ever column of A

• the columns of A are linearly independent

• det(A) does NOT = 0

• 0 is not an eigenvalue of A

A has orthonormal columns if

A^TA = I

A is orthogonally diagonizable if

A = A^T ; a is symmretric

if , A = UΣV T (SVD), what is the size of the original matrix A?

same as the size of Σ