a rectangular array of numbers called entries / elements

m rows x n columns

indexed a_ii

m = n

(# rows = # cols)

diagonal matrix

A and B must be same size

A + B = [aij + bij]

cols of matrix A must equal rows of matrix B

(m x n) * (n x r) -> produces m x r matrix

A * B = [vector ai o vector bj]

AB != BA (not all the time at least)

A(BC) = (AB)C

A(kB) = k(AB)

A(B+C) = AB + AC

(A+B)C = AC + BC

IA = AI = A

A^3 = A * A * A

rowi(A^T) = coli(A)

rotate the such that the nth row becomes the nth col of the new matrix

A = A^T

[aij] = [aji]

think reflecting across the diagonal

(A^T)^T = A

(kA^T) = k(A^T)

(A^r)^T = (A^T)^r (r > 0)

(A + B)^T = A^T + B^T

(AB)^T = (B^T)(A^T)

A is an nxn (square) matrix

A^-1 is an nxn matrix such that

A(A^-1) = (A^-1)A = I

if A is invertible, then A^-1 is unique

null(A) is the set of all vectors x such that Ax = 0

(A times the vector x is the zero vector)

or, null(A) = {x | Ax = 0}

observation:

A is invertible iff null(A) = {0 vector}

a set of vectors that

1) spans S

2) is linearly independent

e.g. a basis for R2 is {[0 1][1 0]} because it is two linearly independent vectors that lets you get around all of R2 through its linear combinations

finding bases for row(A) and col(A)

(the row and col space of matrix A)

row(A) = span({nonzero rows of r.r.e.f. of A})

col(A) = span{(set of columns in *A* that have leading terms in the r.r.e.f. of A)}

WARNING: col(A) != col(r.r.e.f.(A)) (generally)

any matrix performed by performing one elementary row operation on I

1) swap two rows

2) multiply a row by a scalar

3) add a multiple of one row to another

TFAE

a) A is invertible

b) Ax = b has the unique soltuion x = (A^-1)(b) for all b in R^n

c) Ax = b has a unique solution for all b, so x = 0 is the only trivial solution

d) [A | 0] has only the trivial solution (no free variables) [A | 0] = [I | 0]

e) there are elementary row operations R1 ... Rk that transform A into I

let Ei = the elementary matrix of Ri

Ek ... E3E2E1A = I

A = E1^-1(E2^-1)...(En-1^-1)(Ek^-1). thus, 

f) (E1 ... En)^-1 = Ek^-1 ... E1^-1 (invertible)

- 

f) rank(A) = n

g) nullility(A) = 0

h) columns of A are linearly independent

i) columns span R^n

j) columns are a basis of R^n

k) rows of A are linearly independent

l) rows span R^n

m) rows are a basis of R^n

1) 0 vector is in S

2) if u, v in S, then u + v in S (closed under addition)

3) if u in S, c in R, then cu in S (closed under scalar multiplication

thus, a subspace S is a set of vectors such that you can't use linear combinations to get out

equivalent: if u, v i nS and c, d in R then cu + dv in S (closed under linear combinations)

row(A) = space spanned by rows - R^n

col(A) = space spanned by cols - R^m

dim(S) is the number of vectors in a basis of S

therefore, dim(col(A)) == dim(row(A))

because dim(row(A)) = # nonzero rows in rref(A)

                    = # of leading 1s in rref(A)

                    = dim(col(A))

rank(A) = dim(row(A))

        = dim(col(A))

easy proof:

rank(A) = rank(A^T)

nullility(A) = dim(null(A))

observe:

nullility(A) = # free variables in rref(A) aug. with 0 vector

free variables of rref = basis vectors of null space

S is a subspace of Rn with basis B = {v1 ... vk}

For all vectors in S, there is a unique linear combination

c1v1 + c2v2 + ... + ckvk = v

these c's are referred to as what?

the coordinates of v with respect to B

[c1 ... ck] = coordinate vector of v with respect to B

det(A) = a11det(A11) - a12det(A12) +- ... + (-1)^n+1 (a12det(A1n)

A11 = matrix A with row 1 col 1 "struck" out

summation (i = 1 to n) a1i * C1i

^ the cofactor expansion

the determinant of a nxn matrix A = [aij] (n >= 2) can be computed as

det(A) = ai1Ci1 + ai2Ci2 + ... + ainCin

(expansion along row i)

or

det(A) = a1jC1j + a2jC2j + ... + anjCnj

(expansion down column j)

a) if A has an all-0 row or col, det(A) = 0

b) if B comes from swapping two rows of A, then the det(B) = -det(A)

c) if A has two equal rows, then det(A) = 0

d) if B comes from multiplying a row or col by a constant k, then det(B) = k*det(A)

e) if A, B, C are the same except the ith row of column of C is the sum of the ith row/col of A and B then det(C) = det(A) + det(B) (???)

f) if B comes from adding a multiple of row i in A to rowj (i != j) then det(A) = det(B)

theorem:

if E is an elementary matrix in det(EB) = det(E)det(B)

A is invertible iff det(A) != 0