Vectors and Matrices Test 1

i. The span of the columns of A is Rn ii. The columns of A are linearly independent iii. The equation Ax = b has a unique solution for each b in Rn, namely x = A-1b iv. The only solution to Ax = 0 is x = 0 v. Nullity A = 0 vi. Rank A = n vii. R has no zero rows viii. The columns of R are distinct standard vectors in Rn ix. There is a leading 1 in every row of R x. There is a leading 1 in every column of R xi. R = In xii. A is invertible xiii. A can be written as a product of elementary matrices

BIG THEOREM: (Let A be an n x n matrix with RREF matrix R. TFAE )

If column j of R is a linear combination of other columns in R, then column j of A is a linear combination of the corresponding columns of A using the some coefficients, and vice versa.

Column Correspondence Property

an n x n matrix E is called an elementary matrix if we can obtain E from In by a single elementary row operation.

Elementary Matrix

1. The vectors in a vector form of a general solutions to Ax = 0 are linearly independent 2. Standard vectors are linearly independent 3. If 0 vector is in S then S is a linearly dependent set

Examples of linearly independent vectors

1. S = {v} is linearly independent if v != 0 and linearly dependent if v = 0 2. S = {u1,u2} is linearly dependent if u1 = 0 or u2 is in the span of {u1} 3. Suppose S = {u1,...,uk} is linearly independent in Rn, and v is also a vector in Rn a. Then v is not in Span S if and only if {u1,...,uk, v} is linearly independent 4. Every subset containing more than n vectors in Rn must be linearly dependent a. Note: any 2 non-parallel vectors in R2 Span all of R2 5. If S is a subset of Rn and no vectors can be taken out of S without changing the Span S, then S must be linearly independent

Facts about linearly independent and dependent sets

An n x n matrix A is invertible if there exists an n x n matrix B such that AB = BA = In

Inverse Definition

at least one vector is a linear combination of the others.

Linearly Dependent

No one vector can be written as a linear combination of the others.

Linearly Independent

a) Ax = b is consistent b) b is a linear comb of columns of A c) RREF of [A|b] has no row of form [00...0|d], where d does not = 0

Theorem 1.5 a) Ax = b is consistent

i. The span of the columns of A is Rm ii. The equation Ax = b has a least one solution (i.e. is consistent) for each b in Rm iii. Rank A = m iv. R has no zero rows v. There is a leading 1 in every row

Theorem 1.6 (theorem 1.8 and BIG): (Let A be an m x n matrix with RREF matrix R. TFAE)

Ax = b is consistent if and only if b is in the span of the columns of A

Theorem 1.7 Ax = b is consistent IFF

i. The columns of A are linearly independent ii. The equation Ax = b has at most one solution for each b in Rm iii. The only solution to Ax = 0 is x = 0 iv. Nullity A = 0 v. Rank A = n vi. The columns of R are distinct standard vectors in Rm vii. These is a leading 1 in every column of R (additionally, m >= n)

Theorem 1.8 : (Let A be an m x n matrix with RREF matrix R. TFAE)

(AC)T = CT AT IkA = A = AIm

Theorem 2.1: Let A and B be k x m matrices, C be m x n matrix, P and Q be n x p matrices, and let s be a scalar in R

Then there exists an invertible matrix P such that PA = R

Theorem 2.3: let A be an m x n matrix with RREF R.

a) The pivot columns of A are linearly independent (Column j is a pivot column of A if column j of R has a leading 1) b) Each non-pivot column of A is a linear comb of the previous pivot columns, where the coefficients are the entries of the corresponding column of R

Theorem 2.4 Column Correspondence: Let A be a matrix with RREF R

A is invertible if and only if A can be expressed as a product of elementary matrices If A = [ a b ] [ c d ] A↑-1 = (1/ab - bc) [d -b] [-c a]

Theorem 2.4: