Linear Algebra Theorems

Any matrix can be changed to Echelon Form using

1. Interchange 2 rows

2. add a multiple of one row to another

3. Multiply one row by a nonzero constant

Rank vs. Number of Solutions

1. No solution = rank > # of variables

2. Unique Solution = rank = # number of variables

3. Infinitely Many Solutions = rank < # number of variables

Steps to Solve a Linear System

1. Find augmented matrix

2. Find RREF *stop if only one solution

3. Write pivot variables in terms of free variables

4. Parameterize

A subset w of a vector space V is a subspace if and only if

1. x,y E w, then x+y E w

2. For all x E w, all scalers a, axEw

3. 0Ew

Let A₁…A_n be an element of vector space V

The span is a subspace of V

Let M E M(m,n)

The column space of M is a subspace of R^m

A liner system AX=B is solvable iff

B E Column Space(A)

General Solution

Let T be any particular solution to AX=B

Then, the general solution is of the form T+Z where Z is the Nullspace(A)

How to solve Ax=B

1. Find any solution

2. Determine Nullspace(A)

Given a matrix A E M(m,n)

The pivot columns are linearly independent and span the column space of A

n - rows → rank <= n

n + 1 variables → at least one free variable → infinite solutions → at least one is the 0 solution → dependent

Let V be a vector space

Every basis of V has the same cardinality

If A and B E M(m,n) are row equivalent

row space(A) = row space (B)

Dim(row space) vs. Dim(column space)

Dim(row space) = Dim(columb space)

The columns of A are linearly independent iff

rank(A)=n

The rows of A are linearly independant iff

rank(A) = m

AX=B is solvable when

1. rank(A) <= n

2. B E Column Space(A)

Rank-Nullity Theorem

Rank(A)+Nullity(A) = n

A is nonsingular

1. Nullspace(A) = {0}

2. For each B E R^m, AX=B has at most one solution

3. AX=B has a unique solution for all B E R^m

A is nonsingular iff

A is nxn and has rank n