Linear Algebra

# Matrix Fun

$$\begin{pmatrix}2 & 1\\ 1 & 2 \end{pmatrix}$$

$\left[ \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16 \end{array} \right]$

$\left[ \begin{array}{cccc} 1 & 0 & -3 & 5 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right]$

# 1/14/08

Identity Matrix (pg 22)
Rotation Matrices (pg 23)

# 1/23/08

Elementary row operations yield equivilent augmented matrices of a system of linear equations.

## Goal

Transform an augmented matrix into reduced row echelon form so as to easily see the solutions.

## Row Echelon Form

1. Each non-zero row is above every zero row
2. Leading entry of a non-zero row lies in a column to the right of that of the leading entry of any preceding row
3. If a column contains the leading entry of a row, then all entries below it in that column are 0. (follows from #2)

## Reduced Row Echelon Form

1. If a column contains the leading entry of a row, then all other entries of that column (below AND above) are 0
2. Leading entry of each non-zero row is 1

Express leading variables x1,x3,x5 (basic variables) in terms of the others (free variables).
Basic variables are columns containing only a leading entry

# 1/25/08

Every matrix can be transformed into 1 and only 1 matrix in reduced row echelon form by a sequence of elementary row operations

If we have Ax=b
1) Write augmented matrix [A b]

2) Find reduced RE
[Rc] of [Ab]

3) If [Rc] contains a row in which the only non-zero entry lies in the last column, then there's no solution (inconsistent).
Otherwise, there's at least 1 solution.
Then we write the equations corresponding to [Rc] and solve to obtain a general solution of Ax=b.

## Gaussian Elimination

Procedure for obtaining the reduced row echelon form of a matrix

$$\begin{pmatrix}0 & 0 & 2 & 4\\ 0 & 1 & -1 & 1 \\ 0 & 6 & 0 & -6 \end{pmatrix}$$

1) Determine leftmost non-zero column —> Pivot column
Top position in a pivot column is a pivot position
2) In a pivot column interchange rows (if necessary) to have a non-zero entry in the pivot position.

r1<—>r2
$$\begin{pmatrix}0 & 1 & -1 & 1 \\ 0 & 0 & 2 & 4 \\ 0 & 6 & 0 & -6 \end{pmatrix}$$

3) Zero out all entries below pivot position.

-6r1 + r3—>r3
$\bf{A} = $\begin{pmatrix}0 & 1 & -1 & 1 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 6 & -12\end{pmatrix}$$

4) Repeat procedure 1-3 with next pivot column and pivot position.

2 (3rd col, 2nd row) becomes the new pivot position.

-3r2+r3—>r3
$\bf{A} = $\begin{pmatrix}0 & 1 & -1 & 1 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0\end{pmatrix}$$

# 1/28/08

Nothing interesting. Sick. =(

# 1/30/08

• Leading variables are basic variables. (x1,x3,x4,)

Others are free variables (x2,x5) Can be anything (0, 1, or infinite solutions)

• The rank of an m*n matrix A is the number of non-zero rows in the reduced row echelon form of A.
• Nullity of A is n-rank A. For the previous example rank=3; nullity=6(columns)-3=3. 6 is the number of columns.
• Equivalently: Rank equals the number of pivot columns in the close messagematrix. Nullity equals the number of non-pivot columns.
• If an m*n matrix has rank n, then it must be [e1 … en]m*n

If it is square(m*n), then [e1 ..en]m*n == In, identity matrix

Each basic variable corresponds to the leading entry of the augmented matrix Ax=b.

# of basic variables = # of non-zero rows == rank A.

# of free variables of Ax=b is n - # basic variables = n-rank A == nullity of A

• Consistency for Ax=b
1. # of basic variables = rank A.
2. # of free variables = nullity of A.

=> Unique solution iff (if and only if) the nullity of A is 0.

Infinity solutions iff nulity (free variables) of A is positive.

# 02/01/08

## Consistency for Ax=b

• # of basic variables = rank A
• # of free variables = nullity of A (not [Ab])

## Span of a set of vectors

For the non-empty set S={u1 …. uk} of vectors in Rn, the span of S (Span S) is
the set of all linear combinations of u1 … uk in Rn.
Span S = Span {u1 … uk}
Span{0}={0}

# 02/04/08

Test on Wed up to and including 1.4

Span S = set of all linear combinations of S={u1 … uk}
Span{0}={0}
S1={$\bf{A} = $\begin{pmatrix}1 \\ -1 \end{pmatrix}$$}
Span S1 = $\{a (scalar) * $\begin{pmatrix}1 \\ -1 \end{pmatrix}$ \}$
Span3= $\{a$\begin{pmatrix}1 \\ -1 \end{pmatrix}$ + b $\begin{pmatrix} -2 \\ 2 \end{pmatrix}$ + c $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ \}$
S3={u1,u2,u3}={u1,-2u1,u3}=R2

A vector V belongs to the span of S (S={u1 … uk}) if V can be expressed as the linear combination of the vectors u1 … uk in S.

Suppose A=[u1 … uk] in Rn
Then, V is in the span S iff (if and only if) Ax=v is consistent.

x1u1 … = V

# 02/08/08

V "belongs to" Span S = {u1 … uk}

V is a linear combination of the vectors u1 … uk.

• If A = [u1 … uk] then v is in the Span S if + only if Ax=v is consistent.

x1u1 + xkuk = v

Is V= $$\begin{pmatrix}3 \\ 0 \\ -5 \\ 1 \end{pmatrix}$$

# 02/13/08

If one of the u's is 0 then the set is always linear dependant.

The set {u1 .. uk} is linear dependant iff there exists a non-zero solution of _Ax=0__

A=$$\begin{pmatrix}1 & 1 & 1 & 1 \\ 2 & 0 & 4 & 2 \\ 1 & 1 & 1 & 3\end{pmatrix}$$

For an m x n matrix A the following are equivilent:

1. Columns of A are linear independant
2. Ax=b has at most 1 solution for each b in Rm.
3. Nullity of A is 0 (no free vars)
4. Rank=#columns - Nullity
5. Columns of RRE of A are distinct standard vectors in Rm
6. Only solution of Ax=0 is x=0
7. Pivot position in each column of A.

For linear independant/dependant we want to determine whether or not x=0 is the only solution to Ax=0 (homogeneous equation)

If there exist free variables => infinity solutions, linear dependant
*Vectors u1 … uk in Rn are linear dependant iff ui=0 or there exists i>=2 such that ui is a linear combination of the preceding vectors u1 .. ui-1

# 3/10/08

For a linear transf T, w is the range of T iff w (output) = T(v[input])

the range of a linear transformation equals thespan of the columns of its standard matrix A = [T(e1) T …]

A function f Rn —> Rm is onto if its range is all of Rm (range = co-domain)
if every vector in Rm is an image under f

page revision: 34, last edited: 10 Mar 2008 15:47