Linear Algebra

Matrix Fun

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

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

3) Zero out all entries below pivot position.

-6r₁ + r₃—>r₃

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

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 … e_n]_m*n

If it is square(m*n), then [e₁ ..e_n]_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 …. u_k} of vectors in Rⁿ, the span of S (Span S) is
the set of all linear combinations of u₁ … u_k in Rⁿ.
Span S = Span {u₁ … u_k}
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 … u_k}
Span{0}={0}
S₁={}
Span S₁ =
Span₃=
S₃={u₁,u₂,u₃}={u₁,-2u₁,u₃}=R²

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

Suppose A=[u₁ … u_k] 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 … u_k}

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

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

x₁u₁ + x_ku_k = v

Is V=

02/13/08

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

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

A=

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 R^m.
3. Nullity of A is 0 (no free vars)
4. Rank=#columns - Nullity
5. Columns of RRE of A are distinct standard vectors in R^m
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 … u_k in Rⁿ are linear dependant iff u_i=0 or there exists i>=2 such that u_i is a linear combination of the preceding vectors u₁ .. u_i-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 —> R^m is onto if its range is all of R^m (range = co-domain)
if every vector in R^m is an image under f