week6

Standard Linear Transforms

Examples of linear transformation matrices In two-dimensional space R2 linear maps are described by 2 × 2 real matrices. These are some examples:

• rotation by 90 degrees counterclockwise:

\(\mathbf{A}=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\)

• rotation by angle θ counterclockwise:

\(\mathbf{A}=\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\)

• reflection against the x axis:

\(\mathbf{A}=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\)

• reflection against the y axis:

\(\mathbf{A}=\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}\)

• scaling by 2 in all directions:

\(\mathbf{A}=\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}\)

• horizontal shear mapping:

\(\mathbf{A}=\begin{pmatrix}1 & m\\ 0 & 1\end{pmatrix}\)

• squeeze mapping:

\(\mathbf{A}=\begin{pmatrix}k & 0\\ 0 & 1/k\end{pmatrix}\)

• projection onto the y axis:

\(\mathbf{A}=\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}\)

Determine a Transform Matrix based upon results

\(B = LA\)

• L is the transform matrix
• A is the input matrix

Using the unit vector matrix we determine the transform matrix

\(B = L (\begin{pmatrix}1 \\ 0 \end{pmatrix}) = \begin{pmatrix}3 \\ 5 \end{pmatrix}\)

\(B = L (\begin{pmatrix}0 \\ 1 \end{pmatrix}) = \begin{pmatrix}2 \\ -1 \end{pmatrix}\)

\(L = \begin{pmatrix}3 & 2 \\ 5 & -1 \end{pmatrix}\)

Evaluate a transform

Given above calculate
\(B = L (\begin{pmatrix}2 \\ 3 \end{pmatrix}) = ?\)

Method 1

\(B = \begin{pmatrix}3 & 2 \\ 5 & -1 \end{pmatrix} \begin{pmatrix}2 \\ 3 \end{pmatrix}\) \(= \begin{pmatrix}3 & 2 \\ 5 & -1 \end{pmatrix} \begin{pmatrix}2 \\ 3 \end{pmatrix}\)

L=matrix(c(3,5,2,-1),ncol=2)
# does not matter this is "row vector". The multiply matrix will change it 
# to the column vector
A=c(2,3)
B=L%*%A
> A
[1] 2 3
> L
     [,1] [,2]
[1,]    3    2
[2,]    5   -1
> B
     [,1]
[1,]   12
[2,]    7

Method 2

In this case we do not solve for L given the unit vector results and then apply L to the input vector. Here we solve for B directly given the results of scaling the unit vector results.

\(B = L (\begin{pmatrix}2 \\ 3 \end{pmatrix}) = ?\)

\(B = L (2\begin{pmatrix}1 \\ 0 \end{pmatrix} + 3\begin{pmatrix}0 \\ 1 \end{pmatrix}) = ?\)

\(B = 2L(\begin{pmatrix}1 \\ 0 \end{pmatrix}) + 3L(\begin{pmatrix}0 \\ 1 \end{pmatrix}) = ?\)

\(B = 2\begin{pmatrix}3 \\ 5 \end{pmatrix} + 3\begin{pmatrix}2 \\ -1 \end{pmatrix} = ?\)

\(B = \begin{pmatrix}6 \\ 10 \end{pmatrix} + \begin{pmatrix}6 \\ -3 \end{pmatrix} = ?\)

\(B = \begin{pmatrix}12 \\ 7 \end{pmatrix}\)

Solving Linear Equations

Instead of doing the substitution method to solve n-variables and n-equations, arrange the factors in a matrix to solve for a upper triangular matrix.

Example

\(\begin{pmatrix}2 & 4 & -2 \\ 4 & -2 & 6 \\ 6 & -4 & 2 \end{pmatrix} \begin{pmatrix}x_0 \\ x_1 \\ x_2 \end{pmatrix} =\begin{pmatrix}-10 \\ 20\\ 18 \end{pmatrix}\)

We try to form the upper diagonal so that the last row is just one factor times \(x_2\) so its trivial to solve for \(x_2\).

Take each coefficient below the diagonal and divide it by the diagonal to form the factor to reduce that row. Then multiply each term in the diagonal row by the factor and subtract from the existing values in the reducing row.

or

New Row 2 element = Row 2 element minus Row 1 element times Factor.
Factor = Row 2 elimination element / Row 1 keep element

or

Row2,column1 = 4-(2)(4/2) = 4-4=0
Row2,column2 = -2-(4)(4/2) = -2-8=-10
Row2,column3 = 6-(-2)(4/2) = 6+4 = 10
Result2,column1 = 20-(-10)(4/2) = 20+20 = 40

\(\begin{pmatrix}2 & 4 & -2 \\ 0 & -10 & +10 \\ 0 & -16 & 8 \end{pmatrix} \begin{pmatrix}x_0 \\ x_1 \\ x_2 \end{pmatrix} =\begin{pmatrix}-10 \\ 40\\ 48 \end{pmatrix}\)

Repeat for the middle diagonal.

\(\begin{pmatrix}2 & 4 & -2 \\ 0 & -10 & +10 \\ 0 & 0 & -8 \end{pmatrix} \begin{pmatrix}x_0 \\ x_1 \\ x_2 \end{pmatrix} =\begin{pmatrix}-10 \\ 40\\ -16 \end{pmatrix}\)

Representing systems of linear equations as appended matricies

Reducing matrixcies to row echelon form (upper triangle matrix?)

LU Factorization