MAT223 Lecture 17

Activity 1
- $T (\vec{x}) = [\begin{matrix} 3 x + y \\ x + 2 y \end{matrix}]$
- Find $A_{T}$
  - $A_{T} = [\begin{array}{cc} 3 & 1 \\ 1 & 2 \end{array}]$
- Compute $det A_{T}$
  - $det [\begin{array}{cc} 3 & 1 \\ 1 & 2 \end{array}] = (6 - 1) = 5$
- Sketch a visualization of $T$ showing the parallelogram bounded by $T ({\vec{e}}_{1}), T ({\vec{e}}_{2})$ . The fundamental parallegram is the parallelogram bounded by $T ({\vec{e}}_{1}), T ({\vec{e}}_{2})$ .
```
left=-5; right=5;
top=5; bottom=-5;
---
(0,1)
(1,0)
(1,2)
(3,1)
x=0|0<=y<=1
x=1|0<=y<=1
y=0|0<=x<=1
y=1|0<=x<=1
y=2x|0<=x<=1
y=(1/3)x|0<=x<=3
y=2(x-3)+1|3<=x<=4
y=(1/3)(x-1)+2|1<=x<=4
```
- Find the area:
  - $| \vec{u} \times \vec{v} |$
  - $| \begin{matrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 0 \\ 3 & 1 & 0 \end{matrix} |$
  - $= \vec{i} [(2) (0) - (0) (1)] - \vec{j} [(1) (0) - (0) (3)] + \vec{k} [(1) (1) - (2) (3)] = - 5 \vec{k}$
  - $| \vec{u} \times \vec{v} | = 5$
- See that $A_{T} = | \vec{u} \times \vec{v} |$
- We also get that with: $S (\vec{x}) = [\begin{matrix} x + 2 y \\ 2 x - y \end{matrix}]$
  - $det A_{S} = - 5$ but the area is $5$
  - The determinant can be negative but the area is always positive.
Q5
- Suppose $T : R^{n} - R^{n}$ is linear. $\vec{v}$ is an eigenvector for $A_{T}$ with $λ = - 3$
- What happens to $\vec{v}$ when we plug it into the transformation $T$
- $A \vec{v} = λ \vec{v}$
- So it gets stretched by 3 and reflected
6
- $T (\vec{x}) = [\begin{matrix} x + 2 y \\ y - x \\ y \end{matrix}]$
- $S (\vec{a}) = [\begin{matrix} a + b + c \\ 2 a - b - c \end{matrix}]$
- $S \circ T$
- $S (T (\vec{x}))$
- $S (A_{T} \vec{x})$
- $A_{S} A_{T} \vec{x}$
- $A_{S} = [\begin{array}{ccc} 1 & 1 & 1 \\ 2 & - 1 & - 1 \end{array}]$
- $A_{T} = [\begin{array}{cc} 1 & 2 \\ - 1 & 1 \\ 0 & 1 \end{array}]$
- $A_{S} A_{T} = [\begin{array}{cc} (1) (1) + (1) (- 1) & (1) (2) + 1 + 1 \\ (2) (1) + 1 & 4 - 1 - 1 \end{array}]$
- $A_{S} A_{T} = [\begin{array}{cc} 0 & 4 \\ 3 & 2 \end{array}]$