# Matrix representation of function composition

1. May 9, 2016

### Sociomath

Am I on the right path here?

1. The problem statement, all variables and given/known data

i. Prove that $T_{a}$ and $T_{b}$ are linear transformations.
ii. Compose the two linear transformations and show the matrix that represents that composition.

2. The attempt at a solution

i. Prove that $T_{a}$ and $T_{b}$ are linear transformations.
i. $T_{a} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}-x \\ x+y\end{bmatrix}$
$x =\begin{bmatrix}-1\\1\end{bmatrix}+y\begin{bmatrix}0\\1 \end{bmatrix}$
$\begin{bmatrix}-1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}$
$T_{a}$ = Linear transformation.

$T_{b} \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}x+y \\ x -y \end{bmatrix}$
$x \begin{bmatrix}1\\1 \end{bmatrix} + y \begin{bmatrix}1\\-1 \end{bmatrix} = \begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix}x\\y \end{bmatrix}$
$T_{b}$ = Linear transformation.

ii. Compose the two linear transformations and show the matrix that represents that composition.
$T_{a} {\circ} T_{b} = \left[T_{a}\right]\left[T_{e}\right] = \begin{bmatrix}-1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix}$
$= \begin{bmatrix}-1 & -1 \\ 2 & 0 \end{bmatrix}$

2. May 9, 2016

### andrewkirk

That looks correct. However, from the way the question is written, they expect you to not just produce the matrix but also state the the transformation in the same form as that in which the original two were given, ie this form
$$T_{a} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}-x \\ x+y\end{bmatrix}$$

3. May 9, 2016

### Sociomath

$\left[T_{a}\right]\left[T_{b}\right] = \begin{bmatrix}-1 & -1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix}-x -y\\ 2x \end{bmatrix}$

4. May 9, 2016

### andrewkirk

I wouldn't write it like that, because $[T_a]$is the matrix representation of the linear operator $T_a$, rather than the linear transformation itself. The transformation is $T_a\circ T_b$. So writing it the same way as that in which $T_a$ and $T_b$ were presented would be
$$(T_a\circ T_b)\begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix}-x -y\\ 2x \end{bmatrix}$$