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A problem from my Linear Algebra textbook:

Let $T$ be a linear transformation from $M_{2,2}$ into $M_{2,2}$ such that

$T\begin{pmatrix} \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix}\end{pmatrix} = \begin{bmatrix} 1&-1 \\ 0&2 \end{bmatrix}$,  $T\begin{pmatrix} \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}\end{pmatrix} = \begin{bmatrix} 0&2 \\ 1&1 \end{bmatrix}$, 

$T\begin{pmatrix} \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix}\end{pmatrix} = \begin{bmatrix} 1&2 \\ 0&1 \end{bmatrix}$,  $T\begin{pmatrix} \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}\end{pmatrix} = \begin{bmatrix} 3&-1 \\ 1&0 \end{bmatrix}$.

Find $T\begin{pmatrix} \begin{bmatrix} 1&3 \\ -1&4 \end{bmatrix}\end{pmatrix}$

I just need to know how to start this. I tried setting it up like this:

$\begin{bmatrix} a&b \\ c&d \end{bmatrix}$ $\begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix}$ = $\begin{bmatrix} 1&-1 \\ 0&2 \end{bmatrix}$

which gets me to here:

$\begin{bmatrix} a&0 \\ c&0 \end{bmatrix}$ = $\begin{bmatrix} 1&-1 \\ 0&2 \end{bmatrix}$

and then I'm lost...

1 Answers 1

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Use linearity:

$$T\left(\sum_{i=1}^n \alpha_i A_i\right) = \sum_{i=1}^n \alpha_i T(A_i)$$

Also, note that a linear transformation from $M_{2,2}$ to $M_{2,2}$ is not a $2\times 2$ matrix itself.

Finally, note that $$\begin{align*} \pmatrix{1&3\\-1&4} & = \pmatrix{1&0\\0&0} + 3 \pmatrix{0&1\\0&0} - \pmatrix{0&0\\1&0} + 4 \pmatrix{0&0\\0&1} \\ \Rightarrow T\pmatrix{1&3\\-1&4} & = T\pmatrix{1&0\\0&0} + 3T\pmatrix{0&1\\0&0} - T\pmatrix{0&0\\1&0} +4T\pmatrix{0&0\\0&1}\\ & = \pmatrix{1&-1\\0&2} + 3\pmatrix{0&2\\1&1} - \pmatrix{1&2\\0&1} + 4\pmatrix{3&-1\\1&0} \\ & = \pmatrix{12&-1\\7&4} \end{align*}$$