Consider the vector space of,$\mathbb{R}^{2\times 2}$, and the linear operator,$L_{A,B}:\mathbb{R}^{2\times 2}\to\mathbb{R}^{2\times 2}$, given by,$L_{A,B}(X)=(AX)+(XB)$
Where,$A=\begin{bmatrix}
1&3\\
2&4\\
\end{bmatrix}$,and,$B=\begin{bmatrix}
6&5\\
7&8\\
\end{bmatrix}$
Find the matrix representation of,$L_{A,B}$, with respect to the basis:
$(\begin{bmatrix}
1&0\\
0&0\\
\end{bmatrix},\begin{bmatrix}
0&1\\
0&0\\
\end{bmatrix},\begin{bmatrix}
0&0\\
1&0\\
\end{bmatrix},\begin{bmatrix}
0&0\\
0&1\\
\end{bmatrix})$
I defined X as:,$X=\begin{bmatrix}
x_1&x_2\\
x_3&x_4\\
\end{bmatrix}$
Then I found:,$L_{A,B}(X)=x_1\begin{bmatrix}
7&5\\
2&0\\
\end{bmatrix}+x_2\begin{bmatrix}
7&9\\
0&2\\
\end{bmatrix}+x_3\begin{bmatrix}
3&0\\
10&5\\
\end{bmatrix}+x_4\begin{bmatrix}
0&3\\
7&12\\
\end{bmatrix}$
But I have no idea how to go further. I know that there are several examples about this topic, but i do not understand those problems, and I hope that when I solved this problem, I get it. Can someone help me with this problem?
Best Answer
Let's call $\mathcal{B} = (e_1,e_2,e_3,e_4)$ your basis.
In order to find the matrix representation of $L_{A,B}$ all you have to do is calculating $L_{A,B}(e_i)$ for each vector $e_i$ of your basis and then express this in your basis.
Let's give it a try:
$L_{A,B}\left(\begin{bmatrix} 1&0\\ 0&0\\ \end{bmatrix}\right) = \begin{bmatrix} 7&5\\ 2&0\\ \end{bmatrix} = 7 \times \begin{bmatrix} 1&0\\ 0&0\\ \end{bmatrix}+5\times \begin{bmatrix} 0&1\\ 0&0\\ \end{bmatrix}+2\times \begin{bmatrix} 0&0\\ 1&0\\ \end{bmatrix} + 0 \times \begin{bmatrix} 0&0\\ 0&1\\ \end{bmatrix}$
So the first column of $\text{Mat}_{\mathcal{B}}(L_{A,B})$ will be
$$\begin{pmatrix} 7 \\ 5 \\ 2 \\ 0 \\ \end{pmatrix}$$
Do you need more explanation?