Matrix repesentation of a linear operator with respect to its basis 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?
 A: 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?
A: Write your basis as $e_1,e_2,e_3,e_4$. Then $X = x_1e_1+x_2e_2+x_3e_3+x_4e_4$. You've already written out an expression of $L_{A,B}(X)$--from here you just need rewrite the output in terms of the basis, and the coordinate representation of the output of the $i$th basis vector tells us what the $i$th column of the matrix will be. For example, $L_{A,B}(e_1) = 7e_1+5e_2+2e_3+0e_4$. So the first column of the matrix for $L_{A,B}$ with respect to this basis will be $(7,5,2,0)^T$. Applying similar reasoning for the other basis vectors, we get
$$M = \begin{bmatrix}
7 & 7 & 3 & 0 \\
5 & 9 & 0 & 3 \\
2 & 0 & 10 & 7 \\
0 & 2 & 5 & 12
\end{bmatrix}$$
as the matrix representation for $L_{A,B}$. This looks a bit wonky, since our inputs $X$ should be $2\times 2$ matrices, but keep in mind we have to translate $X$ into our basis before applying $M$, turning $X$ into a $4\times 1$ column vector. The dimensions make sense since $M_2(\Bbb R)\cong \Bbb R^4$, so the matrix representation $M$ for $L_{A,B}$ should be $4\times 4$.
A: You have done the correct computation, now it's just a matter of writing this answer in terms of a matrix.
One approach is to build the matrix column-by-column, as I have done on this post. Alternatively, we can build the matrix one row at a time: we have
$$
T(X) = \pmatrix{y_1&y_2\\y_3&y_4},
$$
where for intance $y_1 = 7x_1 + 7x_2 + 3x_3 + 0x_4$. Our goal is to find a matrix $M$ for which
$$
M \pmatrix{x_1\\x_2\\x_3\\x_4} = \pmatrix{y_1\\y_2\\y_3\\y_4}.
$$
We can find the first row of $M$ based on $y_1$. In particular, the matrix of the transformation will have the form
$$
M = \pmatrix{7&7&3&0\\?&?&?&?\\?&?&?&?\\?&?&?&?}
$$
