<s>[INST] Matrix representation of square matrix function with respect to a matrix basis The mapping $f$ from $V$ to $V$ of the vector space on $\mathbb{C}$ formed by the complex square matrices,

\begin{align}
f(x) = \begin{pmatrix}
3& 4&\\
-2& -3
\end{pmatrix}\begin{pmatrix}X\end{pmatrix}\begin{pmatrix}
1&2\\
-1&-1
\end{pmatrix} \qquad\qquad X\in V
\end{align}
Find the matrix representation $A$ of $f$ with respect to the basis $e_1=\begin{pmatrix}1&0\\0&0\end{pmatrix},\,\, e_2=\begin{pmatrix}0&1\\0&0\end{pmatrix},\,\, e_3=\begin{pmatrix}0&0\\1&0\end{pmatrix},\,\, e_4=\begin{pmatrix}0&0\\0&1\end{pmatrix}$ 0f $V$ 
Note: I tried to find the matrix $A$, but stuck after find the $f(e_1),\,f(e_2),\,f(e_3),\, f(e_4)$. How to construct the matrix $A$ (it cannot find with the traditional way like vectors or polynomials because $f(e_i)$`s are also matrices) please give an idea. Thank you
 [/INST] To elaborate a bit on the comment: by calculating $f(e_1),\dots,f(e_4)$, you have done the "hard part".  From there, we use each of these outputs to build the columns of our matrix.
For example, we find that plugging in the second basis element yields
$$
f(e_2) = \pmatrix{-3&-3\\2&2} = -3e_1 + -3e_2 + 2e_3 + 2e_4.
$$
Thus, the matrix $A$ of your transformation $f$ relative to the given basis will have the form
$$
A = \pmatrix{
?&-3&?&?\\
?&-3&?&?\\
?&2&?&?\\
?&2&?&?
}.
$$
Ultimately, you should find that
$$
A = \begin{pmatrix}
3& 4\\
-2& -3
\end{pmatrix}
\otimes
\begin{pmatrix}
1&2\\
-1&-1
\end{pmatrix}^T = \left(\begin{array}{cccc} 3 & -3 & 4 & -4\\ 6 & -3 & 8 & -4\\ -2 & 2 & -3 & 3\\ -4 & 2 & -6 & 3 \end{array}\right).
$$
</s>