How do you find the $4\times 4$ matrix corresponding to the transformation T with respect to the basis?

Question

If the transformation $T$ acting on the vector space $A \in Mat_{2,2}$ is given by $T(A)=CA$, where $ C= \left( \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right) $ how would you find the $4\times 4$ matrix corresponding to $T$ with respect to the basis of $Mat_{2,2}$?

Answer 1

Hint:

If we take the standard basis $\Biggl\{\begin{pmatrix}1&0\\0&0\end{pmatrix}, \begin{pmatrix}0&1\\0&0\end{pmatrix}, \begin{pmatrix}0&0\\1&0\end{pmatrix}, \begin{pmatrix}0&0\\0&1\end{pmatrix}\Biggr\}$ for $M_{2,2}$,

then $T(v_1)=\begin{pmatrix}1&0\\3&0\end{pmatrix}, T(v_2)=\begin{pmatrix}0&1\\0&3\end{pmatrix}, T(v_3)=\begin{pmatrix}2&0\\4&0\end{pmatrix}, T(v_4)=\begin{pmatrix}0&2\\0&4\end{pmatrix}$.

Now write each of these matrices as linear combinations of $v_1,\cdots,v_4$, and use the coefficients to get the columns of the matrix of T with respect to this basis.