Linear transformation: Change of basis

Question

I am given the following linear transformation $L$:

$A=\begin{bmatrix}1&2\\0&3\end{bmatrix} \in \Bbb R^{2 \times 2}$

$L: \space \Bbb R^{2 \times 2} \longrightarrow \Bbb R^{2 \times 2}; \space X \mapsto AX$

I want to find the transformation matrix with respect to the basis

$\mathcal B_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}, \space \mathcal B_2=\begin{bmatrix}0&0\\1&0\end{bmatrix}, \space \mathcal B_3=\begin{bmatrix}0&1\\0&0\end{bmatrix}, \space \mathcal B_4=\begin{bmatrix}0&0\\0&1\end{bmatrix}$

I know the answer is: $M_{\mathcal B}(L)=\begin{bmatrix}1&2&0&0\\0&3&0&0\\0&0&1&2\\ 0&0&0&3\end{bmatrix}$

but I don't know how to get to that matrix.

Usually I would find the new transformation $M$ with respect to a basis $\mathcal B$ by computing:

$$M=C^{-1}AC$$

where $C$ is the matrix that has the alternate basis vectors $b_1,...,b_n$ as its columns. However, in this case, my matrix $C$ would look like this:

\begin{bmatrix}1&0&0&0&0&1&0&0\\0&0&1&0&0&0&0&1\end{bmatrix}

which makes no sense at all. What am I doing wrong here?

Answer 1

The first column of $M_\mathcal{B}(L) = (m_{ij})_{1\le i,j\le 4}$ is defined as the coordinates (with respect to the basis $\mathcal{B}=(\mathcal{B}_1,\mathcal{B}_2,\mathcal{B}_3,\mathcal{B}_4)$) of $A\mathcal{B}_1$.

Since $$\begin{array}{rcl} A\mathcal{B}_1 &=& \begin{bmatrix}1&2\\0&3\end{bmatrix} \begin{bmatrix}1&0\\0&0\end{bmatrix} \\ &=& \begin{bmatrix}1&0\\0&0\end{bmatrix} \\ &=& 1\times \mathcal{B}_1 + 0\times \mathcal{B}_2+ 0\times \mathcal{B}_3+ 0\times \mathcal{B}_4 \\ A\mathcal{B}_1&=& m_{11}\times \mathcal{B}_1 + m_{21}\times \mathcal{B}_2+ m_{31}\times \mathcal{B}_3+ m_{41}\times \mathcal{B}_4. \end{array}$$ This gives us the first column. I let you do the same for the rest of the matrix.