Understanding the solution of a question in chapter 8 of Golan's "Linear algebra" The question is given below:



And a part its solution is given below:



But I do not understand how $\phi_{DD}(\alpha)$ from the calculated values of $\alpha$ on the standard basis. and how the dimension of$\phi_{DD}(\alpha)$ is calculated   could anyone help me understand this, please?
Note: I have asked a similar question here Understanding an example in Golan's "Linear Algebra"  and I know the answer if we have 3 three $\times$ 1 matrices.
 A: To follow from my answer to your previous question, we have $V = W = \mathcal{M}_{2 \times 2}(\Bbb{R})$ equipped with bases
$$B = D = \left(\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}, \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}\right).$$
Note that this is a basis for the space, in that any real $2 \times 2$ matrix can be expressed uniquely as a linear combination:
$$\begin{bmatrix} a & b \\ c & d \end{bmatrix} = a\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} + b\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} + c\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} + d\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}.$$
The dimension of $V$, i.e. the number of vectors in every basis of $V$, is therefore $4$. Any linear transformation from $V$ to itself, when changed to a matrix, is naturally going to be a $4 \times 4$ matrix; that's the reason for the dimensions of $\Phi_{DD}$.
Your solution then follows step 1: to compute $\alpha$ applied to each basis vector. It then brushes over step 2, where these vectors are to be expressed as coordinate column vectors with respect to $D$. Let's fill in one of these steps. For example, we have
$$\alpha\left(\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\right) = \begin{bmatrix}a & b \\ 0 & 0\end{bmatrix} = a\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} + b\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} + 0\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} + 0\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix},$$
hence
$$\left[\alpha\left(\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\right)\right]_D = \begin{bmatrix}a \\ b \\ 0 \\ 0\end{bmatrix}.$$
If you do the same with the other columns, then follow step 3, you'll get the matrix from the solution.
