# 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.

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.