Linear transformation matrix of vector space $\mathbb R^{2x2}$ 
Select a basis B of a vector space $\mathbb R^{2x2}$ and for linear transformation $f:\mathbb R^{2x2}→\mathbb R^{2x2}$ given by $f(X) = \begin{pmatrix}1 & 0 \\ 1 & 0 \end{pmatrix}X+\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}X^T$ compute the matrix relative to the base B.


I selected canonical basis $B=\{\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix},\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix},\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}\}$
Then I computed transformation of each vector in the basis according to the task.
Got $f(B)=\{\begin{pmatrix}2 & 0 \\ 2 & 0 \end{pmatrix},\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 1 \end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 1 \end{pmatrix}\}$, if I'm not wrong.
What should I do now?
 A: If it's indeed the case that
$$f\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} = 2 \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} + 2 \begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}$$
(I didn't check your calculations), then this would tell us that the first row of the matrix is $2, 0, 2, 0$, and so forth.
A: Speedding: your calculations are correct. Another way. If $X=\begin{pmatrix}{x_1}&{x_2}\\{x_3}&{x_4}\end{pmatrix}\in\mathbb{R}^{2\times 2}$ then, $$f(X)=\begin{pmatrix}1 & 0 \\ 1 & 0 \end{pmatrix}X+\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}X^T=\ldots=\begin{pmatrix}{2x_1+x_2}&{x_2+x_3+x_4}\\{2x_1+x_2}&{x_2+x_3+x_4}\end{pmatrix}.$$ But for $M=\begin{pmatrix}{a}&{b}\\{c}&{d}\end{pmatrix}$, its coordinates on $B$ are $(a,b,c,d)^T$ so, we can write $$f\begin{pmatrix}{x_1}\\{x_2}\\{x_3}\\ x_4\end{pmatrix}=\begin{pmatrix}{2x_1+x_2}\\{x_2+x_3+x_4}\\{2x_1+x_2}\\ x_2+x_3+x_4\end{pmatrix}=\underbrace{\begin{pmatrix}{2}&{1}&{0}&0\\{0}&{1}&{1}&1\\{2}&{1}&{0}&0\\{0}&{1}&{1}&1\end{pmatrix}}_{A_B^B}\begin{pmatrix}{x_1}\\{x_2}\\{x_3}\\ x_4\end{pmatrix}.$$
