let $T: M_{2,2} \rightarrow M_{2,2}$ be a linear transformation defined by:
$$T \left(\begin{bmatrix}
a & b\\
c & d\\
\end{bmatrix}\right) = \begin{bmatrix}a + b& b + a \\ c – d&d+b\end{bmatrix}
$$
Find the standard matrix for $S$ by using the standard basis of $M_{2,2}$
This is what i've done so far:
$$T \left(\begin{bmatrix}
1 & 0\\
0 & 0\\
\end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&0\end{bmatrix} = x_1
$$
$$T \left(\begin{bmatrix}
0 & 1\\
0 & 0\\
\end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&1\end{bmatrix} = x_2
$$
$$T \left(\begin{bmatrix}
0 & 0\\
1 & 0\\
\end{bmatrix}\right) = \begin{bmatrix}0&0\\ 1&0\end{bmatrix} = x_3
$$
$$T \left(\begin{bmatrix}
0 & 0\\
0 & 1\\
\end{bmatrix}\right) = \begin{bmatrix}0& 0\\ -1&1\end{bmatrix} = x_4
$$
But I'm not too sure where to go from here. All I know that these need to somehow become part of a larger matrix
Best Answer
Let $v_1,v_2,v_3,v_4$ denote the standard basis. Note that $$ T(v_1) = T \left(\begin{bmatrix} 1 & 0\\ 0 & 0\\ \end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&0\end{bmatrix} = 1v_1 + 1v_2 +0v_3 + 0v_4\\ T(v_2) = T \left(\begin{bmatrix} 0 & 1\\ 0 & 0\\ \end{bmatrix}\right) = \begin{bmatrix}1& 1 \\ 0&1\end{bmatrix} = 1v_1 + 1v_2 + 0v_3 + 1v_4\\ T(v_3) = T \left(\begin{bmatrix} 0 & 0\\ 1 & 0\\ \end{bmatrix}\right) = \begin{bmatrix}0&0\\ 1&0\end{bmatrix} = 0v_1 + 0v_2 + 1v_3 + 0v_4\\ T(v_4) = T \left(\begin{bmatrix} 0 & 0\\ 0 & 1\\ \end{bmatrix}\right) = \begin{bmatrix}0& 0\\ -1&1\end{bmatrix} = 0v_1 + 0v_2 + (-1)v_3 + 1v_4 $$ It follows that the matrix of the transformation is given by $$ S = \pmatrix{ 1&1&0&0\\ 1&1&0&0\\ 0&0&1&-1\\ 0&1&0&1 } $$