<s>[INST] Show that the matrices' linear transformations are similar. Let be $T: M_2(\mathbb{C}) \longrightarrow M_2(\mathbb{C})$ the following linear transformation $$ T\left( \begin{array}{cc}
x & y\\
z & w\\
\end{array} \right) = \left( \begin{array}{cc}
0 & x\\
z-w & 0\\
\end{array} \right) $$
a) Determine the matrix $T$ concerning to canonical basis.
b) Determine the matrix $T$ concerning to the following basis 
$$ B = \left(  \begin{array}{cc} 
1 & 0\\
0 & 1 \end{array} \right), \left(  \begin{array}{cc} 
0 & 1\\
1 & 0 \end{array} \right), \left(  \begin{array}{cc} 
1 & 0\\
1 & 1 \end{array} \right), \left(  \begin{array}{cc} 
0 & 1\\
0 & 1 \end{array} \right) $$
of $M_2(\mathbb{R})$.
c) Show the matrix $M$ such that $[T]_B = M^{-1}[T]_{can}M$.
My attempt:
a) $[T]_{can} = \left( \begin{array}{rrrr}
0 & 0 & 0 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 1 & -1\\
0 & 0 & 0 & 0\\
 \end{array} \right)$
b) $[T]_B = \left( \begin{array}{rrrr}
2 & -1 & 1 & 1\\
1 & 0 & 1 & 0\\
-2 & 1 & -1 & -1\\
0 & 0 & 0 & 0\\
 \end{array} \right)$
c) $[T]_{B,can} = \left( \begin{array}{rrrr}
1 & 0 & 0 & 1\\
0 & 1 & 1 & 0\\
1 & 0 & 1 & 1\\
0 & 1 & 0 & 1\\
 \end{array} \right)$
and $[T]_{B,can}^{-1} = \left( \begin{array}{rrrr}
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
0 & 1 & 1 & 0\\
1 & 0 & 1 & 1\\
 \end{array} \right)$, so $M = [T]_{B,can}$.
My doubt is how find the matrix $M$ without know about eigenvalues and eigenvectors and if my answers in $a$ and in $b$ are correct, thanks in advance!
EDIT: I edited my attempt, I think it's correct now.
 [/INST] Your answer for the first one is correct. I didn't check the second one fully but I think you made some mistakes. For instance, writing $B = (b_1,\ldots,b_4)$, we have $$T(b_1) = T\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}0&1\\-1&0\end{pmatrix} \neq \begin{pmatrix}1&0\\0&1\end{pmatrix} - \begin{pmatrix}1&0\\1&1\end{pmatrix} = b_1 - b_3,$$ so the first column can't possibly be $(1,0,-1,0)$.
For the last one, you are looking for the change of basis matrix from $B$ to $\mathit{can}$. The columns of this matrix simply consist of the coordinates (in the canonical basis) of the basis vectors in $B$. For instance, since $b_1 = e_1 + e_4$, the first column is $(1,0,0,1)$. (Here $\mathit{can} = (e_1,\ldots,e_4)$.)
</s>