Let $A =\begin{pmatrix} a & b \\ c & d\end{pmatrix}$. Then define $T(A)=\begin{pmatrix} b & d \\ a & c\end{pmatrix}$ and $S(A)=\begin{pmatrix} a& d \\ c& b\end{pmatrix}$. Also $T,S \in \mathcal{L}(M_{2 \times 2}(R))$.
Define $ \pi(A) = \begin{pmatrix} tr(A)& (tr\circ T)A \\ (tr\circ S)A & (tr\circ T\circ S)A\end{pmatrix}$, $\pi \in \mathcal{L}(M_{2 \times 2}(R))$
Given the bases matrix, $ \delta= \bigg\{\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}\bigg\} $, I am interested in the matrix form of $[\pi]_\delta^\delta$ and its implication.
$\underline{Attempt:}$
I apply $\pi$ on each of the given basis.
$\pi\begin{pmatrix}1&0\\0&0\end{pmatrix} = \begin{pmatrix}1&0\\1&0\end{pmatrix}$, $\pi\begin{pmatrix}0&1\\0&0\end{pmatrix} = \begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\pi\begin{pmatrix}0&0\\1&0\end{pmatrix} = \begin{pmatrix}0&1\\0&1\end{pmatrix}$, $\pi\begin{pmatrix}0&0\\0&1\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}$
Then : $[\pi]_\delta^\delta = \begin{pmatrix}1 & 0 &0&1 \\ 1&0&1&0 \\ 0&1&1&0 \\0&1&0&1 \end{pmatrix}$
I tried using $ A=\begin{pmatrix}2&3\\4&5\end{pmatrix}$, to check the computation.
$\pi \begin{pmatrix}2&3\\4&5\end{pmatrix} = \begin{pmatrix}7&7\\5&9\end{pmatrix} $
Let $ \begin{pmatrix}1 & 0 &0&1 \\ 1&0&1&0 \\ 0&1&1&0 \\0&1&0&1 \end{pmatrix} = \begin{pmatrix}A_{2\times2} &B_{2\times2} \\ C_{2\times 2} & D_{2\times 2} \end{pmatrix} $
Then $ \pi A^T = \begin{pmatrix}A_{2\times2} &B_{2\times2} \\ C_{2\times 2} & D_{2\times 2} \end{pmatrix} \begin{pmatrix}2&3\\4&5\end{pmatrix}^T = \begin{pmatrix}7&7\\5&9\end{pmatrix} $
Does this attempt make any sense? What is the implication of this transformation? I observe the $[\pi]_\delta^\delta$ is not invertible, will this imply the transformation is not invertible?
Thank you!
$\textbf{Edit}$
Please if you find this post not relevant, I would be so happy to learn from you why this is the case. There were first instances where I typed something different from what was in the note and I have corrected this. I came up with the example case because I found no other numerical example online and I am willing to learn more why this works and the intuitive idea behind it. Numerical errors can be corrected but I am more interested in the why? Thank you so much !
I think what I was looking for was rewriting A in terms of the bases:
$ A = \begin{pmatrix}2&3\\4&5\end{pmatrix} $
as:
$ A = \begin{pmatrix}2&3\\4&5\end{pmatrix} = 2 \begin{pmatrix}1& 0\\0&0\end{pmatrix} + 3\begin{pmatrix}0&1\\0&0\end{pmatrix} + 4 \begin{pmatrix}0&0\\1&0\end{pmatrix} + 5 \begin{pmatrix}0&0\\0&1\end{pmatrix} $
Now, By linearity of $\pi$, it follows that:
$ \pi(A) = \pi \bigg(2 \begin{pmatrix}1& 0\\0&0\end{pmatrix} + 3\begin{pmatrix}0&1\\0&0\end{pmatrix} + 4 \begin{pmatrix}0&0\\1&0\end{pmatrix} + 5 \begin{pmatrix}0&0\\0&1\end{pmatrix} \bigg) $
This is equivalent to (by linearity property):
$ \pi(A) = 2 \cdot \pi\begin{pmatrix}1& 0\\0&0\end{pmatrix} + 3 \cdot \pi \begin{pmatrix}0&1\\0&0\end{pmatrix} + 4 \cdot \pi\begin{pmatrix}0&0\\1&0\end{pmatrix} + 5 \cdot \pi \begin{pmatrix}0&0\\0&1\end{pmatrix} $
Observe that, the $\pi(\cdot)$, in the computation above are the entries of:
$$[\pi]_\delta^\delta = \begin{pmatrix}1 & 0 &0&1 \\ 1&0&1&0 \\ 0&1&1&0 \\0&1&0&1 \end{pmatrix}$$.
This establishes why $\pi(A)$ and $\pi A^T $, gives the same answer.
It is then obvious then why the $[\pi]_\delta^\delta$ is the transformation matrix.