Find the Eigenvectors $T\left(\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\right)=\left[\begin{array}{ll}d & b \\ c & a\end{array}\right]$ $V=$ $M_{2 \times 2}(\mathbb{R})$ y $T\left(\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\right)=\left[\begin{array}{ll}d & b \\ c & a\end{array}\right]$
I think the matrix associated to T is
$A= $$\left[\begin{array}{ll}0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\1 & 0 & 0 & 0 \end{array}\right]$
Then we can get the eigenvalues if we swap $R_4$ to $R_1$
$A`= $$\left[\begin{array}{ll}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{array}\right]$
Then the eigenvalue is 1 but from here Im stuck to find the eigenvectors or is something wrong with my process? thank you!

The matrices $\begin{pmatrix} a & b\\ c & a\end{pmatrix}$ form the eigenspace of dim 3 of eigenvalue 1. The matrix $diag(1,-1)$ is an eigenvector with eigenvalue $-1$.  It spans a 1-dim space of eigenvectors. Since $1+3=4$ there are no other eigenvectors.
