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!
 A: The eigenvalues and eigenvectors of $T$ may be found directly from the given formula
$T \left ( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = \begin{bmatrix} d & b \\ c & a \end{bmatrix}, \tag 1$
for we have
$T^2 \left ( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = T \left ( \begin{bmatrix} d & b \\ c & a \end{bmatrix} \right ) =  \begin{bmatrix} a & b \\ c & d\end{bmatrix}; \tag 2$
thus,
$T^2 = I, \tag 3$
or
$T^2 - I = 0; \tag 4$
if $\mu$ is an eigenvalue of $T$, that is, if
$TZ = \mu Z\tag 5$
for some
$0 \ne Z \in M_{2 \times 2}(\Bbb R), \tag 6$
then
$T^2Z = T(TZ) = T(\mu Z) = \mu TZ = \mu (\mu Z) = \mu^2Z, \tag 7$
whence
$(\mu^2 - 1)Z = \mu^2 Z - Z = T^2 Z - Z = (T^2 - I)Z = 0, \tag 8$
so in light of (6) we have
$\mu^2 - 1 = 0, \tag 9$
which implies
$\mu = \pm 1; \tag{10}$
now if
$\mu = 1, \tag{11}$
$\begin{bmatrix} d & b \\ c & a \end{bmatrix} = T \left (\begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = 
\mu \begin{bmatrix} a & b \\ c & d \end{bmatrix} = 
\begin{bmatrix} a & b \\ c & d \end{bmatrix}, \tag{12}$
which forces
$a = d; \tag{13}$
an eigenmatrix for eigenvalue $1$ thus takes the form
$\begin{bmatrix} a & b \\ c & a \end{bmatrix}, \tag{14}$
where $a, b, c \in \Bbb R$ are arbitrary.  It is now easy to see that the $1$-eigenspace of $T$ is of dimension $3$.  On the other hand, when
$\mu = -1, \tag{15}$
$\begin{bmatrix} d & b \\ c & a \end{bmatrix} = T \left (\begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = 
\begin{bmatrix} -a & -b \\ -c & -d \end{bmatrix}, \tag{16}$
whence,
$a = - d \tag{16}$
$b = -b, \; c = -c ,\Longrightarrow b = c = 0; \tag{17}$
the eigenmatrix thus becomes
$\begin{bmatrix} a & 0 \\ 0 & -a \end{bmatrix}, \; a \in \Bbb R;  \tag{18}$
it is clear that the $-1$ eigenspace of $T$ is of dimension $1$.
Since the sum of the dimensions of the $1$ and $-1$ eigenspaces is
$4 = \dim M_{2 \times 2}(\Bbb R), \tag{19}$
we conclude there are no more eigenvectors/eigenvalues to be had.
A: 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.
