Find the Eigenvectors $Tleft(left[begin{array}{ll}a & b \ c & dend{array}right]right)=left[begin{array}{ll}d & b \ c & aend{array}right]$

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

$T^2 left ( begin{bmatrix} a & b \ c & d end{bmatrix} right ) = T left ( begin{bmatrix} d & b \ c & a end{bmatrix} right ) = begin{bmatrix} a & b \ c & dend{bmatrix}; tag 2$

thus,

$T^2 = I, tag 3$

or

$T^2 - I = 0; tag 4$

if $mu$ is an eigenvalue of $T$, that is, if

$TZ = mu Ztag 5$

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.

The matrices $begin{pmatrix} a & b\ c & aend{pmatrix}$ 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.