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

Question

$V=$ $M_{2 \times 2}(\mathbb{R})$ 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!

Answer 1

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.

Answer 2

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.