# 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!

• This is a pretty special (simple) transformation, so rather than converting this to a problem on $\mathbb R^4$, by playing around with filling out 2x2 matrices with 1s and 0s, I was able to find a bunch of (linearly independent) matrices for which $T(X) = X$. And by adding a -1 into the mix I was able to find another matrix for which $T(X) = -X$. Maybe just try to come up with some yourself? Sep 19, 2020 at 3:29
• Doing row operations changes eigenvectors as this example demonstrates. Sep 19, 2020 at 15:19

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.

• This answer is the same as mine and came several hours later. This is called "plagiarizing". Sep 22, 2020 at 23:24

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.