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

Mathematics Asked on December 20, 2020

$$V=$$ $$M_{2 times 2}(mathbb{R})$$ y $$Tleft(left[begin{array}{ll}a & b \ c & dend{array}right]right)=left[begin{array}{ll}d & b \ c & aend{array}right]$$

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 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.

Answered by Robert Lewis on December 20, 2020

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.

Answered by JCAA on December 20, 2020