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
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