Calculating Eigenvalues from two matrices Let $\alpha$ be the endomorphism given by
$\alpha$:$\mbox{} \left[ \begin{array}{cc}
a & b \\ c & d \end{array} \right]$$\rightarrow$ $\mbox{} \left[ \begin{array}{cc}
 d & -b \\ -c & a \end{array} \right]$
I need to find the eigenvalues and associate eigenspace. Since the determinants are equal, does that mean that the product of the eigenvalues is also the same? And which matrix ought I use to find the eigenvalues? 
 A: It is easy to see that $\alpha^2 = I$, from which it follows that $(\alpha -I)(\alpha + I) = 0$. Hence the set of eigenvalues is $\{\pm 1 \}$.
Choose the basis $e_1 = \pmatrix{ 1 && 0 \\ 0 && 0 }$, 
$e_2 = \pmatrix{ 0 && 1 \\ 0 && 0 }$, $e_3 = \pmatrix{ 0 && 0 \\ 1 && 0 }$, $e_4 = \pmatrix{ 0 && 0 \\ 0 && 1 }$. In this basis, $\alpha$ has the form $A = \pmatrix{ 0 && 0 && 0 && 1 \\ 0 && -1 && 0 && 0 \\0 && 0 && -1 && 0 \\ 1 && 0 && 0 && 0 }$.
The characteristic polynomial is easily computed to be $\det (\lambda I -A) = (\lambda+1)^3 (\lambda-1)$.
Also from $A$ we have $\alpha e_2 = -e_2$, $\alpha e_3 = - e_3$, $\alpha(e_1+e_4) = e_1+e_4$ and $\alpha(e_1-e_4) = -(e_1-e_4)$, which gives all the eigenvectors.
A: maybe relevant
$$
\pmatrix{
1 & 0 \\
0 & 1
}
\to
(+1)
\pmatrix{
1 & 0 \\
0 & 1
}
$$
$$
\pmatrix{
0 & 0 \\
1 & 0
}
\to
(-1)
\pmatrix{
0 & 0 \\
1 & 0
}
$$
$$
\pmatrix{
1 & 0 \\
0 & -1
}
\to
(-1)
\pmatrix{
1 & 0 \\
0 & -1
}
$$
$$
\pmatrix{
0 & 1 \\
0 & 0
}
\to
(-1)
\pmatrix{
0 & 1 \\
0 & 0
}
$$
I would guess that one of the eigenvalues is 1 which corresponds to the 1-dimensional eigenspace
$$
\pmatrix{
x & 0 \\
0 & x
}
$$
and that the other eigenvalue is -1 which corresponds to the 3-dimensional eigenspace
$$
\pmatrix{
y & z \\
w & -y
}
$$
