Possible eigenvalues of a matrix $AB$ Let matrices $A$, $B\in{M_2}(\mathbb{R})$, such that $A^2=B^2=I$, where $I$ is identity matrix. 
Why can be numbers $3+2\sqrt2$ and $3-2\sqrt2$  eigenvalues for the Matrix $AB$? 
Can  be numbers $2,1/2$  the eigenvalues of matrix $AB$? 
 A: Set
$$
A=\left(\begin{matrix}0 & 3-2\sqrt{2} \\ 3+2\sqrt{2} & 0\end{matrix}\right),\quad
B=\left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right).
$$
Then
$$
AB=\left(\begin{matrix} 3-2\sqrt{2} & 0 \\ 0 & 3+2\sqrt{2}\end{matrix}\right).
$$
The eigenvalues of $A,B$ are $\pm 1$, and hence $A^2=B^2=I$, while the eigenvalues of $AB$
are $3\pm2\sqrt{2}$.
Next, set
$$
A=\left(\begin{matrix}0 & 2 \\ 1/2 & 0\end{matrix}\right),\quad
B=\left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right).
$$
Then
$$
AB=\left(\begin{matrix} 2 & 0 \\ 0 & 1/2\end{matrix}\right).
$$
The eigenvalues of $A,B$ are $\pm 1$, and hence $A^2=B^2=I$, while the eigenvalues of $AB$
are $2,1/2$.
In particular, every pair $a,1/a$ can be eigenvalues of $AB$!
A: Note that
$$
\pmatrix{1&0\\0&-1}
\pmatrix{a& b\\-(a^2-1)/b & -a} =
\pmatrix{
a & b\\
(a^2 - 1)/b & a
}
$$
Every such product is similar to a matrix of this form or a triangular matrix with 1s on the diagonal. The associated characteristic polynomial is
$$
x^2 - 2ax + 1
$$
Check the possible roots of this polynomial. The product of two such entries must have eigenvalues of the form
$$
\lambda = a \pm \sqrt{a^2 - 1}
$$
Where $a \in \Bbb C$ is arbitrary. Setting $a=3$ gives you the eigenvalues you mention, and setting $a= 5/4$ gives you $1/2$ and $2$.
