Possible eigenvalues of a matrix $AB$

Question

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 $AB$?

Answer 1

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 $AB$!

Answer 2

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 $2$.