If $A^2=B^2=I$, then what are the eigenvalues of $AB$? Let $A,B\in M_2(\mathbb{R})$. If $A^2=B^2=I$, then what are the eigenvalues of $AB$?
I know that if the eigenvectors of $A$ and $B$ are the same, then the eigenvalues of $AB$ are $+1$ and$-1$. What should I do in general case?
 A: As pointed out in a comment, your question is closely related to Possible eigenvalues of a matrix $AB$ . In fact, Omnomnomnom's answer to that question is also applicable here. But for sake of having an answer, I'll write one below.
First, the spectrum of $AB$ (over $\mathbb C$) must be of the form $\{z, \frac1z\}$. Why? From $A^2=B^2=I$, we get $AABB=I$. Therefore $ABBA=I$, i.e. $(AB)^{-1}=BA$, which is similar to $AB$ because $BA=B(AB)B^{-1}$. So, $(AB)^{-1}$ is similar to $AB$. Hence the eigenvalues of $AB$ occur in reciprocal pairs.
Since nonreal eigenvalues of a real matrix must occur in conjugate pairs, it follows that if the eigenvalues of $AB$ happen to be nonreal, they must also have unit moduli.
We now claim that for every nonzero number $x$ and every complex number $z$ with unit modulus, each of $\{x, \frac1x\}$ and $\{z, \frac1z\}$ is the spectrum of some $AB$ where $A,B$ are real and $A^2=B^2=I$. The aforementioned answer by Omnomnomnom has actually given a unified constructive example that takes care of all possible cases, but I'll construct two examples here, one for a real spectrum and one for a nonreal spectrum.
For any nonzero real number $x$, consider
$$
A=\pmatrix{0&x\\ \frac1x&0},\ B=\pmatrix{0&1\\ 1&0}.
$$
For any nonzero nonreal number $z=e^{i\theta}$, let $R_{\theta/2}$ be the $2\times2$ rotation matrix for angle $\frac\theta2$. Now we may consider
$$
A=R_{\theta/2}\pmatrix{0&1\\ 1&0}R_{\theta/2}^T
\text{ and } B=\pmatrix{0&1\\ 1&0},
$$
so that $\pmatrix{0&1\\ 1&0}R_{\theta/2}^T\pmatrix{0&1\\ 1&0}=R_{\theta/2}$ and $AB=R_\theta$.
