The spectrum of a product of operators Suppose $A,B\in\mathcal{B}(\mathcal{H})$, where $\mathcal{H}$ is an infinite dimensional Hilbert space. In general, we know that there is no relationship between $\sigma(AB)$ and $\sigma(A)$ and $\sigma(B)$ without assuming that perhaps $A$ and $B$ commute. What I've been wondering about is if we can say anything if $A$ and $B$ have finite spectrum - can we conclude that $AB$ has finite spectrum? Or are there examples of bounded operators $A$ and $B$ with finite spectrum but $\sigma(AB)$ is infinite? If nothing can be said in general for this problem, what if $A$ and $B$ are self-adjoint involutions (i.e. $A^* = A$, $B^*=B$ and $A^2 = B^2 = I$)?
I haven't been able to make any serious headway on this due to the complicated nature of a spectrum of a product of operators. I've had a string of if and only ifs for when $AB-\lambda I$ is invertible but nothing enlightening came of it. In the self-adjoint involution case, I would think that the spectral theorem could play a big role but I don't see how to invoke it here in a meaningful way. After some scouring of the internet, I haven't really found any result or even any mention of such a problem.
 A: The answer to your question is no, the spectrum can be infinite, and pairs of self-adjoint involutions are actually a good class of counterexamples because they can be fully described using the spectral theorem. The irreducible pairs of involutions occur in dimensions $1$ and $2$, and the rest are direct integrals of irreducibles. So every such pair is described by the "spectrum of angles" between them with its spectral measure type (i.e. equivalence class) + multiplicity.
Now, an irreducible pair of involutions in dimension $2$ looks like this:
$A_\alpha = \left[\begin{matrix}1\\
 & -1
\end{matrix}\right], B_\alpha = \left[\begin{matrix}\cos\alpha & \sin\alpha\\
\sin\alpha & -\cos\alpha
\end{matrix}\right]$
The product $A_\alpha B_\alpha$ is the unitary operator that rotates by angle $\alpha$, so its spectrum is $\{e^{i\alpha}, e^{-i\alpha}\}$. Clearly, a direct sum of countably many of these, with different $\alpha$'s, can have infinite spectrum.
Incidentally, we see that every unitary operator whose spectral measure type and multiplicity function are symmetric under complex conjugation is a product of two self-adjoint involutions.
