What are the algebraic multiplicities of the eigenvalues of a tensor (i.e. Kronecker) product of linear transformations? We can assume that we are working in finite-dimensional vector spaces over an algebraically closed field.
$\DeclareMathOperator{\tr}{tr}$
Let's denote algebraic multiplicity by $\mu()$. It's easy to see from the matrix definition of trace (i.e., the trace is the sum of the diagonal) that
$$
\tr(A\otimes B) = \tr A \tr B.
$$
Since we also have $\tr(A) = \sum_\lambda \mu_A(\lambda)\lambda$, this strongly suggests to me that
$$
\mu_{A\otimes B}(\lambda) = \sum_{\lambda_A\lambda_B=\lambda}\mu_A(\lambda_A)\mu_B(\lambda_B).
$$
I don't really know to approach this claim, since I can't see a clear way to analyse $\det(A\otimes B – \lambda)$. According to e.g. https://mathoverflow.net/a/188819, it should be true, but I don't know why. It's of course immediately clear that all the products $\lambda_A\lambda_B$ are eigenvalues of $A\otimes B$, but a priori, it's not even clear (to me) that there can't be other eigenvalues.
Update: Since originally posting, an answer has become clear in the comments, thanks to user8675309. I'll go ahead and post an answer based on the comments myself, but other perspectives (especially more abstract ones!) are of course welcome. Other approaches that only get the distinct eigenvalues without multiplicities are also welcome.
Edit note: This question originally also asked for the geometric multiplicities. I've moving that part to another post: What are the geometric multiplicities of the eigenvalues of a Kronecker (or tensor) product?
Best Answer
Credits to user8675309.
Since we're working over an algebraically closed field, we can choose bases for $A$ and $B$ such that their matrices are (upper) triangular. They will have the eigenvalues (with multiplicity) $\{\lambda_{A,1}, \ldots, \lambda_{A,m}\}$ and $\{\lambda_{B,1}, \ldots, \lambda_{B,n}\}$ on the diagonals. The Kronecker product will then also be triangular, with entries $$\left\{\lambda_{A,i}\lambda_{B,j}\mid 1\le i \le m, 1\le j\le n\right\}$$ on the diagonal. (This follows directly from the matrix formula of the Kronecker product). And we are done!
I also wanted to highlight the other nice approach of user8675309. Let $\sigma()$ denote the multiset of eigenvalues with multiplicity. We have, for any non-negative integer $k\ge 0$, $$\DeclareMathOperator{\tr}{tr} \begin{split} \sum_{\lambda\in\sigma(A\otimes B)} \lambda^k &= \tr((A\otimes B)^k)\\ &= \tr(A^k\otimes B^k)\\ &= \tr(A^k)\tr(B^k)\\ &= \sum_{\lambda_A\in\sigma(A)} \lambda_A^k \sum_{\lambda_B\in\sigma(B)} \lambda_B^k\\ &= \sum_{\lambda_A,\lambda_B} (\lambda_A\lambda_B)^k. \end{split}$$ Assuming now that the field has characteristic $0$, then letting $k$ vary shows that the terms on the left- and right-hand sides are equal. (I'm being a biiiit hand-wavy here, but I'm sure the formal argument can't be too bad. :p )