I have a proposal about the eigenvalues of some special matrix but I don't know how to show it is true. I'm pretty sure it is because I can run it numerically with random numbers and find it to be true. I'll state it below.
Let $A\in\mathbb{R}^{c\times c}$ and $U\in\mathbb{R}^{Kc\times p}$. I am interested in the eigenvalues of the $\mathbb{R}^{Kc\times Kc}$ matrix $M$, $$ M=\operatorname{bdiag}_K\{A \} - UU' $$
where $\operatorname{bdiag}_K\{A \}$ is a block diagonal matrix, with $K$ diagonal elements equal to $A$. This matrix will have $Kc$, not necessarily unique, eigenvalues. I propose that in the case when $K>c>p>0$ (but not necessarily only when), $c\times (K - p)$ of the eigenvalues of $M$ can be accounted for by the $c$ eigenvalues of $A$, each replicated $K-p$ times.
Is there anyway I can go about showing this? Or even better, how to find a form for the remaining $cp$ eigenvalues?
I have a previous question similar to this one, How to prove eigenvalues of specific block matrix are as proposed, but this is a more general case. I tried to use a similar method to what was used there in the accepted answer but I'm not able to get it to work (because I cannot nicely use the kronecker products on the right side).
You are assuming $A$ is diagonalizable, i.e. has $c$ linearly independent eigenvectors.
Let $\lambda$ be any eigenvalue of $A$, corresponding to one of these eigenvectors $v$. Consider vectors of the block form $$ V(t) = \pmatrix{ t_1 v\cr t_2 v\cr \ldots \cr t_K v} $$ where $t \in \mathbb R^K$.
The linear map $t \mapsto U' V(t)$ has rank at most $p$, so there are at least $K-p$ linearly independent $t$ for which $U' V(t) = 0$. Then $M V(t) = \lambda V(t)$, so $V(t)$ is an eigenvector of $M$ with eigenvalue $\lambda$.
Considering all $v$ and all such $t$ for each $v$, this gives you $c(K-p)$ linearly independent eigenvectors of $M$ whose eigenvalues are the eigenvalues of $A$.