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I'm working on trying to find a way to get the eigen-values of a complicated matrix but all the original elements themselves are either block-diagonal (as in, all blocks are the same also) or some simple repeated matrix. Computationally, I see a pattern of repeated eigen-values that makes me think I could write them in a more analytical form. I'm however getting into issues because there's a $I_n - D$ like term that's making it troublesome for me. This question is a smaller piece from this though.

Just to be clear on some notation. Let $J_k = 1_k1_k'$ be the square matrix of size $k$ of all 1s. Let $A$ and $B$ be square matrices of size $c$; $A,B\in\mathbb{R}^{c \times c}$.

Is there anyway to rewrite the following sum of kronecker products in such a way that find eigen-values would be "simple"?

$$ \left( I_k \otimes A \right) + \left(J_k \otimes B \right) $$

where all matrix multiplication is possible. I think if it's possible it has something to do with a clever way of making $I_k$ and $J_k$ look more similar to each other but I haven't been able to think of anything.

Specifically in my case $A$ and $B$ have some similar structure in that $A=Z'WZ$ and $B=Z'WCZ$ but any points on the more general problem may be helpful.

There is a "smaller" piece earlier of the form $I_{kc} - J_k \otimes D$, $D\in\mathbb{R}^{c\times c}$, that if I had some other way to rewrite might make the later things simpler.

Edit:

I guess consequently that I'm asking about how to "simplify" the eigen-values of a block-matrix of the form

$$ \begin{pmatrix} A & A + B \\ A + B & A \end{pmatrix} $$

which I don't think exists. Perhaps I've answered myself that the solution is "no" but maybe there's something else I'm missing.

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