In some of my work (statistics), I need the eigenvalues of a very large matrix. As such I would like to reduce it to a simpler problem and it seems entirely possible to me as the matrix has a very clear pattern/design with many things repeated. I think I have found a way to calculate the eigenvalues on smaller matrices that works numerically but I cannot prove to myself why it is true. I rewrote the problem in general below.
Problem: Suppose we have two symmetric matrices $A$ and $B$, both $\in \mathbb{R}^{c\times c}$. Consider a block matrix, $M$, of size $K\geq2$ with $A-B/K$ on the diagonal and $-B/K$ elsewhere. That is, the matrix can be written as $\left(I_K\otimes A\right) - \left(1_K1_K' \otimes \frac{1}{K}B \right)$ where $1_K$ is a column vector of size $K$ with all elements equal to 1 and $\otimes$ is the kronecker product. For example, when $K = 2$,
$$ M = \begin{pmatrix}A - B/2 & -B/2 \\ -B/2 & A- B/2\end{pmatrix} $$
Through some logical guesses and some luck, I believe the eigen-values of $M$ are (in no particular order or care given to the eigen-vectors) given by the $c$ eigen-values of $A - B$, each repeated $K-1$ times, and the $c$ eigen-values of $A$. Combined, this is $c(K-1) + c = cK$ eigen-values.
In my work $c$ is generally small and $K$ becomes very large. I've tried to prove this but haven't been able to. But is seems to work in every numerical situation I try. I also haven't found similar enough questions anywhere else. But I do believe it is true. Maybe I am missing something simple with the block structure of the matrix. I notice that this does not work when $A$ and $B$ are not symmetric.
Below is some R code if you would like to look at it and see it works numerically.
K <- 20
c <- 4
A <- matrix(rnorm(c^2), c, c)
A <- crossprod(A)
B <- matrix(rnorm(c^2), c, c)
B <- crossprod(B)
M <- (diag(K) %x% A) - tcrossprod(matrix(1, K)) %x% B/K
true <- sort(eigen(M, symmetric = TRUE)$values)
one <- eigen(A, symmetric = TRUE)$values
two <- eigen(A - B, symmetric = TRUE)$values
test <- sort(c(rep(one, each = K - 1), two))
all.equal(true, test)
Note that $M (1_K \otimes u) = 1_K \otimes ((A - K B) u)$. Thus the $c$ eigenvalues of $A - K B$ (counted by multiplicity) are eigenvalues of $M$. On the other hand, for $2 \le j \le K$ if $w$ is the column vector with $w_1 = 1$, $w_j = -1$ and $w_i=0$ otherwise, $M (w \otimes u) = w \otimes (Au)$. Thus the $c$ eigenvalues of $A$ are eigenvalues of $M$ with multiplicity $(K-1)$ times their multiplicities in $A$. That makes a total multiplicity of $nK$, so it accounts for all the eigenvalues.