Eigenvalues of a symmetric matrix with blocks that are diagonal matrices

Question

I have the following symmetric matrix of the form (had trouble finding out if there was a name for this kind of special structure): $$A = \begin{bmatrix} X & Y \\ Y & X \end{bmatrix}$$

where $X$ and $Y$ $\in \mathbb{R}^{n \times n}$ and diagonal. How can I find the eigenvalues?

Answer 1

This is actually a block-diagonal matrix in disguise: you can change the order of the rows and columns to rearrange it into a matrix of the form $$ \begin{pmatrix} x_1 & y_1 \\ y_1 & x_1 \\ && x_2 & y_2 \\ && y_2 & x_2 \\ &&&& \ddots \end{pmatrix} $$ (for $1 \leq k \leq n$, send the $k$th row and column to the $(2k)$th, and the $n+k$th to the $(2k+1)$th). Since this is just reordering the basis of the space, it changes no properties of the matrix, so it will have the same eigenvalues, and the eigenvectors are just subject to the same permutation of entries. Now each $2 \times 2$ block can be treated independently. Since each is symmetric, they each have two linearly independent eigenvectors, and so on.

Answer 2

Eigenvalues of such matrix is given by the union of eigenvalues of $X+Y$ and $X-Y$. It is given in book circulant matrices by P.J Davis.