How to find eigenvalues of the following block circulant matrix

Question

I have a block matrix of size PN x PN of the form:

Where A and C are P x P matrices. I would like to find the eigenvalues of the matrix B, that is

where

Answer 1

Progress can be made if $A,C$ are commuting diagonalizable matrices, then one knows that they have a common set of eigenvectors. Call them $v_i$. Then note that the matrix $M$ can be written as a polynomial in the circulant matrix where the top row is $(0,0,...,0,1)$ and each row down you cyclically permute the entries of this vector to the right.

This is an orthogonal matrix whose eigenvectors and eigenvalues are not too difficult to compute. Call this circulant matrix $T$ then the matrix $M$ is equal to $T^{n-1}+T^{n-2}+...+T$. Inserting the eigenvectors of $T$ tensored with the common eigenvectors of $A$ and $C$ you get the eigenvalues of your matrix $B$.

Unfortunately, I do not see anything one could do in the general case.

Answer 2

$$\det(B) = \det(A-C)^{N-1}*\det(A+(N-1)C)$$

I fiddled with numerics until I found this formula. I have no proof other than it works for (lotsa) values I tested.

Using this you can get the eigenvalues via the characteristic equation:

$$0 = \det(B - \lambda I_{PN} ) = \det((A-\lambda I_P)-C)^{N-1}*\det((A-\lambda I_P)+(N-1)C)$$