I have a block matrix of size $3N \times 3N$ of the form
$$B = \begin{bmatrix} A & C & \ldots & C\\ C & A & \ldots & C\\ \vdots & \vdots & \ddots & \vdots\\ C & C & \ldots & A\\ \end{bmatrix}$$
where $A$ and $C$ are $3 \times 3$ matrices. Specifically, $C$ is given by
$$C = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \gamma\\ \end{bmatrix}$$
I would like to find the eigenvalues of the matrix $B$. I have a paper that roughly states that the way to do this is as follows:
First, note that $B = I_N \otimes A + M_N \otimes C$ where $\otimes$ is the Kronecker product, $I_N$ is the $N \times N$ identity, and $M_N$ is the matrix
$$M_N = \begin{bmatrix} 0 & 1 & \ldots & 1 & 1\\ 1 & 0 & \ldots & 1 & 1\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & 1 & \dots & 0 & 1\\ 1 & 1 & \dots & 1 & 0\\ \end{bmatrix}$$
I understand this part. However, the argument proceeds as follows: Let the eigenvalues of $M_N$ be $\mu_1, \dots, \mu_N$. To find the eigenvalues $\lambda$ of $B$ we must use the characteristic equation $\det(B - \lambda) = 0$. Once again, I understand this.
However, I'm confused about the next bit of the argument. The paper states that we can diagonalize $M_N$ and that since this transformation does not affect the identity $I_N$, the characteristic equation for the determinant can be transformed into $N$ equations given by
$$\det(A + \mu_k C - \lambda) = 0, \qquad{} k = 1, \dots, N$$
I don't understand this final transformation, which involves diagonalizing $M_N$. Could someone explain this for me?