# Eigenvalues of block diagonal matrix

I have a block-diagonal matrix of the form \begin{align*} \bf{M} = \begin{bmatrix} \bf{0} & \bf{I} \\ \bf{A} & \bf{B} \end{bmatrix} \end{align*} Can we say anything about the eigenvalues of $$\bf{M}$$ in terms the eigenvalues of the block matrices?

Here, $$\bf{0}$$ is a matrix of zeros and $$\bf{I}$$ is an identity matrix.

This is not in block upper/lower triangular form.

• Welcome to MSE! Please show your attempts. – Culver Kwan Aug 18 at 14:28

In any case, $$\operatorname{Tr} {\bf M}=\operatorname{Tr} {\bf B}$$, and $$\det {\bf M}= - \det {\bf A}$$, so the sum of the eigenvalues of M equals that of those of B, and their product equals minus that of those of A.
I'm not sure of more if A and B do not commute. Splitting the generic eigenvector of M into blocks, $$V\equiv \begin{bmatrix} v\\w \end{bmatrix},$$ you have $$\bf{M} V= \begin{bmatrix} w\\ {\bf A} v + {\bf B} w \end{bmatrix}=\lambda \begin{bmatrix} v\\w \end{bmatrix}$$ so that $$w=\lambda v$$ and $$\bigl ({\bf A} +\lambda ({\bf B} -\lambda { \bf I})\bigr )v=0.$$
If A and B commute, they share eigenvectors, say with respective eigenvalues a and b, so $$\lambda ^2 -\lambda b + a=0 ,$$ and you may proceed conventionally...
You might, or might not, see something useful in the toy example A=3 and B=1 for Pauli matrices, in this case, noncommuting but yielding TrM=0 and detM=a2 with $$\lambda =\pm \sqrt{b^2/2\pm \sqrt{b^4/4+a^2}}$$, but I'd doubt it would contain elements of generality.