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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.

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

But if they do not, I have no insights. Of course, for v null vectors of A, there is no reason these are also eigenvectors of B anymore, so as to furnish eigenvectors V of M.

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.

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