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. 
 A: 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=aσ3 and B=bσ1 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.
