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Let us consider the block matrix

$$M=\left( \begin{matrix} O & I_N\\ A & B \end{matrix} \right) $$

where $I_N$ denotes the $N\times N$ identity matrix and $O,A,B$ have the same dimensions. Do you know if the calculation of eigenvalues of $M$ can be simplified in the above case? Moreover what can we say a priori if we know the eigenvalues of $A$ and $B$?

I am curious if the structure of eigenvalues of $M$ is only determined by the eigenvalues of $B$.

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Multi-posted question. The answer is still and always NO. Clearly $\det(M-\lambda I_{2N})=\det(\lambda^2I_n-\lambda B-A)$. We can conclude when, for example, $A,B$ are simultaneously triangularizable ($AB=BA$ or $AB=0$ or $AB-BA=uA+vB$,....).

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