Eigenvalues of block matrix from the eigenvalues of one block Give a matrix which can be decomposed into $4$ blocks
$$B = \left[\begin{matrix}A &I \\ -I &0\end{matrix}\right]$$
where $I$ denotes the identity matrix and $0$ is a zero matrix.
It's easy to compute the eigenvalues of $A$. Is there a simple way to compute all the eigenvalues of $B$?
 A: Suppose $\begin{pmatrix}u\\ v\end{pmatrix}$ is an eigenvector of $B$ corresponding to an eigenvalue $\lambda$. Then we have $\begin{pmatrix}A &I \\ -I &0\end{pmatrix}\begin{pmatrix}u\\ v\end{pmatrix}=\begin{pmatrix}\lambda u\\ \lambda v\end{pmatrix}$. That is, $Au+v = \lambda u$ and $-u = \lambda v$. Hence $\lambda\not=0$, or else $u=v=0$, contradicting that $\begin{pmatrix}u\\ v\end{pmatrix}$ is an eigenvector. So $Au+v = \lambda u$ and $-u = \lambda v$ imply that $Av = (\lambda+\frac1{\lambda}) v$. Hence each eigenvalue $k$ of $A$ gives rise to a pair of eigenvalues of $B$, which are given by the roots of the equation $\lambda+\frac1{\lambda}=k$.
A: Computing the characteristic polynomial of matrix ${\rm B}$,
$$\begin{aligned} q_{{\rm B}} (s) := \det \begin{bmatrix} s {\rm I}_n - {\rm A} & -{\rm I}_n \\ {\rm I}_n & s {\rm I}_n  \end{bmatrix} &= \det \left( s (s {\rm I}_n - {\rm A}) + {\rm I}_n \right)\\ &= \det \left( s^2 {\rm I}_n - s {\rm A} + {\rm I}_n \right)\\ &= s^n \det \left( \left( s + \frac{1}{s} \right) {\rm I}_n - {\rm A} \right)\end{aligned}$$
where $s \neq 0$. Since $q_{{\rm B}} (0) = 1 \neq 0$, matrix $\rm B$ does not have a zero eigenvalue. Thus, in order to find the $2n$ eigenvalues of matrix $\rm B$, we obtain $n$ equations of the following form
$$s + \frac{1}{s} = \lambda_i ({\rm A})$$

Alternatively, using the Jordan decomposition $\rm A = P J P^{-1}$,
$$q_{{\rm B}} (s) = \det \left( s^2 {\rm I}_n - s {\rm A} + {\rm I}_n \right) = \det \left( s^2 {\rm I}_n - s {\rm J} + {\rm I}_n \right) = \prod_{i=1}^n \left( s^2 - \lambda_i ({\rm A}) \, s + 1\right)$$

matrices block-matrices determinant characteristic-polynomial
