Eigenvalues of block matrix from the eigenvalues of one block

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 $\lambda+\frac1{\lambda}=k$.