Suppose $a,b\in\R$, $B,C\in\Ma(\C,n)$ and let $H\in\Ma(2n,\C)$ be the block-matrix
$$
\left(\begin{array}{cc}
a&B\\
C&d
\end{array}\right)
$$
where, of course, $a$ and $d$ denote $a$ times identity and $d$ times identity, respectively. If $\mu\in\R$ is an eigen-value of $CB$, then any zero of the quadratic polynomial $\l^2-(a+d)\l+ad-\mu$ is an eigen-values of $H$.
Suppose $(x,y)\in\C^n\times\C^n$ is an eigen-vector with eigen-value $\l$, i.e.
$$
ax+By=\l x,
\quad\mbox{and}\quad
Cx+dy=\l y~.
$$
Then $aCx+CBy=\l Cx$. If $CBy=\mu y$, then $\mu y=(\l-a)Cx$ and therefore
$$
\mu Cx=(\l-d)\mu y=(\l-d)(\l-a)Cx~.
$$
i.e. $Cx=0$ or $(\l-d)(\l-a)-\mu=0$; the latter is equivalent to $\l^2-(a+d)\l+ad-\mu=0$.