Eigenvalues of block matrices with zero diagonal blocks Is there any simple relation between the eigenvalues (or the characteristic polynomials) of two matrices $A$ and $B$ with that of matrix $C$ defined as 
$$
C=\begin{bmatrix}
0 & A \\
B & 0
\end{bmatrix}.
$$ 
$A$ and $B$ are invertible $n\times n$ matrices.
 A: We have
$$
\det (C-\lambda I)= \det\begin{bmatrix}-\lambda I&A\\B&-\lambda I\end{bmatrix}
= \det(\lambda^2 I -AB).
$$
Therefore the eigenvalues of $C$ are the square roots of the eigenvalues of $AB$. So, they depend on the eigenvalues of $AB$ rather than the individual eigenvalues of $A$ and $B$.
A: Given two vectors $u,v \in \mathbb{R}^n$ let $w=\begin{bmatrix} u \\ v \end{bmatrix}\in\mathbb{R}^{2n}$ be the vector whose first $n$ components are those of $u$ and whose last $n$ are those of $v$.
The matrix-vector product $C\cdot w$ gives
\begin{equation*}
\begin{bmatrix}
0 & A \\
B & 0
\end{bmatrix}
\begin{bmatrix} u \\ v \end{bmatrix}
=
\begin{bmatrix} Av \\ Bu \end{bmatrix}
\end{equation*}
Therefore, solving $Cw = \lambda w$ reduces to solving the following
\begin{equation*}
\begin{cases}
Av = \lambda u \\
Bu = \lambda v
\end{cases}
\end{equation*}
Since $A$ and $B$ are invertible, it follows that $v=\lambda A^{-1}u$ and $u=\lambda B^{-1}v$:
\begin{equation*}
\begin{cases}
Av = \lambda \cdot \lambda B^{-1}v \\
Bu = \lambda \cdot \lambda A^{-1}u
\end{cases}
\rightarrow
\begin{cases}
BAv = \lambda^2 v \\
ABu = \lambda^2 u
\end{cases}
\end{equation*}
We conclude that $\lambda\in\mathbb{C}$ is an eigenvalue of $C$ iff $\lambda^2$ is an eigenvalue of $AB$ (or $BA$, since they coincide).
(By the way, this solution covers only the case in which $A$ and $B$ are invertible matrices, but as shown by user1551 they need not be)
