Eigenvalues of block matrix I am looking for a good argument that the eigenvalues of the block matrix 
$$\begin{pmatrix} 0 & A \\ -A & 0 \end{pmatrix}$$
where $A = \mbox{diag} (a_1,...,a_n)$ is itself a diagonal matrix, are exactly the diagonal entries of $A$ with both signs multiplied with $i$. By calculating a few explicit examples, I figured out that this might be true, but I don't see a good general argument. If anything is unclear, please let me know.
 A: The vector $v_1=[-ia_1,\underbrace{0,\dots,0}_{n-1},1,\underbrace{0,\dots,0}_{n-1}]^T$ is an eigenvector relative to $ia_1$, because
$$
\begin{bmatrix}
0 & A \\
-A & 0
\end{bmatrix}v_1
=ia_1v_1
$$
Similarly, the vector $v_{n+1}=[ia_1,\underbrace{0,\dots,0}_{n-1},1,\underbrace{0,\dots,0}_{n-1}]^T$ is an eigenvector because
$$
\begin{bmatrix}
0 & A \\
-A & 0
\end{bmatrix}v_{n+1}
=-ia_1v_1
$$
Can you generalize?
A: This is not an answer; I just needed the formatting environments for demonstration.
Just as a simple case, take $A = [a]$ as a $1 \times 1$ matrix. Clearly the eigenvalue of $A$ is just $\lambda_A = a$.
So then your construction of the block matrix, say $B$, would be $$ B = \begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix}.$$ It can be shown quickly that the characteristic polynomial of $B$ is $\lambda^2 + a^2$ and therefore has roots (which are $B$'s eigenvalues) of $\lambda_B = \pm ai $ where $i^2 = -1$.
This hardly agrees with the phrase from your question that "the eigenvalues of $B$...are exactly the diagonal entries of $A$ with both signs" because of the imaginary component. 
A: The characteristic polynomial of the given $2 \times 2$ block matrix is
$$\det \begin{bmatrix} s \,\mathrm I_n & -\mathrm A\\ \mathrm A & s \,\mathrm I_n\end{bmatrix} = \det \left( s^2 \mathrm I_n + \mathrm A^2 \right) = \det \begin{bmatrix} s^2 + a_1^2 & 0 & \dots & 0\\0 & s^2 + a_2^2 & \dots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \dots &  s^2 + a_n^2\end{bmatrix} = \prod_{k=1}^n \left( s^2 + a_k^2 \right)$$ 
and, thus, its eigenvalues are $\pm i a_1, \pm i a_2, \dots, \pm i a_n$.
