Eigenvalues of block matrix

Question

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.

Answer 1

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?

Answer 2

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.