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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.

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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?

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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.

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  • $\begingroup$ you are right, of course you need to multiply by $i$ since the matrix is skew-symmetric, so eigenvalues are always purely imaginary $\endgroup$ – user167575 Sep 6 '15 at 17:09
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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 $\pm i a_1, \pm i a_2, \dots, \pm i a_n$.

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