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I'm trying to show that if the matrix $M=\left( \begin{matrix} A& -\bar{B}\\ B& -\bar{A}\\ \end{matrix} \right)$ has one eigenvalue $\lambda $ then it has another eigenvalue $-\lambda$, where $A$ is a hermitian matrix and $B$ is an antisymmetric matrix, i.e. $B^T=-B$ where I use bar to mean complex conjugate. Hence we have $-\bar{B}=B^{\dagger}$ whence we know $M$ is hermitian with all $\lambda\in \mathbb R$.

I managed to show that when $A=0$ case, i.e. when $M=\left( \begin{matrix} 0& B\\ B^{\dagger}& 0\\ \end{matrix} \right) $, we can construct $U_{total}\equiv\frac{1}{\sqrt{2}}\left( \begin{matrix} U^{\dagger}& V^{\dagger}\\ -U^{\dagger}& V^{\dagger}\\ \end{matrix} \right) $ with $B$ has singular value decomposition $B=U\Sigma V^{\dagger}$, hence we have $U_{total}\left( \begin{matrix} 0& B\\ B^{\dagger}& 0\\ \end{matrix} \right) {U_{total}}^{\dagger}=\left( \begin{matrix} \Sigma& 0\\ 0& -\Sigma\\ \end{matrix} \right) $. Another method is using the Shchur complement to calculate the eigenvalues.

However, when I consider $A\neq 0$ case, I can't see how I can use the method above to show the result.

Any suggestions? Thanks in advance!

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    $\begingroup$ If $(x,y)^T$ is the $\lambda$-eigenvector isn't $(\bar{y},\bar{x})^T$ the $(-\lambda)$ eigenvector? $\endgroup$ Commented Aug 16, 2022 at 12:54
  • $\begingroup$ @ancientmathematician Yes, I didn't think about this answer..! But I want to know if when $A\neq 0$ can we find unitary matrix to diagonalize it. But anyway, you solved my original question. Thank you very much! $\endgroup$
    – Sherlock
    Commented Aug 16, 2022 at 13:01

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