# How to show that the matrix $\left( \begin{matrix} A& -\bar{B}\\ B& -\bar{A}\\ \end{matrix} \right)$ have pairs of eigenvalues $\pm \lambda$?

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.

• If $(x,y)^T$ is the $\lambda$-eigenvector isn't $(\bar{y},\bar{x})^T$ the $(-\lambda)$ eigenvector? Commented Aug 16, 2022 at 12:54
• @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! Commented Aug 16, 2022 at 13:01