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I have a doubt related to eigenvalue of special kind of hermitian matrix

$$H = \begin{pmatrix} 0 & 0 & a_{13} & a_{14}\\ 0 & 0 & a_{23} & a_{24} \\ \overline{a_{13}} & \overline{a_{23}} & 0 & 0 \\ \overline{a_{14}} & \overline{a_{24}} & 0 & 0\end{pmatrix}=\begin{pmatrix} 0_2 & A \\ A^* & 0_2 \end{pmatrix}$$ where, $$A=\begin{pmatrix} a_{13} & a_{14} \\ a_{23} & a_{24} \end{pmatrix}$$ the characteristic polynomial of $H$ is

$$ Ch_H(\lambda)=\lambda^4-(|a_{23}|^2+|a_{14}|^2+a_{14}\overline a_{24}-a_{23}a_{13})\lambda^2-(|a_{13}|^2|a_{24}|^2-a_{13}a_{24}\overline a_{14}\overline a_{23}-|a_{23}|^2|a_{14}|^2-a_{13}a_{24}\overline a_{23}\overline a_{14})=\det(\lambda^2-AA^*) $$ the $Ch_H(\lambda)$ contains even degree terms in $\lambda$.

Result There is a result for symmetric matrix of the above kind which says non zero eigenvalues of the matrix are symmetric with respect to y-axis, that is if $\lambda$ is an eigenvalue with multiplicity $k$ then $-\lambda$ is also an eigenvalue of the matrix with multiplicity $k.$

Doubt We know the eigenvalues of hermitian matrix are real. But my intuition is the above Result is also holds for hermitian matrix $H$?

Thanks in advanced.

Student of Mathematics
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  • This follows immediately from the fact that $\det\left(\lambda I-\pmatrix{0&A\ A^\ast&0}\right)=\det(\lambda^2-AA^\ast)$. It relies on the commutativity between the sub-blocks. Therefore the analogous result does not hold for a general Hermitian matrix $H$. – user1551 Dec 28 '24 at 13:43
  • You can also get to the result by arguing that $\text{trace}\big(H^k\big)=0$ for odd $k$ and then applying Newton's Identities – user8675309 Dec 29 '24 at 01:10

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