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We have a $2N\times2N$ complex block matrix $H$ in the form:

$H=\left(\begin{array}{cc} \alpha & \beta\\ -\beta^{*} & -\alpha^{*} \end{array}\right)$,

where $\alpha$ is a hermitian matrix, $\beta$ is a symmetric matrix, and $\alpha^\ast$ denotes the complex conjugate (but not the conjugate transpose, which is denoted as $\alpha^{\dagger}$) of a matrix $\alpha$.

Is that possible to find a unitary matrix $U$ such that $UHU^{\dagger}=-H$ ?

At first, considering the simplest case where $N=1$, I can use the general form of unitary matrix $U$ to prove that there doesn't exist such $U$. But there is no general form of unitary matrix for higher dimension, so this method fails.

I wonder whether the non-existence is always true for higher dimension?

Discussions for the condition of a matrix $T$ unitary equivalent to $-T$ is also helpful.

phchen
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  • (Ignore my answer, it was horribly wrong). – hmakholm left over Monica May 20 '16 at 12:11
  • This might be a possible way to find a general counterexample: Since similar matrices have the same spectrum, $UHU^{\dagger}=-H$ implies that $\lambda \in \sigma(H)$ iff $-\lambda \in \sigma(H)$. Is the spectrum of this $H$ always symmetric w.r.t. $0$? (For $\beta = 0$ it is!) – Roland May 20 '16 at 12:21
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    @Roland, yes, the spectrum of this $H$ truly always symmetric with respect to 0. The proof can be found in this question – phchen May 23 '16 at 08:00

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