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Fix $p,q \in \mathbb{N}$ and let $n = p+q$. Denote by $V$ the subset of $\mathbb{C}^{n \times n}$ defined by:

$$ V =\left\{ \begin{pmatrix} 0_{p \times p} & X \\ \overline{X}^T & 0_{q \times q} \end{pmatrix} \in \mathbb{C}^{n \times n} \ \middle| \ X \in \mathbb{C}^{p \times q}\right\}.$$

Is it somehow possible to find all possible eigenvalues of matrices in $V$? Is it maybe true, that the only two conditions would be, that the sum of the eigenvalues is equal to zero $\sum_{j=1}^n \lambda_j =0$, since the trace of all the matrices in $V$ is zero and that the eigenvalues are all real numbers, since the matrices are all hermitian?

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Since hermitian matrices only have real eigenvalues $(\lambda_1, ... ,\lambda_n)$ and $trace(A)=\sum_{i=1}^n\lambda_i$ your assumption should be correct.

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  • $\begingroup$ but couldn't there be some "hidden" restrictions? $\endgroup$ – Olorin May 6 '16 at 15:13
  • $\begingroup$ @Olorin The hidden restrictions are If $\lambda_m = x \neq 0$ is an eigenvalue, there exits and eigenvalue $\lambda_n = -\lambda_m$ and if n is odd, at least one eigenvalue is 0. I am really searching for an other hidden restriction... but I can't see any other restriction... That's all just some basic stuff. $\endgroup$ – Patrick Abraham May 6 '16 at 15:29

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