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Let $A \in \mathbb{R}^{m \times m}$ be a nonsymmetric zero diagonal matrix such that $A^k, k=2n+1, n\in\mathbb{N}_0$ has also zero diagonal.

Is there an easy proof or reference that the spectrum of $A$ is symmetric with respect to the imaginary axis?

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  • $\begingroup$ Notably, $\operatorname{Tr}(A^k) = \sum_{i}\lambda_i^k$. If this comes out to $0$ for all odd $k$, maybe we could do something using Newton's identities. $\endgroup$ – Omnomnomnom Aug 17 '17 at 23:18
  • $\begingroup$ Also, it is notable that in the complex matrix case, the real condition is symmetry with respect to the origin. That is, $\lambda$ is an eigenvalue implies that $-\lambda$ is an eigenvalue. $\endgroup$ – Omnomnomnom Aug 17 '17 at 23:21
  • $\begingroup$ There's a relevant discussion here. $\endgroup$ – Astor Aug 17 '17 at 23:35
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Using that $Tr(A^k)=\sum_i \lambda_i^k = 0$ for all odd $k$, the sets of complex numbers having this property cancel each other in pairs as proven here.

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