Non-zero eigenvalues of a skew-hermitian matrix are purely imaginary as shown here. If the matrix is additionally hollow, that is, all diagonal elements are zero, do the non-zero eigenvalues come in conjugate pairs? If yes, how can one proof this?
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I don't know why this question was closed, it is a clear mathematical question, and, of course potentially helpful and relevant for others to see the answers. – Karsten Leonhardt Oct 21 '24 at 12:12
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1The eigenvalues of $\pmatrix{0&i&i\ i&0&i\ i&i&0}$ are $-i,-i$ and $2i$. – user1551 Oct 21 '24 at 14:33
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@user1551 Thank you, such a counterexample is exactly what I needed. – Karsten Leonhardt Oct 21 '24 at 14:45
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By the Schur-Horn theorem, for any $\lambda_1 \ge \cdots \ge \lambda_n$ such that $\lambda_1 + \cdots + \lambda_n = 0$, there is an $n\times n$ hermitian matrix $A$ whose diagonal entries are zeroes and eigenvalues are $\lambda_k$’s. Then $iA$ is a skew-hermitian matrix whose diagonal entries are zeroes and eigenvalues are $i\lambda_k$’s. Clearly, we can pick $\lambda_k$’s to be not symmetric about $0$, so they don’t necessarily come in conjugate pairs.
Pranay
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