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Normally, content related to my question proves the converse. For example, if a matrix is hermitian, then its eigenvectors corresponding to different eigenvalues are orthogonal.

If the eigenvectors of a square matrix under (separately) both $\mathbb{R}$ and $\mathbb{C}$ are orthonormal (or unitary), however, can we infer something about the matrix?

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Complex case: If $M[e_1 ... e_n] = [e_1 ... e_n] \mbox{ diag } (\lambda_1 ... \lambda_n)$, or $MP = P\Lambda$ then $M=P\Lambda P^{-1} = P\Lambda P^*$. This is Hermetian iff $\Lambda$ is real (i.e. only real eigenvalues). In the real case if you assume that there are $n$ real eigenvectors then implicitly you have that all eigenvalues are real so the same argument shows that $M$ is automatically symmetric in this case.

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