Non-Hermitian operator with real eigenvalues?

Question

So we know that in Quantum Mechanics we require the operators to be Hermitian, so that their eigenvalues are real ($\in \mathbb{R}$) because they correspond to observables.

What about a non-Hermitian operator which, among the others, also has real ($\mathbb{R}$) eigenvalues? Would they correspond to observables? If no, why not?

Answer 1

For Hermitian matrices eigenvectors corresponding to different eigenvalues are orthogonal. This guarantees that not only are the eigenvalues real, expectation values are too.