quantum mechanics - If all the eigenvalues of an operator are real, is the operator Hermitian?

How do I prove or disprove the following statement?

The eigenvalues of an operator are all real if and only if the operator is Hermitian.

I know the proof in one way, that is, I know how to prove that if the operator is Hermitian, then the eigenvalues must be real. It's the other direction that I'm not sure.

Answer

We shall assume that the vector space $V$ (where the linear operator $A:V\to V$ acts) is a complex vector space (as opposed to a real vector space), and that $V$ is equipped with a sesquilinear form $\langle\cdot,\cdot \rangle:V\times V \to \mathbb{C}$. (We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.)

Since OP says he already knows how to prove that a Hermitian operator $A$ has real eigenvalues, he is essentially asking

If all the eigenvalues $\lambda_i$ are real, is the operator $A$ Hermitian?

The answer is No, only if $A$ is diagonalizable in an orthonormal basis.

In other words, the eigenspaces $\ker(A-\lambda_i {\bf 1})\subseteq V $ should be mutually orthogonal and together span the whole vector space $V$.

A version of the Spectral Theorem says that $A$ is orthonormally diagonalizable iff $A$ is a normal operator.