Diagonalisability of Self-Adjoint Operators for Non-Symmetric Metrics

Question

Let $V$ be a finite dimensional vector space and $(\cdot,\cdot)$ a non-degenerate bilinear form. When $(\cdot,\cdot)$ is symmetric, every self-adjoint operator on $V$ is diagonalisable. What happens when $(\cdot,\cdot)$ is not necessarily symmetric?

Answer 1

The inner product over complex numbers, for example, is not symmetric, although it has the conjugate symmetry axiom: $\langle x,y\rangle=\overline{\langle y,x\rangle}$. That didn't prevent people to prove the diagonalisability of self-adjoint operators. Actually, self-adjoint operators are only defined for such inner-products.