Is it possible to define an inner product such that an arbitrary operator is self adjoint?

Given a vector space $V$ (possibly infinite dimensional) with inner product $(.,.)$. We say an operator $A$ is self adjoint if $(Af,g)=(f,Ag)$.

The definition as stated require us to start with an inner product $(.,.)$ in $V$ and check if the operator $A$ satisfies the equality.

My question is:

If we start with an operator $B$ on a vector space $W$ what are the necessary and sufficient conditions such that we can define an inner product such that $B$ is self adjoint with that inner product?

Solution 1:

I will address the case when $V$ is finite dimensional (in infinite dimensions, things may be more subtle). Let me also assume that the base field is either $\mathbb{R}$ or $\mathbb{C}$.

I claim that the following is a necessary and sufficient condition for the existence of an inner product for which $B$ is self-adjoint: There exists a basis of $V$ consisting of eigenvectors of $B$ with real eigenvalues (i.e., $B$ is diagonalizable with real eigenvalues).

This condition is necessary by the spectral theorem.

It is also sufficient: given a basis of eigenvectors, we can construct an inner product by declaring that basis to be orthonormal, and extending the inner product (sesqui-)linearly. If the eigenvalues are real, then $B$ will be self-adjoint with respect to this inner product.