On self adjoint operator having discrete spectrum

Question

If $T$ is a self adjoint operator in $\mathcal{H}$, having discrete spectrum, is it true that set of eigenvectors form an orthonormal basis for $\mathcal{H}$? Under which condition it will form onb for $\mathcal{H}$.

Answer 1

• Assume that $\mathcal{H}$ is separable. Then any subspace of $\mathcal{H}$ admits a countable orthogonal basis.

Under this additional assumption your statement is (more or less) true. In fact, that the eigenspaces span the entire Hilbert space follows from the spectral theorem.

What is left to prove is the orthogonality of the eigenfunctions. Note that eigenfunctions corresponding to different eigenvalues are orthogonal: this follows from a classical argument (you can google it e.g.).

The question is what happens to eigenfunctions associated to the same eigenspace. These in general are not orthogonal, but in view of our additional assumption we can choose an orthogonal basis for every eigenspace, and this will form an orthogonal basis for the entire $\mathcal{H}$.