# Eigenbases in the infinite dimensional case

Let $$H$$ be a Hilbert space and $$T \in \mathcal{L}(H)$$ be self-adjoint. I know that this is a basic question, but I do not understand the infinite dimensional case of linear algebra very well. My question is, when does there exist a Hilbert basis for $$H$$ consisting entirely of eigenvectors of $$T$$? In the finite dimensional case, we only had to show that the matrix representation of $$T$$ in any basis was similar to a diagonal matrix. But in the infinite dimensional case, it is not clear that matrix representations even make any sense.

1. Example: let $$H=L^2[0,1]$$ and let $$T:H \to H$$ be defined by $$(Tf)(t)=tf(t)$$. Then $$T$$ is self-adjoint but has no eigenvalues.

2. If $$A:H \to H$$ is compact and self-adjoint, then there exists an orthonormal basis of $$H$$ consisting of eigenvectors of $$A$$. (See: https://en.wikipedia.org/wiki/Compact_operator_on_Hilbert_space).

• Ah, thank you! I was looking for a characterization like this. Nov 30 '18 at 9:12