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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.

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  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).

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  • $\begingroup$ Ah, thank you! I was looking for a characterization like this. $\endgroup$
    – J_Psi
    Nov 30 '18 at 9:12

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