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If I am not misunderstanding anything: by the spectral theorem, Hermitian operators that act upon finite-dimensional Hilbert space as well as compact Hermitian operators that act upon infinite-dimensional Hilbert spaces have eigenvectors that form a complete set. I would like an example of a Hermitian operator whose eigenvectors do not do so.

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For example, multiplication by $x$ on $L^2[0,1]$. It has continuous spectrum $[0,1]$ and no eigenvectors.

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  • $\begingroup$ My apologies--I think I misstated the question, I meant to ask about a complete set of eigenvectors, not a complete set of eigenvalues. $\endgroup$ – Andrew Ho Oct 14 '13 at 19:47
  • $\begingroup$ Yes, that's what I thought you meant. $\endgroup$ – Robert Israel Oct 14 '13 at 20:39

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