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The spectral theorem says that if $A$ is a self-adjoint operator on a Hilbert space $H$ with compact inverse, then the eigenvectors of $A$ form a complete orthonormal basis of $H$. Furthermore each eigenvalue is real and $|\lambda_n|\to\infty$ as $n\to\infty$.

Now let us slightly change the conditions of the theorem. Let $A$ be a self-adjoint operator with increasing sequence of eigenvalues $0<\lambda_1<\lambda_2<...$ Is it true that the eigenvectors of $A$ form a complete orthonormal basis of $H$?

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  • $\begingroup$ Do you mean "eigenvalue" in the strict sense of $Ax = \lambda x$, or do you mean it in a more general sense (e.g. $A - \lambda I$ fails to be invertible)? $\endgroup$ – Omnomnomnom May 28 '18 at 14:49
  • $\begingroup$ @Omnomnomnom In the strict sense: $Ax=\lambda x$. $\endgroup$ – Oleg May 28 '18 at 14:50
  • $\begingroup$ I suspect that there exists a modification of the $\phi(t) \mapsto t\phi(t)$ map described here which has an increasing sequence of eigenvalues whose eigenvectors fail to span the space. $\endgroup$ – Omnomnomnom May 28 '18 at 14:59
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Let $H=\ell^2(\mathbb N)\oplus L^2[0,1]$. Write $\{e_n\}$ for the canonical basis of $\ell^2(\mathbb N)$. Now define $A$ by $$ Ae_n=\lambda_ne_n, $$ and on the second component let $(Af)(t)=tf(t). $ Now the eigenvectors of $A$ are $\{e_n\}$, which do not span $H$; in fact, their orthogonal complement is infinite-dimensional. You can tweak the second component to make the spectrum of $A$ almost anything you want.

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  • $\begingroup$ Martin, thank you very much for a very nice example! It means that this compact inverse condition in spectral theorem cannot be replaced by my suggested condition. $\endgroup$ – Oleg May 28 '18 at 15:08
  • $\begingroup$ Indeed. Basically, you cannot learn much structure from just the eigenvalues. $\endgroup$ – Martin Argerami Jun 4 '18 at 19:01

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