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I relate to this one question about corollary 4.10.2 pag. 198 of "Introduction to Hilbert Spaces - Debnath, Mikusinki" third edition, that states

Let $A$ be a compact self-adjoint operator on an infinite-dimensional Hilbert space $\mathcal{H}$. Then $\mathcal{H}$ has a complete orthonormal system $\{v_k\}$ consisting of eigenvectors of $A$

So I interpret it like "if an Hilbert space admits a compact self-adjoint operator, it will be separable" but this is not true (also this).

So what are the authors really telling, that I'm not understanding, with that proposition?

Little note:

$\ker{A}$ is sequentially closed subspace of $\mathcal{H}$, because $A$ is linear and compact (hence bounded, hence continuos), so by decomposition theorem $$\mathcal{H} = \ker{A} \oplus \ker{A}^{\perp\mathcal{H}}$$ Now, Hilbert-Schmidt theorem basically says that exists a sequence of eigenvectors $u_k$ of $A$ associated to non-null eigenvalues, such that $$ \text{cl}\,\text{Span}\{u_k\} = \ker{A}^{\perp\mathcal{H}} $$ but nothing is said about $\ker{A}$ that may also be non-separable, so that's why I cited $\ker{A}$ in the title

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    $\begingroup$ What about the case where $A=0$ on a non-separable space? $\endgroup$ Jul 27, 2022 at 5:26
  • $\begingroup$ @DisintegratingByParts Exactly! $A=0$ is linear, bounded, compact and self-adjoint with $\ker{A}=\mathcal{H}$, right? So what are the authors telling? $\endgroup$
    – Rob Tan
    Jul 27, 2022 at 10:38
  • $\begingroup$ If the orthonormal system has to be countable then the statement is false, as the case $A=0$ on any non-separable Hilbert space shows. $\endgroup$
    – daw
    Jul 27, 2022 at 14:15

2 Answers 2

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It has nothing to do with separability.

Every element of $\ker A$ is an eigenvector for $A$ with eigenvalue $0$. So you just choose an orthonormal basis of $\ker A$, countable or not, and you put it together with an orthonormal basis of $(\ker A)^\perp$ made out of eigenvectors, to get a full basis of eigenvectors.

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  • $\begingroup$ Very clear, thanks. I miss one part: if an hilbert space is non-separable is it always possible to find a non-countable orthonormal basis? Sorry for the question, I'm not a mathematician and in the book I'm reading orthonormal bases and bases in general are always considered in separable spaces. How can I find a proof? Thank you. $\endgroup$
    – Rob Tan
    Jul 27, 2022 at 21:21
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    $\begingroup$ The proof is the same. In both cases you need to find a maximal orthonormal set. The existence of maximal objects is usually guaranteed by Zorn's Lemma, which is equivalent to the Axiom of Choice. So in the usual ZFC framework of set theory, this works. If you want to do things formally you cannot avoid set theory, as you need a framework (axioms) to work with infinite objects. $\endgroup$ Jul 28, 2022 at 2:23
  • $\begingroup$ Thank you very much, I'll give a look to the subjects, very interesting $\endgroup$
    – Rob Tan
    Jul 28, 2022 at 7:21
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It is certainly true that the orthogonal complement to the kernel decomposes as (the closure of) a countable direct sum of finite-dimensional eigenspaces, since the non-zero eigenvalues' only possible limit point is $0$. But (as alluded-to in @MartinArgerami's answer), this says nothing about the separability-or-not of the kernel itself. Yes, this does say that a compact operator on a non-separable Hilbert space must have a very large (=inseparable) kernel. Meanwhile, on separable Hilbert spaces, $0$ need not be an eigenvalue at all (though it is inevitably in the spectrum).

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  • $\begingroup$ I understand, very clear, I just don't get what you say in the last phrase, sorry $\endgroup$
    – Rob Tan
    Jul 27, 2022 at 21:12
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    $\begingroup$ Well, just to repeat, on an infinite-dimensional separable Hilbert space, a compact self-adjoint operator can possibly have only non-zero eigenvalues. Yes, since the spectrum is a closed, and eigenspaces are finite-dimensional, $0$ must be a limit point, so is definitely in the spectrum. There are examples where $0$ is an eigenvalue, and, also, where it is not. That is, there are examples where the kernel is trivial... $\endgroup$ Jul 27, 2022 at 21:20

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