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Since a compact operator $K$ on an infinite dimensional separable Hilbert space $H$ can't be invertible, the spectrum of $K$ must contain zero $0\in \mbox{sp}(K)$. However, $K$ could be injective (e.g. if $K$ is strictly positive). Hence, all the elements in the spectrum $\mbox{sp}(K)$ are eigenvalues of $K$, except perhaps zero. Is this true? ...I'm a little confused about this, since I've always heard that all the elements in the spectrum of a compact operator must be eigenvalues.

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closed as off-topic by Bill Johnson, paul garrett, David Handelman, Pace Nielsen, Chris Godsil Jan 23 at 0:26

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    $\begingroup$ Yes, this is true. The spectrum of a compact operator is the union of the set of eigenvalues and the point 0. $\endgroup$ – Fedor Petrov Jan 22 at 18:11
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    $\begingroup$ ... but/and this should be moved to Math Stack Exchange... $\endgroup$ – paul garrett Jan 22 at 20:01
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All the non-zero points of the spectrum of a compact operator are eigenvalues, yes, and, yes, in an infinite-dimensional Hilbert space $0$ is always in the spectrum, being an accumulation point of the non-zero spectrum.

$0$ may or may not be an eigenvalue, depending on the operator. The operator on $\ell^2$ that sends $e_n\to e_n/n$ is compact, and $0$ is not an eigenvalue, though $0$ is in the spectrum. And it is also easy to make compact operators with $0$ as an eigenvalue.

(... but this should be moved to Math Stack Exchange...)

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