# Kernel of compact operators [closed]

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.

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

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• Yes, this is true. The spectrum of a compact operator is the union of the set of eigenvalues and the point 0. – Fedor Petrov Jan 22 at 18:11
• ... but/and this should be moved to Math Stack Exchange... – paul garrett Jan 22 at 20:01

## 1 Answer

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...)