# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – paul garrett, David Handelman, Pace Nielsen, Chris Godsil
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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

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.