Does each compact operator have a non–zero eigenvector?

Question

I have learned that 0 must be an eigehvalue for a compact operator. But I do not know if there exists a compact operator that have no non–zero eigenvector. Any hint would be most appreciated.

Answer 1

First, it must be said that $0$ is not always an eigenvalue of a compact operator. What is true is that $0$ is always in the spectrum of a compact operator, when the space it acts on is infinite-dimensional. For example, the operator $T\in\mathcal B(\ell^2)$ defined by $(Tx)(n)=\frac{1}{n}x(n)$ is compact, but $0$ is not an eigenvalue of $T$.

As for the main question, note that the zero operator is compact, and the spectrum of this operator is $\{0\}$.

EDIT I may have read the question wrong. One can (relatively) easily construct a compact operator with empty point spectrum, hence no eigenvectors. One example is given by the operator $S\in\mathcal B(\ell^2)$ defined by $(Sx)(n)=\frac{1}{n}x(n-1)$ for $n>1$ and $(Sx)(1)=0$. It is the composition of the above mentioned compact operator $T$ (hence $S$ is compact), with a shift operator. One can show that $\sigma(S)=\{0\}$, but $0$ is not an eigenvalue of $S$.