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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.

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closed as off-topic by Namaste, Nils Matthes, Mostafa Ayaz, Chris Custer, Adrian Keister Jul 14 '18 at 0:05

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    $\begingroup$ $0$ is always in the spectrum of a compact operator (i.e. the set of $\lambda$-s such that $T-\lambda I$ is not bijective), but it needs not be an eigenvalue of $T$ (i.e. $T-\lambda I$ may be injective). The correct result is that, if $T$ is compact, then $\lambda\in\Bbb R\setminus\{0\}$ is in the spectrum $\sigma(T)$ if and only if it is a non-zero eigenvalue. Moreover, a terminological issue: eigenvectors are always non-zero. $\endgroup$ – Saucy O'Path Jun 21 '18 at 4:24
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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 $S$.

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