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I know of a theorem that tells me, that every compact linear operator on an infinitedimensional Hilbert space has to have the eigenvalue $0$. On the other hand I have the operator \begin{eqnarray*} & T:\ell^{2}\rightarrow\ell^{2}\\ & \left(x_{1},x_{2},\ldots\right)\mapsto\left(\lambda_{1}x_{1},\lambda_{2}x_{2},\ldots\right), \end{eqnarray*} where $\left(\lambda_{n}\right)_{n}$ is a sequence of real nonnegative numbers, tending to $0$. Then this mapping can't have $0$ as an eigenvalue, since if that were the case, there had to be a $\left(y_{1},y_{2},\ldots\right)\in\ell^{2}$ with not all $y_{n}$'s being zero, such that $\lambda_{n}y_{n}=0$ for all $n\in\mathbb{N}$. Since $\lambda_{n}\neq0$, that would imply that all $y_{n}$'s are there.

Where is my error ? The operator $T$ is compact and $\ell^{2}$ is infinitedimensional, so this should be a counterexample to the theorem above.

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    Unlike in finite dimensions, an operator can have a residual and continuous spectrum. The operator may be injective but not surjective.2012-07-28

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$0$ being in the spectrum means that $T$ isn't invertible, which in infinite-dimensional space no longer means that it's not injective. You should be able to show that $T$ isn't surjective.

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    @user36675: Related info: Compact operators on infinite dimensional spaces are never surjective (alt$h$oug$h$ they can be injective as in your example). All *nonzero* elements o$f$ the spectrum of a compact operator are eigenvalues (e.g. see the [Fredholm alternative](http://en.wikipedia.org/wiki/Fredholm_alternative)).2012-07-29