# Has this operator $0$ as an eigenvalue / where is my error?

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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The theorem does not say that $0$ is an eigenvalue, only that the spectrum contains $0$. See Cocopuff's answer. – Erick Wong Jul 28 '12 at 5:36
Unlike in finite dimensions, an operator can have a residual and continuous spectrum. The operator may be injective but not surjective. – copper.hat Jul 28 '12 at 5:54

$0$ 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.
@Cocopuffs So just to gmake sure I understood this properly: $0$ is in the spectrum, since $T$ isn't surjectiv, but since $0$, as I have shown, isn't an eigenvalue, $T$ has to be injective ? – pink_pyjamas Jul 28 '12 at 12:45