Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 4 down vote accepted

$0$ 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.

share|improve this answer
1  
You might want to replace the word bijective by the word invertible. –  Rasmus Jul 28 '12 at 9:28
1  
@Rasmus Alright. They should be the same in Banach spaces, though. –  Cocopuffs Jul 28 '12 at 9:50
    
Right, this theorem helps us out. –  Rasmus Jul 28 '12 at 12:02
    
@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 ? –  user36675 Jul 28 '12 at 12:45
1  
@user36675: Related info: Compact operators on infinite dimensional spaces are never surjective (although they can be injective as in your example). All nonzero elements of the spectrum of a compact operator are eigenvalues (e.g. see the Fredholm alternative). –  Jonas Meyer Jul 29 '12 at 2:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.