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Can anybody please give me an example of a quasinilpotent operator $T$, i.e. an operator such that $\sigma(T)=\{0\}$ on $l_2$ such that it has finite dimensional but non-trivial kernel and is not compact?

This is probably easy and well known but I just can't figure it out it and I am getting frustrated.

Thanks!

2 Answers 2

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Take a quasinilpotent operator with trivial kernel and a finite Jordan Block and glue them together.

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    Could you please expand on your answer, perhaps by giving a more concrete example? Why is the kernel finite dimensional? Thanks2012-08-15
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    Sorry, I was in a hurry and made a fundamental typo. I edit the answer.2012-08-15
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    Apologies abatkai, could you please be more explicit what you mean by "glue them together"?2012-08-16
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    @Theo Gluing = taking direct sum. Say, you have a quasinilpotent non-compact operator $T:\ell_2\to \ell_2$ with trivial kernel. Let $H=\ell_2\oplus \mathbb R$. Define $\tilde T:H\to H$ by $\tilde T(v,x)=(Tv,0)$. This operator has 1-dimensional kernel.2012-08-17
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    @LeonidKovalev Ahh....so simple and clear. Thank you.2012-08-17
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    Oh, thank you. It is summertime, I am not sitting here very often...2012-08-17
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    @Theo: Do you know a quasinilpotent non-compact injective operator to finish the answer?2012-08-19
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    @JonasMeyer I realize I don't. Perhaps a weighted left shift? I will think about it.2012-08-21
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    @Theo: Depends on what you mean by "left " I guess, but weighted shifts are a good idea. E.g., a countable direct sum of copies of the operator $(a_0,a_1,a_2,a_3,\ldots)\mapsto (0,a_0,a_1/2,a_2/3,a_3/4,\ldots)$.2012-08-23
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If your definition of a quasinilpotent element is just the following: $$T \in B(H) \quad \text{is quasinilpotent if} \quad \sigma(T)=\{0\}$$ then a nice, non-trivial example of a quasinilpotent element is $T:l^2 \rightarrow l^2$ given by $$T(x_1,x_2,...)=(0,\frac{x_1}{2},\frac{x_2}{4},...,\frac{x_n}{2^n},...)$$ Why is this operator quasinilpotent?

Recall that spectral radius of $T$, denoted $r(T)$, is given by $$r(T)=\lim_{n \rightarrow \infty} \|T^n\|^{\frac{1}{n}}=\text{inf} \, \|T^n\|^{\frac{1}{n}}.$$ Since spectrum of an operator is always non empty, it is enough to show that $r(T)=0$. Now notice that for any $x=(x_1,x_2,...) \in l^2$ such that $\|x\|=1$, we have \begin{equation} \begin{split} \|T(x_1,x_2,...)\| & = \left\|(0,\frac{x_1}{2},\frac{x_2}{4},...,\frac{x_n}{2^n},...)\right\|\\ & = \frac{1}{2}\left\|(0,x_1,\frac{x_2}{2},...,\frac{x_n}{2^{n-1}},...)\right\|\\ & \leq \frac{1}{2}\|x\|\\ & = \frac{1}{2} \end{split} \end{equation} \begin{equation} \begin{split} \|T^2(x_1,x_2,...)\| & = \left\|T(0,\frac{x_1}{2},\frac{x_2}{4},...,\frac{x_n}{2^n},...)\right\|\\ & = \left\|(0,0,\frac{x_1}{2^{(1+2)}},\frac{x_2}{2^{(2+3)}},...)\right\|\\ & = \frac{1}{2^3}\|(0,0,x_1,\frac{x_2}{2^2},...)\|\\ & \leq \frac{1}{2^3}\|x\|\\ & = \frac{1}{2^3}, \end{split} \end{equation} and for an arbitrary $n$, \begin{equation} \begin{split} \|T^n(x_1,x_2,...)\| & = \left\|T(0,0,...,0,\frac{x_1}{2^{(1+2+...+n)}},...)\right\|\\ & = \left\|T(0,0,...,0,\frac{x_1}{2^\frac{n(n+1)}{2}},...)\right\|\\ & = \frac{1}{2^\frac{n(n+1)}{2}}\|T(0,0,...,0,x_1,...)\|\\ & \leq \frac{1}{2^\frac{n(n+1)}{2}}\|x\|\\ & = \frac{1}{2^\frac{n(n+1)}{2}}. \end{split} \end{equation} Since $x$ was an arbitrary element of norm $1,$ this implies that $$\|T^n\| \leq \frac{1}{2^\frac{n(n+1)}{2}},$$ which in turn implies that $$\|T^n\|^{\frac{1}{n}} \leq \left(\frac{1}{2^\frac{n(n+1)}{2}}\right)^{\frac{1}{n}}=\frac{1}{2^{\frac{(n+1)}{2}}}.$$ Thus, since $\frac{1}{2^{\frac{(n+1)}{2}}}$ goes to zero as $n$ increases, $\|T^n\|^{\frac{1}{n}}$ goes to zero as $n$ increases. Therefore, $$r(T)=\text{inf} \, \|T^n\|^{\frac{1}{n}}=0,$$ as desired.

(example taken from http://www.polishedproofs.com/example-of-a-quasinilpotent-element)

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    Note that this is a weighted shift with weights converging to zero, and thus is compact. That said, this operator won't be too helpful for the purposes of this question. Also note that this operator is easily seen to be quasinilpotent. It is a weighted shift, and thus its spectrum has circular symmetry, yet it's spectrum must also be a sequence converging to 0 as the operator is compact. Put these together to deduce that its spectrum is $\{0\}$.2017-03-22
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    @ZackCramer You are correct. Should I delete my answer then? Or should I leave it just as an example of a quasinilpotent element?2017-03-23
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    I'd leave it up. It's still a nice example 2017-03-23
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    @ZackCramer I'll do that! Thank you for your input!2017-03-23