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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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    @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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    @ZackCramer I'll do that! Thank you for your input!2017-03-23