# Example of a quasinilpotent operator

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!

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

• Could you please expand on your answer, perhaps by giving a more concrete example? Why is the kernel finite dimensional? Thanks
– Theo
Aug 15, 2012 at 16:48
• Sorry, I was in a hurry and made a fundamental typo. I edit the answer. Aug 15, 2012 at 17:14
• Apologies abatkai, could you please be more explicit what you mean by "glue them together"?
– Theo
Aug 16, 2012 at 2:53
• @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.
– user31373
Aug 17, 2012 at 1:59
• @LeonidKovalev Ahh....so simple and clear. Thank you.
– Theo
Aug 17, 2012 at 2:39

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 $$$$\begin{split} \|T(x_1,x_2,...)\| & = \left\|(0,\frac{x_1}{2},\frac{x_2}{4},...,\frac{x_n}{2^n},...)\right\|\\ & = \frac{1}{2}\left\|(0,x_1,\frac{x_2}{2},...,\frac{x_n}{2^{n-1}},...)\right\|\\ & \leq \frac{1}{2}\|x\|\\ & = \frac{1}{2} \end{split}$$$$ $$$$\begin{split} \|T^2(x_1,x_2,...)\| & = \left\|T(0,\frac{x_1}{2},\frac{x_2}{4},...,\frac{x_n}{2^n},...)\right\|\\ & = \left\|(0,0,\frac{x_1}{2^{(1+2)}},\frac{x_2}{2^{(2+3)}},...)\right\|\\ & = \frac{1}{2^3}\|(0,0,x_1,\frac{x_2}{2^2},...)\|\\ & \leq \frac{1}{2^3}\|x\|\\ & = \frac{1}{2^3}, \end{split}$$$$ and for an arbitrary $$n$$, $$$$\begin{split} \|T^n(x_1,x_2,...)\| & = \left\|T(0,0,...,0,\frac{x_1}{2^{(1+2+...+n)}},...)\right\|\\ & = \left\|T(0,0,...,0,\frac{x_1}{2^\frac{n(n+1)}{2}},...)\right\|\\ & = \frac{1}{2^\frac{n(n+1)}{2}}\|T(0,0,...,0,x_1,...)\|\\ & \leq \frac{1}{2^\frac{n(n+1)}{2}}\|x\|\\ & = \frac{1}{2^\frac{n(n+1)}{2}}. \end{split}$$$$ Since $$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.

• 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 $\{0\}$. Mar 22, 2017 at 19:25
• @ZackCramer You are correct. Should I delete my answer then? Or should I leave it just as an example of a quasinilpotent element? Mar 23, 2017 at 2:38
• I'd leave it up. It's still a nice example 🙂 Mar 23, 2017 at 2:38
• @ZackCramer I'll do that! Thank you for your input! Mar 23, 2017 at 2:40
• Tipp: There's an easier way to find the norms even on the nose: Use the C*-identity $\|T\|^2=\|T^*T\|$. Applying this for (the powers of) your shift T=S^n you obtain diagonal operators whence the norm may be easily and on the nose read off as the sup of the diagonal values, i.e. the weights. Apr 13, 2020 at 18:10