Example of a quasinilpotent operator

Question

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!

Answer 1

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

Answer 2

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.