Are there non nilpotent operators with spectrum 0? If a linear operator on a vector space V is nilpotent, then its spectrum is 0. Makes me wonder, are there also operators with spectrum 0 that are not nilpotent? 
Necessarily such an operator is not invertible, but I can't construct any examples and was hoping to see one.
 A: You'll have to go into infinite dimensional spaces. For example
$$V:=\left\{\,\{x_n\}_{n\in\Bbb N}\subset\Bbb R\right\}$$
with the usual operations of sum and scalar multiplcation, and the operator
$$R:V\to V\;\;,\;\;\;R\{x_1,.x_2,\ldots\}:=\{0,x_1,x_2,\ldots\}$$
has zero in its spectrum, but it is not nilpotent.
A: DonAntonio gave an algebraic example. If we put a norm on $V$, then the keyword is: quasinilpotent which, by definition means that the spectrum $=\{0\}$, ie the spectral radius $=0$. Note this is no longer algebraic since now invertible means bijective and bi-bounded.
This is equivalent to nilpotent in finite dimension (this follows readily from the Jordan normal form, with or without norm since boundedness is automatic in finite dimension), not in infinite dimension.
See the Volterra operator here for a counterexample. See also this related fact.
A: If your vector space $V$ is a Hilbert space (which we will denote $H$)
then a nice, non-trivial example of an element of spectrum consisting only of $0$ 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 (i.e. its spectrum is $\{0\})$?
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)
