Past exam of the analysis course of the University of Oxford / 2009 / A / 3 / iii / Solution

The weighted right shift $T$ is injective, so zero is not an eigenvalue. If $Ta=\lambda a$ with $\lambda \ne 0$, the first coordinate forces $a_1=0$ and subsequent coordinates force every $a_j=0$. Thus its point spectrum is empty.

For $m\ge 1$,since the largest column sum is attained at $j=1$. Therefore the spectral radius formula gives $\nu (T)={\lim }_m(m!{)}^{-1/m}=0$. The limit follows, for example, from $m!\ge (m/2{)}^{m/2}$ for large $m$. Zero is in the spectrum because the output first coordinate is always zero, so $T$ is not onto. HenceThis is a nonzero quasinilpotent element, specifically the factorial weighted right shift.