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

First give the general definitions and radius facts. The spectrum of a bounded operator is $\sigma (A)=\{\lambda :A-\lambda I\text{ has no bounded inverse}\}$, and the point spectrum is $p\sigma (A)=\{\lambda :\ker (A-\lambda I)\ne 0\}$. If $|\lambda |>\Vert A\Vert$, the Neumann series inverts $I-A/\lambda$, hence also $A-\lambda I=-\lambda (I-A/\lambda )$. Thus $\sigma (A)$ is contained in the closed disc of radius $\Vert A\Vert$ and is bounded.

The spectral radius is $\nu (A)={\sup }_{\lambda \in \sigma (A)}|\lambda |$. The spectral radius formula isFor the left shift $R$, an eigenvector satisfies $a_{j+1}=\lambda a_j$, so $a_j={\lambda }^{j-1}{a}_1$. A nonzero such sequence is in ${\ell }_1$ exactly when $|\lambda |<1$. Since $\Vert R\Vert =1$, its spectrum is contained in the closed unit disc. It contains the open unit disc as point spectrum and therefore its closure by closedness of the spectrum. Hence

For the diagonal operator $S$, the equation $Sa=\lambda a$ gives $(j^{-1}-\lambda )a_j=0$ for every $j$. Thus a nonzero eigenvector exists exactly for $\lambda =1/j$, with eigenvector $e_j$; zero is not an eigenvalue because all diagonal entries are nonzero.

If $\lambda$ lies outside $\{0\}\cup \{1/j:j\ge 1\}$, its distance from this closed set is positive. Dividing each coordinate by $j^{-1}-\lambda$ gives a bounded inverse of $S-\lambda I$. Zero lies in the spectrum: $\Vert S{e}_j{\Vert }_1=1/j\to 0$, excluding a bounded inverse. Therefore

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.