$\sigma(T)=\{\lambda:T-\lambda I\text{ is not invertible}\}$ and $r(T)=\sup_{\lambda\in\sigma(T)}|\lambda|$, well-defined because the <spectrum> is nonempty compact. If $|\lambda|>\|T^n\|^{1/n}$, factor $T^n-\lambda^nI$ and use a Neumann <series>; spectral mapping then excludes $\lambda$. Thus $r(T)\le\inf_n\|T^n\|^{1/n}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 21I / a / Solution

 Solution (source code)