Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 1 / 41C / a / Solution

Since $H$ is diagonalizable, write $H=V\Lambda V^{-1}$. ThenIf the spectral radius $\rho (H)<1$, every eigenvalue satisfies $|{\lambda }_i|<1$, so ${\Lambda }^k\to 0$ and hence $z^{(k)}\to 0$ for every initial vector. Conversely, taking $z^{(0)}$ to be an eigenvector of any eigenvalue $\lambda$ shows that convergence for every initial vector forces ${\lambda }^k\to 0$, hence $|\lambda |<1$. Therefore