= Solution
Since $H$ is diagonalizable, write $H=V\Lambda V^{-1}$. Then
$$
z^{(k)}=H^kz^{(0)}=V\Lambda^kV^{-1}z^{(0)}.
$$
If the <spectral radius> $\rho(H)<1$, every eigenvalue satisfies $|\lambda_i|<1$, so $\Lambda^k\to0$ and hence $z^{(k)}\to0$ for every initial vector. Conversely, taking $z^{(0)}$ to be an eigenvector of any eigenvalue $\lambda$ shows that convergence for every initial vector forces $\lambda^k\to0$, hence $|\lambda|<1$. Therefore
$$
\boxed{z^{(k)}\to0\text{ for all }z^{(0)}
\quad\Longleftrightarrow\quad \rho(H)<1}.
$$
Back to article page
