Limit of powers of matrix if spectral radius < 1

Question

I want to show $\lim_{k\rightarrow \infty}A^k=0$ iff $\sigma(A)<1$. I have $0\leq ||A^k||\leq ||A||^k=||Ax||^k=||\lambda_{max}x||^k=|\lambda_{max}|^k<1$ if we choose $x$ s.t. $||x||<1$ and $Ax=\lambda_{max}x$ for the backwards direction and $\lim_{k\rightarrow\infty}||A^kx||=|\lambda_{max}|^k ||x||= 0\Rightarrow \sigma(A)<1$ if we choose $x$ in the same way for the forwards direction. Is this sufficient?

Answer 1

If $A$ is nonsymmetric, it need not be the case that there exists an eigenvector $x$ such that $\|A\| = \|Ax\| = |\lambda_{\max}| \|x\|$. For take

\begin{equation} A = \begin{pmatrix} r & 1 \\ 0 & r \end{pmatrix}.\tag{1} \end{equation}

Then the only eigenvalue of $A$ is $r$ but $\|A\| > r$ since $\|Ae_2\| = \sqrt{1+r^2}$, where $e_2 = \begin{pmatrix} 0 & 1 \end{pmatrix}^T$. (Here, $\|\cdot\|$ is the vector 2-norm and matrix 2-norm $\|A\| = \max_{\|x\|\le 1} \|Ax\|$.) However, by direct computations of powers of $A$, we observe

$$ A^k = \begin{pmatrix} r^k & kr^{k-1} \\ 0 & r^k \end{pmatrix} \to 0 \textrm{ as } k\to\infty \textrm{ provided } |r| < 1. $$

One way to tackle this problem is to compute a Jordan normal form for your matrix, $A = VJV^{-1}$, which block diagonalizes $A$ into matrices which look like (1). In which case you observe, $0\le\|A^k\|= \|VJ^k V^{-1}\| \le \|V\| \|J\|^k \|V^{-1}\|$. From here, the analysis reduces to showing that if $|\lambda| < 1$, then a Jordan block $J$ with eigenvalue $\lambda$ satisfies $J^k \to 0$ as $k\to\infty$.

Answer 2

This is wrong. An $x$ that maximizes $\|Ax\|/\|x\|$ is not necessarily an eigenvector. For example, try $$ A= \pmatrix{0 & 2\cr 0 & 0\cr}$$ where the only eigenvalue is $0$, but $\|A\| = 2$. Here $x = \pmatrix{0\cr 1}$ satisfies $\|A x\| = 2$ and $\|x\| = 1$. By looking at $\|A\|$ alone you wouldn't know that $\lim_{n\to \infty} A^n = 0$ (in fact in this case $A^2 = 0$).