TITLE: Limit of powers of matrix if spectral radius < 1
QUESTION [1 upvotes]: 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?

REPLY [1 votes]: 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 $A^2 = 0$).
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