Use Schur to prove that for any norm $\lim_{n \rightarrow \infty} ||A^n|| = 0 \text{ iff } \rho(A) < 1$ where $\rho$ denotes spectral radius. I know I need to prove both directions of the biconditional, but am having some trouble in both ways.
First, looking left to right, If we assume the limit equals zero, it means that the $A$ matrix keeps getting smaller each time we apply $A$. Thus, since eigenvalue represent the effect of the matrix the maximum would have to be less than one but this doesn't seem very rigorous.
Then, I'm not sure how to proceed by assuming the spectral radius is less than 1 except for doing the backwards reasoning of what I talked about above. Any help is appreciated.

$\rho(A)=\lim\sup ||A^{n}\|^{1/n}$. If $\rho(A) <1$ and $r =\frac {1+\rho (A)} 2$ then $\|A\|^{n}<r^{n}$ for $n$ sufficiently large and $r^{n} \to 0$ since $r<1$.
Suppose $\|A^{n}\| \to 0$. There exists $k$ such that  $\|A^{k}\|< \frac 1  2$. If $\lambda $ is an eigen value of $A$ then $\lambda ^{k}$ is an eigen value of $A^{k}$ so $|\lambda^{k} |<1$ which implies $|\lambda|<1$. Hence, $\rho(A) <1$.
