Let $\|\cdot\|$ denote the Euclidean operator norm when applied to matrices, and the $L2$ norm when applied to vectors. Let $\rho(\cdot)$ denote the spectral radius (maximum magnitude of $A$'s eigenvalues). $\newcommand{\R}{\mathbb{R}}$
Given $A \in \R^{n \times n}$ and $x \in \R^n$, by definition $ \| A x \| \leq \|A\| \|x\|, $ and therefore $$ \| A^k x \| \leq \|A\|^k \|x\|. $$ On the other hand, if $\rho(A) < 1$, then $$ \lim_{k \to \infty} \|A^k\| = 0. $$ So in the case where $\rho(A) < 1,\ \|A\| > 1$, the $\|A\|^k$ bound is loose to the point of being useless.
The difficulty in bounding $\|A^k x\|$ is that in the sequence
$
x_1 = Ax,\ x_2 = A^2 x,\ \dots ,
$
it's possible that $\|A x\| > \|x\|,\ \|A^2 x\| > \|x\|$, and so on.
The sequence of $\|x_n\|$ might look something like this:
The Kreiss matrix theorem states that if $\rho(A) < 1$, there exists a constant $C$ (details omitted) such that $$ \sup_k \|A^k\| < C. $$ However, a constant bound like this does not help my application.
Question: When $\rho(A) < 1$ and $\|A\|>1$, does there exist a bound $B$ of the form $$ \|A^k x\| < B(A, k) \|x\| $$ such that $\lim_{k \to \infty} B(A, k) = 0$?