Given a non-singular matrix $A \in \mathbb{R}^{n\times n}$, let $\rho(A)$ be its spectral radius. Then, $\forall k \in \mathbb{N_0}$, can we find a constant $a > 0$ such that the following inequality holds?
$$\|A^k\| \leq a k^{n-1} \rho^k(A),$$
where $\|\cdot\|$ is an induced matrix norm.
Thanks!
PS: I guess this inequality is correct since it holds for any Jordan block; for a general $A = Q^{-1} J Q$, where $J = \bigoplus_i J_i$, the Jordan block $J_i$ corresponding to $\rho(A)$ dominates $\|A^k\| = \|Q^{-1} J^k Q\|$.
We may assume that $\rho(A)=1$. Let $PJP^{-1}$ be a “Jordanisation” of $A$. As all norms are equivalent on a finite-dimensional vector space, it suffices to prove the inequality for the norm defined by $\|X\|=\|P^{-1}XP\|_\infty$, where $\|\cdot\|_\infty$ denotes the induced supremum norm (i.e., the maximum absolute row sum norm). Let $J=J_{m_1}(\lambda_1)\oplus\cdots\oplus J_{m_s}(\lambda_s)$. Then \begin{align} \|A^k\|=\|J^k\|_\infty &=\max_i\|J_{m_i}(\lambda_i)^k\|_\infty\\ &\le\max_i\sum_{r=0}^{\min(m_i-1,k)}\binom{k}{r}|\lambda_i|^r\\ &\le\max_i\sum_{r=0}^{\min(m_i-1,k)}\binom{k}{r}\\ &\le\sum_{r=0}^{\min(n-1,k)}\binom{k}{r}.\tag{1}\\ \end{align} When $k\le n-1$, the sum on $(1)$ inside the summation sign is just $2^k\le 2^n$. When $k\ge n$, the sum is $$ 1+k+\frac{k(k-1)}{2!}+\cdots+\frac{k(k-1)\cdots(k-(n-2))}{(n-1)!}, $$ which is bounded above by $nk^{n-1}$. Therefore there exists a constant $C$ such that $\|A^k\|\le Ck^{n-1}$ for all $k\ge1$. In particular we may choose $C=\max\left(\max_{1\le k<n}\frac{2^k}{k^{n-1}},\,n\right)$ or a looser estimate $C=\max(2^n,n)$.