Asymptotics of Gelfand's formula

Question

In the following paper, it is stated that for for any matrix norm, $n \in \mathbb{N}$ and $A \in \mathbb{C}^{d \times d}$, the following holds:

$\rho(A) \ge \gamma^{(1+\ln n)/n}\|A^{n}\|^{1/n}$

for some constant $\gamma \in (0,1)$ and $\rho(A)$ is the spectral radius of $A$. Unfortunately, I could not find the source of this claim. In what way does $\gamma$ depend on $d$ and the specific norm? I am interested in the $\infty$-norm, i.e., $\|A\| = \max_{i}\sum_{j=1}^{d}|A_{i,j}|$.

Thank you very much.

Answer 1

Unless $\gamma$ also depends on $A$, I don't think the claim is true. With $n=1$, the claim boils down to $\rho(A) \ge \gamma \|A\|$ for all $A$, which is demonstrably false if you consider the maximum absolute row sum norm and $A=\pmatrix{1&x\\ 0&1}$. We have $\rho(A)=1$ but $\|A\|=|x|+1\to+\infty$ when $x\to\infty$.

Answer 2

A somewhat similar bound can be acheived by considering the Cauchy integral definition for $A^{k}$. If one consider the pseudo spectrum $\Lambda_{\epsilon}(A)=\{z \in \mathbb{C}:\| (zI-A)^{-1} \| \ge \epsilon^{-1}\}$ and $\rho_{\epsilon}(A)=\sup_{z \in \Lambda_{\epsilon}(A)}|z|$:

$$ \|A^{k}\| \le \frac{\rho_{\epsilon}(A)^{k+1}}{\epsilon} $$

See, for example, here.