# Recover approximate monotonicity of induced norms

Let $$A$$ some square matrix with real entries. Take any norm $$\|\cdot\|$$ consistent with a vector norm.

Gelfand's formula tells us that $$\rho(A) = \lim_{n \rightarrow \infty} \|A^n\|^{1/n}$$.

Moreover, from , for a sequence of $$(n_i)_{i \in \mathbb{N}}$$ such that $$n_i$$ is divisible by $$n_{i-1}$$, we also know that the sequence $$\|A^{n_i}\|^{1/n_i}$$ is monotone decreasing and converges towards $$\rho(A)$$. I am interested in what happens when this divisibility property is not verified.

1. If the matrix has non-negative entries, it seems the general property holds: For integers $$n$$ and $$m$$ such that $$m > n$$, it is the case that $$\|A^m\|^{1/m} \leq \|A^n\|^{1/n}$$.

2. If the matrix can have positive and negative entries, this more general observation does not seem to hold. I am trying to understand why it fails, how worse can the inequality become, and if it is possible to recover an inequality up to some function of $$A$$: $$\|A^m\|^{1/m} \leq f(A)\cdot\|A^n\|^{1/n}$$.

Any references to 1., or pointers for understanding 2. would be much appreciated.

 Yamamoto, Tetsuro. "On the extreme values of the roots of matrices." Journal of the Mathematical Society of Japan 19.2 (1967): 173-178.

• Note that if (1) holds, then it holds also for all matrices that are similar to one with non-negative entries (since $v \mapsto \|Xv\|$ is a norm). Aug 18, 2020 at 7:04

This is not a complete answer: If you allow positive and negative entries then this monotonicity will not hold in general. Consider $$A = \left[\begin{matrix} 0 & 1 & -1 & 0 & 0 \\ 0& 0 & 1&1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & -1 \\ 0& 0&0 &0&0 \end{matrix}\right]$$ then $$A^2 = \left[\begin{matrix} 0 & 0 & 1 & 0 & 0 \\ 0& 0 & 0&1 & 0 \\ 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 0& 0&0 &0&0 \end{matrix}\right] \ \ \ \textrm{and} \ \ \ A^3 = \left[\begin{matrix} 0 & 0 & 0 & 1 & 1 \\ 0& 0 & 0&0 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0& 0&0 &0&0 \end{matrix}\right].$$ Thus, $$\|A^3\|^{1/3} > 1= \|A^2\|^{1/2}$$.