# 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 [1], 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.

[1] 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, $$\|A^3\|^{1/3} > 1= \|A^2\|^{1/2}$$.