MathOverflow Asked by ippiki-ookami on November 3, 2020
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.
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 $|A^m|^{1/m} leq |A^n|^{1/n}$.
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.
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, $|A^3|^{1/3} > 1= |A^2|^{1/2}$.
Answered by Chris Ramsey on November 3, 2020
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