How bad could $\|A^k\|$ be when $\rho(A) < 1-\delta$ [closed]

Question

(Sorry, I do hate editing this many many times but let me try the last time)

Gelfand's formula says that

$$\lim_{k\rightarrow \infty} \|A^k\|^{1/k} = \rho(A)$$

I am wondering whether there is any way to make this non-asymptotic. for example, I would like to have a set $S$ of matrices so that for any matrix $A\in S$, $\|A^k\|$ goes to 0 with some exponential rate. (A candidate might be, for any matrix $A$ with $\rho(A) < 1/2$ and $\|A\|\le T$, $\|A^k\| \le T^{100}(2/3)^k$. Although I haven't thought through whether there is any trivial counterexample for this statement. )

Answer 1

To get a hang of the behaviour of matrix powers, you should consider powers of Jordan blocks: $$ J_k(\lambda)^n = \begin{bmatrix} \lambda^n & \binom{n}{1}\lambda^{n-1} & \binom{n}{2}\lambda^{n-2} & \cdots & \cdots & \binom{n}{k-1}\lambda^{n-k+1} \\ & \lambda^n & \binom{n}{1}\lambda^{n-1} & \cdots & \cdots & \binom{n}{k-2}\lambda^{n-k+2} \\ & & \ddots & \ddots & \vdots & \vdots\\ & & & \ddots & \ddots & \vdots\\ & & & & \lambda^n & \binom{n}{1}\lambda^{n-1}\\ & & & & & \lambda^n \end{bmatrix} $$ From this case, you can generalize to see what happens if you vary the magnitude of the off-diagonal entries $$ \begin{bmatrix} \lambda & \mu & \\ & \lambda & \mu & \\ & & \ddots & \ddots \\ & & & \ddots & \ddots \\ & & & & \lambda & \mu\\ & & & & & \lambda \end{bmatrix}^n = \begin{bmatrix} \lambda^n & \binom{n}{1}\lambda^{n-1}\mu & \binom{n}{2}\lambda^{n-2}\mu^2 & \cdots & \cdots & \binom{n}{k-1}\lambda^{n-k+1}\mu^{k-1} \\ & \lambda^n & \binom{n}{1}\lambda^{n-1}\mu & \cdots & \cdots & \binom{n}{k-2}\lambda^{n-k+2}\mu^{k-2} \\ & & \ddots & \ddots & \vdots & \vdots\\ & & & \ddots & \ddots & \vdots\\ & & & & \lambda^n & \binom{n}{1}\lambda^{n-1}\mu\\ & & & & & \lambda^n \end{bmatrix} $$ As you can see, the off-diagonal elements can grow arbitrarily before the exponential convergence kicks in. You can generalize further to cases in which the $\mu$ terms are in a different diagonal (the trick to get these expansions is always using the binomial theorem on $(\lambda I + \mu Z)^n$, where $Z$ contains the desired diagonal).

If, instead of an exact Jordan block, you have bounds on the magnitude of diagonal and off-diagonal entries, you can turn these identities into estimates.

Once you leave the realm of nonnormal matrices (for which $||A^n||=||A||^n$), this is what is going to happen in general. The Schur factorization tells you that every matrix is essentially triangular (up to an orthogonal factor which does not affect norms), so the only possible bounds that I can foresee are the ones that are derived from the above technique.

To get something more tame, you could try restricting to sets of matrices for which the off-diagonal part of the Schur factorization is small with respect to its diagonal (small "distance from normality"). If I recall correctly, there is some discussion of this phenomenon of growth of powers of nonnormal matrices in the famous paper nineteen dubious ways to compute the matrix exponential, but at the moment I can't get behind the SIAM paywall.