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If a matrix $A$ has spectral radius $\rho(A)<1$, it is well-known that $A^n\to0$ as $n\to\infty$, or equivalently $\lVert A^n\rVert\to 0$ for some matrix norm $\lVert\cdot\rVert$; however, it may increase a fair amount first. Are there any known results on when this sequence begins to decrease? I'm using the induced $\infty$-norm primarily, but results in any norm are welcome.

For the context of where this arose, I am attempting to find a uniform bound on the sequence $\lVert A^n\rVert$ between several different matrices each with spectral radius less than 1, and finding an upper bound on when the sequences of norms start to decrease would be sufficient to solve the problem at hand.

For normal matrices, the bound is of course $n=0$, but the matrices I'm interested are not normal. If a matrix is a Jordan matrix, then it's pretty straightforward to show that this occurs no later than $$n=\max_{\lambda\in\sigma(A)}\frac{d_\lambda}{1-|\lambda|}$$ where $d_\lambda$ is the degeneracy of the eigenvalue, since by then each entry of the matrix is decreasing in magnitude. The Jordan canonical form suggests that the best bound probably looks like this usually; however, from some numerical experimentation I have found that if the original matrix $A$ is very close to one with an eigenvalue with higher degeneracy, then the effects of that other matrix will be felt in the bound. For instance, this happens with the matrix $\left(\begin{smallmatrix}9/10 & 1 \\ 1/1000 & 9/10\end{smallmatrix}\right)$, which achieves its maximum norm at $n=9$ despite being diagonalizable, because the eigenvectors are nearly colinear. This suggests that the pseudospectrum of $A$ might be relevant.

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This is really more of an extended comment. Any bound for $n$ must take account of not only the spectral radius of $A$ but the size of matrix entries. Consider the 2-by-2 matrix $A(k)$ which has 0.9 in the diagonal entries, $k$ in the top right and $10^{-k}$ in the bottom left (for $k\geq 1$ say). Then $A(k)$ is diagonalizable and its spectral radius is less than $r$ for some $0.9<r<1$ independent of $k$.

On the other hand, it is intuitively clear that as $k$ increases, one will need to take larger and larger powers to make the modulus of the top-right corner of $A(k)$ less than $1$, and any reasonable norm on $M_2({\bf C})$ dominates the function "take the modulus of the top-right corner". (I have not done the calculation but I expect that computer algebra software will easily show what the correct asymptitic behaviour is.)

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