(I was asking a similar question before but maybe this is the best way to put it: )

I am wondering whether there exists constant $C$ such that for any $n$, there exists a square matrix $A$ of dimension $n\times n$ such that the spectral radius $\rho(A)$ of $A$ is bounded away from 1 (say, $\rho(A) < 1-1/n$ or $\rho(A) < .99$), and in the meantime, there exists a $k\ge 0$ such that the operator norm of $A^k$ is very big, say $\|A^k\|\ge n^{C}$.

Thanks to the comments such bad situation could happen. Although my point here is not really to solve a brain-teaser but to try to build some constructive theory here. For example, my plan for rescuing the situation is as follows:

Could one assume additionally some matrix norm bound on $A$ so that this doesn't happen. That is, whether there exists a matrix norm so that for any matrix with some matrix norm $\le poly(n)$ (with a fixed poly) and $\rho(A) < 1-\delta$, it is true that $\|A^k\|\le poly(n)$ for any $k$. (Or slightly stronger, I would like to have $\|A^k\|\le poly(n) (1-\delta)^k$.

Thanks a lot! Any pointers/references would be most appreciated!

(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. )

(I was asking a similar question before but maybe this is the best way to put it: )

I am wondering whether there exists constant $C$ such that for any $n$, there exists a square matrix $A$ of dimension $n\times n$ such that the spectral radius $\rho(A)$ of $A$ is bounded away from 1 (say, $\rho(A) < 1-1/n$ or $\rho(A) < .99$), and in the meantime, there exists a $k\ge 0$ such that the operator norm of $A^k$ is very big, say $\|A^k\|\ge n^{C}$.

I feel like there should probably exists such bad situations. If situation like this do exist, a further question would be, could one add some matrix norm bound on $A$ so that this doesn't happen. That is, whether there exists a matrix norm so that for any matrix with some matrix norm $\le poly(n)$ (with a fixed poly) and $\rho(A) < 1-\delta$, it is true that $\|A^k\|\le poly(n)$ for any $k$. (Or slightly stronger, I would like to have $\|A^k\|\le poly(n) (1-\delta)^k$.

Thanks a lot! Any pointers/references would be most appreciated!

(I was asking a similar question before but maybe this is the best way to put it: )

I am wondering whether there exists constant $C$ such that for any $n$, there exists a square matrix $A$ of dimension $n\times n$ such that the spectral radius $\rho(A)$ of $A$ is bounded away from 1 (say, $\rho(A) < 1-1/n$ or $\rho(A) < .99$), and in the meantime, there exists a $k\ge 0$ such that the operator norm of $A^k$ is very big, say $\|A^k\|\ge n^{C}$.

Thanks to the comments such bad situation could happen. Although my point here is not really to solve a brain-teaser but to try to build some constructive theory here. For example, my plan for rescuing the situation is as follows:

Could one assume additionally some matrix norm bound on $A$ so that this doesn't happen. That is, whether there exists a matrix norm so that for any matrix with some matrix norm $\le poly(n)$ (with a fixed poly) and $\rho(A) < 1-\delta$, it is true that $\|A^k\|\le poly(n)$ for any $k$. (Or slightly stronger, I would like to have $\|A^k\|\le poly(n) (1-\delta)^k$.

Thanks a lot! Any pointers/references would be most appreciated!