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Consider any matrix $A \in \mathbb R^{n \times n}$ with the p-norm $||A||_{p} < 1$.

  1. I would like to show that $\lim\limits_{k \rightarrow \infty}{A^k} = 0$.
  2. Consider the reverted scenario. Let $\lim\limits_{k \rightarrow \infty}{A^k} = 0$,

I actually have two questions regarding number 1. It's not stated which p-norm, does this mean it has to be valid for all p-norms? And then I'm a little lost on to how to prove this. Does this mean all elements of the matrix are smaller than 1?

For number 2 I have a hunch it's false. Consider a matrix with $1$ on the top right corner, $A^k$ will be $0$ but it's $p=1$ norm is 1. (and therefore also all other norms?)

Many thanks in advance!

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Hint: $\|AB\|\le\|A\|\|B\|$ for all $p$-norm. – ziyuang Apr 24 '12 at 16:42
You are correct for #2. If your matrix is all zeroes except for the top right corner $A_{1,n} = 1$, then you have $A e_n = e_1$, and $||e_1||_p = 1$ for all the $p$-norms, hence all induced $p$-norms are at least 1. – copper.hat Apr 24 '12 at 16:47
thanks @copper.hat! – Clash Apr 24 '12 at 16:50
A useful (standard) result is that the limit is $0$ iff all eigenvalues lie in the open unit ball of $\mathbb{C}$. Another useful result is that all eigenvalues lie in the open unit ball of $\mathbb{C}$ iff there exists an induced norm of value less than 1. – copper.hat Apr 24 '12 at 16:50
up vote 3 down vote accepted

From $||A^k||_{p} \le ||A||_{p}^k=c^k$, with $c<1$. We know $ \lim\limits_{k \rightarrow \infty}{\|A^k\|_p} =0$, implying what you want.

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well that was easy! I didn't really remember/think showing the norm is 0 means the matrix is also 0. thanks guys! – Clash Apr 24 '12 at 16:49

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