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Hello how to show the following fact:

When the spectral radius of a matrix is less than $1$ then $B^n\to0$ as $n$ goes to infinity?

Thank you!

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Are you aware of this formula? – Jonas Meyer Jan 24 '13 at 5:15
up vote 0 down vote accepted

One way to see this fact (without using Gelfand's formula) is the following. Let $B=SJS^{-1}$, where $J$ is the (complex) Jordan form of $B$. Let $D=\operatorname{diag}(1,\varepsilon,\varepsilon^2,\ldots,\varepsilon^{n-1})$, where $\varepsilon>0$. You can make all off-diagonal entries of $D^{-1}JD$ arbitrarily small if $\varepsilon$ is small enough. Since $\rho(B)<1$, you can pick $\varepsilon$ such that $\|D^{-1}JD\|_\infty<1$, where $\|\cdot\|_\infty$ is the maximum absolute row sum norm (i.e. $\|X\|_\infty=\max_i\sum_j|x_{ij}|$). Define $\|X\|=\|D^{-1}S^{-1}XSD\|_\infty$. Since $\|\cdot\|_\infty$ is submultiplicative, so is $\|\cdot\|$. Also, $\|B\|=\|D^{-1}JD\|_\infty<1$. Hence $0\le\|B^n\|\le\|B\|^n\to0$ as $n\to\infty$. Since all norms on a finite dimensional vector space are equivalent, we conclude that $B^n\to0$ as $n\to\infty$.

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There is a proof on the Wikipedia page for spectral radius.

Also there you will find the formula $\lim\limits_{n\to\infty}\|B^n\|^{1/n}$ for the spectral radius, from which this fact follows. However, the Wikipedia article's author(s) used the result in your question to prove the formula.

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