Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
2  
Are you aware of this formula? –  Jonas Meyer Jan 24 '13 at 5:15

2 Answers 2

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 $n\to\infty$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.