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Let $B$ be a $n\times n$ matrix and let $X$ be the set of all eigenvalues of $B$. Prove that if $|m|<1$ then $\lim \limits_{k\rightarrow\infty}B^k=0$, where $m=\max X$.

Thanks.

Actually, there isn't a order involved. Sorry. The correct question is:

Let $B$ be a $n\times n$ matrix, $X$ be the set of all eigenvalues of $B$ and $|X|=\{|x|;x\in X\}$. Prove that if $|m|<1$ then $\lim \limits_{k\rightarrow\infty}B^k=0$, where $m=\max |X|$.

Pedro
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    Using $\max (X)$ assumes there is an order involved. What about if there are complex eigenvalues? – Git Gud Jun 02 '13 at 20:14
  • More generally, in a unital Banach algebra, we have the spectral radius formula $\rho(b)=\lim_{k\rightarrow +\infty}|b^k|^\frac{1}{k}$. So $\rho(b)<1$ implies easily $\lim |b^k|=0$. Regardless of the norm (as long as it is Banach algebra norm and satifies in particular $|ab|\leq |a||b|$). – Julien Jun 02 '13 at 20:59

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In my answer I'll be assuming that the OP means that $X$ is the set of absolute values of the eigenvalues of $B$, (as suggested by the title), which makes $m$ be $B$'s spectral radius.

Consider the Jordan Normal Form, this and this.

Let $J$ be a jordan matrix for $B$. Then $B=P^{-1}JP$ and $(B^k)_{k\in \Bbb N}$ converges if, and only if, $(J^k)_{k\in \Bbb N}$ converges.

Now consider the powers of $J$. The above links should make it clear that if $\color{grey}{m=\rho (J)=}\rho (B)<1$, then $\lim \limits_{k\to +\infty }(J^k)=0$ and the result follows.

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Git Gud
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