$A_{n\times n}$ is a matrix having each row sum $<1$ and its largest eigenvalue is also $<1$. I need to show $A^t\to 0,\text{ i.e } a^t_{ij}\to 0\forall i,j\text{ as } t\to\infty$ given that $0<a_{ij}<1$.

Well, suppose $0<\lambda<1$ be the largest eigenvalue of $A$ then $A^tx=\lambda^tx\to 0\text{ as } t\to\infty\Rightarrow A^t\to0\because x\ne0$, is my argument works? Well $\lambda$ may be negative? Thanks.

up vote 2 down vote accepted

The maximum absolute row sum of a matrix is a matrix norm, the $\infty$-norm.

Therefore, if $\|A\|_{\infty}<1$, then $A^n \to 0$ since $\|A^n\|_{\infty} \le \|A\|_{\infty}^n\to 0$.

This argument uses the hypothesis that $a_{ij}>0$ but not the hypothesis on the largest eigenvalue.

  • where are we using the fact that its largest eigenvalue is $<1$?, Is that a redundant information to prove the fact? – Markov Dec 6 at 16:31
  • @Wow, I think the extra information is redundant. – lhf Dec 6 at 16:37

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