# $A^t\to 0$ when its row sum is strictly less than one?

$$A_{n\times n}$$ 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.

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.

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? – Marso Dec 6 '18 at 16:31