91Ó°ÊÓ

We have given an n × n matrix A whose entries are positive or zero and all column sums of A are less than 1. Let r be the largest column sum ofA. Then

$$\begin{gathered} 0≤\mathrm{r}≤1 \\ \mathrm{Thus}\mathrm{we}\mathrm{have} \\ A=\underset{p=1}{\overset{m}{∑}}{a}_{ip}{a}_{pj}≤r \\ A≤r \\ {A}^m≤r^m \end{gathered}$$

This shows that the entries of $A^m$are less than or equal to r$r^m$, for all positive integers m.

Since from part (a) we get

$A^m≤r^m$ for all positive integer m and $0≤r≤1$

$$\begin{gathered} ⇒\underset{m→∞}{\lim }{A}^m≤\underset{m→∞}{\lim }{r}^m=0\left(âˆ\mu 0≤r<1\right) \\ ⇒\underset{m→∞}{\lim }{A}^m=0 \end{gathered}$$

From part (a) we have

$A^m≤r^m$ for all positive integer m and $0≤r<1$,

$⇒A≤r,A^2≤r^2,...A^m≤r^m$

Therefore,

$l_n+A+A^2+...+A^m+...≤1+r+r^2+...+r^m+...$

Since we know that$1+r+r^2+...+r^m+...$is a geometric series with common ratio r.

And the geometric series converges for r < 1and diverges for$râ‰\yen 1$.

Here we have$0≤r<1$, therefore the series$1+r+r^2+...+r^m+...$converges and hence the series$l_n+A+A^2+...+A^m+...$also converges.

We have to find the product $(l_n-A)(l_n+A+A^2+...+A^m)$

localid="1659868348236" $$\begin{gathered} (l_n-A)(l_n+A+A^2+...+A^m)=l_n^2-A{l}_n+A{l}_n-A^2+A^2{l}_n-A^3+...A_m{l}_n-A^3+...+A^m{l}_n-A^{m+1} \\ =l_n-A+A-A^2+A^2-A^2-A^3+...+A^m-A^{m+1} \\ =l-A^{m+1} \\ ⇒(l_n-A)(l_n+A+A^2+...+A^m)=l_n-A^{m+1} \end{gathered}$$