Question

$$\begin{gathered} \mathbf{If}\mathbf{}\mathbf{we}\mathbf{}\mathbf{have}\mathbf{}\mathbf{an}\mathbf{}\mathit{n}\mathbf{\times }\mathit{n}\mathbf{}\mathbf{matrix}\mathbf{}\mathit{A}\mathbf{}\mathbf{whose}\mathbf{}\mathbf{entries}\mathbf{}\mathbf{are}\mathbf{}\mathbf{positive}\mathbf{}\mathbf{or}\mathbf{}\mathbf{zero}\mathbf{}\mathbf{and}\mathbf{}\mathbf{all}\mathbf{}\mathbf{column}\mathbf{}\mathbf{sums}\mathbf{} \\ \mathbf{of}\mathbf{}\mathbf{A}\mathbf{}\mathbf{are}\mathbf{}\mathbf{less}\mathbf{}\mathbf{than}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{}\mathbf{Let}\mathbf{}\mathbf{r}\mathbf{}\mathbf{be}\mathbf{}\mathbf{the}\mathbf{}\mathbf{largest}\mathbf{}\mathbf{column}\mathbf{}\mathbf{sum}\mathbf{}\mathbf{of}\mathbf{}\mathit{A}\mathbf{.}\mathbf{}\mathbf{Then} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ \mathbf{a}\mathbf{.}\mathbf{}\mathbf{}\mathbf{The}\mathbf{}\mathbf{entries}\mathbf{}\mathbf{of}\mathbf{}{\mathit{A}}^{\mathbf{m}}\mathbf{}\mathbf{are}\mathbf{}\mathbf{less}\mathbf{}\mathbf{than}\mathbf{}\mathbf{or}\mathbf{}\mathbf{equal}\mathbf{}\mathbf{to}\mathbf{}{\mathit{r}}^{\mathbf{m}}\mathbf{,}\mathbf{}\mathbf{for}\mathbf{}\mathbf{all}\mathbf{}\mathbf{positive}\mathbf{}\mathbf{integers}\mathbf{}\mathbf{m}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ \mathbf{b}\mathbf{.}\mathbf{}\mathbf{}\underset{\mathbf{m}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{A}}^{\mathbf{m}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{(}\mathbf{meaning}\mathbf{}\mathbf{that}\mathbf{}\mathbf{all}\mathbf{}\mathbf{entries}\mathbf{}\mathbf{of}\mathbf{}{\mathit{A}}^{\mathbf{m}}\mathbf{}\mathbf{approach}\mathbf{}\mathbf{zero}\mathbf{)}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ \mathbf{c}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{The}\mathbf{}\mathbf{infinite}\mathbf{}\mathbf{series}\mathbf{}{\mathit{l}}_{\mathbf{n}}\mathbf{+}\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{A}}^{\mathbf{m}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{converges}\mathbf{}\mathbf{(}\mathbf{entry}\mathbf{}\mathbf{by}\mathbf{}\mathbf{entry}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ \mathbf{d}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{The}\mathbf{}\mathbf{product}\mathbf{}\mathbf{(}{\mathit{l}}_{\mathbf{n}}\mathbf{-}\mathit{A}\mathbf{)}\left(l_n+A+A^2+...+A^m+...\right)\mathbf{=}\mathit{l}\mathbf{-}{\mathit{A}}^{\mathbf{m}\mathbf{+}\mathbf{1}}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{If}\mathbf{}\mathbf{}\mathbf{‘}\mathbf{m}\mathbf{’}\mathbf{}\mathbf{go}\mathbf{}\mathbf{to}\mathbf{}\mathbf{infinity}\mathbf{,}\mathbf{}\mathbf{then}\mathbf{}\mathbf{} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{(}{\mathbf{l}}_{\mathbf{n}}\mathbf{-}\mathbf{A}\mathbf{)}}^{\mathbf{-}\mathbf{1}}{\mathit{l}}_{\mathbf{n}}\mathbf{+}\mathit{A}\mathbf{+}{\mathit{A}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{A}}^{\mathbf{m}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Short answer

$$\begin{gathered} \mathrm{If}\mathrm{we}\mathrm{have}\mathrm{an}n\times n\mathrm{matrix}\mathrm{A}\mathrm{whose}\mathrm{entries}\mathrm{are}\mathrm{positive}\mathrm{or}\mathrm{zero}\mathrm{and}\mathrm{all}\mathrm{column}\mathrm{sums} \\ \mathrm{of}\mathrm{A}\mathrm{are}\mathrm{less}\mathrm{than}1.\mathrm{Let}\mathrm{r}\mathrm{be}\mathrm{the}\mathrm{largest}\mathrm{column}\mathrm{sum}\mathrm{of}\mathrm{A}.\mathrm{Then} \\ \mathrm{a}.\mathrm{The}\mathrm{entries}\mathrm{of}{A}^m\mathrm{are}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}{r}^m,\mathrm{for}\mathrm{all}\mathrm{positive}\mathrm{integers}\mathrm{m}. \\ \mathrm{b}.\underset{m\to \infty }{lim}{A}^m=0(\mathrm{meaning}\mathrm{that}\mathrm{all}\mathrm{entries}\mathrm{of}{A}^m\mathrm{approach}\mathrm{zero}). \\ \mathrm{c}.\mathrm{The}\mathrm{infinite}\mathrm{series}{l}_n+A+A^2+...+A^m+...\mathrm{converges}(\mathrm{entry}\mathrm{by}\mathrm{entry}) \\ \mathrm{d}.\mathrm{The}\mathrm{product}(l_n-A)\left(l_n+A+A^2+...+A^m+...\right)=l-A^{m+1} \\ \mathrm{If}‘\mathrm{m}’\mathrm{go}\mathrm{to}\mathrm{infinity},\mathrm{then} \\ {(l_n-A)}^{-1}{l}_n+A+A^2+...+A^m+... \end{gathered}$$

Step-by-step solution

 To prove part (a)

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

$$\begin{gathered} 0\le \mathrm{r}\le 1 \\ \mathrm{Thus}\mathrm{we}\mathrm{have} \\ A=\sum _{p=1}^m{a}_{ip}{a}_{pj}\le r \\ A\le r \\ {A}^m\le r^m \end{gathered}$$

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

 To prove part (b)

Since from part (a) we get

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

$$\begin{gathered} \Rightarrow \underset{m\to \infty }{\lim }{A}^m\le \underset{m\to \infty }{\lim }{r}^m=0\left(\because 0\le r<1\right) \\ \Rightarrow \underset{m\to \infty }{\lim }{A}^m=0 \end{gathered}$$

To prove part (c)

From part (a) we have

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

$\Rightarrow A\le r,A^2\le r^2,...A^m\le r^m$

Therefore,

$l_n+A+A^2+...+A^m+...\le 1+r+r^2+...+r^m+...$

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$r\ge 1$.

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

 To prove part (d)

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

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

 Final Answer

$$\begin{gathered} \mathrm{If}\mathrm{we}\mathrm{have}\mathrm{an}n\times n\mathrm{matrix}\mathrm{A}\mathrm{whose}\mathrm{entries}\mathrm{are}\mathrm{positive}\mathrm{or}\mathrm{zero}\mathrm{and}\mathrm{all}\mathrm{column}\mathrm{sums} \\ \mathrm{of}\mathrm{A}\mathrm{are}\mathrm{less}\mathrm{than}1.\mathrm{Let}\mathrm{r}\mathrm{be}\mathrm{the}\mathrm{largest}\mathrm{column}\mathrm{sum}\mathrm{of}\mathrm{A}.\mathrm{Then} \\ \mathrm{a}.\mathrm{The}\mathrm{entries}\mathrm{of}{A}^m\mathrm{are}\mathrm{less}\mathrm{than}\mathrm{or}\mathrm{equal}\mathrm{to}{r}^m,\mathrm{for}\mathrm{all}\mathrm{positive}\mathrm{integers}\mathrm{m}. \\ \mathrm{b}.\underset{m\to \infty }{lim}{A}^m=0(\mathrm{meaning}\mathrm{that}\mathrm{all}\mathrm{entries}\mathrm{of}{A}^m\mathrm{approach}\mathrm{zero}). \\ \mathrm{c}.\mathrm{The}\mathrm{infinite}\mathrm{series}{l}_n+A+A^2+...+A^m+...\mathrm{converges}(\mathrm{entry}\mathrm{by}\mathrm{entry}) \\ \mathrm{d}.\mathrm{The}\mathrm{product}(l_n-A)\left(l_n+A+A^2+...+A^m+...\right)=l-A^{m+1} \\ \mathrm{If}‘\mathrm{m}’\mathrm{go}\mathrm{to}\mathrm{infinity},\mathrm{then} \\ {(l_n-A)}^{-1}{l}_n+A+A^2+...+A^m+... \end{gathered}$$