Exercise 5.3.16

Answers

If $A$ is a (not necessarily square) matrix with row vectors $A_{1}, A_{2}, \dots, A_{r}$ , we have the fact

A^{t} u = A_{1}^{t} + A_{2}^{t} + \dots + A_{r}^{t} .

This vector is equal to $u$ if and only if the sum of entries is $1$ for every column of $A$ . Using this fact we’ve done the proof of Theorem 5.15.

For the Corollary after it, if $M$ is a transition matrix, we have

{(M^{k})}^{t} u = {(M^{t})}^{k} u = {(M^{t})}^{k - 1} u = \dots = u .

And if $v$ is probability vector,

{(Mv)}^{t} u = v^{t} M^{t} u = v^{t} u = (\begin{matrix} 1 \end{matrix}) .

By Theorem 5.15 we get the conclusion.

jephianlin

2011-06-27 00:00