Chapter 6.3, Problem 29E

Let A be the PageRank transition matrix and let $x_k$ be a vector in the Markov chain with starting probability vector $x_0.$ Since n is very large, the direct multiplication $x_{k+1}=A{x}_k$ is computationally intensive. However, the computation can be simplified dramatically if we take advantage of the structured components of A given in equation (5). Because M is sparse, the multiplication $w_k=M{x}_k$ is computationally much simpler. Show that if we set
$b=\frac{1-p}{n}e$
then
$E{x}_k=e$ and $x_{k+1}=p{w}_k+b$
where M, E, e, and p are as defied in equation (5).