I want to find the expression for the PageRank of a webpage defined as in the original paper of Sergey and Larry (The Anatomy of a Large-Scale Hypertextual Web Search Engine).
Consider a directed graph of $n$ vertices with adjacency matrix $A$. Choose $\alpha\in[0,1]$ and obtain the matrix $A'$ as follows:
If row $i$ of $A$ is composed only by zeros, replace each entry of the row by the number $\frac{1}{n}$, where $n$ is the number of vertices in the graph. Otherwise, divide each entry of the row by $C_{T_{i}}$, the number of non zero entries of the row. Define
$$ P = (1-\alpha)A' + \frac{\alpha}{n}{1_{N}} $$
Where $1_{N}$ is the square matrix of dimension $N$ where every entry is $1$.
Now it is easy to prove that P is an stochastic matrix. Also, we can show that this matrix has a unique stationary distribution $\Pi_{0}$. Furthermore, we can set the following equation to find $\Pi_{0}$:
$$ \Pi_0 = \Pi_0 P $$
Then each entry $\pi_i$ of $\Pi_0$ satisfies
$$ \begin{equation}\label{eq1} \pi_i = \sum_{j=1}^{n}{\pi_j\left(\frac{1-\alpha}{C_{T_j}}+\frac{\alpha}{n}\right)} \end{equation} $$
Considering the previous expression, is it possible to get the following equation?
$$ PR(A) = \alpha + (1-\alpha)\sum_{j=1}^{n}{\frac{PR(W_j)}{C_{T_j}}} $$
According to the expression deduced by first principles we can get
$$ \begin{align}\label{eq2} \pi_i =& (1-\alpha)\sum_{j=1}^{n}{\frac{\pi_j}{C_{T_j}}+\frac{\pi_j\alpha}{n}} \\ =&\sum_{j=1}^{n}\frac{\pi_j\alpha}{n} + (1-\alpha)\sum_{j=1}^{n}{\frac{\pi_j}{C_{T_j}}} \\ =& \frac{\alpha}{n} + (1-\alpha)\sum_{j=1}^{n}{\frac{\pi_j}{C_{T_j}}} \end{align} $$
Is there any way to get rid of that $\frac{1}{n}$?
Thank you
