Finding limiting probabilities thorough the $\lambda = 1$ eigenvector

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I'm practicing the Google PageRank algorithm and so far I have the transition matrix

$$P_{\alpha}=\left(\begin{array}{cccc} 3/80 & 3/80 & 37/80 & 37/80\\ 3/80 & 37/80 & 3/80 & 37/80\\ 71/80 & 3/80 & 3/80 & 3/80\\ 77/240 & 3/80 & 77/240 & 77/240 \end{array}\right)$$

I need to find the vector of limiting probabilities via the eigenvector of the eigenvalue $\lambda = 1$ (for educational purposes), in other words $$\underset{k\rightarrow\infty}{\lim}\left(P_{\alpha}^{k}X\right)=g$$

I transposed that matrix and now I have

$$P_{\alpha}^{T}=\left(\begin{array}{cccc} 3/80 & 3/80 & 71/80 & 77/240\\ 3/80 & 37/80 & 3/80 & 3/80\\ 37/80 & 3/80 & 3/80 & 77/240\\ 37/80 & 37/80 & 3/80 & 77/240 \end{array}\right)$$

Then I look for the eigenvector corresponding to $\lambda = 1$. It is $$\mathbf{v}=\left[ \begin{array}{c} \frac{123506 t}{104721} \\\\ \frac{7520 t}{34907} \\\\ \frac{31711 t}{34907} \\\\ t \end{array} \right]=\left[ \begin{array}{c} \frac{123506}{104721} \\\\ \frac{7520}{34907} \\\\ \frac{31711}{34907} \\\\ 1 \end{array} \right] t$$

And here's where I got confused.

I can't find a decent description of looking for stationary probabilities through the eigenvector and I thought that the eigenvector's coordinates will be the numbers I'm looking for. However, the sum of the eigenvector's coordinates is not 1.

Question: What do I do with this eigenvector next? Do I have to normalize it or do I do something else?

I appreciate any help.