Show that the vector of ones, e, is an eigenvector of the Google matrix transpose

Asked May 06 '19 at 18:01

Active May 06 '19 at 18:01

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In the Google matrix where $$G=\alpha A+(1-\alpha)\frac{1}{n}ee^T$$

and $e$ is a vector of ones, how do I show that $e$ is the eigenvector of $G^T$ corresponding to the eigenvalue of 1

I need to show that $G^Te=e$ right?

I get $G^Te=(\alpha A^T+(1-\alpha)\frac{1}{n}e^Te)e$, not sure how to proceed from there.

asked May 06 '19 at 18:01

user61871

If $e^T = (1,\ldots,1) \in \mathbb{R}^n$, then $e^Te = n$;
$\ G^T e = (\alpha A^T + (1-\alpha))e = e \iff A^Te = e$
– dcolazin May 06 '19 at 18:09
But the same missing step is $(\alpha A^T+(1-\alpha))e = e$, right? – user61871 May 06 '19 at 18:12
isn't $A^T$ the normalized adjacency matrix? – dcolazin May 06 '19 at 18:14

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