# How does scaling rows to sum to 1, of a positive matrix change the perron vector?

Reposting from math.sx due to lack of response. Let $A$ be a $N\times N$ positive matrix such that $A_{ij}>0$. By Perron-Frobenius theorem, there is a unique positive left eigenvector called Perron vector $x$ corresponding to the largest eigenvalue and also it sums to 1. Call it $x$. Let $D$ be a diagonal matrix such that $B=DA$ is a stochastic matrix with rows summing to 1. Call $y$ the Perron vector of $B$. How are $x$ and $y$ related?

• 1) Can you edit in a link to the m.se question? 2) Have you tried doing this in detail for $N=2$ to see what happens there? – Gerry Myerson Oct 12 '15 at 4:34
• 1) Link is added 2) Yes, I should be doing that first!!. Thanks. – dineshdileep Oct 12 '15 at 4:46
• I am afraid that the reason why you got no answer on math.se is because there is no simple relation between them. – Federico Poloni Oct 12 '15 at 7:33
• A minimal remark: Wlog you can always assume that $A$ is the adjacency matrix of a weighted, undirected graph and then $D$ is the matrix of its (weighed, directed) degrees. I agree with @FedericoPoloni that there should be no simple relation between these Perron eigenvectors, but perhaps this graph approach might give you some keywords to look after. – Delio Mugnolo Oct 12 '15 at 9:12
• Yes, I will take a look. Isn't it a directed graph since $\mathbf{A}$ may or may not be symmetric? – dineshdileep Oct 13 '15 at 4:10