Suppose $A$ is an entrywise nonnegative matrix that is symmetric, irreducible and has a zero diagonal. By Perron-Frobenius theorem, the spectral radius $\rho(A)$ is an eigenvalue and $A$ has, up to scaling, a unique entrywise positive eigenvector $v$ (i.e. the Perron vector) for this eigenvalue.
Let $r_i$ denotes the $i$-th row sum of $A$. In general, the maximum row sum of $A$ and the maximum entry of $v$ occur in different positions. For example, when $$ A=\pmatrix{0&0&0&2\\ 0&0&3&2\\ 0&3&0&3\\ 2&2&3&0}, $$ the Perron vector is $(0.2067,\ 0.5236,\ 0.5908,\ 0.5780)^\top$. Therefore the maximum entry of the Perron vector is the third one, but the maximum row sum of $A$ occurs in the fourth row. However, in the computer experiments that I have carried out, this kind of exceptional cases are relatively infrequent. So, it seems that by imposing some mild conditions, one may force $\arg\max_i r_i=\arg\max_i v_i$. Now my question is:
Under what conditions can we conclude that $\arg\max_i r_i=\arg\max_i v_i$?
(Let's ignore the pathological cases where $\arg\max_i r_i$ or $\arg\max_i v_i$ are not unique.) This question arose when I was pondering what could happen if we apply Google style ranking to a weighted undirected graph. If $\arg\max_i r_i=\arg\max_i v_i$, there is no need to compute the Perron vector if I only want to find the highest-ranked member.
