Conjecture: Let $A$ be a $4 \times 4$ row-stochastic, primitive matrix. Let $p_{i}$ be four probabilities such that $p_1 + p_2 + p_3 + p_4 = 1$. Let \begin{align} V &= \text{diag}_i\{p_{i + 1} + p_{i + 3} - p_i \} \\ &= \text{diag}\{p_2 + p_4 - p_1, p_1 + p_3 - p_2, p_2 + p_4 - p_3, p_1 + p_3 - p_4\}. \end{align} Then, $$\rho(A + V) > \rho(A) = 1$$ where $\rho(X)$ denotes the spectral radius of matrix $X$.
Some notes:
- $\rho(A) = 1$ follows directly from Perron-Frobenius.
- $\text{Tr}(V) = 1$.
- $\rho(A + V)$ is a convex function of $V$ (see Cohen 1981.)
- If $p_{i + 1} + p_{i + 3} - p_i > 0$ for all $i$ then the conjecture is true since the derivative of the Perron root with respect to any element is positive (see Theorem 2 of Vahrenkamp 1976).
- If $A$ is doubly-stochastic, the conjecture is true, in fact we have the stronger bound $\rho(A + V) > 1.25$. The proof is the same idea as Lemma 1 of Johnson 1994. I can't quite see how it would generalize to the row-stochastic case since the right-Perron eigenvector will be $\mathbf{1}$, but not necessarily the left-Perron eigenvector.
I have some numerical evidence to support the conjecture, but that's about it.
This is not true. Let $$ B=\pmatrix{1&0&0&0\\ 1&0&0&0\\ 1&0&0&0\\ 1&0&0&0\\ }, \ V=\pmatrix{2s-1\\ &1-2s\\ &&s\\ &&&1-s}, $$ where $V$ is generated from the probability vector $p=(1-s,\,s,0,0)$ where $0<s<\frac12$. Then $\rho(B+V)=\max\{2s,\,1-2s,\,s,\,1-s\}<1$. It follows that if $A$ is a positive (hence primitive) stochastic matrix that is close to $B$, then $\rho(A+V)<1$. A more concrete counterexample can be obtained by putting $A=(1,1,1,1)^T(0.97,\,0.01,\,0.01,\,0.01)$ and $s=0.1$. Numerically we have $\rho(A+V)=0.92865$.