# Spectral radius of a row-stochastic matrix plus a certain diagonal matrix

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 $$0. Then $$\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$$.