How to prove each entry in $(x\mathbf{I} - \mathbf{A})^{-1}$ is non-increasing non-negative This is at page 265 of the On the Perron roots of principal submatrices of co-order one of irreducible nonnegative matrices by Savchenko.

Given an $n\times n$ symmetric nonnegative matrix $\mathbf{A}$, let $\lambda$ be its largest root. Then each entry in $(x\mathbf{I} - \mathbf{A})^{-1}$ is a non-increasing non-negative function on the interval $x\in (\lambda, \infty)$.

For $x>\lambda$, the matrix $x\mathbf{I}-\mathbf{A}$ is an M-matrix and therefore its inverse $(x\mathbf{I}-\mathbf{A})^{-1}$ must be a nonnegative matrix. But I'm not sure how to prove that each entry is nonincreasing on the interval.
The paper by Savchenko cited Matrix theory by Gantmacher, but I don't have a copy of this book.
 A: Can't one expand $(x\mathbf I - \mathbf A)^{-1})$ in powers of $1/x$:
$$(x\mathbf I - \mathbf A)^{-1} = \frac 1 x \sum_{n\ge0}  \frac 1 {x^n}\mathbf A^n,$$
convergent for all $x>\lambda$?  The $i,j$ entry is a series in powers of the decreasing function $1/x$, the coefficients of which are $(\mathbf A^n)_{ij}$, which are nonnegative because all the entries in $\mathbf A$ are nonnegative.
I don't see why the symmetry is needed here, though.  Maybe it simplifies the proof of the convergence of the power series, by allowing one to assume all the eigenvalues of $\mathbf A$ are real.  But any $x>\rho(\mathbf A)$ should, it seems to me,  make the series converge. Write, if need be, $x>u>\lambda$, so that $\rho(\mathbf A/u)<1$. Then $(\mathbf A/u)^n\to0$ and the series is dominated by $\sum_{n\ge0} (u/x)^n$.
A: Since you have already noted that $(xI-A)^{-1}$ is non-negative for $x > \lambda$ the result follows from entry by entry differentiation. 
We have for $x \in (\lambda,\infty)$
$$ (xI-A)^{-1}(xI-A) = I.$$ Differentiating entry by entry we get
$$\left(\dfrac{d}{dx}(xI-A)^{-1}\right)(xI-A) +(xI-A)^{-1} = 0,$$ i.e.,
$$\dfrac{d}{dx}(xI-A)^{-1} = -(xI-A)^{-1}(xI-A)^{-1},$$ 
because all elements of the RHS are non-positive the result follows.
