I am reading the proof of Lemma 8.7.1 of Algebraic Graph Theory written by Godsil and Royle, and have the following question:
Let $A$ be an $n\times n$ nonnegative irreducible matrix. Here I guess it is sufficient to know that "$A$ being irreducible" implies that $(I+A)^{n-1}$ has all its entries positive. Detailed definition of irreducibility will be in attachment.
For every nonnegative vector $x\neq0$, define $F(x)=\min_{i:x_i\neq0} \frac{(Ax)_i}{x_i}$, as usual the underscript $i$ indicates row index. What makes me confused is that the authors claim that $F((I+A)^{n-1}x)\geq F(x)$ for all $x$.
All I know is that, for $x\neq0$, $(I+A)^{n-1}x$ has all its entries positive, therefore, it is equivalent to show that $\min_{i} \frac{(A(I+A)^{n-1}x)_i}{((I+A)^{n-1}x)_i}\geq \min_{i:x_i\neq0} \frac{(Ax)_i}{x_i}$.
I would be much indebted for any clues!
Given a non-negative $x \neq 0$.
Write $i_0 := \arg\min_{i: x_i \neq 0} \frac{(Ax)_i}{x_i}$.
Then, $$\begin{align*} F((I + A)^{n - 1}x) &= \min_{i: ((I + A)^{n - 1}x)_i\neq 0} \frac{(A(I + A)^{n - 1}x)_i}{((I + A)^{n - 1}x)_i} \\ &= \min_{i} \frac{(A(I + A)^{n - 1}x)_i}{((I + A)^{n - 1}x)_i} \\ &= \min_{i} \frac{((I + A)^{n - 1}(Ax))_i}{((I + A)^{n - 1}x)_i} \\ &= \min_{i} \frac{\sum_{j=1}^n((I + A)^{n - 1})_{ij}(Ax)_j}{\sum_{j=1}^n((I + A)^{n - 1})_{ij}x_j} \\ &\geq \min_{i} \frac{\sum_{j: x_j \neq 0}((I + A)^{n - 1})_{ij}(Ax)_j}{\sum_{j: x_j \neq 0}((I + A)^{n - 1})_{ij}x_j} \\ &\geq \min_{i} \frac{\sum_{j: x_j \neq 0}((I + A)^{n - 1})_{ij}(Ax)_{i_0}}{\sum_{j: x_j \neq 0}((I + A)^{n - 1})_{ij}x_{i_0}} \\ &= \frac{(Ax)_{i_0}}{x_{i_0}} \\ &= F(x). \end{align*}$$