Trouble Proving Invertability I have this theorem. 
So if A is a nonnegative matrix, then the following conditions are equivalent.


*

*$(I-A)^{-1}$ exists and is non-negative

*There is a non-negative vector $\vec{x}$ so that $(I-A)\vec{x}$ is positive (that is, all entries of $(I-A)\vec{x}$ are positive)


I have proved that 1 implies 2, but I am getting stuck on 2 implies 1.
Here's what I have for 1 implies 2. First we will start with 1 $\rightarrow$ 2. So we assume the first condition to be true, meaning that $(I-A)^{-1}$ exists and is non-negative. By definition $\vec{x}$ can be solved for using the equation $\vec{x} = (I-A)^{-1}\vec{b}$. Additionally, it is known that $\vec{b}$ cannot contain negative values as the external demand can only be positive. Using these two pieces of information, it can be seen that $\vec{x}$ must be positive, as only positive numbers are being used in its calculation. Now, the equation can be modified to solve for $\vec{b}$, giving $(I-A)\vec{x}=\vec{b}$ since we know that $\vec{x}$ and $\vec{b}$ are non-negative we can conclude that there is some non-negative vector $\vec{x}$ such that all the values of $(I-A)\vec{x}$ are positive.
Any help on 2 implies 1 would be greatly appreciated.
 A: A proof of $1 \implies 2$:
Let $e = (1,\dots,1)^T$. First, we show that there is necessarily a positive vector $x'$ for which $(I - A)x'$ is positive. In particular, we note that $\lim_{t \to 0^+}(I - A)(x + te)$ is positive. It follows that there exists a $t > 0$ for which $x' = x + te$ is such that $(I - A)x'$ is positive, and for this $t$, $x'$ is clearly positive.
Let $D = \operatorname{diag}(x')$. We have
$$
0 < D^{-1}(I - A)x' = D^{-1}(I - A)D(D^{-1}x') = (I - D^{-1}AD)(D^{-1}x') =  (I - B)e
$$
where $B = D^{-1}AD$.  We see that $B$ is a matrix for which $(I - B)e$ is positive. That is, we have
$$
0 < (I - B)e \implies Be < e.
$$
Thus, $B$ is a matrix for which all row-sums are at strictly less than $1$. We note that
$$
\rho(B) = \lambda_{\max}(B) \leq \|B\|_\infty = \max_{i=1,\dots,n} \sum_{j=1}^n |b_{ij}| < 1.
$$
Here, $\rho$ denotes the spectral radius and $\|\cdot\|_\infty$ denotes the induced $\infty$-norm.  Because $\rho(B) < 1$, it follows that the "Neumann series" $(I - B)^{-1} = \sum_{i=0}^\infty B^i$ converges. We then see that because $B$ is non-negative, this sum and hence $(I - B)^{-1}$ must be non-negative.
We can conclude that $(I - A)^{-1} = D(I - B)^{-1}D^{-1}$ is also non-negative, which was what we wanted.
