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.