# Trouble Proving Invertability

I have this theorem.

So if A is a nonnegative matrix, then the following conditions are equivalent.

1. $$(I-A)^{-1}$$ exists and is non-negative
2. 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.

• An observation: with diagonal similarity, the case where $x$ is positive can be further reduced to the case where $x = (1,\dots,1)$. May 20, 2020 at 19:06
• Note also that because of the Perron Frobenius theorem and because of the infinite series $$(I - A)^{-1} = I + A + A^2 + \cdots,$$ it would suffices to show that $A$ has maximal positive eigenvalue less than $1$. May 20, 2020 at 19:10
• In your proof, how did you conclude that $x = (I - A)^{-1}b$ is positive as opposed to merely non-negative? And what do you mean by "external demand"? May 20, 2020 at 22:11
• Here is a correct proof for $1 \implies 2$: take $b = (1,\dots,1)$ and $x = (I - A)^{-1}b$. $b$ is positive, and because $(I - A)^{-1}$ is non-negative, $x = (I - A)^{-1}b$ must be non-negative. Also, we indeed have $(I - A)x = b$. May 20, 2020 at 22:31
• All we need to do to make my proof work is pick a $b$ with positive (as opposed to non-negative) entries. $b = (1,\dots,1)$ is one such vector. May 21, 2020 at 0:53

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.