In the article of Evans and Murthy (1977) the following lemma is given:
If $A,b$ satisfy the relationship $$\sum_{i=0}^r a_iA^ib=0 \quad \quad a_i > 0 \quad i = 0,1,\dots,r$$ then any vector $x$ in the subspace $V \leq \mathbb{R}^n$ spanned by $A^ib$, $i=0,1,\dots,r$ can be expressed as a linear combination of $A^ib$, $i=0,1,\dots,r$ with positive coefficients. That is, for any $x \in V$ there exist (nonunique) $c_i > 0, i=0,1,\dots,r$ such that $$x = \sum^r_{i=0} c_iA^ib$$
There is no proof given in the article for this lemma, but I would like to find out the proof for why this holds. I think that, since $a_i >0$, $A^ib \leq 0$ for some $i$. But I don't get any further... Can someone help me with a hint or something?
To illustrate, let $y = -A^1 b + A^2b$. This representation is not valid since the coefficient for $A^1 b$ is negative. Since we have $$a_1 (A^1 b) + \cdots + a_r (A^r b) = 0$$, we can write $$ A^1 b = -\frac{a_2}{a_1} (A^2 b) - \cdots -\frac{a_r}{a_1} (A^r b) $$; the coefficient of $A^j b$ in RHS are all negatives. So we can write $y$ as \begin{align*} y &= - \left( -\frac{a_2}{a_1} (A^2 b) - \cdots -\frac{a_r}{a_1} (A^r b) \right) + A^2 b \\ & = \left( \frac{a_2}{a_1} (A^2 b) + \cdots +\frac{a_r}{a_1} (A^r b)\right) + A^2 b \end{align*} whose coefficients are now all positive.
This special case is just illustrative for general case; all negative coefficients can be flipped into positives via put the following equality to negatively coefficiented terms: $$A^k b = -\frac{1}{a_k}\left( a_1 A^1b + \cdots + a_{k-1}A^{k-1} b + a_{k+1}A^{k+1} b +\cdots + a_r A^r b\right)$$ This is from the fact that $A^i b$ are linealy dependent with positive coefficients.