Consider the matrix:
$A = \pmatrix{-1&1 \\1&-1 \\ 0& 0}$
For this matrix, it is easy to see that no linear combination of its columns will yield a positive vector $b$ (namely, where $b_i > 0, \forall i$). For other matrices, it is easy to see that there exists some linear combination that will yield a desired $b$.
I am interested in the general case: How do I prove whether or not a given matrix $A \in \mathbb{R}^{m \times n}$ ($m > n$) can yield a linear combination that is positive? Note that there are no constraints on the linear combination - I am looking for any $x$ such that $Ax = b, \forall i, b_i > 0$
The Motzkin transposition theorem seems relevant, but I can't find exactly how to exploit it for this use case.