Representing a graph's vertices as linear combinations of paths

Question

I have a (directed graded) graph $G=(V,E)$ with two distinguished subsets of $V$: sources and destinations. I am to show that the set of indicator functions of paths starting in a source and ending in a destination generate the linear spaсe $\mathbb R^V$.

I've found an ad hoc proof for my case but while at it I got the feeling that this is, in general, a pretty natural problem. Are there are any general results of this type?

Clarification. My graph $G$ is a very particular graph which I came across in my research and which turned to have this property: vertices are expressible as linear combinations of paths. I, on the other hand, would like to know now whether this actually holds for any broad enough classes of graphs.

Answer 1

In full generality you have the condition that a collection $\mathcal C$ of vertex subsets fails to span if and only if there is some (signed) weighting of the vertices such that every $C \in \mathcal C$ has zero weight (other than the constant zero weighting). This is just a restating of the algebraic fact that a proper subspace is orthogonal to at least one non-zero vector, but the phrasing makes it easier to choose suitable vectors combinatorially. For example, if the sources $S$ and destinations $T$ are disjoint independent sets then giving the elements of $S$ weight $1$ and the elements of $T$ weight $-1$ witnesses that the $S$–$T$ paths don't span, as every path contains exactly one element of $S$ and one element of $T$.

Depending on what exactly you mean by a graded directed graph, that structure might make it easy to find bad weightings (e.g. if every path passes exactly once between a fixed pair of adjacent levels).