# Positive vector in the kernel of an skew-symmetric incidence matrix

Let $$G=(V,A)$$ be an oriented graph, stronlgy connected with $$n\in\mathbb{N}^*$$ vertices. Let $$M\in\mathcal{M}_n(\mathbb{R})=(m_{i,j})$$ be an skew-symmetric matrix of size $$n$$ and rank $$r$$, such that for all $$i,j\in V$$: $$i\longrightarrow j\Longleftrightarrow m_{i,j}>0$$ (hence $$M$$ is a skew-symmetric incidence matrix of $$G$$). Is there any necessary and sufficient condition on $$G$$ and $$r$$ ensuring that there exists a positive vector in $$\text{Ker}(M)$$?

I've tried several matrices for $$n\in\{3,4,5\}$$ and it seems that, given $$G$$ and $$r$$, the fact that there exists a positive vector in $$\text{Ker}(M)$$ does not depend on the exact values of the coefficients $$m_{i,j}$$, but I do not know how to predict that such a vector exists. I have just noticed that given $$G$$, the answer of the question can vary with $$r$$.

• Sounds like a manifestation of (a weighted) Kastelyn's Theorem. $Det(M)$ is the square of the Pfaffian. However in some cases the latter is a sum of terms corresponding to matchings, involving the weights $m_{ij}$ and all contributing with the same sign. See, e.g., arxiv.org/abs/1409.4631 Jun 22, 2020 at 13:49