Given $E$ a finite subset of a real vector space $V$, a **circuit** of the associated matroid is a minimal linearly dependent subset of $E$. For each circuit $\underline C$, a minimal linear dependence
$$\sum\_{v\in \underline C}\lambda\_vv=0$$ gives rise to a **signed circuit** $C = (C^+,C^-)$ defined by $$C^+= \{v\in \underline C: \lambda\_v>0\}$$ and $$C^-=\{v\in \underline C:\lambda\_v<0\}.$$ (So $(C^-,C^+)$ is also a signed circuit.)


Given an arbitrary linear dependence of the form
$$\sum\_{v\in S\subset E}\lambda\_v v = 0$$ with $\lambda\_v\neq 0$ for all $v\in S$, by definition there is a circuit $\underline C$ with $\underline C\subset S$.


My question is: if we define $S^+ = \{v\in S:\lambda\_v>0\}$ and $S^-=\{v\in S:\lambda\_v<0\}$, must there exist a *signed* circuit $C$ with $C^+\subset S^+$ and $C^-\subset S^-$? Certainly not every circuit $\underline C\subset S$ has such a signature, but it seems easy enough to find one in particular examples by ad hoc means.


If it's useful or necessary, for my purposes I am only interested in situations where all coefficients are integers.


Thanks.



Circuits in a linear oriented matroid
Yes. If the dependence is minimal, then $(S^+, S^-)$ is a signed circuit. Now suppose that the dependence is not minimal. Then there is a subset $T\subsetneq S$ and a dependence 
$$\sum\_{v\in T} \mu\_v v =0$$
with $\mu\_v \neq 0$ for all $v \in T$. Define $\mu\_v=0$ for $v\in S\setminus T$. Let $t$ be the smallest positive real number such that $\lambda\_v+t\mu\_v=0$ or $\lambda\_v-t\mu\_v=0$ for some $v\in S$. Without loss of generality, suppose that the first case happened. Let $\rho\_v=\lambda\_v+t\mu\_v$ for all $v\in S$. Then 
$$\sum\_{v\in T} \rho\_v v =0,$$
there is a $u\in S\setminus T$ such that $\rho\_u\neq 0$, there is a $w\in T$ such that $\rho\_w=0$, and for all $v \in S$, we have $\lambda\_v\rho\_v \ge 0$. In other words, we obtained a smaller dependence $S\_1 \subsetneq S$, where $S\_1^+\subseteq S^+$ and $S\_1^-\subseteq S^-$. A signed circuit is then obtained by induction.


