Corollary 1 (section 4)

Proof. Let $P$ be the characteristic polynomial of $M$; by hypothesis, $P$ has a root $\alpha \in \Re^{*}_{+}$, which means that $\det (M-\alpha I)=0$. Since $M$ is not degenerate, it has a decomposition in circuits. Let us suppose that $M$ has no positive circuit. $M-\alpha I$ possesses the same decompositions in circuits as $M$: diagonal terms of $M$ were negative and we substracted a positive number, so if they were not 0 in $M$ they are not 0 in $M-\alpha I$; other terms are untouched. If we apply lemma 1 to matrix $M-\alpha I$, which is degenerate but has a decomposition in circuits, we find that $M-\alpha I$ has a positive circuit, and so does $M$. We thus arrive to a contradiction which concludes our proof. $\qedsymbol$