Equivalent condition for the stability of mutualistic Lotka-Volterra systems

<div style="clear:both;"></div>Based on numerical observations <a href="http://users.fmg.uva.nl/hvandermaas">Han</a> conjectured the following equivalences, of which here I give a proof (as far as they actually hold): (<a title="if you don't want to install the plugin, here's a pdf" href="http://users.fmg.uva.nl/rgrasman/worknotes/equivalence.pdf">pdf</a>) <latex height=700> \begin{lemma} Let $M \in \mathbb{R}^{K\times K}$ be a non-negative matrix, then the following are equivalent (actually only the first two are equivalent): \renewcommand\theenumi{\roman{enumi}} \newcommand{\abs}{\mathrm{abs}} \begin{enumerate} \item[i] $\Re\,\lambda_i(M-I) < 0$, for $i=1,2,\ldots$---i.e., all eigenvalues of $M$ are negative \label{negRe} \item[ii] $|\lambda_i(M)|<1$ for $i=1,2,\ldots$---i.e., all eigenvalues of $M$ lie within the unit circle of the complex plane\label{unitCircle} \item[iii] $0 < (-1)^K |M-I| $ \label{detCriterion} \end{enumerate} \end{lemma} Although the third statement follows trivially from the first, it is in fact not equivalent with the other two: \begin{proof}\hfill \begin{itemize} \item[(ii) $\Rightarrow$ (i)] %[(\ref{unitCircle} $\Rightarrow$ \ref{negRe})] Clearly $|\lambda(M)|<1 \Leftrightarrow (\Re\;\lambda(M))^2 + (\Im\;\lambda(M))^2 < 1$ which implies that $\Re\;\lambda(M) < 1 \Leftrightarrow \Re\;\lambda(M)-1<0 \Leftrightarrow \Re\;\lambda(M-I) < 0$. \item[(ii) $\Leftarrow$ (i)] %[(\ref{unitCircle} $\Leftarrow$ \ref{negRe})] If $M$ is nonnegative, then $M+\epsilon^2 11'$ is positive for all $\epsilon\in \mathbb{R}$. Let $$\rho(\epsilon) \equiv \max_{i} |\lambda_i(M + \epsilon^2 1 1')|,$$ then clearly $\rho(\epsilon)>0$, and by Theorem 7.44 of Schott (1997) \cite{Schott97}, as indicated above, the corresponding eigenvector $x(\epsilon)$ satisfies $(M + \epsilon^2 11')\,\abs[x(\epsilon)] = \rho_M(\epsilon) \,\abs[x(\epsilon)]$, and thus $\rho_M(\epsilon)$ is a real eigenvalue of $M+\epsilon^2 1 1'$. Therefore (\ref{negRe}) implies $\Re\;\rho_M(\epsilon) < 1$, but $\Re\;\rho_M(\epsilon) = \rho_M(\epsilon) > 0$, and so $0 < \rho_M(\epsilon) < 1$. %Theorem 7.4.6 of Schott (1997) implies that By definition of $\rho_M(\epsilon)$ all eigenvalues of $M+\epsilon^2 1 1'$ satisfy $|\lambda_i(M+\epsilon^2 1 1')| \leq \rho_M(\epsilon)$ and hence $|\lambda_i(M+\epsilon^2 1 1')| < 1$ for all $i=1,2,\ldots$. Continuity of the eigenvalues in $\epsilon$ implies that if $\rho_M(0)<1$ we can always find an $\epsilon$ small enough such that $\rho_M(\epsilon)<1$, and this implies (\ref{unitCircle}) for all nonnegative $M$. \item[(i) $\Rightarrow$ (iii)] %[(\ref{negRe} $\Rightarrow$ \ref{detCriterion})] Trivial. [(\ref{negRe}) implies $(-1)^K |M-I| = (-1)^K (-1)^K \prod |\lambda(M-I)|>0$] \item[(i) $\not\Leftarrow$ (iii)] %[(\ref{negRe} $\not\Leftarrow$ \ref{detCriterion})] A simple counter example shows that (\ref{detCriterion}) does not in general imply (\ref{unitCircle}) or (\ref{negRe}): if we set $M = \mathrm{diag}(3,2)$, then $|M-I| = (3-1)(2-1) = 2 > 0,$ but $\lambda_1(M)=3$ and $\lambda_2(M) = 2$---i.e., both are greater than one in absolute value. \end{itemize} \end{proof} </latex><div style="clear:both; padding-bottom:0.25em"></div>