# Why does the trace of the Jacobian and its eigenvalues determine equilibrium stability?

I am a high school student studying differential equations! I just cannot understand exactly why the trace of the Jacobian Matrix and its eigenvalues determine equilibrium stability (I encountered the Jacobian when learning about the Lotka-Volterra equations). Could anybody offer an explanation without invoking topology or Lyapunov stability? Thank you.

• "Why does the trace of the Jacobian and its eigenvalues determine equilibrium stability?" In general they don't, so it is impossible to give an answer to your question. – John B Feb 24 '18 at 13:30
• That's interesting! Most sources I have found claim that whether the eigenvalues of the Jacobian at the equilibrium point(s) are purely imaginary, real, or have both real and imaginary parts determines the kind of stability in a system. I certainly might be wrong so could you please point to a source that opposes that? Thank you very much. – George Orf. Feb 24 '18 at 13:35
• The condition you describe in your comment is not the condition you describe in your question (and in the title). Before asking for references, perhaps make up your mind? – Did Feb 24 '18 at 13:54
• Consider this, math.stackexchange.com/questions/546155/…. But e.g. the trace of $\mathrm{diag([-3,1])}$ is $-2$, however unstable, the trace of $\mathrm{diag([-1,-1])}$ is $-2$, and corrsponding system is stable. – Carlos Feb 24 '18 at 14:20
• I sincerely apologize for any potential discrepancy in the title and comment @Did and would certainly be open to editing the title. I am new on StackExchange and open to constructive feedback. My question is possibly better phrased in my comment above. – George Orf. Feb 24 '18 at 14:32

A system is stable if all eigenvalues $\lambda _i$, satisfy $$Re(\lambda _i) < 0.$$ Otherwise the positive real part of eigenvalue will generate an exponential function which diverges and cause the equilibrium to be unstable.
• Solutions of linearized system are linear combinations of $e^{\lambda _it}$ where $\lambda _i$s are eigenvalues. The linearized system approximates the original system. – Mohammad Riazi-Kermani Feb 24 '18 at 14:42