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 LotkaVolterra equations). Could anybody offer an explanation without invoking topology or Lyapunov stability? Thank you.
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1"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

1That'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

1The 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, https://math.stackexchange.com/questions/546155/proofthatthetraceofamatrixisthesumofitseigenvalues. 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

http://www.scholarpedia.org/article/Equilibrium – Dadep Oct 29 '18 at 09:54
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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.
Mohammad RiaziKermani
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I see! Could you perhaps elaborate on the generation of a divergent exponential function? Thank you! – George Orf. Feb 24 '18 at 14:25

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 RiaziKermani Feb 24 '18 at 14:42

@MohammadRiaziKermani Thank you for your explanation, sir! Can you recommend any sites/books about the linearization of systems? – George Orf. Feb 24 '18 at 14:49

The differential equation book by Polking,Boggess, and Arnold published by Pearson is a down to earth undergraduate book which explains systems very well. – Mohammad RiaziKermani Feb 24 '18 at 14:57


By the way, that the real parts are all nonpositive certainly does not guarantee stability. In dimension $1$, consider $$\dot x=x^3$$ – Did Feb 24 '18 at 17:02

@Did Sure, the linearized system is an approximations anyway. The linearized system for $ \dot x=x^3$ is $ \dot x=0$ which is stable. Whether or not the actual system is stable depends on how well the linearization approximates the actual system. – Mohammad RiaziKermani Feb 24 '18 at 17:57

@MohammadRiaziKermani Yeah, but approximation can be useless as in case of $\dot{x} =x^3$. What is known is that approximation is useful (in terms of stability) when equilibrium has no eigenvalues on imaginary axis. – Evgeny Feb 24 '18 at 18:17

2The question is not whether the linearized system is un/stable but if the original system is. Instead of arguing about trivialities, you could correct the answer. – Did Feb 24 '18 at 18:34

Was not the question how the trace of the Jacobian is related to stability? The answer is clearly that there is no relationship in this direction. – Carlos Feb 24 '18 at 19:05


@Did i just mentioned that the title of this question is about the relationship of the trace and stability, however, it’s not possible to conclude stability or instability by considering the trace of the Jacobian – Carlos Mar 07 '18 at 08:46