Stability for higher dimensional dynamical systems

Question

I remember learning that in order for a steady state to be locally stable in a system of two equations, it is sufficient for the Jacobian evaluated at a steady state to have: $$Tr(J)<0$$ $$Det(J)>0$$

Is this true for systems of n dimensions?

EDIT: Changed from "must" to "it is sufficient" after Hans and Evgeny

Answer 1

I think that @RobertIsrael wanted to say that this is only a sufficient condition, not necessary. If equilibrium has Jacobian with such properties then it's hyperbolic (no zero or purely imaginary eigenvalues) and all its eigenvalues lie in the left halfplane, therefore it's asymptotically Lyapunov stable. But there are examples of systems that disobey this conditions and still can be Lyapunov stable or even Lyapunov asymptotically stable:

Example 1.
\begin{array}{ccc} \dot{x} & = & \omega y \\ \dot{y} & = & - \omega x \end{array} Eigenvalues are $\pm i \omega$, this is linear system with center equilibrium. It's Lyapunov stable.

and another one:

Example 2.
\begin{array}{ccc} \dot{x} & = & \omega y - x (x^2+ y^2) \\ \dot{y} & = & - \omega x - y(x^2+y^2) \end{array} Eigenvalues are $\pm i \omega$ again, but this is nonlinear nonhyperbolic focus. If you consider $\frac{d(x^2+y^2)}{dt} = -2(x^2+y^2)^2$ and $\frac{d}{dt}\left ( \arctan \frac{y}{x} \right) = - \omega$ then you clearly see that this equilibrium is asymptotically stable.

If you want similar criterion for higher dimensional systems, then take a look at Routh-Hurwitz stability criterion. It's the same from three points of view: it guarantees the hyperbolicity of equilibrium, it guarantees that eigenalues lie if left halfplane and it describes sufficient conditions for stability in terms of characteristic polynomial coefficients.

Answer 2

Obtaining the eigenvalues of the system can also gives us the trace and the determinant of the Jacobian matrix. $$ \begin{aligned} \text{det} (J) = \lambda_1 \cdot \lambda_2 \cdot \lambda_3 \cdots \lambda_n \\ Tr(J) = \lambda_1 + \lambda_2 + \lambda_3 + \cdots + \lambda_n \end{aligned} $$ The sign of each eigenvalue would gives us the stability of the corresponding eigenvector. But the determinant of the Jacobian can no longer state the local stability of the fixed points.

Say you have 3 dimension with negative eigenvalues for each dimension. So you the condition of stability along each dimension satisfied. But if you rely on the determinant fo the Jacobian matrix: $$ \text{det}(J) = \lambda_1 \lambda_2 \lambda_3 < 0 $$ This tells you that you have a saddle-node fixed point which is not correct. I think the best would be to check each eigenvalue individually but if someone knows better ways, I would be happy to hear.