Specifically, if the eigenvalues all have real parts that are negative, then the system is stable near the stationary point. If any eigenvalue has a real part that is positive, then the point is unstable. If the largest real part of the eigenvalues is zero, then the Jacobian matrix does not allow for an evaluation of the stability.

Yes—for a continuous-time autonomous system

x ˙ =f(x),f(x ∗ )=0,

this is the standard linearization criterion, with J=Df(x ∗ ):

If every eigenvalue of J has strictly negative real part, then x ∗ is locally exponentially asymptotically stable. If at least one eigenvalue has strictly positive real part, then x ∗ is unstable. If no eigenvalue has positive real part but at least one has real part 0, linearization is generally inconclusive. Nonlinear terms or a center-manifold analysis are needed.

The last case really can go either way. For example, all three scalar equations below have Jacobian J=0 at x=0:

x ˙ =−x 3 , x ˙ =x 3 , x ˙ =0.

Yet 0 is respectively asymptotically stable, unstable, and neutrally stable.

A slightly more precise wording is therefore:

If the spectral abscissa

α(J)= λ∈σ(J) max

Reλ

is negative, the equilibrium is locally exponentially stable. If α(J)>0, it is unstable. If α(J)=0, the Jacobian test is inconclusive.

This criterion concerns the Jacobian matrix of a dynamical system at an equilibrium; it is unrelated to the “constant Jacobian determinant” condition in the Jacobian conjecture.