Trace-Determinant Plane I'm a bit confused about why a spiral sink or spiral source in a linearization guarantees that the original, nonlinear system also has a spiral sink or spiral source, respectively.  Let's state the rational for why a center in a linearization is inconclusive on the type of the critical point in the original, nonlinear system.


  If the center in the linearization is the point in the
  trace-determinant plane ($0$, $D_L$), then the associated point in the
  nonlinear system may not be ($0$, $D_L$), since the linearization is
  only an approximation; it may be a nearby point ($T_N$, $D_N$)
  instead, where $T_N \neq 0$.


My question is, why isn't it possible for there to be a different system of ODEs where ($T_N$, $D_N$) is the linearized point and ($0$, $D_L$) is the nonlinearized point, in which case the spiral sink or spiral source in the linearization would also be inconclusive?
First of all, it's misleading (and maybe actually the cause of your confusion?) to say that the nonlinear system corresponds to another point in the trace-determinant plane, since “trace” and “determinant” refers specifically to properties of linear systems (only). The nonlinear system corresponds to the same point in the trace-determinant plane as its linearization; it's just that you have additional nonlinear terms which are not taken into account by looking only at the trace and the determinant.

But anyway, the intuitive explanation goes as follows:

If the trace of the linearized system is zero, then the linear system has closed orbits, which is a very delicate property; the slightest nonlinear perturbation might cause the orbit to not return exactly where it started, and instead go into a spiral.

But if the linearized system is a spiral sink, then its orbits are spirals going inwards (the orbits return a bit inwards from where they started, after one lap), and if the perturbation is sufficiently small (which it is if you start close enough to the equilibrium point), then the orbits of the nonlinear system will do the same (maybe they will return a bit further out, but still inwards from where they started, so they too go into a spiral).
First of all, it's misleading (and maybe actually the cause of your confusion?) to say that the nonlinear system corresponds to another point in the trace-determinant plane, since “trace” and “determinant” refers specifically to properties of linear systems (only). The nonlinear system corresponds to the same point in the trace-determinant plane as its linearization; it's just that you have additional nonlinear terms which are not taken into account by looking only at the trace and the determinant.

But anyway, the intuitive explanation goes as follows:

If the trace of the linearized system is zero, then the linear system has closed orbits, which is a very delicate property; the slightest nonlinear perturbation might cause the orbit to not return exactly where it started, and instead go into a spiral.

But if the linearized system is a spiral sink, then its orbits are spirals going inwards (the orbits return a bit inwards from where they started, after one lap), and if the perturbation is sufficiently small (which it is if you start close enough to the equilibrium point), then the orbits of the nonlinear system will do the same (maybe they will return a bit further out, but still inwards from where they started, so they too go into a spiral).