Jacobian linearization, does it need to be around a hyperbolic fixed-point? Everything that I read about Jacobian linearization of systems of nonlinear equations is about approximations near hyperbolic fixed-points (cf. the Hartman-Grobman theorem). 
It seems to me that even nonlinear systems, provided that they have a solution, qualify for Taylor expansions. I don't see why a Jacobian linearization of a surface would not hold when evaluated in the neighborhood of any point. What am I not seeing?
 A: It's not that a nonhyperbolic system doesn't have a Taylor expansion around the equilibrium. The question is what conclusions one can draw by looking only at the linear terms in that expansion.
For example, think of the linear system
$$
\begin{pmatrix} \dot x \\ \dot y \end{pmatrix}
=
\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix}
.
$$
(Eigenvalues $\pm i$, so not hyperbolic.)
The orbits are periodic (concentric circles around the origin). Now, for an orbit to exactly return where it started is a very delicate thing, so if you perturb the system by adding higher-order terms, you can easily break this property and get a nonlinar system whose flow is qualitatively very different from the flow of its linear part (for example, the orbits might be spirals towards the origin).
The hyperbolicity condition is a "robustness" condition which is necessary in order to be able to guarantee that the nonlinear system really behaves like its linear part (near the equilibrium, of course).
