Our instructor requires us to memorize this table for our differential equations exam. So I wonder if anyone has some deeper intuition or observation to help with this?
For example, I noticed that whenever eigenvalues $< 0$, the stability (for linear and locally linear systems) is stable. Why is this?
@snarski I'm terrified too for the exam. I write about the most general eigenvalue, when $r = \lambda \pm iu$. Solutions to $\mathbf{ x' = Ax }$ are, from BOYCE p161, $x(t) = e^{\lambda t}(\cos(ut) + i\sin(ut)).$ So when $\lambda <0$, then $\lim_{t \to \infty} x(t) = \lim_{t \to \infty} e^{\lambda t}(..) = 0$ ?
The imaginary part only adds oscillations to the general magnitude behavior. So the result only depends on the real part of the eigenvalues. Negative real parts have falling eigenfunctions, positive parts have eigenfunctions that grow over all bounds.
To have (asymptotic) stability, you will need all real parts negative.
If only one real part is positive, there is at least one infinitely growing solution starting close to the fixed point, so the fp is instable.
If some real parts are zero (and all the other negative), the growth behavior depends on the multiplicity of the eigenvalue and, in the non-linear case, the higher order terms of the Taylor expansion.
It makes not really much sense to separate this discussion into real and complex eigenvalues, since the important point is their location on the positive or negative real-part half-plane. Of course, the shape of the two-dimensional phase pictures is different for complex conjugate pairs and pairs of unrelated real eigenvalues.