Consider the linear diffeq $$\dot{x} = Ax$$
The general solution has the form $$x(t) = c_{1}e^{\lambda_{1}t}v_{1}+c_{2}e^{\lambda_{2}t}v_{2}$$ if $A = V \Lambda V^{-1}$ and so is diagonalizable.
An observation one makes after looking at a few graphs is that the trajectories satisfy the following properties:
Assuming $0>\lambda_{1}>\lambda_{2}$
$x(t)$ becomes parallel to the direction with smallest absolute eigenvalue as $t\rightarrow \infty$,
and $x(t)$ becomes parallel to the direction with largest absolute eigenvalue as $t\rightarrow -\infty$.
More specifically, how do I show the above statement?