I am trying to understand why the eigenvectors are the $A$ invariant subspaces of a phase portrait.
An A-invariant subspace is defined by the relation $AV \subseteq V$ where $V$ is a subspace and $A$ is some operator
Then there exist a theorem that says given a point $x_o \in V$, $e^{At}x_o \in V$
So suppose we have
Then let $x_o \in L_1$, then $\lim\limits_{t\to\infty} e^{At}x_o = 0$, $0 \in L_1$, so $L_1$ is an invariant subspace..
What about $L_2$? $x_o \in L_1$, then $\lim\limits_{t\to\infty} e^{At}x_o = \infty$ (aside from $x_o = 0$). Can we say that $L_2$ is also an invariant subspace since the line extends to infinity?
Also why are eigenvectors the only invariant subspaces? And does it matter if the eigenvector is convergent or not?
Let $x$ be an eigenvector of $A$, i.e. $A x = \lambda x$, then $V = \{\alpha x | \alpha \in \mathbb{C}\}$ is an $A$-invariant subspace, because $A (\alpha x) = \alpha \lambda x \in V$.
Now observe that $e^{At} x = e^{\lambda t}x$ for all $t \in \mathbb{R}$ (Spectral Mapping Theorem). Therefore if you select $x_0 = x$, the trajectory will be in $V$ forever and it doesn't matter if it converges or not.
This is also true for all linear combinations of eigenvectors of $A$. This is why all the trajectories are also $e^{At}$-invariant.