Eigenvectors and Invariant Sets

Question

Consider a system $\dot x=Ax$
Suppose $A\nu = \lambda\nu,\, \nu\ne0,\,\lambda\in \Bbb R$
Then how do I show that the line $\{ t\nu\,|\,t\in\Bbb R\}$ is invariant?

Answer 1

Hint: $x(t) = C e^{\lambda t} \nu$ satisfies the ODE.

Answer 2

Let $x = t \nu$. Then, $Ax = (A)(t\nu) = t(A)(\nu) = t\lambda \nu \in \{t \nu \mid t \in \Bbb R\}$.

Answer 3

Let $x(0) = tv,\,v$ is an eigenvector of the system $\dot x = Ax, \, t\in\Bbb R$
Therefore, $tv$ is also and eigenvector of $A$
$x(t)=e^{\lambda t}tv$ (Solution of $\dot x = Ax$ with $t_0=0,\, x(t_0) = tv$)
$x(t)=(te^{\lambda t})v$
$x(t)=Tv,\,T=te^{\lambda t},\,T\in\Bbb R$
So, for all $t>t_0,\,x(t)$ lies in the set.