Let $A \in Mat(n, \mathbb{R})$ be a square matrix. Consider the vector field $Z_A$ on $\mathbb{R}^n$ given by $ Z_A(\vec{x}) = A \vec{x}$ for all $\vec{x} \in \mathbb{R}^n$.
I found the flow of this vector field as $$ \phi : \mathbb{R} \times \mathbb{R}^n \rightarrow \mathbb{R}^n : (t, \vec{x}) \mapsto \exp(tA) \vec{x}. $$ Since this is a global flow, $Z_A$ is a complete vector field.
However, I'm now asked to prove the following:
Problem: For any non-zero vector $v \in \mathbb{R}^n$, denote by $\gamma_v$ the integral curve of $Z_A$ with $\gamma_{v} (0) = v$. Prove the following: the image of $\gamma_v$ is contained in the ray $\left\{t v : t > 0 \right\}$ if and only if $v$ is an eigenvector of $A$.
I'm not sure how to do this. I assume $Im(\gamma_v) \subset \left\{t v : t > 0 \right\}$. Then I need to show $Av = \lambda v$ for some $\lambda \in \mathbb{R}$. Do I need to use the uniqueness of integral flows maybe? I know that $Z_A(v) = Av$. Not sure how to find the $\lambda$.
Also the other direction is not clear to me. Help is appreciated!
You know $\gamma_v(t)=e^{tA}v$, since you already know the flow of $Z$.
Now suppose $Im \gamma_v\subset \{tv|t>0\}$. Then the derivative of $\gamma_v$, $\gamma_v'(t)=e^{tA}Av$ is a multiple of $v$, $\forall t$. In particular $\gamma_v'(0)=Av=\lambda v$ for some $\lambda \in \mathbb{R}$. So, $v$ is eigenvector of $A$.
On the other hand, if $v$ is eigenvector of $A$, $Av=\lambda v$ for some $\lambda \in \mathbb{R}$. Then $\gamma_v'(t)=e^{tA}Av=\lambda e^{tA}v=\lambda e^{\lambda t}v$ (for this last inequality just expand $e^{tA}$ and evaluate in $v$). Integrating both sides we get $\gamma_v(t)=(e^{\lambda t}+C)v$ but making $t=0$ gives us $C=0$. So that $\gamma_v(t)=e^{\lambda t}v$. Then, $Im \gamma_v\subset \{tv|t>0\}$, the curve going from left to right if $\lambda > 0$.
$\textbf{Added}:$ if $\lambda <0$, then also $Im \gamma_v\subset \{tv|t>0\}$, but the curve runs from right to left in this ray.
If $\lambda=0$, $\gamma_v(t)=v, \forall t$.