Image of this curve is contained in the ray if and only if v is an eigenvector of A

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 $$\lambda=0$$, $$\gamma_v(t)=v, \forall t$$.
• Sorry why is the derivative of $\gamma_v$ a multiple of $v$? How do you know this. – Kamil Nov 9 '18 at 8:00
• Also, it is not clear to me how for the other inclusion, from the fact that $\gamma_v^{'} (t) = \lambda e^{\lambda t } v$ you conclude that the image is contained in the ray. – Kamil Nov 9 '18 at 8:05
• If the curve is contained in the ray, its tangent vector is in $span(v)$ because that's exactly the tangent space to the ray at any of its points. Geometrically it should be clear, but here it goes: $\gamma_v(t)=f(t)v, \forall t$, where $f$ is differentiable. Now write the definition of the derivative as the limit of a quotient. – Laz Nov 9 '18 at 13:12