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 $\lambda=0$, $\gamma_v(t)=v, \forall t$.

  • $\begingroup$ Sorry why is the derivative of $\gamma_v$ a multiple of $v$? How do you know this. $\endgroup$ – Kamil Nov 9 '18 at 8:00
  • $\begingroup$ 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. $\endgroup$ – Kamil Nov 9 '18 at 8:05
  • $\begingroup$ 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. $\endgroup$ – Laz Nov 9 '18 at 13:12
  • $\begingroup$ I've edited my answer fixing a little bug it had, and also clarifying your second doubt. $\endgroup$ – Laz Nov 9 '18 at 14:28

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.