Given a vector $\vec{u} \in \mathbb{R}^n$. For what functions $\psi(t)$ can $\vec{x}(t) = \psi(t)\vec{u}$ be a solution of $\dot{\vec{x}} = A \vec{x}$ for some $n \times n$ matrix $A$?

I'm trying to prove that $\psi(t)$ has to be of the form $e^{\lambda t}$ for some constant $\lambda$, but I'm not sure about that and I do not know exactly how to prove that.


1 Answer 1


We have $$\dot \psi u= A \psi u$$

Suppose that $\psi(t_0)\neq0$ for some $t_0$. Then by continuity, in some neighbourhood

$$ A u = \frac{\dot\psi}\psi u$$

(and hence $\dot \psi/\psi$ is an eigenvalue of $A$ assuming $u\neq0$.) But the left-hand side is a constant; therefore so is the other, and the result follows.

  • $\begingroup$ The fraction cannot switch continuously between different values since the right-hand side of $$ A u = \frac{\dot\psi}\psi u$$ is constant. It follows that $\frac{\dot\psi}\psi$ is constant and therefore, $\psi(t) = e^{\lambda t}$ for some $\lambda$, right? $\endgroup$
    – Edward
    Commented May 22, 2013 at 1:37
  • $\begingroup$ Indeed! I've simplified that statement to make it clearer. $\endgroup$ Commented May 22, 2013 at 1:41
  • $\begingroup$ I meant left-hand side, of course. Thank you! $\endgroup$
    – Edward
    Commented May 22, 2013 at 1:46

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