Prove or give a counterexample to:
Let $X(t) $ be the solution of the problem of initial value $\dot X=AX$ with $X(t_0)=X_o $ with $X_0\ne 0$. If there exists $t_1 \ne t_0$ s.t. $X(t_1)=X_0$ then A has an eigenvalue $0$ or purely imaginary.
Well I tried proving it and I'm quite lost. Can I get a hint?
I think the answer to this is yes, at least, assuming $A$ has constant entries and the maximal solution is not trivial. This follows from the general theory of ODE's. That is, if $\varphi:I\to \mathbb{R}^n$ is the maximal solution of $x'=Ax$, then either
$1).\ \varphi $ is injective;$\quad $ or $\quad 2).\ \varphi $ is constant; $\quad $or $\quad3).\ \varphi $ is periodic
for, if $\varphi$ is not injective then there are $r<s\in I$ such that $\varphi(r)=\varphi(s):=x_0$.
Set $\psi(t)=\varphi(t+(s-r))).$ Then, $\psi(r)=\varphi(s)=\varphi(r)$ so both $\varphi$ and $\psi$ are solutions to the initial value problem $x'=Ax;\ x(r)=x_0.$ By uniqueness, $\varphi=\psi\Rightarrow \varphi(t)=\varphi(t+(s-r))$ and so $\varphi$ is either periodic or constant.
Of course, if $\varphi$ is injective, then the hypothesis of your question does not hold. So we are left with the remaining two cases.
If $\varphi$ is constant, then $Ax=0$ so either $A$ is singular and so has $0$ as an eigenvalue or $x=0$ which we are excluding.
If $\varphi$ is periodic, then since the solutions of $x'=Ax$ are all of the form $ve^{\lambda t}$ (because $A$ has constant entries) where $v\in \mathbb R^n$ and $\lambda$ is an eigenvalue of $A,\ \lambda$ must be pure imaginary.
