$x'=Ax$ has one periodic solution. Prove that all solutions are periodic. I want to prove the following:
1) Suppose
$$A_{2,2}=\begin{pmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \end{pmatrix}$$
 real
   and suppose the system of differential equations
$$\begin{matrix}
        x'=a_{1,1}x+a_{1,2}y\\
        y'=a_{2,1}x+a_{2,2}y
 \end{matrix}$$
has at least one periodic solution $\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} f(t) \\ g(t) \end{pmatrix}$. 
   Show that in this case, all solutions are periodic. 
2) Suppose $A \in M_{n,n}(\mathbb{R})$ is invertible with $n$ odd. Show that there exists a solution to the system of equations $x'=Ax$ that is not periodic.
I came up to the following. 
We proved in lecture that an ODE of this kind has a solution of the form $x(t)=\exp(At)\,x(0)$. As our solution is periodic of period $\omega$, we know: 
$$
\begin{pmatrix} f(t) \\ g(t) \end{pmatrix}
=exp(At)\,\begin{pmatrix} f(0) \\ g(0) \end{pmatrix}=\exp(A(t+\omega))\,\begin{pmatrix} f(0) \\ g(0) \end{pmatrix}=\begin{pmatrix} f(t+\omega) \\ g(t+\omega) \end{pmatrix}$$
My ideas was now to show that $\exp(At)=\exp(A(t+\omega))$ and then deduce that all solutions have to be periodic. But I kinda stuck by how I should prove this. 

Hints: (1) the complex conjugate of a non-real eigenvalue is an eigenvalue.
(2) there is a real eigenvalue.
