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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.

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    $\begingroup$ (1) should say it has at least one nontrivial periodic solution: $x = y = 0$ is always a solution, and it is periodic. $\endgroup$ – Robert Israel May 5 '14 at 14:49
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Hints: (1) the complex conjugate of a non-real eigenvalue is an eigenvalue.

(2) there is a real eigenvalue.

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