Periodic Solutions 2x2 Linear System "Let $A \in M_{2x2}(\Bbb R)$ such that $\dot x =Ax$ has a non trivial periodic solution $x(t)$. Let $T$ be the period of such solution. Prove all the solutions to the system are $T$ periodic."
I need help writing a proper proof. I know that periodic solutions to this planar linear system come from the solutions associated to complex eigenvalues, and since the matrix is 2x2, if one eigenvalue is complex, then the other one is as well, so all solutions are periodic, but I don't know how to write it properly.
 A: As you know, the matrix $A$ must have $u+iv$ as an eigenvalue for some real $u,v$ with $v$ non-zero. That means that $\lambda=u+iv$ is a solution to the characteristic equation $$\det(A-\lambda I_2)=0,$$ which (as $A$ is a $2\times 2$ matrix with real entries) is necessarily a quadratic equation with real coefficients. In particular, it has the form $$-\lambda^2+b\lambda +c=0\tag{1}$$ for some real $b$ and $c,$ which I leave to you to verify. As copper.hat points out in the comments above, if you recall that non-real solutions to polynomial equations with real coefficients come in conjugate pairs, then we immediately conclude that $\lambda=u-iv$ is another (indeed, the other) solution, but it isn't difficult to deduce that in this case, anyway, as follows:

Making the substitution $\lambda=u+iv$ yields $$-(u+iv)^2+b(u+iv)+c=0\\-u^2-2iuv+v^2+bu+biv+c=0\\v^2-u^2+bu+c+(b-2u)iv=0,$$ whence $$v^2-u^2+bu+c=-(b-2u)iv.$$ However, the left-hand side of this last equation is real, meaning the right-hand side must be, too. Since $v$ is non-zero real, we must therefore have have $b=2u,$ so $$v^2-u^2+2u^2+c=0\\u^2+v^2+c=0,$$
whence $c=-u^2-v^2,$ and so our characteristic equation has the following (equivalent) forms: $$-\lambda^2+2u\lambda-u^2-v^2=0\\\lambda^2-2u+u^2+v^2=0\\(\lambda-u)^2-(-1)v^2=0\\(\lambda-u)^2-i^2v^2=0\\(\lambda-u)^2-(iv)^2=0\\(\lambda-u-iv)(\lambda-u+iv)=0$$ so readily, $\lambda=u\pm iv$ are the solutions to the characteristic equation.

As copper.hat also pointed out in the comments above, the two eigenvalues of $A$ are necessarily purely imaginary (that is, $u=0$). Again, though, this isn't too tricky to show.

Now, having these eigenvalues at our disposal, take any non-zero vector $x_1$ such that $$Ax_1=(u+iv)x_1,$$ and consider $$x=e^{(u+iv)t}x_1=e^{ut}\bigl(\cos(vt)+i\sin(vt)\bigr)x_1.$$ Readily, we can verify that $\dot x=Ax.$
Similarly, if $x_2$ is a non-zero vector such that $Ax_2=(u-iv)x_2$ and we let $x=e^{ut}\bigl(\cos(vt)-i\sin(vt)\bigr)x_2,$ then $\dot x=Ax.$ Consequently, our general solution to $\dot x=Ax$ is of the form $$x=d_1e^{(u+iv)t}x_1+d_2e^{(u-iv)t}x_2,$$ where $d_1$ and $d_2$ are constants. We can instead rewrite this as $$x=e^{ut}\cos(vt)\bigl(d_1x_1+d_2x_2\bigr)+ie^{ut}\sin(vt)\bigl(d_1x_1-d_2x_2\bigr),$$ which makes it clear that, unless $u=0,$ there won't be any nontrivial periodic solutions! Thus, we have $u=0,$ so that the solutions have the form $$x=d_1e^{ivt}x_1+d_2e^{-ivt}x_2,$$ or $$x=d_1\bigl(\cos(vt)+i\sin(vt)\bigr)x_1+d_2\bigl(\cos(vt)-i\sin(vt)\bigr)x_2,$$ which can be verified to be periodic regardless of our choices of $d_1,d_2.$ Moreover, they all have the same fundamental period: $T=2\pi.$

