Given Hill's equation $\ddot{x}+k(t)x=0$ such that $k$ is T-periodic ($k(t)=k(t+T)$). We can easily see that if $x(t)$ is a solution of this equation so is $x(t+T)$ (although this doesn't mean $x(t)$ is periodic with $T$)
given two solution $x_1(t)$ and $x_2(t)$ we can write $x_1(t+T)=Ax_1(t)+Bx_2(t)$, or in vector form:
$$ \begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix} x_1(t) \\x_2(t) \end{pmatrix}=\begin{pmatrix} x_1(t+T) \\x_2(t+T) \end{pmatrix}=T \begin{pmatrix} x_1(t) \\x_2(t) \end{pmatrix} \tag1$$
Now a statement in my textbook says that we can write the eigenvectors of matrix $T$ such that $$Ty(t)=e^qy(t)=y(t+T)\tag2$$ in the following way:
$$y(t)=e^{\frac{qt}{T}}u(t)\tag3$$
where $u(t)$ is a periodic function! $$u(t+T)=u(t)\tag4$$
why is this so? How can one show that we can write our function $y(t)$ in this way?
ps: It can also be shown that $\det(T)=1$, I don't know if this is of any help tho.