Find a periodic orbit for the differential equation Let $B=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, $P=\begin{pmatrix} 3 & 1 \\ 5 & 2 \end{pmatrix}$, and $A=P^{-1}BP$.
Find a periodic orbit for the differential equation $\dot{x}=Bx$
Find a periodic orbit for the differential equation $\dot{x}=Ax$.
So what I did was calculate eigenvalues of matrix $B$ which are $\lambda _1=i$, $\lambda _2=-i$ and its eigenvectors $\begin{pmatrix} i \\ 1\end{pmatrix}$, $\begin{pmatrix} -i \\ 1 \end{pmatrix}$. Then calculated matrix exponential $e^{Bt}=\begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}$. But then I don't know how to proceed afterwards. Somehow the period seems to be $2\pi$.
I know that the solution is $x(t)=e^{Bt}x_0$ which can be written as $\begin{pmatrix}x(t) \\ y(t)\end{pmatrix}=\begin{pmatrix}\cos t & -\sin t \\ \sin t & \cos t\end{pmatrix}\begin{pmatrix}x_0 \\ y_0\end{pmatrix}$.
How can periodic orbit for $\dot{x}=Bx$ be calculated from this?
I guess same approach can be followed for finding periodic orbit for $\dot{x}=Ax$ by calculating eigenvalues and eigenvectors of $P^{-1}AP$ then finding $e^{At}$. But that calculation is tedius. There must be easier way to find periodic orbit for $\dot{x}=Ax$ from matrix $B$. Any thoughts?
 A: For the equation
$\dot x = Bx, \tag 1$
with
$B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \tag 2$
we recall the general solution is given by
$x(t) = e^{B(t - t_0)} x(t_0), \tag 3$
as is easily checked: since
$\dfrac{d}{dt}(e^{B(t - t_0)}) = Be^{B(t - t_0)}, \tag 4$
(3) yields
$\dot x(t) = \dfrac{d}{dt}(e^{B(t - t_0)})x(t_0) =  Be^{B(t - t_0)}x(t_0) = Bx(t). \tag 5$
We observe that
$x(t + 2\pi) =  e^{B(t + 2\pi - t_0)} x(t_0) = e^{(2\pi B) + B(t - t_0)}x(t_0) = e^{2\pi B}e^{B(t - t_0)}x(t_0),  \tag 6$
and since
$B^2 = -I, \tag 7$
it follows that
$e^{Bt} = I\cos t + B\sin t, \tag 8$
in parallel with the well-known Euler identity
$e^{it} = \cos t + i \sin t, \tag 9$
which is easily seen via the power series expansion
$e^{it} = \displaystyle \sum_0^\infty \dfrac{(it)^n}{n!} \tag{10}$
and comparing it term-by-term to
$e^{Bt} = \displaystyle \sum_0^\infty \dfrac{(Bt)^n}{n!} \tag{11}$
using (7); from (8),
$e^{2\pi B} = I \cos 2\pi + B\sin 2\pi = I, \tag{12}$
and
$e^{Bt} \ne I \tag{13}$
for
$0 < t < 2\pi, \tag{14}$
which shows that $e^{Bt}$ is of period $2\pi$; thus
$x(t + 2\pi) = e^{2\pi B}e^{B(t - t_0)}x(t_0) = e^{B(t - t_0)}x(t_0) = x(t),  \tag{15}$
which shows that every non-trivial orbit of (1) is periodic.
We turn to the equation
$\dot x = Ax. \tag{16}$
Given that
$A = P^{-1}BP, \tag{17}$
with
$P = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}, \tag{18}$
we write (16) in the form
$\dot x = P^{=1}BPx, \tag{19}$
whence
$(Px)' = P\dot x = P(P^{-1}BP)x = (BP)x = B(Px), \tag{20}$
which shows that $Px$ satisfies (1); as such, $Px$ may be written in accord with (3):
$(Px)(t) = e^{B(t - t_0)} (Px)(t_0), \tag{21}$
and then via (8):
$(Px)(t) = (I\cos(t - t_0)  + B\sin(t - t_0))(Px)(t_0), \tag{22}$
which we left multiply by $P^{_1}$:
$x(t) = P^{-1}(I \cos(t - t_0) + B \sin(t - t_0)) (Px)(t_0), \tag{23}$
and algebraically re-arrange:
$x(t) = P^{-1} (Px)(t_0) \cos(t - t_0) + P^{-1}B(P x)(t_0)\sin(t - t_0)$
$= (P^{-1} P)x(t_0) \cos(t - t_0) + (P^{-1}BP) x(t_0)\sin(t - t_0)$
$= (I\cos(t - t_0) + A\sin(t - t_0))x(t_0), \tag{24}$
and since via (17) we have
$A^2 = P^{-1}BPP^{-1}BP = P^{-1}B^2P$
$= P^{-1}(-I)P = -P^{-1}IP = -I, \tag{25}$
we may infer exactly as we did in the above discussion of equation (1) that
$e^{A(t - t_0)} = I\cos(t - t_0) + A\sin(t - t_0) \tag{26}$
and so we may write the solution to (16) as
$x(t) = e^{A(t - t_0)} x(t_0). \tag{27}$
