$$
\operatorname{det} \Pi\left(t, t_{0}\right)=1
$$
and hence the characteristic equation for the monodromy matrix
$$
M\left(t_{0}\right)=\left(\begin{array}{cc}
c\left(t_{0}+T, t_{0}\right) & s\left(t_{0}+T, t_{0}\right) \
\dot{c}\left(t_{0}+T, t_{0}\right) & \dot{s}\left(t_{0}+T, t_{0}\right)
\end{array}\right),
$$
is given by
$$
\rho^{2}-2 \Delta \rho+1=0
$$
where
$$
\Delta=\frac{\operatorname{tr}\left(M\left(t_{0}\right)\right)}{2}=\frac{c\left(t_{0}+T, t_{0}\right)+\dot{s}\left(t_{0}+T, t_{0}\right)}{2} .
$$
If $\Delta^{2}>1$ we have two different real eigenvalues
$$
\rho_{\pm}=\Delta \pm \sqrt{\Delta^{2}-1}=\sigma \mathrm{e}^{\pm T \gamma}
$$
with corresponding eigenvectors
$$
u_{\pm}\left(t_{0}\right)=\left(\begin{array}{c}
1 \
m_{\pm}\left(t_{0}\right)
\end{array}\right)
$$
where
$$
m_{\pm}\left(t_{0}\right)=\frac{\rho_{\pm}-c\left(t_{0}+T, t_{0}\right)}{s\left(t_{0}+T, t_{0}\right)}=\frac{\dot{s}\left(t_{0}+T, t_{0}\right)}{\rho_{\pm}-\dot{c}\left(t_{0}+T, t_{0}\right)}
$$
MATH4068 COURSE NOTES ：
$u_{1}, \ldots, u_{m}$ is a basis for $V_{1}$ and $u_{m+1}, \ldots, u_{n}$ is a basis for $V_{2}$, implies that $A$ is transformed to the block form
$$
U^{-1} A U=\left(\begin{array}{cc}
A_{1} & 0 \
0 & A_{2}
\end{array}\right)
$$
in these new coordinates. Moreover, we even have
$$
U^{-1} \exp (A) U=\exp \left(U^{-1} A U\right)=\left(\begin{array}{cc}
\exp \left(A_{1}\right) & 0 \
0 & \exp \left(A_{2}\right)
\end{array}\right)
$$
Hence we need to find some invariant subspaces which reduce $A$. If we look at one-dimensional subspaces we must have
$$
A x=\alpha x, \quad x \neq 0
$$
