Consider matrix ODE: $latex \dot{\phi}(t)=A(t)\phi(t)$ Where $latex A(t)$ is a given periodic matrix with period $latex T$, i.e. $latex A(x)=A(x+T), \forall x\in R$. Then the solution $\phi(t)$ satisfied identity: $latex \phi(t+T)=\phi(t)\phi^{-1}(0)\phi(T)$. This could be explained as $latex \phi^{-1}\phi(T)=\int_{0}^T\phi(t)$. Now we consider to solve the equation: $latex e^{TB}=\phi^{-1}(0)\phi(T)$. At least formally it could be solved: $latex B=\frac{1}{T}log(\frac{\phi(T)}{\phi(0)})$. (Unfo…
