Let $A$ be a continuous function, periodic of period $T > 0$. We are interested in the homogeneous differential system $$\forall t \in \mathbb { R } , \quad X ^ { \prime } ( t ) = A ( t ) X ( t ) \tag{IV.1}$$ We fix $t _ { 0 } \in \mathbb { R }$. We denote by $U$ and $V$ the two solutions of (IV.1) satisfying $U \left( t _ { 0 } \right) = \binom { 1 } { 0 }$ and $V \left( t _ { 0 } \right) = \binom { 0 } { 1 }$, and set $E ( t ) = [ U ( t ) , V ( t ) ]$. Let $t _ { 1 } \in \mathbb { R }$ and $W = \binom { w _ { 1 } } { w _ { 2 } } \in \mathbb { C } ^ { 2 }$. Assume that $E \left( t _ { 1 } \right) W = \binom { 0 } { 0 }$. Show that the function $Y : \left\lvert \, \begin{aligned} & \mathbb { R } \rightarrow \mathbb { C } ^ { 2 } \\ & t \mapsto E ( t ) W = w _ { 1 } U ( t ) + w _ { 2 } V ( t ) \end{aligned} \right.$ is zero. Deduce that for all real $t , E ( t )$ is invertible.
Let $A$ be a continuous function, periodic of period $T > 0$. We are interested in the homogeneous differential system
$$\forall t \in \mathbb { R } , \quad X ^ { \prime } ( t ) = A ( t ) X ( t ) \tag{IV.1}$$
We fix $t _ { 0 } \in \mathbb { R }$. We denote by $U$ and $V$ the two solutions of (IV.1) satisfying $U \left( t _ { 0 } \right) = \binom { 1 } { 0 }$ and $V \left( t _ { 0 } \right) = \binom { 0 } { 1 }$, and set $E ( t ) = [ U ( t ) , V ( t ) ]$.
Let $t _ { 1 } \in \mathbb { R }$ and $W = \binom { w _ { 1 } } { w _ { 2 } } \in \mathbb { C } ^ { 2 }$. Assume that $E \left( t _ { 1 } \right) W = \binom { 0 } { 0 }$. Show that the function $Y : \left\lvert \, \begin{aligned} & \mathbb { R } \rightarrow \mathbb { C } ^ { 2 } \\ & t \mapsto E ( t ) W = w _ { 1 } U ( t ) + w _ { 2 } V ( t ) \end{aligned} \right.$ is zero. Deduce that for all real $t , E ( t )$ is invertible.