Recently I posted a book exercises about fundamental matrices I managed to solve. The following one is a little harder imo:
Let $G(x)$ and $H(x)$ be fundamental matrices of the homogeneous linear system of equations $$ y^{\prime}=A(x) y. $$ Show: $G(x)^{-1} H(x)$ does not depend on $x$.
I played a little bit around with the equations:
If $G(x)$ and $H(x)$ are fundamental matrices, then $$ G^\prime(x)=A(x)G(x) $$ and $$ H^\prime(x)=A(x)H(x). $$ Then $$ G^\prime(x)G(x)^{-1}=A(x)=H^\prime(x)H(x)^{-1} $$ and $$ G(x)^{-1}H(x)=A(x)H(x)G^\prime(x)^{-1}=H^\prime(x)G^\prime(x)^{-1}. $$
But it got me nowhere and I wanted to ask for some help. Thanks in advance for any kind of hint or help.
Take the derivative with respect to x
$f(x)=G(x)^{-1}H(x) \Rightarrow f'(x)=\frac{d~(G(x)^{-1})}{dx}H(x)+G(x)^{-1}H'(x)$
$\frac{d~(G(x)^{-1})}{dx}=-G(x)^{-1}G'(x)G(x)^{-1}$
$\Rightarrow f'(x)=-G(x)^{-1}G'(x)G(x)^{-1}H(x)+G(x)^{-1}H'(x) = -G(x)^{-1}A(x)H(x)+G(x)^{-1}H'(x) = -G(x)^{-1}H'(x)+G(x)^{-1}H'(x)=0 \Rightarrow f(x)=constant$