Lately I was faced by a problem and I can't think some way to solve it:
Let $A(t)$ be an antisymmetryc matrix, for all $t \in I$. Prove that $\phi(t,t_{0})$ (the state-transition matrix) is an orthogonal matrix $\forall t \in I$ in the ODE $$ u'(t)=A(t)u(t). $$
Then, one must prove, using this result, the fundamental theorem of local theory of curves, the same described by Manfredo.
I've tried to write $\phi(t,t_{0}) = \phi(t)(\phi(t_{0}))^{-1} = \phi(t_{0},t)$, but I can't related this to the transpose. Any help will be very appreciated! Thanks in advance!
You know that $ϕ(t_0,t_0)=I_d$, which is, among other things, orthogonal. Now consider the evolution of the orthogonality condition $$ \frac{d}{dt}[ϕ(t,t_0)^Tϕ(t,t_0)-I_d]=[A(t)ϕ(t,t_0)]^Tϕ(t,t_0)+ϕ(t,t_0)^T[A(t)ϕ(t,t_0)] $$ and use the anti-symmetry of $A$.
Another approach to the same calculation is to take a fixed vector $C$ and compute the derivative of the norm square $\|ϕ(t,t_0)C\|_2^2$.