We assume from ODE theory that given a smooth A: I → gl(n;R) there exists a
unique smooth solution F : I → gl(n;R), defined on the same interval I on which
A is defined, of the initial value problem F' = FA and F(t0) = F0 ∈ gl(n;R) given.
(i) Show that two solutions Fi : I → GL(n;R) of the ODE F' = FA satisfy
F2 = CF1 for a constant invertible matrix C ∈ GL(n;R).
(ii) Show that for A a constant matrix F(t) = exp(tA) is a solution of F' = FA
(iii) If A: I → gl(n;R) is not constant, why is F(t) = exp(∫A(s)ds [from t0 to t]) not
solving F' = FA, or is it? Explain.
F' = FA then...
|F|' = tr(A)*|F| where |...| signifies the determinant
The Attempt at a Solution
i) I recall once I needed to show the uniqueness of the solutions for a standard complex valued ODE and the solution was given as something like z(t) = z0exp(of an integral) and I wanted to show that w(t) was also a solution. If I recall correctly someone told me to differentiate the ratio of the two (z(t)/w(t)) and I found that it ended up equaling zero or something like that and that showed there was only a constant difference between them or something like that? Can anyone help here.. I'm not sure if this is the right way to show F1 and F2 are actually separate solutions.
ii) For this part I would assume that since we are given the solution we can just plug it into the DE and show that it works, correct?
iii) Would this be because when A is constant it can be taken out of the exp in a nice and easy way when exp(...) is differentiated? I have a feeling there's more to this and ii) but I can't think of anything more complex than just trying the solution.