Hmm, this seems wrong but fixable. Namely, exp(A) is close to (I+A/n)^n, so raising both sides of det(exp(A))=exp(tr(A)) to the power of 1/n gives something like what we want. Still a bit too algebraic though, I wonder if we can do better.
Another thing to say is if A(0)=I then
ddxdet(A(x))∣∣x=0=tr(A′(0)).
Hmm, this seems wrong but fixable. Namely, exp(A) is close to (I+A/n)^n, so raising both sides of det(exp(A))=exp(tr(A)) to the power of 1/n gives something like what we want. Still a bit too algebraic though, I wonder if we can do better.
Another thing to say is if A(0)=I then