My intuition for exp is that it tells you how an infinitesimal change accumulates over finite time (think compound interest). So the above expression is equivalent to det(I+εA)=1+εtr(A)+O(ε2). Thus we should think ‘If I perturb the identity matrix, then the amount by which the unit cube grows is proportional to the extent to which each vector is being stretched in the direction it was already pointing’.
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.
Interesting, can you give a simple geometric explanation?
My intuition for exp is that it tells you how an infinitesimal change accumulates over finite time (think compound interest). So the above expression is equivalent to det(I+εA)=1+εtr(A)+O(ε2). Thus we should think ‘If I perturb the identity matrix, then the amount by which the unit cube grows is proportional to the extent to which each vector is being stretched in the direction it was already pointing’.
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