Re: derivative of the matrix log

Pouya D. Tafti wrote:
Hello everyone,

In one of the exercises for a noncredit course that I am
attending it has been asked to show the following:

d logA = A^-1 dA. (1)

Here A is a nonsingular matrix and d supposedly denotes
differentiation -- I am interpreting it as being with
respect to a scalar parameter, but I may be wrong.

I don't think it is true unless A and dA commute. So I think the restrictions in your proof below are appropriate.



Below is a description of my incomplete attempt at deriving
(1). As you will see, I could certainly benefit from some
advice.

First of all, I remember (somewhat vaguely) from undergrad
school that matrix exponentials may be defined by the
absolutely convergent series

I + sum_i>0 A^i / i!.

I don't remember having seen matrix logarithms before, but
if A is symmetric positive-definite with eigen-decomposition
USU', then

logA := U logS U'

poses itself as an agreeable definition, as it satisfies

exp logA = log expA = A.

For arbitrary A this definition may not work; but then
again, regarding log as the inverse of exp one may generally
write

A = I + sum_i>0 (logA)^i / i!. (2)

Now if for some matrix X, X and dX commute, one can show
that

d(X^n) = n X^(n-1) dX.

Using this and assuming that (2) can be differentiated
term-by-term (by some extension of the corresponding scalar
result), I can derive (1), but only if logA and d(logA)
commute. Is this extra condition really necessary for (1)
to hold? If yes, how restrictive is it? If no, how can one
prove (1) without it?

Thanks very much,
.