Explicit proof of the derivative of a matrix logarithm

Solution 1:

A simple expression can be derived by manipulating the Taylor series $\ln X = \sum_{n=1}^\infty -\frac{(-1)^n}{n}(X-1)^n$ with the result $$\frac{d}{ds}\ln X(s) = \int_0^1 \frac{1}{1-t\,(1-X(s))} X'(s) \frac{1}{1-t\,(1-X(s))}\, dt\ .$$ While not in closed form, this formula can be easily computed numerically, for example. In the above expressions, 1 is the unit matrix.

To derive: $$\frac{d}{ds}\ln X(s) = -\sum_{n=1}^\infty \frac{(-1)^n}{n}\sum_{a=0}^{n-1}(X-1)^a X' (X-1)^{n-1-a}\\ =-\sum_{a=0}^\infty \sum_{n=a+1}^\infty \frac{(-1)^n}{n}(X-1)^a X' (X-1)^{n-1-a}\\ = -\sum_{a=0}^\infty\sum_{b=0}^\infty\frac{(-1)^{a+b+1}}{a+b+1}(X-1)^a X' (X-1)^{b}\\ = \sum_{a=0}^\infty\sum_{b=0}^\infty \int_0^1 dt\, t^{a+b}(1-X)^a X' (1-X)^{b}\ . $$ On performing the sums over $a$ and $b$ one gets the formula stated above.

Solution 2:

You can write $d\log X = dX\,X^{-1}$ if and only if $X$ and $dX$ commute. In that case, of course: $$ dX\,X^{-1} = X^{-1}dX. $$

In the general case they do not commute, and there is no simple rule for the derivative of the logarithm. Even though the expressions $dX\,X^{-1} $ and $X^{-1}dX$ are called "logarithmic derivatives", as they share some properties with the actual derivatives of the logarithm, they are not.

The reason behind this is that, for general matrices: $$ e^A\,dA\ne d(e^A) \ne dA\,e^A, $$

unless $A$ and $dA$ commute. This can be seen from the definition by the Taylor series: $$ d(e^A) = d \left( 1 + A + \frac{1}{2}A^2 +\dots \right) = 0 + dA + \frac{1}{2}A\,dA + \frac{1}{2}dA\,A +... $$

which is not equal to: $$ dA + dA\,A +...= dA (1+A+...) = dA\,e^A, $$

because $\frac{1}{2}(dA\,A+A\,dA)\ne dA\,A$ in general.

You might feel that if $dA$ is "small", then the commutator is "small". That is a dangerous assumption, the truth is that the commutator is the same order as $dA$, so it matters.