TITLE: How to find the Frechet derivative at $A\rightarrow A^{-1}$ mapping?
QUESTION [6 upvotes]: I am reading on my own the Lectures on the Geometry of Manifolds (http://nd.edu/~lnicolae/Lectures.pdf ) , and got stuck in solving the exercise 1.1.3 (b) . 
The 1.1.3 (b) is :
Let F: $U\rightarrow U$ be defined as $A\rightarrow A^{-1}$.  Show that $%
D_{A}F(H)=-A^{-1}HA^{-1}$ for any $n\times n$ matrix $H.$
$D_{A}F(H)$ is the Frechet derivative of F at A.  H I guess should be the small "drifting", so that it is actually trying to calculate the derivative of $(A+tH)^{-1}$
I tried to expand $(A+H)^{-1}$ as $(I+A^{-1}H)^{-1}A^{-1}=I-A^{-1}H+\frac{%
A^{-2}H^{2}}{2!}-\frac{A^{-3}H^{3}}{3!}+...,$ but it doesn't looks like $%
-A^{-1}HA^{-1}$
Some one can give me a hint?

REPLY [9 votes]: You are almost correct, you made a mistake in your expansion:
Assuming $\|A^{-1}H\| < 1$, we have
$(A+H)^{-1} = (A(I+A^{-1}H))^{-1} = (I+A^{-1}H)^{-1}A^{-1} = \sum_{k=0}^\infty (-1)^k(A^{-1}H)^k A^{-1}$. This gives $(A+H)^{-1} = A^{-1}-A^{-1} H A^{-1} + o(H)$, from which it follows that $ D F(A)(H) = -A^{-1} H A^{-1}$.
(It doesn't matter here, but you can't assume that $A^{-1}$ and $H$ commute. And you have factorials in your expansion, which are incorrect.)