TITLE: Inversion of matrices is a diffeomorphism.
QUESTION [6 upvotes]: I am having problems showing that the function
$$ \operatorname{inv}:G\rightarrow G$$
$$A\rightarrow A^{-1}$$
where $G$ is the set of all invertible $n\times n$ matrices, is a diffeomorphism. I have already shown that such function is a homeomorphism, and its inverse is itself, but I don't know how I can show that this function is differentiable.
The exercise also tells us that the derivative of $\operatorname{inv}$ in $A$ is the linear mapping $M\rightarrow M$ such that $X\rightarrow -A^{-1}\cdot X\cdot A^{-1}$.
Can anybody give me a hint?

REPLY [6 votes]: Another approach would be to note that if $\|X\|<1$, then $(I+X)^{-1} = I-X+X^2-\cdots$. Then consider the expression $(A+X)^{-1}$, where $X$ is a suitably small perturbation. A small computation shows that $(A+X)^{-1} = (I+A^{-1}X)^{-1} A^{-1}$. Then expand using the above series, and look at the linear term of the expansion. Both differentiability and the form of the derivative follow from this expansion.