# Inversion of matrices is a diffeomorphism.

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?

-
How did you prove the $\text{inv}$ is a homeomorphism ? – Fardad Pouran Mar 8 '15 at 12:26

Here is a hint: what does Cramer's rule tell you about the matrix entries of the inverse in terms of the original matrix?

-
Does this means that, since every entry of the inverse are determinants (that is, the adjunt is the transpose of the matrix of the cofactors, which are determinants), the inversion is a multilinear function times the $\dfrac{1}{det A}$ which are both differentiable, then it is differentiable? – Marra Apr 6 '12 at 0:55
What it means is that since $A^{-1}$ is just $\displaystyle \frac{1}{\det(A)}\text{adj}(A)$ the inverse entries are just rational functions in the entries of the original matrix--thus trivially smooth. – Alex Youcis Apr 6 '12 at 2:11
@GustavoMarra : If you write \det A rather than det A, that not only prevents det from being italicized but also results in proper spacing before and after det, thus: $a\det A$. And it's standard usage. – Michael Hardy Apr 6 '12 at 17:47
Thanks @Michael, I didn't know that det was a LaTeX function! – Marra Apr 6 '12 at 20:09

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.

-
I'll check this. Can I always make this expansion $(I+X)^{-1}=I-X+X^2-...$ or is it possible only when $||X||<1$? – Marra Apr 6 '12 at 0:58
If you take $X \in \mathbb{R}$, you can see that the series fails to converge if $|X| \geq 1$. – copper.hat Apr 6 '12 at 4:17