Prove that matrix multiplication and inversion is differentiable.

Question

Let $f:GL(n,\mathbb{R})\rightarrow GL(n,\mathbb{R})$ be defined by $f(x)=x^{-1}$ and let $g:GL(n,\mathbb{R})\times GL(n,\mathbb{R}) \rightarrow GL(n,\mathbb{R})$ be defined by $g(x,y)=xy$ where we take the relative topologies from $\mathbb{R}^{n^2}$ and $\mathbb{R}^{n^2}\times \mathbb{R}^{n^2}$ respectively. Prove that $f$ and $g$ are differentiable when considered as functions on open subsets of $\mathbb{R}^{n^2}$ and $\mathbb{R}^{n^2}\times \mathbb{R}^{n^2}$ respectively mapping into $\mathbb{R}^{n^2}$.

So far I've managed to prove that $GL(n,\mathbb{R})$ is open in $\mathbb{R}^{n^2}$ and disconnected. I've also shown that $f$ and $g$ are continuous, but I'm not sure how to show that differentiable follows. Any help would be greatly appreciated.

Answer 1

Hint: For the product, make sure that $(AB)_{ij}=\sum_{k=1}^{n}A_{ik}B_{kj}$, in other words g is polynomial and therefore differentiable. Now, consider $F: GL(n,ℝ)×GL(n,ℝ) \to GL(n,ℝ)×GL(n,ℝ)$ given by $(A,B) \to (A, AB)$ and proof that F is diffeomorphism (User Inverse Theorem), then $F^{-1}(A,B)=(A,AB ^{-1})$ is differentiable, and therefore $f(A)=F^{-1}(Id,A)$.