Invertibility condition

Question

if linear application v is C1 then I can conclude that: -if det(Jacobian of v)=/0 and v is bijective then the inverse of v exists.

is this correct?

Answer 1

Sure you can but this has no interest at all, you assumed that the linear map is bijective, hence invertible. Even though you drop this assumption this is still meaningless. Indeed, if $\ell$ is linear, then for all $x$, $\mathrm{d}_x\ell=\ell$. Therefore, the invertibility of $\textrm{Jac}_x\ell$ is equivalent to the invertibility of $\ell$. In particular, the inverse function theorem has no role to play here.

Maybe I'll give some insight to clarify the situation. Recall that if $f\colon U\rightarrow\mathbb{R}^n$ is a smooth map on an open subset of $\mathbb{R}^n$, then $\mathrm{d}_xf+f(x)$ is the affine map of $\mathbb{R}^n$ that gives the best approximation of $f$ in a neighbourhood of $x$ in the following sense: $$f(x+h)=f(x)+\mathrm{d}_xf\cdot h+o(h).$$ If $f$ is already linear, then $\mathrm{d}_xf=f$, this correspond to the above intuition. With this interpretation, the inverse function theorem means that if the affine map that approximates the better $f$ near $x$ is invertible, then so is $f$ in a neighbourhood of $x$, this is a kind of a transfer. The inverse function theorem can truly be thought in the following fashion: if the linear obstruction to be invertible vanishes, then the map is locally invertible.

For the record, here is the rigorous statement of the inverse function theorem:

Theorem. Let $f\colon U\rightarrow\mathbb{R}^n$ be a $C^1$-mapping on an open subset of $\mathbb{R}^n$ and let $x\in U$. If $\mathrm{d}_xf\in\textrm{GL}(\mathbb{R}^n)$, then there exists $V$ an open neighborhood of $x$ in $U$ such that $f\colon V\rightarrow f(V)$ is a $C^1$-diffeomorphism.

Sketch of a proof. It is a Newton's method applied to $w\mapsto f(w)-y$ for $y$ in a neighbourhood of $f(x)$. The convergence is ensured by the mean value theorem. In other words, one seeks a fixed point for $w\mapsto w-{\mathrm{d}_xf}^{-1}\cdot(f(w)-y)$. $\Box$