The Inverse and Implicit Function Theorems

A linear system $Ax=y$ has a unique solution for every $y$ exactly when the matrix $A$ is nonsingular, $\det A\ne 0$. The several-variable calculus asks the same question of a nonlinear system

and the answer, locally, is governed by the same determinant condition applied to the derivative. The object playing the role of $\det A$ is the Jacobian determinant $\det Df(x)$. Where it is nonzero, the best linear approximation $Df(x)$ is invertible, and the inverse function theorem guarantees that $f$ itself is invertible on a small neighborhood.

The contraction mapping principle

Both theorems rest on one fixed-point result. A map $T$ on a complete metric space that uniformly shrinks distances has exactly one fixed point, reached as the limit of iterating $T$ from any starting point.

This is the same principle that produces solutions of differential equations by Picard iteration; here it produces a local inverse instead. The iterates form a Cauchy sequence because consecutive gaps shrink geometrically, $d(f_{n+1},f_n)\le {\lambda }^{n}d(f_1,f_0)$, and completeness supplies the limit, which continuity of $T$ pins as the fixed point.

The inverse function theorem

If $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is differentiable at $x_0$ with $\det Df(x_0)\ne 0$, then $Df(x_0)$ is an invertible linear map. Geometrically, the best linear approximation to $f$ at $x_0$ is a linear isomorphism, and the theorem says the nonlinear $f$ inherits invertibility on a neighborhood.

The derivative formula is forced by the chain rule: differentiating $f^{-1}(f(x))=x$ gives $D{f}^{-1}(f(x))\cdot Df(x)=I$, so $D{f}^{-1}$ is the matrix inverse of $Df$. This is the several-variable form of the one-variable identity $(f^{-1}{)}^{\prime }(y)=1/f^{\prime }(x)$ from the inverse function theorem in one variable: reciprocal becomes matrix inverse.

Local, not global. The theorem is local. Two failures show why.

• Vanishing Jacobian. If $\det Df(x_0)=0$, the theorem says nothing, and $f$ may fail to be invertible: $f(x)=x^2$ on $\mathbb{R}$ has $f^{\prime }(0)=0$ and is not injective near $0$. (A nonzero derivative is sufficient but not necessary: $f(x)=x^3$ is invertible near $0$ despite $f^{\prime }(0)=0$.)
• Nonsingular everywhere, still not global. For $f(x,y)=(e^x\cos y,e^x\sin y)$, the Jacobian determinant is $e^{2x}\ne 0$ at every point, so $f$ is locally invertible everywhere. But $f$ is $2\pi$-periodic in $y$, hence not injective on ${\mathbb{R}}^2$. Local invertibility does not add up to global invertibility.

Constructing the inverse by contraction

The proof reduces to a fixed-point problem. After a linear change of variables one may assume $x_0=0$, $f(x_0)=0$, and $Df(0)=I$. To solve $f(x)=y$ for a given small $y$, define

A point $x$ solves $g_y(x)=x$ if and only if it solves $f(x)=y$. Writing $g(x)=x-f(x)$, the hypothesis $Df(0)=I$ gives $Dg(0)=0$, and continuity of $Dg$ makes $g$ shrink distances by a factor $\frac{1}{2}$ on a small ball $B(0,r)$. Then $g_y$ maps $B(0,r)$ into itself for $\Vert y\Vert <r/2$ and is a contraction, so it has a unique fixed point — the sought $x=f^{-1}(y)$. Continuity and then differentiability of $f^{-1}$ follow from the bound $\Vert x_1-x_2\Vert \le 2\Vert f(x_1)-f(x_2)\Vert$ and the fact that matrix inversion is a smooth operation on the open set of invertible matrices.

The two supporting facts about the space of matrices are used repeatedly and worth stating.

Openness holds because $\det$ is continuous and $GL({\mathbb{R}}^n)$ is the preimage of the nonzero reals; smoothness of inversion holds because $B^{-1}=(\det B{)}^{-1}[\mathrm{Cof}B{]}^{\mathsf{T}}$ has entries that are polynomials in the entries of $B$ divided by the nonvanishing $\det B$. These are what let the proof conclude that $[Df(x){]}^{-1}$ varies continuously with $x$, so $f^{-1}$ is genuinely $C^1$.

The implicit function theorem

The inverse function theorem solves $f(x)=y$ for all the unknowns at once. Often one instead has a single relation among several variables and wants to solve for some of them in terms of the rest — to know when a level set is locally the graph of a function. That is the implicit function theorem.¹

Consider $F:{\mathbb{R}}^n\times {\mathbb{R}}^m\to {\mathbb{R}}^m$, written $F(x,y)$ with $x\in {\mathbb{R}}^n$ and $y\in {\mathbb{R}}^m$, and the solution set $\{(x,y):F(x,y)=0\}$. Split the Jacobian of $F$ into the block $D_{x}F$ of partials in the $x$-variables and the square block $D_{y}F$ of partials in the $y$-variables.

The proof is the inverse function theorem in disguise: apply it to the auxiliary map $\Phi (x,y)=(x,F(x,y))$, whose Jacobian is invertible exactly when $D_{y}F$ is, and read off $g$ from the inverse. The derivative formula then comes from differentiating the identity $F(x,g(x))=0$ with the chain rule: $D_{x}F+D_{y}F\cdot Dg=0$.

The geometric reading: $D_{y}F(a,b)$ invertible means the level set is not vertical in the $y$-directions at $(a,b)$, so locally it projects one-to-one onto the $x$-space and is the graph of $g$. At a point where $D_{y}F$ is singular, a vertical tangent, the level set may fold back and fail to be a graph, exactly as a circle cannot be written $y=g(x)$ near its leftmost and rightmost points.

Summary

The two theorems share one hypothesis and one tool: a nonsingular Jacobian, the linear-algebra fact that a matrix is invertible, and the contraction mapping principle, the analysis fact that shrinking maps have fixed points. Wherever the derivative is invertible, the map is locally invertible, and the nonlinear system can be solved for the corresponding variables.²