# Implicit Function Theorem, an application

Let $f: \mathbb{R}^{k+n} \to\mathbb{R}^n$ be of class $C^1$; suppose that $f(\mathbf{a})=0$ and that $Df(\mathbf{a})$ has rank n. Show that if $\mathbf{c}$ is a point of $\mathbb{R}^n$ sufficiently close to $\mathbf{0}$, then the equation $f(\mathbf{x}) = \mathbf{c}$ has solution.

So this sounds almost like a reinstatement of the theorem. Here I will recite it for this question

Let $\mathbf{a} = (a_1,a_2, \dots, a_n) \in \mathbb{R}^{n + k}$; I somehow have to show that $$\det \frac{\partial f}{\partial \mathbf{x}} (\mathbf{a}) \neq 0$$

All I know is that the derivative matrix is invertible, so does this also imply $\det Df(\mathbf{a}) \neq 0$?

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Note that $Df(a)$ is not square. So it does not have a determinant. –  1015 Mar 4 '13 at 3:00
You need to 'split' the variable $a$ into $a_1,a_2$ such that $\frac{\partial f ((a_1, a_2))} {\partial a_1}$ consists of the $n$ linearly independent columns of $Df(a)$. –  copper.hat Mar 4 '13 at 3:29
@copper.hat, would that Jacobian give me a 1 x n matrix? Still not square –  hosun Mar 4 '13 at 3:38

Since $\operatorname{rk} Df(a)=n$, there exists indices $i_1,...,i_n$ such that the columns $\{ Df(a) e_{i_j} \}_{j=1}^n$ are linearly independent ($e_j$ is the $j$th unit vector in $\mathbb{R}^{n+k}$). Let $\pi$ be a permutation of $\{1,...,n+k\}$ such that $\pi_j = i_j$ for $j=1,...,n$, and let $\Pi=\begin{bmatrix} e_{\pi_1} \cdots e_{\pi_{n+k}} \end{bmatrix}$ be the corresponding permutation matrix.
Let $\binom{\hat{x}}{\hat{y}} = \ \Pi^{-1} a$, where $\hat{x} \in \mathbb{R}^n$, $\hat{y} \in \mathbb{R}^k$. Define $\phi:\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$ by $\phi(x,c) = f(\Pi \binom{x}{\hat{y}}) -c$ (note that $\hat{y}$ is 'fixed'), note that $\phi$ is $C^1$ and $\phi(\hat{x},0) = 0$. Furthermore, $\frac{\partial \phi(\hat{x},0)}{\partial x}= \begin{bmatrix} Df(a)e_{\pi_1} \cdots Df(a)e_{\pi_{n}} \end{bmatrix}$, and so is invertible. Hence by the implicit function theorem, there exists neighborhoods $V$ of $\hat{x}$, $U$ of $0$ and a function $\eta:U \to V$ such that $\phi(\eta(c),c) = 0$ for $c \in U$.
Hence $f(\Pi \binom{\eta(c)}{\hat{y}}) =c$, for $c \in U$.