Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\det Df(\mathbf{a}) \neq 0$?

share|cite|improve this question
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

The main complication here is rearranging the indices of the function argument.

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 $c \in U$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.