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Let ${\bf f}:U\to \mathbb R^{n-k}$ be a continuously differentiable function. Then ${\bf f}^{-1}(0)$ is a manifold if $[{\bf D}{\bf f}(x)]$ is surjective at all $x$. This is equivalent to the condition that $[{\bf D}{\bf f}(x)]\neq 0$ for all $x\in M$.

Given an $M$ described in this way, is it always possible to find an ${\bf f}$ such that $\{\nabla f_1,\dots,\nabla f_{n-k}\}$ is a linearly independent set of partial derivatives?

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Thank you, I knew I was probably being stupid about something. Make that an answer and I'll give it to you. – user21725 Apr 25 '13 at 0:38
Maybe I'm misunderstanding your question, but I'm not sure how my comment answers your question... – Jason DeVito Apr 25 '13 at 0:42
@JasonDeVito, I'm confused actually. It's not true that $[Df(x)]$ is square. The onto condition should mean that the columns are linearly independent, no? Isn't this the conclusion that I require? – user21725 Apr 25 '13 at 0:44
You're right that my comment is wrong - give me a second... – Jason DeVito Apr 25 '13 at 0:46
Ok, in my case the matrix $[Df(x)]$ is $(n-k)\times k$, with each column the $k$ partial derivatives. The onto condition is equivalent to asking that the rows be linearly independent, I think. Which is the same as asking that the $\nabla f_i$'s are linearly independent. – user21725 Apr 25 '13 at 0:49
up vote 1 down vote accepted

The answer is that the partial derivatives will certainly not always be independent, but the gradients will be. This since the gradients will form the rows of the matrix (in order to satisfy the surjectivity condition the matrix $[{\bf D f}]$ must be a wide matrix with full row rank. The partials form the columns of this matrix, and the gradients form the rows. The conclusion follows.

[Note that my statement that the set of gradients was a linearly independent set of partial derivatives is nonsense.]

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Looks like you got it ;-) - sorry to be confusing earlier. – Jason DeVito Apr 25 '13 at 1:00
Thanks for helping me work through it. :) – user21725 Apr 25 '13 at 1:02

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