Let's say we have a $C^1$-function $f:X\to\mathbb{R}^m$ ($X\subset\mathbb{R}^{n+m}$ an open set) and the rank of the matrix $Df(x)$ is $m.$ We'll let $Z=\lbrace x\in X:f(x)=0\rbrace$ and take some $x_0\in Z.$ In addition, we let $T$ be the set of $u\in\mathbb{R}^{n+m}$ such that there is an open $Y\subset\mathbb{R}$ with $0\in Y$ and continuously differentiable $c$ with $c(Y)\subset Z,$ $c(0)=x_0$ and $Dc(0)=u.$

I aim to show that $T$ is an $n$-dimensional subspace of $\mathbb{R}^{n+m},$ and I've got to a point where I'm a bit stuck, so I'm looking for help on where to go next. I wanted to use the implicit function theorem, first taking the $m$ rows that were linearly dependent to the end WLOG, so that they make an $m\times m$ (sub)matrix which must be invertible. Then I could use the implicit function theorem and find some set $X'$ containing $x_0,$ and a set $U\subset\mathbb{R}^n,$ with a $C^1$ $g:V\to\mathbb{R}^m$ so that $Z\cap X'=\lbrace (v,g(v)):v\in V\rbrace.$ The fact that $T$ should be a vector space is more or less straightforward but I can't quite prove that the dimension should be $n,$ but I believe the implicit function theorem can be used for this.

I've also shown, since $T$ looks a lot like a tangent space at $x_0$ or something of the sort, that $\langle Df(x_0),u\rangle=0$ for any $u.$ I'm kind of tempted to let $u$ run over unit vectors or something like that, but I'm not entirely sure if it would be useful here, it would tell me more about the orthogonal complement. I get the feeling I'm quite close to completing the proof, but I just need some help knowing which way to go from here (ie. hints greatly appreciated).


First of all, by the way you defined it, $T$ is indeed the tangent space to $Z$ at the point $x_0$ (if indeed $Z$ is a $C^1$ manifold in the first place), because you construct $T$ as the set of tangent vectors of curves in $Z$ at the point $x_0$. So you have to prove that $Z$ is an $n$-dimensional $C^1$ manifold.

You can use the implicit function theorem by choosing new coordinates with $x_0$ at the origin and the first $m$ coordinates chosen such that the total derivative of $$ \tilde{x}_1,\ldots,\tilde{x}_m \mapsto \tilde{f}(\tilde{x}_1,\ldots,\tilde{x}_m,0,\ldots,0) $$ is still surjective, hence invertible. Here $\tilde{f}$ is simply $f$ represented in the new coordinates.

Then you can apply the implicit function theorem to find $Z$ as the graph of the implicit function $\tilde{g}\colon \mathbb{R}^n \mapsto \mathbb{R}^m$ of the remaining $n$ coordinates.


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