I want to prove the following:

Let $U\subseteq\mathbb{R}^{n}$ be open and $\Phi:U\rightarrow\mathbb{R}^{n}$ be a $C^{1}$ mapping. Let $x\in U$. These are equivalent:
(i) $\Phi$ is regular at $x$
(ii) $\mbox{det}J(\Phi)\Big\vert_{x}\ne0$
(iii) $\text{rank}(D\Phi(x))$ is maximal, where the rank of a linear mapping (whose image is finite-dimensional) is defined to be precisely the dimension of the image. This coincides with the number of linearly independent columns of its matrix, which also coincides with the number of linearly independent lines.

Unfortunately, I am quite shaky in these topics. I would like you to check my proof so far and to aid me in the last part.

Proof. $(i)\Rightarrow(ii):$ Suppose that $\Phi$ is regular at $x$ . In particular, $D\Phi(a)$ is injective. Now, we know that, if $D\Phi(a)$ is an endomorphism (and in this case it is), then it is injective if and only if the determinant of its matrix is nonzero. Using this, we get that $\mbox{det}(J(\Phi)\Big\vert_{a})\ne0$.

$(ii)\Rightarrow(iii):$ Suppose that $\mbox{det}(J(\Phi)\Big\vert_{a})\ne0$ . We know that if the rank of a matrix $n\times n$ is smaller than $n$ , then at least two lines are linearly dependent, and so the determinant of the matrix is zero. By the contrapositive of this result, we get that the rank of $J(\Phi)\Big\vert_{a}$ is $n$ , which is the maximal value for the rank of $D\Phi(x)$ (since its target set is $\mathbb{R}^{n}$).

$(iii)\Rightarrow(i):$ How?

Note: By $\Phi$ is regular at $x$, I mean that $D\Phi(x)\in\text{Aut}.(\mathbb{R^n})$

  • $\begingroup$ Given basic facts of linear algebra there is nothing to prove here. By definition $\Phi:\ {\mathbb R}^n\to{\mathbb R}^m$ is regular at $x$ if $d\Phi(x)$ has maximal rank, which is $=\max\{n,m\}$. $\endgroup$ – Christian Blatter Feb 16 '17 at 10:26
  • $\begingroup$ @Christian, while it doesn't matter for this particular problem, we should probably point out that $\Phi$ is regular at $x$ if $d\Phi(x)$ has rank $m$, not $\max\{n,m\}$. The latter number prohibits any map $\Phi\colon\mathbb{R}^3\to\mathbb{R}^2$, for example, from having any regular points, and this isn't what we want. It also allows a map $\Phi\colon\mathbb{R}\to\mathbb{R}^2$ to have regular points, though such a map has none. $\endgroup$ – 211792 Feb 16 '17 at 14:58
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    $\begingroup$ @AustinChristian: I don't agree, but never mind. E.g., we talk about regular parametrizations of curves or surfaces all the time. $\endgroup$ – Christian Blatter Feb 16 '17 at 15:03
  • $\begingroup$ @Christian: That's a very fair point, and I imagine the difference has to do with settings. I was thinking in terms of things such as Sard's theorem. $\endgroup$ – 211792 Feb 16 '17 at 15:15

Most sources (with which I'm familiar) would say that $x\in U$ is a regular point for $\Phi\colon U\to\mathbb{R}^n$ if $D\Phi(x)$ is a surjective map rather than an injective one. An important insight for this problem is that it doesn't actually matter which definition of regular we use. Can you explain why, in this case, $D\Phi(x)$ is surjective if and only if it is injective?

Using your definition of regular, your argument that (i) implies (ii) is fine. It still works if we use the surjective definition instead, because we're in a situation where surjective and injective are the same thing.

Your argument that (ii) implies (iii) also works. Here's a different argument, for the sake of variety: since $\det(J(\Phi)(x))\neq 0$, the matrix $J(\Phi)(x)$ has a trivial kernel. By the rank-nullity theorem, this matrix must have rank $n$.

Finally, we can show that (iii) implies (i) by pointing out that if the linear map $D\Phi(x)$ has full rank, then it is surjective, since its image has the greatest possible dimension. According to the surjective definition of regularity, we're finished -- $\Phi$ is regular at $x$. If we use the definition of regularity you've provided, then we point out that since $D\Phi(x)$ is surjective, it must also be injective (why?), so $\Phi$ is regular at $x$.

The linchpin here is the fact that $D\Phi(x)$ is injective if and only if it is surjective. Can you prove this?

  • $\begingroup$ I added the definition I used of regular to my question. It's from Kolt's book. $\endgroup$ – Soap Feb 16 '17 at 10:09
  • $\begingroup$ To answer your (very pertinent (!)) question: The dimensions of the domain and codomain are finite and equal (both are equal to $n$). In these cases, $f$ is injective iff bijective iff surjective. $\endgroup$ – Soap Feb 16 '17 at 10:22
  • $\begingroup$ One question: you said "then it is surjective, since its image has the greatest possible dimension." This is precisely what I did not know if I could conclude: couldn't the image be a subset of $\mathbb{R}^n$ of dimension $n$? Or any vector subspace of $\mathbb{R}^n$ with dimension $n$ must be $\mathbb{R}^n$ itself? $\endgroup$ – Soap Feb 16 '17 at 10:29
  • $\begingroup$ Yes, any $n$-dimensional subspace $W$ of an $n$-dimensional vector space $V$ must be $V$ itself. If $W$ were properly contained inside $V$, we could take a basis $w_1,\ldots,w_n$ of $W$ and then choose some $v\in V$ not in $W$. Then $w_1,\ldots,w_n,v$ would give us a collection of $n+1$ linearly independent vectors in $V$, and this can't happen. $\endgroup$ – 211792 Feb 16 '17 at 13:38

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