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Question: Let $f: E \rightarrow \mathbb{R}^{m}$ and $E \subset \mathbb{R}^{n}$ And let $g: U \rightarrow \mathbb{R}^{n}$ with $U \subset \mathbb{R}^{m}$. And: $g(f(x)) = x \space \space \space \forall x \in E$. Prove if $m \ge n$ then $Df(a)$ is a matrix of rank $n$.

My attempt:

Case 1: if $m = n$ $Df(a)$ is $m \times n$ matrix, but $m = n$, So $Df(a)$ is a $n \times n$ matrix. Its rank is then $n$.

Case 2: if $m > n$

$D(g \cdot f) = D(g(f(a)) \cdot D(f(a))$ But $g(f(a)) = a$, so $D(g \cdot f) = D(g(a)) \cdot D(f(a))$

The information I have is that $Df(a)$ is a $m \times n$ matrix and

$Dg(a)$ is a $n \times m$ matrix

And then I got stuck

illegalsh
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1 Answers1

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You're on some good tracks here, but you're making a couple of mistakes.

Your Case 1 doesn't work. You show $Df(a)$ is a $n \times n$ matrix, but that only proves that the rank is $\le n$. Fortunately, we can just disregard this case and solve the problem using your "Case 2" approach even if $m = n$.

Your Case 2 is the right strategy here. First, use Chain Rule to write $$D(g \circ f)(a) = Dg(f(a)) \cdot Df(a).$$ For the left side: we know $g \circ f(x) = x$ for all $x \in E$, so $D(g \circ f)(a) = I_n$, the $n \times n$ identity matrix. For the right side: $Dg(f(a))$ is some $n \times m$ matrix, and $Df(a)$ is some $m \times n$ matrix.

Up to this point, I'm just writing a slightly more careful version of your own work. To finish from here we don't need any more calculus; we only need to remember some linear algebra.

  • If we have matrices $A, B$ and $\text{rank}(AB) = n$, then we must have $\text{rank}(A) \ge n$ and $\text{rank}(B) \ge n$. In our problem we know $\text{rank}(I_n) = n$, so we see $\text{rank}(Df(a)) \ge n$.
  • On the other hand, $Df(a)$ is an $m \times n$ matrix, and the rank can't be bigger than the smallest dimension, so we have $\text{rank}(Df(a)) \le n$.

Therefore $\text{rank}(Df(a)) = n$ as required.

David Clyde
  • 5,251