Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$S\subseteq \mathbb{R}^N, U\subseteq \mathbb{R}^k, \phi\in C^1(U,\mathbb{R}^N) \ \text{ s.t.} \ \phi(U)=S.$$

$J_\psi$ is the Jacobian matrix. Let $V\subseteq\mathbb{R}^k$ be open, and let $\psi:V\to S$ a bijection so that $h:=\psi^{-1}\circ\phi\in C^1(U,V)$ and $h^{-1}=\phi^{-1}\circ\psi\in C^1(V,U)$

I'm trying to get to the end statement that $J_\psi(h(x))J_h(x)=J_\phi(x)$ for all $x\in U$.

I'm thinking along the lines of chain rules, but the notation is strange as it is. I think that $J_\phi(x)$ would be the values of $\phi$ with respect to the partial derivatives $x_i$, but it's sending me in circles.

Ideas?

share|improve this question
    
Don't be misled: the jacobian is a ttrick to express a linear operator as a matrix. Try thinking in terms of derivatives as linear operators, and apply the chain rule. –  Siminore Oct 6 '12 at 12:39
    
I don't quite follow. If we generalise, say $U\subset \mathbb{R}^n$ is open, $f:U\to \mathbb{R}^m, V\subset \mathbb{R}^m$ open and $g:V\to \mathbb{R}^k$ with $f(U)\subset V$. Letting $x\in U$, the definition of the chain rule is $$d(f\circ g)(x)=dg(f(x))\circ df(x)$$ But I still don't understand how that would fit into this notation. Are $\psi$ and $\phi$ the linear operators?! –  Ronan Oct 6 '12 at 12:51
    
If so, this leads to $$d(\phi\circ \psi)(x)=d\psi (\phi(x))\circ(d\phi(x))$$ If we can say that $d\psi$ is equal to $J_\psi$, and my brain is still undecided on that front, then $$d(\phi\circ \psi)(x)=J_\psi(\psi\circ h(x))\circ d(\psi\circ h(x))$$ Now, if we can expand, and again, not certain that we can: $$d(\phi\circ \psi)(x)=J_\psi(\psi\circ h(x))\circ J_\psi\circ J_h(x)$$ Someone please come along and tell me I'm wrong, this feels non-rigorous. –  Ronan Oct 6 '12 at 13:07

1 Answer 1

Think first of the differentials. If $\rm can$ denotes the canonical basis, in all cases, we have $\left[{\rm d}\phi(x)\right]_{\rm can} = J_\phi(x)$, and so on. And the matrix of the composition if the product of the matrices.

For starters, for all $x \in U$, since $h = \psi^{-1}\circ \phi$, we have: $${\rm d}h(x) = {\rm d}(\psi^{-1}\circ \phi)(x) \color{red}{=} {\rm d}(\psi^{-1})(\phi(x))\circ {\rm d}\phi(x), $$where we used the chain rule in $\color{red}{=}$. On the other hand, take $h(x) \in V$. Since $\psi^{-1}\circ \psi = {\rm id}$ and the identity is linear, we have: $$ {\rm d}({\rm id})(h(x)) = {\rm d}(\psi^{-1}\circ \psi)(h(x)) \color{red}{=} {\rm d}(\psi^{-1})(\psi(h(x)))\circ {\rm d}\psi(h(x)) = {\rm d}(\psi^{-1})(\phi(x))\circ {\rm d}\psi(h(x)),$$where again I used the chain rule in $\color{red}{=}$. By my initial remark: $$\begin{cases} {\rm d}h(x) = {\rm d}(\psi^{-1})(\phi(x))\circ {\rm d}\phi(x) \\ {\rm id} = {\rm d}(\psi^{-1})(\phi(x))\circ {\rm d}\psi(h(x))\end{cases} \implies \begin{cases} \left[{\rm d}h(x)\right]_{\rm can} = \left[d(\psi^{-1})(\phi(x))\circ {\rm d}\phi(x)\right]_{\rm can} \\ \left[{\rm id}\right]_{\rm can} = \left[{\rm d}(\psi^{-1})(\phi(x))\circ {\rm d}\psi(h(x))\right]_{\rm can}\end{cases} $$Proceeding: $$\begin{cases} \left[{\rm d}h(x)\right]_{\rm can} = \left[{\rm d}(\psi^{-1})(\phi(x))\right]_{\rm can}\left[{\rm d}\phi(x)\right]_{\rm can} \\ \left[{\rm id}\right]_{\rm can} = \left[{\rm d}(\psi^{-1})(\phi(x))\right]_{\rm can} \left[{\rm d}\psi(h(x))\right]_{\rm can}\end{cases} \implies \begin{cases} J_h(x) = J_{\psi^{-1}}(\phi(x))J_\phi(x) \\ {\rm id} = J_{\psi^{-1}}(\phi(x)) J_\psi(h(x))\end{cases} $$ From here, we have $(J_{\psi^{-1}}(\phi(x)))^{-1} = J_\psi(h(x))$, so: $$J_h(x) = J_{\psi^{-1}}(\phi(x))J_\phi(x) \implies (J_{\psi^{-1}}(\phi(x)))^{-1}J_h(x) = J_\phi(x) \implies J_\psi(h(x))J_h(x)=J_\phi(x).$$

share|improve this answer

Your Answer

 
discard

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.