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Let's say we're in $\mathbb{R}^n \times \mathbb{R}^n$ and we have the identity mapping $f:\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n \times \mathbb{R}^n$, $f(x,y) = (x,y)$.

What I want to do is find the Jacobian determinant, but when starting I get a $2\times n$ matrix, not a square one. So, I have a basic misconception about how to find the Jacobian, I thought it would be $$J = \frac{\partial(x,y)}{\partial(x_1,x_2,\ldots,x_n)}$$, but that doesn't seem to really be making sense.

Since it's a conceptual error, it's probably fine to just do $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$, $f(x) = x$.

Hang on a second, I think I'm being dumb and am forgetting that component functions are a thing. I should probably sleep more.


So, to spell it out more, when $n=1$ we have:

$$f(x,y) = \big(f_1(x,y),f_2(x,y)\big),\text{ where } f_1(x,y)=x,f_2(x,y)=y $$ $$ \Rightarrow J=\frac{\partial(f_1,f_2)}{\partial(x,y)}=I_2$$

When $n=2$ we have:

$$f(x,y) = \big(f_1(x,y),f_2(x,y)\big),\text{ where } \\f_1(x,y)=x, f_2(x,y)=y$$

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  • $\begingroup$ Do you still need help with this? $\endgroup$ – Git Gud Apr 24 '14 at 14:01
  • $\begingroup$ Something's not clicking. $\endgroup$ – al92 Apr 24 '14 at 14:01
  • $\begingroup$ @alg I would start by replacing $n$ with $1$, there seems to be part of what's generating your confusion. $\endgroup$ – Git Gud Apr 24 '14 at 14:02
  • $\begingroup$ I added a bit more to my answer. I'm not sure what to do with tuples of tuples. $\endgroup$ – al92 Apr 24 '14 at 14:10
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    $\begingroup$ No, you weren't. Everything is fine with your notation (at the start). I'm now going to tell you something and I hope it might help you. If after this you don't understand what is going on, I will type an answer. It shouldn't be $\dfrac{\partial(x,y)}{\partial(x_1,x_2,\ldots,x_n)}$, but rather $\dfrac{\partial(x,y)}{\partial(x_1,x_2,\ldots,x_n, y_1, \ldots , y_n)}$. $\endgroup$ – Git Gud Apr 24 '14 at 14:38
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As noted in the comments, I'll deal with the case $\varphi\colon \mathbb R^m\to \mathbb R^m, x\mapsto x$.

Any $x\in \mathbb R^m$ equals $(x_1, \ldots , x_m)$, for some $x_1, \ldots ,x_m\in \mathbb R$.

For each $i\in \{1, \ldots, m\}$, let $\varphi_i$ denote the scalar map $\colon \mathbb R^m\to \mathbb R, x\mapsto x_i$.

The jacobian matrix $J_\varphi$ is thus the $m\times m$ matrix that follows: $$\begin{bmatrix} \dfrac{\partial \varphi _1}{\partial x_1}& \ldots &\dfrac{\partial \varphi_1}{\partial x_m}\\ \vdots & \ddots & \vdots\\ \dfrac{\partial \varphi_m}{\partial x_1} & \ldots & \dfrac{\partial \varphi}{\partial x_m}\end{bmatrix}.$$

Since $\dfrac{\partial \varphi _i}{\partial x_j}$ is the null function whenever $i\neq j$ and it is the map $x\mapsto 1$, whenever $i=j$, the jacobian matrix is the identity matrix.

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  • $\begingroup$ The above can be translated to your question which essentially is $m=2n$. But the setting in your question is more troublesome and I see no advantage in using it over the one in my answer. $\endgroup$ – Git Gud Apr 24 '14 at 15:14
  • $\begingroup$ Thanks for the thorough answer! $\endgroup$ – al92 Apr 24 '14 at 15:26
  • $\begingroup$ @al92 You're welcome. $\endgroup$ – Git Gud Apr 24 '14 at 15:28

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