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:

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.