Let us consider the inversion in $\mathbb R^2$ with respect to the unit circle : $$ f(x,y) = \left(\frac{x}{(x^2+y^2)^{1/2}}, \frac{y}{(x^2+y^2)^{1/2}}\right)$$
I found the Jacobian matrix (and wolfram agree with me) given by $$ \frac{1}{(x^2+y^2)^{3/2}} \begin{pmatrix} y^2 & -xy \\ -xy & x^2 \end{pmatrix}$$
However this has determinant zero which is absurd ... Where is the mistake ? Thanks !
On
I'm having a hard time believing your map is the inversion. In polar, inversion through the unit circle is represent as follows:
$$ (r ,\theta) \to ( \frac{1}{r}, \theta)$$
Now, let's compute the jacobian:
$$ \frac{ \partial( \frac{1}{r}, \theta)}{\partial(r ,\theta)} = \begin{vmatrix} -\frac{1}{r^2} & 0 \\ 0 & 1\end{vmatrix}= - \frac{1}{r^2}$$
On
Let me consider inversion suggested in @peek-a-boo comment and write $\xi = \frac{x}{x^2+y^2}$ and $\eta=\frac{y}{x^2+y^2}$. Then $\frac{\partial x}{\partial \xi} = - \frac{\partial y}{\partial \eta} = \frac{\eta^2-\xi^2}{(\eta^2+\xi^2)^2}$ and $\frac{\partial x}{\partial \eta} = - \frac{\partial y}{\partial \xi} = -\frac{2\eta\xi}{(\eta^2+\xi^2)^2}$. Then Jakobian $$\frac{D(x,y)}{D(\xi,\eta)}=-\frac{1}{(\eta^2+\xi^2)^2}\lt 0$$
On
Your transformation is normalization (transforming a vector into its normalized version). It's not inversion with respect to unit circle which is:
$$(x,y) \ \mapsto \ \left(\dfrac{x}{x^2+y^2},\dfrac{y}{x^2+y^2}\right)$$
See this answer for a detailed computation and geometrical interpretation: https://math.stackexchange.com/q/2621344
That's the case because your map has not inversion. You're taking all elements of $\mathbb{R}^2$ to the unit circle. Thus you can regain the direction of your point, but you cannot regain the distance from $(0,0)$:
We have $f((1,0)) = (1,0)$ and $f((2,0)) = (1,0)$. Thus you have no isomorphism and no invertible map and that means that the Jacobian has a determinant of $0$.