Let
$$f: \mathbb{R}^n \setminus \{ 0\} \to \mathbb{R}^n \setminus \{ 0\}, x \mapsto \frac{x}{\| x \|^2}$$
be the inversion of the unit sphere, which is differentiable. I want to show, that the Jacobian-Matrix of $f$ is proportional to an orthogonal matrix, i.e. that $$\forall x \in \mathbb{R} \setminus \{ 0\}: J_f(x)^\top \cdot J_f(x) = c \cdot I_n$$ for some constant $c \in \mathbb{R}$.
How should I approach this? I tried just calculating it but this turned out to be rather complicated. I tried to calculate the Jacobian-Matrix and found it to be:
$$(a_{ij})_{i,j = 1,…,n} := J_f(x) = \begin{cases}
\cfrac{\|x\|^2 – 2x_j^2}{\|x\|^4} & \text{if } i = j \\
\cfrac{-2x_ix_j}{\|x\|^4} & \text{if } i \not= j
\end{cases}$$
I'm very thankful for any kind of advice!
Best Answer
EDIT: I think I found a ridiculously simple solution: we have $f \circ f (x) = x$ (inversion is an involution) so letting $y=f(x)$ and taking derivatives we have $J_f(y)\cdot J_f(x) = I_n$. Since $y=\frac{x}{\|x\|^2}$ and every entry of $J$ is homogenous of degree $-2$, we have $J_f(y)=\|x\|^4J_f(x)$. Also $J_f$ is a symmetric matrix as your computation shows, and so $J_f=J_f^T$.
So $$J_f(x)^TJ_f(x) = \frac{1}{\|x\|^4}J_f(y)\cdot J_f(x) = \frac{1}{\|x\|^4}I_n$$
Below is my earlier less elegant proof
Here is the explicit calculation that $J^TJ=cI$ (where the constant $c$ is allowed to depend on $x$ as even the case $n=1$ requires). By symmetry, we can say that the norm of each columns of the Jacobian are the same. So it just remains to show that the $(i,j)$ entry of $J^TJ$ is $0$ when $i\neq j$ (midway through I omitted the denominator of $\|x\|^4$)
end of proof.
Here is what I tried when trying unsuccessfully to come up with a elegant solution.
Applying the derivative $\partial/\partial x_i$ to $f(x)^Tx=1$ and using product rule gives $$(\frac{\partial f}{\partial x_i})^Tx + f(x)^T\cdot e_i=0$$ where $e_i$ is standard basis vector. Since $f(x)=c_xx$ for $c_x=-1/|x|^2$ putting all the partial derivatives together to form the Jacobian matrix we hvae $$J_f(x)^Tx=-f(x)=-c_xx$$ But I couldn't figure out where to go from here because as $x$ varies, then $J_f(x)$ varies