Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be an isometric transfomation. Then $J^TJ=I_3$, where $J $ is the jacobian matrix of $f$ , and $T$ refers to the transpose of a matrix.
Can some one help me see how this holds?
Note: this is not a homework at all, I was reading this article in computer vision where they state this proposition. No proof is provided, I try to search or derive it by myself, but I fail.
If $f$ preserves distances, then for every $x$ and every vector $h$, the difference $f(x+h)-f(x)$ has norm $\|h\|$. Let $J$ be the Jacobian matrix at $x$, which by definition has the property $$ Jh = \lim_{t\to 0} \frac{f(x+ht)-f(x)}{t} $$ Since $\|f(x+ht)-f(x)\| = |t|\|h\|$, it follows that $\|Jh\|=\|h\|$ for all $h$. In particular, the matrix norm of $J$ is $1$.
Recall the relation of norm and dot product: $$ \|h\|^2 = \|Jh\|^2 = \langle Jh, Jh\rangle = \left\langle J^TJh, h\right\rangle \le \| J^TJh\| \|h\| \le \|h\|^2 $$ which also uses the Cauchy-Schwarz inequality and the fact that $\|J^T\|=\|J\|$. Since we started and ended with $\|h^2\|^2$, equality holds everywhere here, including the Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality turns into equality only when $J^TJh$ is a scalar multiple of $h$. This multiple must be $1$ for the rest of things to be equal. So $J^TJh=h$ as claimed.