Field of Deformation

Question

Let $y(x)$ be a field of deformation s.t. $$ || y(x_1) - y(x_2) || = ||x_1 - x_2 || \ \ \forall x_1, x_2 \in \mathbb R^d$$

Can anybody show me why $y(x) = Qx + a$ holds true for appropriate $ Q \in \mathbb R^{d \times d}$ and $a \in \mathbb R^d$ ?

Answer 1

If

$$ || y(x_1) - y(x_2) || = ||x_1 - x_2 || \ \ \forall x_1, x_2 \in \mathbb R^d$$

then exist $Q, a$ such that $y(x) = Q x + a$ because making $u = x_1-x_2$ and $x = x_2$

$$ || y(x+u) - y(x) || = ||u ||$$

and then

$$ || Q(x+u) - Q x || = ||Q u || = ||u||$$

as for instance $Q = I$ or any rotation matrix because if $R$ is a rotation then $||R u|| = ||u||$ and finally

$$ y(x) = R x + a $$