why preserving norm is equivalent to preserving inner product in rigid body transformation

Solution 1:

This is a consequence of the Theorem of Mazur-Ulam: Every bijective isometry $f: E \to F$ between normed spaces is affine.

You have in the link I provided a proof of Mazur-Ulam theorem. In your case (rigit body motion), $g$ is indeed bijective.

Solution 2:

Another way to show it without using the Mazur-Ulam theorem:

$v = V - 0$

$u = U - 0$

$g_*(v) = g(V) - g(0)$

1. Using the rigid-body motion property $||g_*(x)|| = ||x||$ we have:

$||g_*(-v)|| = ||-v|| = ||v|| = ||g_*(v)|| = ||-g_*(v)||$

2. Using the rigid-body motion property $||X - Y|| = ||g(X) - g(Y)||$ we have:

$||-g_*(-v) + g_*(v)|| = ||-(g(-V) - g(0)) + (g(V) - g(0))|| = ||g(V) - g(-V)|| = ||V - (-V)|| = 2||v||$

1 + 2 imply:

$g_*(-v) = -g_*(v)$

Then:

$||u + v|| = ||(U - 0) + (V - 0)|| = ||U - (-V)|| = ||g(U) - g(-V)|| = ||(g(U) - g(0)) - (g(-V) - g(0))|| = ||g_*(u) - g_*(-v)|| = ||g_*(u) + g_*(v)||$

Also with the rigid-body motion property $||g_*(x)|| = ||x||$ we have:

$\sqrt{||g_*(u)||^2 + ||g_*(v)||^2 + 2<u, v>} = \sqrt{||u||^2 + ||v||^2 + 2<u, v>} = ||u + v|| = ||g_*(u) + g_*(v)|| = \sqrt{||g_*(u)||^2 + ||g_*(v)||^2 + 2<g_*(u), g_*(v)>}$

That implies: $<u, v> = <g_*(u), g_*(v)>$