Despite the long physical introduction, I swear this is a mathematical question. While introducing the motion of a point seen by two observers O and O' (with respective Euclidean spaces E3 and E3'), my Professor of Analytical Mechanics stated that there is an isometry between the two spaces (which is a postulate of kinematics) and then claimed that the function P = f(P', t) - which expresses the position of P with respect to O as a function of its position with respect to O' - is an isometry iff it can be written as f(P') = f(W) + Q(P' - W) ∀ P' ∈ E3', where W is an arbitrary point of E3' and Q: E3' → E3 is an orthogonal linear application which does not depend on W (E3' and E3 are the vector spaces related to E3' and E3). The proof of the sufficient condition was made of 4 steps: Q preserves the norms, Q preserves the inner products, Q is linear and Q does not depend on W. In particular, the end of the second step
u′ · v′ = 1/2 [|v′ − u′|^2 − |u′|^2 − |v′|^2] = 1/2 [|v − u|^2 − |u|^2 − |v|^2] = u · v, where u = Q(u') and v = Q(v')
relies on the first step, as |u′| = |Q(u')| and |v′| = |Q(v')|, but I do not think we can say |v′ − u′| = |Q(v') - Q(u')| before proving linearity. What is the reason u' · v' = Q(u') · Q(v') then?