# Showing that a rigid motion preserves distances

For a linear map $\Phi : \mathbb{R}^n \to \mathbb{R}^m$ which has the form $[ x \mapsto Ax]$ for an $n \times m$ matrix $A$.

A rigid motion is an affine map where $A$ is orthogonal. How would you show that a rigid motion preserves distances?

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ok I've changed that, I'm still learning how to use this website. – user53076 Jan 2 '13 at 12:17

Let $x,y \in \mathbb{R}^n$ and let your affine transformation be $f : v \mapsto Av + b$ for some $b \in \mathbb{R}^m$. Then $$\lVert f(x) - f(y) \rVert = \lVert (Ax+b)-(Ay+b) \rVert = \lVert A(x-y) \rVert$$ Now use a familiar property of orthogonal matrices in a vector space to show that this is equal to $\lVert x-y \rVert$.
Hint: If $v$ and $w$ are vectors then $v \cdot w = v^Tw$, and $\lVert v \rVert = \sqrt{v \cdot v}$.