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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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2 Answers 2

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 $\lVert v \rVert = \sqrt{v \cdot v}$.

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length of a vector x is x^t*x under trasformation it becomes (Ax)^t*(Ax)=x^tA^tAx=x^tx so norm remains unchanged

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