Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $\lVert v \rVert = \sqrt{v \cdot v}$.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.