1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Existence of some special kind of isometry in R^n-1

  1. Apr 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Is there a theorem that states the following?

    Let [itex]P= \{ P_{1}, . . . , P_{n} \}[/itex] be the set of n distinct points in [itex]\mathbb{R}^{n-1}[/itex] and [itex]P'= \{ P'_{1}, . . . , P'_{n} \}[/itex] also a set of points in [itex]\mathbb{R}^{n-1}[/itex]. If for all [itex]i,j[/itex] [itex]|P_{i} - P_{j}|=|P'_{i} - P'_{j}|[/itex] then there is an isometry [itex]f: \mathbb{R}^{n-1} \to \mathbb{R}^{n-1}[/itex] such that for all [itex]i[/itex] [itex]f(P_{i})=P'_{i}[/itex].

    Or at least is it ture? If true how to prove it?


    2. Relevant equations



    3. The attempt at a solution
    Geometrically, it seems plausible especially in some cases where such an isometry is a rotation, a translation, a reflection, or a combination of them.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Existence of some special kind of isometry in R^n-1
  1. Vector Basis of R^n (Replies: 0)

Loading...