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Trying to find an isometry T(X)=MX

  1. Oct 19, 2009 #1
    Hi, I have a question:
    I am trying to find an isometry such that T(aU+bV)≠aT(U)+bT(V).
    I have tried so many possibilities. I gave T(X)=MX given that M is a matrix that doesn't have an inverse. But i still can't find a nice matrix that will make the proposition possible.
    Help please
     
  2. jcsd
  3. Oct 19, 2009 #2

    DrGreg

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    Re: Isometry

    If T is given by a matrix, T(X)=MX, then T(aU+bV)=aT(U)+bT(V) is always true, isometry or not. You might like to search for the Mazur-Ulam theorem.
     
  4. Oct 20, 2009 #3
    Re: Isometry

    Maybe a translation? T(x) = x+w for some fixed vector w. It is an isometry. It is not linear (as you request). But it is affine.
     
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