Trying to find an isometry T(X)=MX

  • Thread starter viviane363
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  • #1
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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
 

Answers and Replies

  • #2
DrGreg
Science Advisor
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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
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.
 
  • #3
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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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