Trying to find an isometry T(X)=MX

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The discussion centers on finding an isometry defined by the transformation T(X) = MX, where M is a non-invertible matrix. The user seeks an example where T(aU + bV) ≠ aT(U) + bT(V), which is not achievable with a matrix transformation due to the linearity of matrix operations. The suggestion to explore the Mazur-Ulam theorem and consider affine transformations, such as T(x) = x + w for a fixed vector w, provides a viable alternative that maintains isometric properties without linearity.

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viviane363
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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
 
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viviane363 said:
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.
 


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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