Trying to find an isometry T(X)=MX

[viviane363]

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

 

[DrGreg]

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.

 

[g_edgar]

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.