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    Hi

    I have attached the following transformation
    I need to classify it (I believe it to be a rotation around z axis by 90 degrees, followed by a translation....so overall rotation)
    and also prove that it is an isometry. How do i show this?

    I know an isometry transformation is distance preserving...

    Anyone have any ideas relating to this?

    Thanks in advance
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    (Original post by number23)
    Hi

    I have attached the following transformation
    I need to classify it (I believe it to be a rotation around z axis by 90 degrees, followed by a translation....so overall rotation)
    and also prove that it is an isometry. How do i show this?

    I know an isometry transformation is distance preserving...

    Anyone have any ideas relating to this?

    Thanks in advance
    Assuming you're allowed to quote certain facts about isometries showing that the rotation matrix is othogonal should be sufficient, as a translation is certainly an isometry and the combination of two isometries is an isometry.
    If you can't quote this, then more work would need to be done.
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    Well f is an affine transformation which are of the form x \mapsto Ax + b, where A is some linear transformation and b is a vector.

    Now, you want to prove that f is indeed an isometry, then you must prove that |x - y| = |f(x) - f(y)|, which in terms of the matrix and vector translates to |x - y| = |(Ax + b) - (Ay + b)| = |Ax - Ay|. So you only need to check that A is an isometry via the usual Euclidean metric.
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    thanks for that. i think that level of depth should be fine, since im on a placement abroad so theres a slight language barrier.
    So I could just multiply the matrix by its transpose, and if its the identity then its orthogonal... hence distance preserving transfomation
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    (Original post by 0x2a)
    Well f is an affine transformation which are of the form x \mapsto Ax + b, where A is some linear transformation and b is a vector.

    Now, you want to prove that f is indeed an isometry, then you must prove that |x - y| = |f(x) - f(y)|, which in terms of the matrix and vector translates to |x - y| = |(Ax + b) - (Ay + b)| = |Ax - Ay|. So you only need to check that A is an isometry via the usual Euclidean metric.
    Thanks for the answer. For the final part on the RHS you just multiply it all out, and you end up with modulus of a vector in Rn?
    The standard euclidean metric.. is that squaing each element... adding them all up.. then square rooting?
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    (Original post by number23)
    Thanks for the answer. For the final part on the RHS you just multiply it all out, and you end up with modulus of a vector in Rn?
    The standard euclidean metric.. is that squaing each element... adding them all up.. then square rooting?
    Yes

    If you do that then it's obvious that the values of |x - y| and |Ax - Ay| are the same since you're taking away the same values in the rows of the vectors.
 
 
 
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