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I Why is this Isometry a rotation?

  1. Jun 27, 2016 #1
    Hello,

    i need a little help. Did someone have an idea how to prove this?

    Thanks in advance.

    Be ##\Phi## an direct isometry of the euclidean Space ##\mathbb{R}^3## with

    ##\Phi (\begin{pmatrix} 2\\0 \\1 \end{pmatrix})##=##\begin{pmatrix} 2\\1 \\0 \end{pmatrix}## and ##\Phi (\begin{pmatrix} 1\\1 \\0 \end{pmatrix}) ##=## \frac{1}{3} \begin{pmatrix} 1\\4 \\1 \end{pmatrix}##

    Why is ##\Phi## a rotation?
     
  2. jcsd
  3. Jun 27, 2016 #2

    berkeman

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    Staff: Mentor

    All schoolwork-type question should be posted in the Homework Help forums, and you need to show your efforts toward solving the question. Advanced schoolwork questions are sometimes allowed in the technical forums, but only if the poster shows substantial work.

    Can you say what level class this is for, and please show your thoughts for how to proceed?
     
  4. Jun 27, 2016 #3

    fresh_42

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    Is it? Have you calculated the angles between the vectors before and after? Or the matrix wrt the basis (1,0,0), (0,1,0) and (0,0,1)?
     
  5. Jun 27, 2016 #4

    micromass

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    It might also be a good idea to start with defining what you understand under "rotation" and "direct isometry".
     
  6. Jun 28, 2016 #5

    mathwonk

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    you might ask yourself whether every linear isometry might actually be an isometry, and if not why not. If you know the basic theory there is only one number to compute, (and that only up to sign!).
     
  7. Jul 6, 2016 #6

    chiro

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    Hey Doradus.

    In addition to the above advice I'd write down the conditions for a rotation matrix.

    Hint - Think about determinants and other matrix properties of rotation matrices.
     
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