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Rotation matrix

  1. Feb 5, 2013 #1
    For a mathematician, how is rotation defined in the most general sense?

    Question arose to me because it occured to me that an essential property of the rotation matrix is that it preserves lengths. Is this the only mapping (if not please give me a counterexample) that has this property (identity is also a special case of a rotation of course) and can this be shown?

    Might seem like a very lose question and indeed it is, so speak freely about anything that you think could improve my understanding of the fundamental mathematics behind rotation.
  2. jcsd
  3. Feb 5, 2013 #2
    I would personally define a rotation as everything in [itex]SO_n(\mathbb{R})[/itex]. So the linear maps which preserve length and which preserve the orientation.

    We had some threads about this before though. I remember Fredrik say that he would consider [itex]O_n(\mathbb{R})[/itex] as the rotations for convenience. So my answer is definitely not the only possible answer you can give.

    I'll see if I can dig up some old threads.
  4. Feb 5, 2013 #3
  5. Feb 5, 2013 #4


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    I have seen at least two conventions in the literature:

    1. Members of O(n) are called rotations (or orthogonal transformations), and members of SO(n) are called proper rotations.
    2. Members of O(n) are called orthogonal transformations, and members of SO(n) are called rotations.

    In one of those threads I said that I prefer convention 1. Lately I've been thinking that I prefer convention 2.
  6. Feb 6, 2013 #5


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    Here is a simple concrete example in ##\mathbb{R}^2##: let ##T:\mathbb{R}^2 \rightarrow \mathbb{R}^2## be the map defined by
    $$T\left(\begin{bmatrix}x \\ y \end{bmatrix} \right) = \begin{bmatrix}y \\ x \end{bmatrix}$$
    i.e. we interchange the ##x## and ##y## axes. This is a linear map which preserves lengths, but it is not a rotation in the geometric sense. (There is also a "flip" involved.) Note that the matrix representation of ##T## with respect to the usual basis is
    $$[T] = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$
    so the determinant is -1, which shows that ##T \in O(\mathbb{R}^2)## but ##T \not\in SO(\mathbb{R}^2)##.
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