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I Dimension of the group O(n,R) - How to calc?

  1. Jul 6, 2016 #1
    Hi, I want to find the number of parameters needed to define an orthogonal transformation in Rn.
    As I suppose, this equals the dimension of the orthogonal group O(n,R) - but, correct me if I'm wrong.

    I haven't been able to figure out how to do this yet. If it helps, I know that an orthogonal matix should have n(n+1)/2 "free components".
    That said, I'd appreciate any hint from this point.
     
  2. jcsd
  3. Jul 6, 2016 #2

    micromass

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    An orthogonal tranformation ##A## satisfies ##AA^T = I##. Those are initially ##n^2## constraints, but some coincide since the matrix ##AA^T## always is symmetric even if the matrix is orthogonal. So the amount of constraints consist of an upper diagonal matrix, which has ##\frac{n(n+1)}{2}## entries.

    The amount of freedom is then ##n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}##, which is the dimension.
     
  4. Jul 6, 2016 #3
    Thanks, very concise.
     
  5. Jul 7, 2016 #4

    mathwonk

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    you can also use induction and geometry. such a matrix consists of n orthogonal unit length columns. so just ask how many ways there are to choose these. If n = 1, only two choices exist of unit vectors, so the dimension is zero. then to go from n to n+1, we begin by choosing a unit vector in n+1 space, i.e. a point on the n dimensional sphere in n+1 space. so the inductive step giving the answer fore n+1 space, just adds n to the answer for n space, so just add n to n(n-1)/2 and you get (n+1)n/2.
     
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