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Isometries Proofs

  1. Jan 14, 2010 #1
    Have been revising geometry today and have came across some proofs that I can't seem to find in books, but I can't get through either. Any help would be great.

    Let A be a 3x3 orthogonal matrix and let x and y be vectors in R^3

    a) Show that detA = +/- 1

    b) Show that the length of Ax is the same as that of x and that x and y are orthogonal iff Ax and Ay are orthogonal
    Suppose further that A represents a rotation through angleθ , with axis of rotation along the unit vector n, show that if m is a unit vector orthogonal to n, then n.m^Am = sinθ

    attempt at a)

    from defn of orthogonal matrix (Atr.A = I)

    det(Atr . A) = det(I) = 1
    using standard results such as det(A.B) = detA. detB and det(Atr) = det(A)
    we have det(Atr).det(A) = det(A)^2
    therefore det A = +\- 1

    b) struggling to start.

    Thank you in advance
  2. jcsd
  3. Jan 15, 2010 #2


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    For b) if you have a vector v with transpose tr(v), how do you calculate the length of v?

    You also might want to know that tr(x y) = tr(y) tr(x).
  4. Jan 15, 2010 #3
    The only way I know how to find the length of a vector is by squaring and square-rooting. I have never used the transpose to find the length. Ahh im very stuck!
  5. Jan 15, 2010 #4


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    Can you convince yourself that
    ||v||2 = v . vT
    where the dot denotes the inner product?
  6. Jan 15, 2010 #5
    I can sort of see it, but working on a proof right now. Will check back later if I get one out. For some reason I just don't see how this answers the question! Sorry for being a pain and thank you for your advice so far.
  7. Jan 15, 2010 #6
    Your proof of part a is correct.

    For part b.

    Try using the formula <x,Ay> = <Atrx,y> which works because A is orthogonal.

    In the rotation question, you are looking at the area of the parallelogram spanned by the two unit vectors m and Am.
    Last edited: Jan 15, 2010
  8. Jan 15, 2010 #7


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    Well, if the formula I gave is valid for any vector, then in particular it holds for Ax.

    So if you calculate (Ax) . (Ax)T you should get precisely the length x . xT back.
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