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Continuity argument?

  1. Sep 28, 2005 #1
    What's a continuity argument? For example, a question asks to prove that the determinant of a rotation matrix is always 1 using a continuity argument?
     
  2. jcsd
  3. Sep 28, 2005 #2
    Anyone? How would i prove in general that the det of a rotation matrix is 1?
     
  4. Sep 28, 2005 #3

    Tom Mattson

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    I am unsure of what you mean by a continuity argument as it pertains to matrices and determinants. But you can straightforwardly show that the determinant of a rotation matrix is 1 by writing down the matrix and taking its determinant.

    For instance, a clockwise rotation by an angle [itex]\theta[/itex] about the [itex]z[/itex] axis is described by the following matrix:

    [tex]R_z(\theta)=\left[\begin{array}{ccc}\cos(\theta)&sin(\theta)&0 \\ -\sin(\theta)&\cos(\theta)&0 \\ 0&0&1 \end{array}\right ][/tex]

    It's a piece of cake to show that [itex]det(R_z(\theta))=1[/itex].
     
    Last edited: Sep 28, 2005
  5. Sep 28, 2005 #4
    I'm not sure what the question means by continuity argument either. That's why I asked.

    I know could just take the determinant of the rotation matrix. Since by euler's rotation theorem any rotation matrix M can be expressed as rotations over three perpendicular axis, det M = det R1 * det R2 * det R3 = 1, where R1, R2, and R3, are just the general rotation matrices for rotating over x,y, and z axis. But I wasn't sure if this 'proof' is sufficiently rigorous and general.
     
  6. Sep 30, 2005 #5

    lurflurf

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    When asking questions like this one it would be nice to define what you are talking about. In this case a rotation matrix. The definition
    Rotation matrix: A matrix for which det(A)=1
    would be very helpful

    Another definition is
    Rotation matrix: An isometric orientation preserving linear transform
    by isometric I mean the inner product defined by A matches the one defined by I (idenity matrix)
    (Ax)'Ay=x'y
    where ' is the adjoint (conjugate transpose or if A is real transpose)
    immediately we have
    A'A=I
    This only insures abs(det(A))=1
    so we turn to the orientation preserving bit
    That is fancy talk, but it means that I is our prototype rotation
    that is
    T(x,y,z)=(-x,y,z)
    would be a "bad" or improper rotation
    we want for A a rotation and B any matrix
    det(AB)=det(B)
    which again does not teach us much
    I said we want I to be a prototype
    det(I)=1 but again we want to go further
    say h is a small number and we wish to construct an almost rotation
    I+Ah
    we have
    (I+Ah)'=I+A'h
    and
    (I+Ah)^-1~I-Ah
    thus we require
    A'=-A
    we could also make better almost rotations
    I+Ah+A^2h^2/2+A^3h^3/6+...
    The ultimate result being the actual rotation
    exp(Ah)
    and
    of course
    A'=-A->tr(A)=0
    det(exp(At))=exp(tr(At))=exp(0)=1
    Thus rotation matricies are of the form exp(A) where A'=-A thus have det=1
     
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