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Improper rotation matrix

  1. Dec 7, 2008 #1
    In Goldstein there is a problem asking to find a vector representation for a reflection in a plane of a unit normal [tex]\mathbf{\hat{n}}[/tex]. I find it to be

    [tex] \mathbf{r'} = \mathbf{r} - 2(\mathbf{r\cdot \hat{n}})\mathbf{\hat{n}}[/tex]

    and it has a corresponding transformation matrix with elements

    [tex]A_{ij} = \delta_{ij} - 2l_i l_j[/tex]

    where [tex]l_i\, , i=1,2,3[/tex] are the direction cosines for the orientation of the plane. Goldstein then asks to show that this matrix is an improper orthogonal one. I can show orthogonality by simply noting that [tex]A^T = A[/tex], and then I multiply [tex]A^2 = I[/tex], which shows that [tex]A^T = A^{-1}[/tex], which is the condition for orthogonality.

    However, the improper nature of the matrix is unclear to me. If I compute the determinant, by explicitly writing out the form of the matrix, the result I obtain is +1, instead of -1:

    [tex] \text{det}(A) = \begin{vmatrix}
    1-l_1^2 & -l_1 l_2 & -l_1 l_3\\
    -2l_1 l_2 & 1-2l_2^2 & -2l_2 l_3\\
    -2l_1 l_3 & -2l_2 l_3 & 1-2l_3^2\end{vmatrix} = 1[/tex]

    Does it mean that the matrix is a proper one? Or is there an error in the problem statement, or am I missing something?
     
  2. jcsd
  3. Dec 7, 2008 #2

    Avodyne

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    Science Advisor

    If I set l1=l2=0, and l3=1, your matrix is diagonal with entries 1, 1, -1, which obviously has det = -1. So I think you must have made a mistake taking the determinant.
     
  4. Dec 8, 2008 #3
    Oh yeah, that's right. Conclusion: don't drink behind the table.
     
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