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Proving a certain orthogonal matrix is a rotation matrix

  1. Jul 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Let ##U## be a ##2\times 2## orthogonal matrix with ##\det U = 1##. Prove that ##U## is a rotation matrix.

    2. Relevant equations


    3. The attempt at a solution
    Well, my strategy was to simply write the matrix as
    $$U = \begin{pmatrix}
    a & b\\
    c & d
    \end{pmatrix}$$
    and using the given properties to solve for ##a,b,c## and ##d##. I have 4 equations:
    1. Determinant = 1
    2. ##U^TU = I## where ##I## is identity
    where the last property gives me 3 equations (one of the entries is redundant). Thus, I have 4 equations in 4 unknowns. When I solve them, I get 1 free variable, and my matrix turns out to be of the form:
    $$U = \begin{pmatrix}
    x & \mp \sqrt{1-x^2}\\
    \pm \sqrt{1-x^2} & x
    \end{pmatrix}$$
    and since all entries must be real (orthogonal matrix) we have the constraint on ##x##:
    $$-1≤x≤1$$
    Clearly, these properties are satisfied if we let ##x=\cos\theta## or ##x=\sin\theta##, thus obtaining a rotation matrix in its standard notation. However, I do not see how to prove that these trigonometric functions are the only possible solutions. Also, how does one define formally and rigorously a rotation matrix? Only as a matrix of cosines and sines (with the appropriate values of determinant etc.)?

    Any suggestions/comments will be greatly appreciated!
     
  2. jcsd
  3. Jul 18, 2015 #2

    ShayanJ

    User Avatar
    Gold Member

    There is no need for that. You've done all things needed. You have shown that for all allowed values of x, there exists a ##\theta## that either ##x=\sin\theta ## or ##x=\cos\theta##. It means for all allowed values of x, this matrix corresponds to a rotation. It may correspond to many other things, but that doesn't matter. All we care about now, is that it corresponds to a rotation. So you're done with this question.
     
  4. Jul 18, 2015 #3
    Ah, I think I get it. Basically, the question can be rephrased as "Prove the ##U## can be represented as a rotation matrix."? Thus my proof will end as:
    ##x=\cos\theta \ ∀x## for some angle ##\theta## and thus, ##U## is always a rotation matrix about some angle ##\theta##.

    Thanks for the reply!
     
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