Proving a certain orthogonal matrix is a rotation matrix

  • #1
ELB27
117
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Homework Statement


Let ##U## be a ##2\times 2## orthogonal matrix with ##\det U = 1##. Prove that ##U## is a rotation matrix.

Homework Equations

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!
 
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  • #2
ELB27 said:
Clearly, these properties are satisfied if we let x=cosθ or x=sinθ, 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.)?
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.
 
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  • #3
Shyan said:
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.
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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