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Math Help - 2 x 2 matrix representing a rotation

  1. #1
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    2 x 2 matrix representing a rotation

    I'm trying to work through the following theorem in Brannan et al.'s book "Geometry" (you might wish to skip ahead to the bold text "The Source of My Problem", so that you don't waste your time if the context is not necessary):

    "A  2 \times 2 matrix  \mathbf{P} represents a roation of  \mathbb{R}^2 about the origin if and only if it satisfies the following two conditions:

    (a)  \mathbf{P} is orthogonal;
    (b)  \mathrm{det} \, \mathbf{P} = 1 ."

    So, the proof proceeds as follows:

    A matrix  \mathbf{P} represents a rotation if and only if it is of the form

     \left( \begin{array}{cc}<br />
\cos\theta & - \sin\theta \\<br />
\sin\theta & \cos\theta<br />
\end{array} \right) . (*)

    I'm accustomed to, and am happy with, this fact (well, the 'if' part), and I am happy that it is easy to verify that if it is of this form, then  \mathbf{P} satisfies conditions (a) and (b).

    Next, the proof procees to let  \mathbf{P} = \left( \begin{array}{cc}<br />
a & b \\<br />
c & d<br />
\end{array} \right) be a matrix that satisfies conditions (a) and (b).

    Then, since  \mathbf{P} is orthgonal, the vector  \left( \begin{array}{c}<br />
a \\<br />
c<br />
\end{array} \right) has length 1; that is,  a^2 + c^2 = 1 . Thus, there is a number  \theta for which

     a = \cos\theta and  c = \sin\theta .

    Also, since  \mathbf{P} is orthogonal, the vectors  \left( \begin{array}{c}<br />
a \\<br />
c<br />
\end{array} \right) = \left( \begin{array}{c}<br />
\cos\theta \\<br />
\sin\theta<br />
\end{array} \right) and  \left( \begin{array}{c}<br />
b \\<br />
d<br />
\end{array} \right) are orthogonal; that is,  \left( \begin{array}{cc}<br />
\cos\theta & \sin\theta<br />
\end{array} \right) <br />
\left( \begin{array}{c}<br />
b \\<br />
d<br />
\end{array} \right) = 0 (all fine by me so far) or

    *******************************
    The Source of My Problem:

     \cos\theta \cdot b + \sin\theta \cdot d = 0.

    Then the proof goes on to say,

    "So there exists some number  \lambda , say, such that

     b = - \lambda \sin\theta and  d = \lambda \cos\theta ."

    Would somebody be able to explain the rationale behind this? Why must  b and  d be of this form? I can see from the rest of the proof that it would be very helpful if they would be - but I'm not sure why they have to be. To try to explain my confusion, I think I would be happy saying that there must be some numbers,  \lambda and  \mu , say, such that

     b = - \lambda \sin\theta and  d = \mu \cos\theta .

    But I can't think of any reason for which  \lambda should equal  \mu .

    Thanks in advance for your help.

    -----
    For anyone interested, the rest of the proof is as follows:

    "Then since  \mathrm{det} \, \mathbf{P} = 1 , we have

     1 = ad - bc = \lambda \cos ^2 \theta + \lambda \sin ^2 \theta ,

    so that  \lambda = 1 . It follows that  \mathbf{P} must be of the form (*), and so represent a rotation of  \mathbb{R} ^2 about the origin."
    Last edited by Harry1W; September 20th 2009 at 03:23 PM.
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  2. #2
    Senior Member
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    Paris
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    If \cos\theta =0 or \sin\theta =0, then b=-\sin\theta,\ d=\cos\theta is the only solution (because of (a) and (b)).

    Assume both \cos\theta and \sin\theta are different from 0.

    Then any pair of real (b,d) can be written under the form (-\lambda\sin\theta, \mu\cos\theta) for some \lambda,\mu \in\mathbb{R}.

    Since \cos\theta\sin\theta(\mu-\lambda) = \cos\theta .b +\sin\theta .d=0 and \cos\theta \sin\theta \neq 0, we have \lambda =\mu\ .
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  3. #3
    Junior Member
    Joined
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    Thanks so much for that. However, I'm struggling to see the reason why,

    Quote Originally Posted by clic-clac View Post
    If \cos\theta =0 or \sin\theta =0, then b=-\sin\theta,\ d=\cos\theta is the only solution (because of (a) and (b)).
    I'm sorry to be slow on the uptake, but I'd really appreciate it if you could spell this out for me. Thanks again.
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  4. #4
    Senior Member
    Joined
    Nov 2008
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    Paris
    Posts
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    No problem. There may be a better proof, but that's what I was thinking when writting this one:
    if a=\cos\theta=0, then c=\sin\theta\in\{-1,1\} and -cb=\det\mathbf{P}=1 leads to b=-c=-\sin\theta. Then orthogonality imposes that d=0=\cos\theta.

    A similar argument for the case b=\sin\theta=0 ends the proof.
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  5. #5
    MHF Contributor

    Joined
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    Quote Originally Posted by Harry1W View Post
    I'm trying to work through the following theorem in Brannan et al.'s book "Geometry" (you might wish to skip ahead to the bold text "The Source of My Problem", so that you don't waste your time if the context is not necessary):

    "A  2 \times 2 matrix  \mathbf{P} represents a roation of  \mathbb{R}^2 about the origin if and only if it satisfies the following two conditions:

    (a)  \mathbf{P} is orthogonal;
    (b)  \mathrm{det} \, \mathbf{P} = 1 ."

    So, the proof proceeds as follows:

    A matrix  \mathbf{P} represents a rotation if and only if it is of the form

     \left( \begin{array}{cc}<br />
\cos\theta & - \sin\theta \\<br />
\sin\theta & \cos\theta<br />
\end{array} \right) . (*)

    I'm accustomed to, and am happy with, this fact (well, the 'if' part), and I am happy that it is easy to verify that if it is of this form, then  \mathbf{P} satisfies conditions (a) and (b).

    Next, the proof procees to let  \mathbf{P} = \left( \begin{array}{cc}<br />
a & b \\<br />
c & d<br />
\end{array} \right) be a matrix that satisfies conditions (a) and (b).

    Then, since  \mathbf{P} is orthgonal, the vector  \left( \begin{array}{c}<br />
a \\<br />
c<br />
\end{array} \right) has length 1; that is,  a^2 + c^2 = 1 . Thus, there is a number  \theta for which

     a = \cos\theta and  c = \sin\theta .

    Also, since  \mathbf{P} is orthogonal, the vectors  \left( \begin{array}{c}<br />
a \\<br />
c<br />
\end{array} \right) = \left( \begin{array}{c}<br />
\cos\theta \\<br />
\sin\theta<br />
\end{array} \right) and  \left( \begin{array}{c}<br />
b \\<br />
d<br />
\end{array} \right) are orthogonal; that is,  \left( \begin{array}{cc}<br />
\cos\theta & \sin\theta<br />
\end{array} \right) <br />
\left( \begin{array}{c}<br />
b \\<br />
d<br />
\end{array} \right) = 0 (all fine by me so far) or
    One of the conditions for an orthogonal matrix has been dropped here: we must also have b^2+ d^2= 1

    *******************************
    The Source of My Problem:

     \cos\theta \cdot b + \sin\theta \cdot d = 0.

    Then the proof goes on to say,

    "So there exists some number  \lambda , say, such that

     b = - \lambda \sin\theta and  d = \lambda \cos\theta ."

    Would somebody be able to explain the rationale behind this? Why must  b and  d be of this form? I can see from the rest of the proof that it would be very helpful if they would be - but I'm not sure why they have to be. To try to explain my confusion, I think I would be happy saying that there must be some numbers,  \lambda and  \mu , say, such that

     b = - \lambda \sin\theta and  d = \mu \cos\theta .

    But I can't think of any reason for which  \lambda should equal  \mu .

    Thanks in advance for your help.

    -----
    For anyone interested, the rest of the proof is as follows:

    "Then since  \mathrm{det} \, \mathbf{P} = 1 , we have

     1 = ad - bc = \lambda \cos ^2 \theta + \lambda \sin ^2 \theta ,

    so that  \lambda = 1 . It follows that  \mathbf{P} must be of the form (*), and so represent a rotation of  \mathbb{R} ^2 about the origin."
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