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Math Help - Rotation matrix.

  1. #1
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    Rotation matrix.

    Explain what it means for a matrix M to be orthogonal.
    With respect to the standard basis in cartesian plane, the anticlockwise rotation through angle \theta around the origin is represented by:

    {M}_{\theta } =

    ( \cos \theta -\sin \theta )
    ( \sin \theta \cos \theta )

    Supposed to be 2x2 matrix sorry I don't know the tex for it?

    Verify that {M}_{\theta } is orthogonal and verify the identity {M}_{\theta + \varphi} = {M}_{\theta}{M}_{\varphi}
    Last edited by mr fantastic; June 12th 2011 at 03:49 PM.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by TeaWithoutMussolini View Post
    Explain what it means for a matrix M to be orthogonal.
    With respect to the standard basis in cartesian plane, the anticlockwise rotation through angle \theta around the origin is represented by:

    {M}_{\theta } =

    ( \cos \theta -\sin \theta )
    ( \sin \theta \cos \theta )

    Supposed to be 2x2 matrix sorry I don't know the tex for it?

    Verify that {M}_{\theta } is orthogonal and verify the identity {M}_{\theta + \varphi} = {M}_{\theta}{M}_{\varphi}
    I will find all the orthogonal matrices at M_2(\mathbb{R}).

    Let A=\begin{pmatrix}a & b\\ c & d\end{pmatrix} orthogonal matrix .

    Hence,  AA^*=AA^T=I.

    In other words:

    A^T=A^{-1}=\frac{1}{|A|}\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}.

    Now, from AA^T=I we get:

    1=|I|=|AA^T|=|A||A^T|=A^2

    Hence |A|=\pm1 .

    Suppose that |A|=1, then:


    \begin{pmatrix}a & c\\ b & d\end{pmatrix}=\begin{pmatrix}d & -b\\ -c & a\end{pmatrix}

    So:  b=-c, a=d.

    A^{+}=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}<br />

    In the case in which |A|=-1 we will get:

    \begin{pmatrix}a & c\\ b & d\end{pmatrix}=\begin{pmatrix}-d & b\\ c & -a\end{pmatrix}<br />

    So:  b=s, a=-d

    A^{-}=\begin{pmatrix}a & b\\ b & -a\end{pmatrix}

    In both cases we can deduce from orthogonality that a^2+b^2=1.

    |a|\leq 1 so exists angel 0\leq \theta \leq 2\pi for which  a=cos\theta and then b=sin\theta .

    Finally we can write he follwing:


    A^+=\begin{pmatrix}cos\theta  & sin \theta \\-sin\theta & cos\theta\end{pmatrix}

    A^-= \begin{pmatrix}cos\theta  & sin \theta \\sin\theta & -cos\theta\end{pmatrix}
    Last edited by mr fantastic; June 12th 2011 at 03:50 PM.
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  3. #3
    Behold, the power of SARDINES!
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    For the first bit here is a link. Orthogonal matrix - Wikipedia, the free encyclopedia

    The LaTex code is

    \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \\ \end{bmatrix}

    For the last part verify the definitions that you have or the one's in the link and I would start with the left hand side of the equality. Use matrix multiplication and then use the sum identities for sine and cosine.

    Good luck
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