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Math Help - [SOLVED] Tricky little matrix problem

  1. #1
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    [SOLVED] Tricky little matrix problem

    This exercise has been troubling me. It seems like it should be simple, but I can't seem to crack it.

    Let A=\left[\begin{array}{cc} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta}\end{array} \right]

    Find A^k, where k\in\mathbb{N}

    To solve this, I tried to find a pattern in the following:

    A^2=\left[\begin{array}{cc} \cos^2{\theta}-\sin^2{\theta} & 2\sin{\theta}\cos{\theta} \\ -2\sin{\theta}\cos{\theta} & \cos^2{\theta}-\sin^2{\theta}\end{array} \right]

    A^3=\left[\begin{array}{cc} \cos^3{\theta}-3\sin^2{\theta}\cos{\theta} & 3\sin{\theta}\cos^2{\theta}-\sin^3{\theta} \\ -3\sin{\theta}\cos^2{\theta}+\sin^3{\theta} & \cos^3{\theta}-3\sin^2{\theta}\cos{\theta}\end{array} \right]

    A^4=\left[\begin{array}{cc}\cos^4{\theta}-6\sin^2{\theta}\cos^2{\theta}+\sin^4{\theta}&4\sin  {\theta}\cos^3{\theta}-4\sin^3{\theta}\cos{\theta}\\-4\sin{\theta}\cos^3{\theta}+4\sin^3{\theta}\cos{\t  heta}&\cos^4{\theta}-6\sin^2{\theta}\cos^2{\theta}+\sin^4{\theta}\end{a  rray}\right]

    From this I observe that

    A^k=\frac{1}{k^2}\left[\begin{array}{cc}kf'(\theta)&k^2f(\theta)\\f''(\th  eta)&kf'(\theta)\end{array}\right]

    But I can't seem to generalize f(\theta) in terms of k. Any ideas?
    Last edited by hatsoff; September 1st 2008 at 02:50 PM.
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  2. #2
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    Did you notice that A^2  = \left[ {\begin{array}{cc}<br />
   {\cos (2x)} & {\sin (2x)}  \\<br />
   { - \sin (2x)} & {\cos (2x)}  \\<br />
\end{array}} \right]?
    I don't know if that helps?
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  3. #3
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    Quote Originally Posted by hatsoff View Post
    This exercise has been troubling me. It seems like it should be simple, but I can't seem to crack it.

    Let A=\left[\begin{array}{cc} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta}\end{array} \right]

    Find A^k, where k\in\mathbb{N}
    Think geometrically.
    This is a rotation matrix. Meaning if \bold{x}\in \mathbb{R}^2 then A\bold{x} represent rotation the vector \bold{x} counterclockwise by \theta (around the origin).

    Now A^2\bold{x} = A(A\bold{x}) is rotating \bold{x} twice by \theta i.e. 2\theta. And in general A^k is a rotation by k\theta. The matrix for the rotation is given by,
    A^k = \left[ \begin{array}{cc}\cos k\theta & \sin k\theta \\ - \sin k\theta & \cos k\theta \end{array} \right]
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  4. #4
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    Quote Originally Posted by hatsoff View Post
    This exercise has been troubling me. It seems like it should be simple, but I can't seem to crack it.

    Let A=\left[\begin{array}{cc} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta}\end{array} \right]

    Find A^k, where k\in\mathbb{N}

    To solve this, I tried to find a pattern in the following:

    A^2=\left[\begin{array}{cc} \cos^2{\theta}-\sin^2{\theta} & 2\sin{\theta}\cos{\theta} \\ -2\sin{\theta}\cos{\theta} & \cos^2{\theta}-\sin^2{\theta}\end{array} \right]

    A^3=\left[\begin{array}{cc} \cos^3{\theta}-3\sin^2{\theta}\cos{\theta} & 3\sin{\theta}\cos^2{\theta}-\sin^3{\theta} \\ -3\sin{\theta}\cos^2{\theta}+\sin^3{\theta} & \cos^3{\theta}-3\sin^2{\theta}\cos{\theta}\end{array} \right]

    A^4=\left[\begin{array}{cc}\cos^4{\theta}-6\sin^2{\theta}\cos^2{\theta}+\sin^4{\theta}&4\sin  {\theta}\cos^3{\theta}-4\sin^3{\theta}\cos{\theta}\\-4\sin{\theta}\cos^3{\theta}+4\sin^3{\theta}\cos{\t  heta}&\cos^4{\theta}-6\sin^2{\theta}\cos^2{\theta}+\sin^4{\theta}\end{a  rray}\right]

    From this I observe that

    A^k=\frac{1}{k^2}\left[\begin{array}{cc}kf'(\theta)&k^2f(\theta)\\f''(\th  eta)&kf'(\theta)\end{array}\right]

    But I can't seem to generalize f(\theta) in terms of k. Any ideas?
    A thread of related interest: http://www.mathhelpforum.com/math-he...wer-100-a.html
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