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Thread: Orthogonal Matrix

  1. #1
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    Orthogonal Matrix

    Find the value of for which the matrix is orthogonal.
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  2. #2
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    Quote Originally Posted by mr_motivator View Post
    Find the value of for which the matrix is orthogonal.
    An Orthogonal matrix is one whos transpose is its inverse.

    In other words:

    $\displaystyle AA^T = I = A^TA $

    For your matrix, the transpose is:

    $\displaystyle \left( \begin{array}{ccc}
    \frac{\sqrt{3}}{2} & \frac{1}{2} \\
    m & -\frac{\sqrt{3}}{2}
    \end{array} \right)$

    So multiply it by the original matrix
    $\displaystyle \left( \begin{array}{ccc}
    \frac{\sqrt{3}}{2} & \frac{1}{2} \\
    m & -\frac{\sqrt{3}}{2}
    \end{array} \right)\left( \begin{array}{ccc}
    \frac{\sqrt{3}}{2} & m \\
    \frac{1}{2}
    & -\frac{\sqrt{3}}{2}
    \end{array} \right)
    $$\displaystyle = \left( \begin{array}{ccc}
    1 & 0 \\
    0
    & 1
    \end{array} \right)$

    $\displaystyle \left( \begin{array}{ccc}
    \bigg(\frac{\sqrt{3}}{2}\bigg)^2+\bigg(\frac{1}{2} \bigg)^2 & \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} \\
    \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} & m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2
    \end{array} \right)
    $$\displaystyle = \left( \begin{array}{ccc}
    1 & 0 \\
    0
    & 1
    \end{array} \right)$

    $\displaystyle \left( \begin{array}{ccc}
    1 & \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} \\
    \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} & m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2
    \end{array} \right)
    $$\displaystyle = \left( \begin{array}{ccc}
    1 & 0 \\
    0
    & 1
    \end{array} \right)$

    Hence $\displaystyle \frac{\sqrt{3}}{2}m - \frac{\sqrt{3}}{4} =0$ and $\displaystyle m^2+\bigg(\frac{\sqrt{3}}{2}\bigg)^2 = 1 $
    Last edited by Mush; Feb 17th 2009 at 04:06 AM.
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