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Math Help - Orthogonal Matrix

  1. #1
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    Orthogonal Matrix

    A=\begin{bmatrix}1&2&2\\2&1&-2\\2&-2&1\end{bmatrix}

    I have found the eigenvalues and vectors to be \lambda=3,3,-3 and eigenvectors \left\{\begin{bmatrix}1\\1\\0\end{bmatrix},\begin{  bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}-1\\1\\1\end{bmatrix}\right\}.

    Now I need to find an orthogonal matrix such that P^tAP=D.
    D being my diagonal matrix.

    I tried Gram-Schmidting my eigenspace but that didn't work. What do I need to do?
    Last edited by dwsmith; November 29th 2011 at 01:04 PM.
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  2. #2
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    Re: Orthogonal Matrix

    the problem lies with your eigenvectors for 3, as both are orthogonal to (-1,1,1). so apply gram-schmidt to just those 2.

    for example if we take u1 = v1 = (1,1,0), then gram-schmidt gives:

    u2 = v2 - [(u1.v1)/(u1.u1)]u1 = (1,0,1) - (1/2)(1,1,0) = (1/2,-1/2,1).

    by construction, u2 is orthogonal to u1, and (1/2,-1/2,1).(-1,1,1) = -1/2 + -1/2 + 1 = 0.
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  3. #3
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    Re: Orthogonal Matrix

    That gives me the Diagonal matrix

    \begin{bmatrix}6&0&0\\0&\frac{9}{2}&0\\0&0&-9\end{bmatrix}

    If I am not mistaken, shouldn't D have entries corresponding to the eigenvalues?
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  4. #4
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    Re: Orthogonal Matrix

    normalize the eigenvectors.
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  5. #5
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    Re: Orthogonal Matrix

    Quote Originally Posted by Deveno View Post
    normalize the eigenvectors.
    Normalizing didn't help. Here is the output:
    Attached Thumbnails Attached Thumbnails Orthogonal Matrix-normalized.jpg  
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  6. #6
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    Re: Orthogonal Matrix

    \frac{1}{\sqrt{\frac{3}{2}}} = \frac{2}{\sqrt{6}}

    not \frac{\sqrt{6}}{2}.

    re-check your matrix product with the proper matrix, because i get:

    P^TAP = \begin{bmatrix}3&0&0\\0&3&0\\0&0&-3\end{bmatrix}

    as expected.
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