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Math Help - find an othonormal set of eigenvectors

  1. #1
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    find an othonormal set of eigenvectors

    Find an orthonormal set of eigenvectors for the following symmetric matrix.

    [ 10 0 2 ]
    [ 0 6 0 ]
    [ 2 0 7 ]

    that's a 3x3 matrix, just don't know how to put it all in one.
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    Ok, where should you start? What have you done?
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    well i tried to find the characteristic equation,

    and i got (x= lamda)
    x^3 - 23x^2 - 168x - 396.. but then couldn't factorise it
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    Double-check your signs. I agree with the magnitudes of the coefficients, but I think you have a sign error in there somewhere.
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    ok, now i have , (x= lamda)

    x^3 - 23x^2 + 168x - 396..

    and i have (x-11)( x^2 -12x+36),

    i think the second term is made up of two complex factors, but i forget how to do it?
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  6. #6
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    Looking better. The quadratic factor you have there factors over the reals. In fact, it factors fairly straight-forwardly over the reals. You could always use the quadratic formula. What do you get?
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    lmao wow i can't believe i actually just did that. ( i swear i'm not this stupid, haha)
    well, i have the eigeevalues now, i'l see how i go from here, thanks
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    Ok, let me know how it goes.
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  9. #9
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    Quote Originally Posted by linalg123 View Post
    Find an orthonormal set of eigenvectors for the following symmetric matrix.

    [ 10 0 2 ]
    [ 0 6 0 ]
    [ 2 0 7 ]
    If you expand the determinant \left|\begin{array}{ccc}10-\lambda & 0 & 2 \\ 0 & 6-\lambda & 0\\ 2 & 0 & 7-\lambda\end{array}\right| on the middle row you get (6- \lambda)\left|\begin{array}{cc} 10- \lambda & 2 \\ 2 & 7- \lambda\end{array}\right| = (6- \lambda)((10-\lambda)(7- \lambda)- 4)= (6- \lambda)(66- 17\lambda+ \lambda^2) which obviously has \lambda= 6 as a root.
    Since the expansion you give obviously has 11 as a root, try factoring that quadratic as (\lambda^2- 17\lambda+ 66)= (\lambda- 11)(\lambda- 6) so that 6 is a double eigenvalue and 11 is an eigenvalue. Now start looking for eigenvectors.

    (Double click on the formulas to see the LaTex code.)

    that's a 3x3 matrix, just don't know how to put it all in one.
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  10. #10
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    ok, i have two linearly independant eigenvectors. Is that a problem that i can't get a third?
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    Show us your work. It isn't necessarily a problem. It depends on whether you're required to get a basis of orthonormal eigenvectors. If you're just asked to get a set of orthonormal eigenvectors, then who cares how many there are in the set?
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    for \lamda =11

    \left|\begin{array}{ccc}1 & 0 & -2 \\ 0 & 5 & 0\\ 2 & 0 & 4\end{array}\right|

    y=0
    x=2z eigenvector \left|\begin{array}{ccc}2 & 0 & 1\end{array}\right|

    for \lamda=6

    \left|\begin{array}{ccc}-4 & 0 & -2 \\ 0 & 0 & 0\\ -2 & 0 & -1\end{array}\right|

    y=0
    x=-2z eigenvector \left|\begin{array}{ccc}1 & 0 & -2\end{array}\right|

    that look ok? so then i just make them orthonormal and i'm done?
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    that didn't work =S
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  14. #14
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    It looks like you've correctly found two eigenvectors. (You should clean up the way that post looks by enclosing your LaTeX code with math tags.) However, you're missing an eigenvector.
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    for \lamda =11
    <br />
\left|\begin{array}{ccc}1 & 0 & -2 \\ 0 & 5 & 0\\ 2 & 0 & 4\end{array}\right|

    y=0
    x=2z eigenvector \left|\begin{array}{ccc}2 & 0 & 1\end{array}\right|

    for \lamda=6

    \left|\begin{array}{ccc}-4 & 0 & -2 \\ 0 & 0 & 0\\ -2 & 0 & -1\end{array}\right|

    y=0
    x=-2z eigenvector \left|\begin{array}{ccc}1 & 0 & -2\end{array}\right|

    that look ok? so then i just make them orthonormal and i'm done?
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