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Math Help - Help with final step for eigenvector problem

  1. #1
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    Help with final step for eigenvector problem

    B = \left(\begin{array}{ccc}4&0&-1\\0&4&1\\-1&1&3\end{array}\right)

    Find all the other eigenvalues and eigenvectors.

    I have found the eigenvalues for this matrix being \lambda = 2, \lambda = 4 & \lambda = 5


    I have subtracted your eigenvalues from the diagonal to get the following matricies

    for \lambda = 2 \left(\begin{array}{ccc}2&0&-1\\0&2&1\\-1&-1&1\end{array}\right)
    for \lambda = 5 \left(\begin{array}{ccc}-1&0&-1\\0&-1&1\\-1&-1&-2\end{array}\right)

    My problem is that I am stuck on where to go from here. I have used the elimination method but I always end up with a row of zeros and am unsure on where to go from there.

    Any help would be appreciated.
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  2. #2
    Super Member Random Variable's Avatar
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    You should get a row of zeroes. For every eigenvalue there are infinitely-many associated eigenvectors.

    For example, the eigenvectors associated with  \lambda = 2 are vectors of the form  \alpha \left(\begin{array}{c} 1 \over 2 \\ -\frac{1}{2} \\ 1\end{array}\right)
    Last edited by Isomorphism; June 4th 2009 at 08:06 PM.
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  3. #3
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    Ok. Im still confused as to how you got that answer. Are you able to show me the working out that you did?
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  4. #4
    Super Member Random Variable's Avatar
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    The reduced row echelon form of  \left(\begin{array}{ccc}2&0&-1\\0&2&1\\-1&1&1\end{array}\right) is  \left(\begin{array}{ccc}1&0&-\frac{1}{2}\\0&1&1 \over 2\\0&0&0\end{array}\right)

    Since x_{3} can be anything, just call it  \alpha

    Then  x_{2} + \frac {1}{2} x_{3} = 0 which implies that x_{2} = -\frac{\alpha}{2}

    And  x_{1} - \frac {1}{2} x_{3} =0 which implies that  x_{1} = \frac {\alpha}{2}

    so solutions are of the form  \left(\begin{array}{c} \alpha \over 2 \\ -\frac{\alpha}{2} \\ \alpha\end{array}\right)

    and then you can factor out the  \alpha
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