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Thread: Help with final step for eigenvector problem

  1. #1
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    Help with final step for eigenvector problem

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

    Find all the other eigenvalues and eigenvectors.

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


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

    for $\displaystyle \lambda = 2$$\displaystyle \left(\begin{array}{ccc}2&0&-1\\0&2&1\\-1&-1&1\end{array}\right)$
    for $\displaystyle \lambda = 5$$\displaystyle \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 $\displaystyle \lambda = 2 $ are vectors of the form $\displaystyle \alpha \left(\begin{array}{c} 1 \over 2 \\ -\frac{1}{2} \\ 1\end{array}\right) $
    Last edited by Isomorphism; Jun 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 $\displaystyle \left(\begin{array}{ccc}2&0&-1\\0&2&1\\-1&1&1\end{array}\right) $ is $\displaystyle \left(\begin{array}{ccc}1&0&-\frac{1}{2}\\0&1&1 \over 2\\0&0&0\end{array}\right) $

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

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

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

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

    and then you can factor out the $\displaystyle \alpha $
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