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Thread: repeated eigenvalue general solution problem

  1. #1
    Newbie
    Joined
    Dec 2006
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    repeated eigenvalue general solution problem

    Hello all,

    I am having trouble solving this repeated eigenvalue problem.

    $\displaystyle
    x'=\left|\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0 \end{array} \right|{x}$

    solving for the eigenvalues i get
    $\displaystyle
    \left|\begin{array}{ccc}-\lambda&1&1\\1&-\lambda&1\\1&1&-\lambda \end{array} \right|=-(\lambda+1)^2(\lambda-2)=0$
    which means
    $\displaystyle \lambda_1 = -1, \lambda_2 = 2 $

    solving for the non-repeated solution using the eigenvalue
    $\displaystyle \lambda_1 = 2$
    $\displaystyle
    \left|\begin{array}{ccc}-2&1&1\\1&-2&1\\1&1&-2 \end{array} \right|
    $
    which gives me the eigenvector and solution
    $\displaystyle
    x_1(t)=\left|\begin{array}{ccc}1\\1\\1\end{array} \right| e^{2t}
    $

    however when I try to solve for the repeated value
    $\displaystyle \lambda = -1$
    i get the matrix
    $\displaystyle
    \left|\begin{array}{ccc}1&1&1\\1&1&1\\1&1&1 \end{array} \right|
    $

    I dont know how to find a LI eigenvector from this matrix. I have the solution from the back of my textbook, but even after just arbitrarily taking the same eigenvector that they chose, I cannot come up with the same solution. I believe the repeated solution should be of the form
    $\displaystyle x^{(2)}(t) = u.te^\lambda_2t + w.e^\lambda_2t$
    where u and w are vectors, and u is actually the eigenvector from the matrix acquired using $\displaystyle \lambda = -1$

    I am stuck here, any help would be greatly appreciated.

    Thanks,
    Nick
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  2. #2
    TD!
    TD! is offline
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    The system comes down to the equation x+y+z = 0, from which you can get two linearly independent vectors, e.g. (1,0,-1) and (1,-1,0).
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