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Another eigenvector problem

  1. Nov 26, 2006 #1
    I need to solve the differential equation
    [tex]\mathbf{x'} = \left(
    3 & 0 & -1\\
    0 & -3 & -1\\
    0 & 2 & -1
    \right) \mathbf{x} [/tex]

    solving for the eigenvalues by taking the determinate and using the "basketweave" yields

    [tex] (3 - \lambda)(-3-\lambda)(-1-\lambda) + 2(3-\lambda) = 0 [/tex]

    and further simplification shows that

    [tex] -\lambda^3 - \lambda^2 +7\lambda +15 = 0 [/tex]

    guessing roots, I found that 3 is one and divided the polynomial by the root

    [tex](\lambda - 3)(-\lambda^2 - 4\lambda -5)=0[/tex]

    so the eigenvalues are (solving for both eqns in the brackets)
    [tex]\lambda = 3[/tex]
    [tex]\lambda = 2 + i[/tex]
    [tex]\lambda = 2 - i[/tex]

    The thing is that when I put in the real eigenvalue, it doesn't seem to have an eigenvector. Is this right? I think it would be because I don't have any idea how to get the solution for a mixture of real and complex eigenvalues. The solutions I know how to solve are when the eigenvectors are of the form v1,v2=p±iq. Though I imagine that if the real eigenvalue made an eigenvector I could just take the general solution of that and add it to the general solution of the complex eigenvectors.

    Basically what I want to know is, what do I do know?
  2. jcsd
  3. Nov 26, 2006 #2
    there's some miscalculation there.
    eigenvalues are 3, (-2 + i), (-2 -i)

    you can just look at the matrix and
    pick the eigenvector that gives 3 as
    the eigenvalue. (1, 0, 0).
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