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Homework Help: Defective eigenvalues?

  1. Dec 6, 2005 #1
    if you have a differential equation of the form
    x' = Ax
    where A is the coefficient matrix, and you get a triple eigenvalue with a defect of 1. (meaning you get v1 and v2 as the associated eigenvector). How do you get v3 and how do you set up the solutions?
    I tried finding v3 such that [itex](\mathbf{A}-\lambda \mathbf{I})^2 \mathbf{v_3} = \mathbf{0}[/itex] but not exactly sure what to do after that. I got another v_2 (called it v_2') by using [itex](\mathbf{A}-\lambda \mathbf{I}) \mathbf{v_3} = \mathbf{v_2'}[/itex] but again not sure how to set up solution.
    my answer was
    [tex]\mathbf{x_1(t)} = \mathbf{v_1} e^{\lambda t}[/tex]
    [tex]\mathbf{x_2(t)} = \mathbf{v_2'} e^{\lambda t}[/tex]a
    [tex]\mathbf{x_3(t)} = (\mathbf{v_2'} + \mathbf{v_3}t)e^{\lambda t}[/tex]
    but it is not what is in the back of the book. I also tried using the original v_2:
    [tex]\mathbf{x_1(t)} = \mathbf{v_1} e^{\lambda t}[/tex]
    [tex]\mathbf{x_2(t)} = \mathbf{v_2} e^{\lambda t}[/tex]a
    [tex]\mathbf{x_3(t)} = (\mathbf{v_2} + \mathbf{v_3}t)e^{\lambda t}[/tex]
    Can anyone help?
    Thanks in advance.
  2. jcsd
  3. Dec 6, 2005 #2


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    Staff Emeritus
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    Gold Member

    Well, it's awfully brutish, but if you think that the general solution can be written in terms of your [itex]v_1[/itex], [itex]v_2[/itex], and [itex]v_3[/itex], possibly with an extra factor of t on some terms, then why not just make the most general candidate possible:

    [tex](C_1 v_1 + C_2 t v_1 + C_3 v_2 + C_4 t v_2 + C_5 v_3 + C_6 t v_3)e^{\lambda t}[/tex]

    and plug it into the original equation, to see what relations must hold between the various coefficients?
  4. Dec 8, 2005 #3


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    Homework Helper

    Maybe it would help to examine the solutions to a simpler version of the same sort of problem. For example:

    [tex]A = \left(\begin{array}{ccc}2&0&0\\0&2&1\\0&0&2\end{array}\right)[/tex]

    The difficulty is the same as for this problem:

    [tex]A = \left(\begin{array}{cc}2&1\\0&2\end{array}\right)[/tex]

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