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Thread: Eigenvectors of a Jacobian matrix

  1. #1
    Senior Member chella182's Avatar
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    Eigenvectors of a Jacobian matrix

    Sorry, if this is the wrong subforum, firstly.

    So I have a dynamical system with fixed points $\displaystyle (0,0)$, $\displaystyle (1,0)$ and $\displaystyle (-1,0)$.

    The bit I'm stuck on is this; I have the Jacobian matrix for $\displaystyle (1,0)$ and $\displaystyle (-1,0)$ as $\displaystyle A=\left( \begin{array}{cc} 0 & 1 \\ -2 & 1 \end{array}\right)$ (so the trace $\displaystyle \tau=1$ and the determinant $\displaystyle \delta=2$) and the eigenvalues as $\displaystyle \lambda_1=\frac{1}{2}(1_+7i)$ and $\displaystyle \lambda_2=\frac{1}{2}(1-7i)$.

    To find the eigenvectors $\displaystyle \mathbf{U}$ and $\displaystyle \mathbf{V}$ I know I need to solve

    $\displaystyle A=\left( \begin{array}{cc} 0 & 1 \\ -2 & 1 \end{array}\right) \left( \begin{array}{c} U_1 \\ U_2 \end{array} \right)=\frac{1}{2}(1+7i)\left( \begin{array}{c} U_1 \\ U_2 \end{array} \right)$

    and

    $\displaystyle A=\left( \begin{array}{cc} 0 & 1 \\ -2 & 1 \end{array}\right) \left( \begin{array}{c} V_1 \\ V_2 \end{array} \right)=\frac{1}{2}(1-7i)\left( \begin{array}{c} V_1 \\ V_2 \end{array} \right)$

    and this is what I'm stuck on. I've found both lines of the simultaneous equations to be

    $\displaystyle U_2=\frac{1}{2}(1+7i)U_1$
    $\displaystyle -2U_1+U_2=\frac{1}{2}(1+7i)U_2$

    but I can't get anywhere with it. I did give it a go and got $\displaystyle U_1=0$ but I get the feeling that's not right this was never my strong point when we initially learned it, but I think it's the complex number that's confusing me. Any ideas? Cheers in advance.
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  2. #2
    Senior Member
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    Finding the eigenvector

    This seems to be an eigenvector problem. If I call $\displaystyle \lambda$ = one of your $\displaystyle \lambda_1$ or $\displaystyle \lambda_2$, then I bring the RHS of your matrix equation to the left, I get:

    $\displaystyle
    \left[
    \begin {array}{ccc}
    {0-\lambda}&1\\
    \noalign{\medskip}
    -2&{1-\lambda}
    \end {array}
    \right]
    \left[
    \begin {array}{ccc}
    U1\\
    \noalign{\medskip}
    U2
    \end {array}
    \right]=
    \left[
    \begin {array}{ccc}
    0\\
    \noalign{\medskip}
    0
    \end {array}
    \right]
    $

    I can then choose U1=1 and U2=$\displaystyle \lambda$ .
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  3. #3
    Senior Member
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    To find eigenvectors always solve the system
    $\displaystyle
    (A - \lambda I)v = 0.
    $
    The result is the eigenvector $\displaystyle v$ corresponding to the eigenvalue $\displaystyle \lambda$.
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  4. #4
    Senior Member chella182's Avatar
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    See, I knew all of this, but that wasn't what the problem was, it was the fact there was complex numbers there, I needed walking through it a little. S'alright though, I handed my thing in now, so we'll see how I did.
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  5. #5
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    The complex numbers should just be treated as any scalar. Sorry, I didn't understand completely your question.
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