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Math Help - 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 (0,0), (1,0) and (-1,0).

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

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

    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

    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

    U_2=\frac{1}{2}(1+7i)U_1
    -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 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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    Thanks
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    Finding the eigenvector

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

    <br />
\left[<br />
\begin {array}{ccc}<br />
{0-\lambda}&1\\<br />
\noalign{\medskip}<br />
-2&{1-\lambda}<br />
\end {array}<br />
\right]<br />
\left[<br />
\begin {array}{ccc}<br />
U1\\<br />
\noalign{\medskip}<br />
U2<br />
\end {array}<br />
\right]=<br />
\left[<br />
\begin {array}{ccc}<br />
0\\<br />
\noalign{\medskip}<br />
0<br />
\end {array}<br />
\right]<br />

    I can then choose U1=1 and U2= \lambda .
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  3. #3
    Senior Member
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    To find eigenvectors always solve the system
    <br />
(A - \lambda I)v = 0.<br />
    The result is the eigenvector v corresponding to the eigenvalue \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
    Senior Member
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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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