Eigenvectors of a Jacobian matrix

chella182 · #1 $Old$ November 19th, 2009, 02:08 PM

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.

qmech · #2 $Old$ November 20th, 2009, 06:13 PM

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:

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

I can then choose U1=1 and U2= $\lambda$ .

lvleph · #3 $Old$ November 20th, 2009, 07:44 PM

To find eigenvectors always solve the system
$(A - \lambda I)v = 0.$
The result is the eigenvector $v$ corresponding to the eigenvalue $\lambda$ .

chella182 · #4 $Old$ November 21st, 2009, 04:07 AM

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.

lvleph · #5 $Old$ November 21st, 2009, 09:00 AM

The complex numbers should just be treated as any scalar. Sorry, I didn't understand completely your question.