# DIFFEQ - Complex eigenvectors

1. Nov 13, 2005

### FrogPad

I'm having trouble with this problem. Actually I'm having trouble with all of this set of problems (when the eigenvectors are complex). I must not be finding these things correctly, because nothing is matching up with the book. Any help would be awesome.
$$\vec {x}\,' = \left( \begin{array}{cc} 1 & 2 \\ -5 & -1 \end{array} \right) \vec x$$

So...
$$\left| \begin{array}{cc} 1-\lambda & 2 \\ -5 & -1-\lambda \end{array} \right| = (1-\lambda)(-1-\lambda)+10=0=\lambda^2+9$$

$$\lambda = 3i$$

So then:
$$\left( \begin{array}{cc} 1-3i & 2 \\ -5 & -1-3i \end{array} \right) \left(\begin{array}{cc}x1\\x2 \end{array}\right) =\vec 0$$

Which is the NULLSPACE of the $$\vec A - r\vec I$$ matrix.
Gaussian-Jordan reduction yields:
$$\left( \begin{array}{cc} 1 & \frac{1}{5}+\frac{3}{5}i \\ 0 & 0 \end{array} \right)$$

$$x1=-\left(\frac{1}{5}+\frac{3}{5}i\right)\alpha$$
$$x2=\alpha$$
$$\vec V = \left( \begin{array}{cc} \frac{-1}{5}-\frac{3}{5}i \\1 \end{array} \right)$$
$$\vec V = \left( \begin{array}{cc} \frac{-1}{5} \\ 1 \end{array} \right) + i \left( \begin{array}{cc} \frac{-3}{5} \\ 0 \end{array} \right)$$
So, I'm pretty sure I'm making the mistake in here. But where? Isn't everything I'm doing legit?

Last edited: Nov 13, 2005
2. Nov 14, 2005

### HallsofIvy

Staff Emeritus
You never did say exactly what it was you were trying to do but, yes, that is an eigenvector corresponding to the eigenvalue 3i.

3. Nov 14, 2005

### FrogPad

woops... I definitely could have been more specific. I basically just wanted confirmation that I was finding the eigenvectors correctly. I've taken linear algebra, and I remember just using an algorithmic approach when solving for the nullspace, but in diffeq they seem to be setting certain constants to 'certain' values, and my answers are just NOT matching up with the books... and I can't figure out in what way they are different. Like, their answers all seem to be flipped for complex roots. But, anyways... thanks for the confirmation. I'll just go with what I have then. :)

4. Nov 14, 2005

### HallsofIvy

Staff Emeritus
I don't know what you mean by "flipped". One possible source, not of error, but of confusion, is that there exist an infinite number of eigenvectors for any eigenvalue (the constitute a subspace, of course).

Yes,
$$\vec V = \left( \begin{array}{cc} \frac{-1}{5}-\frac{3}{5}i \\1 \end{array} \right)$$
is an eigenvector corresponding to eigenvalue 3i.
But since multiplying or dividing by a number still gives an eigenvector
$$\vec V = \left( \begin{array}{cc}1\\ \frac{-1}{2}+\frac{3}{2}i \end{array} \right)$$
is also an eigenvector corresponding to eigenvalue 3i. (I divided both components by $-\frac{1}{5}-\frac{3}{5}i$.)

Don't forget to find an eigenvalue corresponding to -3i!

5. Nov 14, 2005

### FrogPad

Dude you seriously rock. Thank you. You cleared it up.
hehe... sorry about saying "flipped" (I know how technical that sounds). You did correctly reason out the problem I was having though.
I was not thinking in terms of the factor itself being a complex number. Instead I was assuming that the factor had to be a real. So I was like how the hell are they changing:
$$\vec V = \left( \begin{array}{cc} \frac{-1}{5}-\frac{3}{5}i \\1 \end{array} \right)$$
into
$$\vec V = \left( \begin{array}{cc}1\\ \frac{-1}{2}+\frac{3}{2}i \end{array} \right)$$
But of course. The factor is a complex number. So that's how they flip it .

oh, and yeah... I got the other eigenvalue. I figured it wasn't necessary to explain my problem.

Last edited: Nov 14, 2005
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