Finding the eigenvectors from complex eigenvalues

[gtse]

Homework Statement

This isn't really a question in particular.
I am doing my first Differential Equations course, and in the complex eigenvalues part, I am getting confused as to how to find the eigenvectors.

Example:
Solve for the general solution of:
x' = (1 -1)x (don't know how to type a matrix using latex sorry)
(5 -3)

Homework Equations

I know how to find the eigenvectors if there were real eigenvalues, since I've taken Linear Algebra and know that you can just simply reduce the matrix into Gauss-Jordan form.

The Attempt at a Solution

The eigenvalues are -1 +/- 2i (this is the easy part)
What confuses me is the next step:
The examples usually only plug in one eigenvalue (say -1+2i), which I don't know why.

And so the matrix will now look something like this;
(2-2i -1)
( 5 -2-2i)

What happens next I don't understand, usually the example would take
(2-2i)v1 - v2 = 0 if (v1,v2) was one of the eigenvectors
And it is around here I get the most confused.
- do I use v1 or v2 as the free variable?
- do I use the top or bottom row (5v1 + (-2-2i)v2 = 0) to find v1 and v2?

I have tried reading the books and the examples but they never show what they exactly do.

 

[hgfalling]

First off, check your eigenvalues. Remember that the product of the eigenvalues has to equal the determinant of the matrix, but (-1 + 2i)(-1 - 2i) != 2.

Now you plug in one of the eigenvalues, and obtain some matrix (not exactly the one you reproduced, but close). You now solve the matrix equation (A - \lambda I)x = 0. Suppose you start with the first row, you get some kind of linear relationship between v₁ and v₂, which is normal. Now considering the second row, you will see that it's just a constant multiple of the first row, so you won't get any more information from that. You can use whatever free variable you like: if (3+i,1) is an eigenvector, so is (10,3-i), right?

Then you can do the same thing with the other eigenvalue; or you can just rely on the fact that complex eigenvalues come in pairs, so if a + bi is an eigenvalue, so is a - bi. And if (d+ti,c) is an eigenvector associated with the eigenvalue a + bi, then (d - ti, c) is an eigenvector associated with the eigenvalue a - bi. That's usually easier.

 

[NeoDevin]

For matrices in LaTeX try this:

[NOPARSE][tex]\left( \begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i \end{array} \right)[/tex][/NOPARSE]

\left( \begin{array}{ccc}<br /> a &amp; b &amp; c \\<br /> d &amp; e &amp; f \\<br /> g &amp; h &amp; i \end{array} \right)

(Or you can click on the LaTeX images, and it shows you the code used to create it)