# Finding the eigenvectors from complex eigenvalues

1. Apr 15, 2010

### gtse

1. The problem statement, all variables and given/known data
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)
2. Relevant 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.

3. 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.

2. Apr 15, 2010

### 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.

3. Apr 15, 2010

### NeoDevin

For matrices in LaTeX try this:

Code (Text):

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

$$\left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right)$$

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