# How do you find this eigenvector?

1. Oct 18, 2012

### msell2

[0 1]
[-2 -2] This is the 2x2 matrix.

[λ -1]
[2 λ+2] This is the matrix that equals λI - A.

Here are the eigenvalues I found:
λ = -1 + i, -1 - i

I am really confused at what to do next to find the eigenvectors. I keep looking online for help but I still can't figure it out...

Thanks!

2. Oct 18, 2012

### Dick

Now put one of your eigenvalues into λI - A and try to find a vector v such that (λI - A)v=0. Then v would be an eigenvector of λ. It's just a set of linear equations to solve with a free parameter.

3. Oct 18, 2012

### LCKurtz

For $\lambda$ equal an eigenvalue, your two equations are dependent, so you only need to solve one and you have one free variable. The first one is the same as $\lambda x - y =0$ or $y = \lambda x$. So let $x = 1$ and what do you get for $y$? Do that for each value of $\lambda$ and you will have two [x,y] eigenvectors.

4. Oct 18, 2012

### msell2

I get what I need to do in theory, I just don't actually know how to do it. How do you find v such that (λI-A)v=0?

5. Oct 18, 2012

### Dick

Just try it! What is the matrix (λI-A) when λ=-1+i? Write that matrix times a vector (x,y), set it equal to zero and try to find a solution for x and y.