# Eigenvectors - stuck on the basics

1. Dec 9, 2011

### Kaldanis

In the last 2 weeks we've begun learning about eigenvalues/vectors. It will come up in my exam in January so I'm trying hard to get my head around this. I've tried various different sources to learn this but I'm hoping someone here can offer a different view on it. Basically, I can work out the eigenvalues no problem but that's as far as I can get.

basic example:

$$A= \begin{bmatrix}2 & 0 \\ 0 &-1\end{bmatrix}$$
Using det(λI-A) I get that the eigenvalues are λ = 2 and λ = -1.

Then: $$\begin{bmatrix}2 & 0 \\ 0 &-1\end{bmatrix} \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} = \lambda \begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}$$

After this I'm not sure how to solve it or how to write it. I thought I had to start with λ = 2 and substitute it into the equation above, giving: $$2x_{1} + 0x_{2} = 2x_{1}$$$$0x_{1} - 1x_{2} = 2x_{2}$$ but from here I really don't know what to do. Can someone please tell me if I'm along the right tracks and explain simply what to do next?

2. Dec 9, 2011

### Quinzio

Well, you found that any value of $x_1$ will fulfill the system, while $x_2=0$
So any vector of the form (any value, 0) is an eigenvector.

An eigenvector is a vector which retains the same direction when a linear transformation is applied. The lenght is not necessarily retained.
This is beautifully explained with the Monalisa picture here: http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors.
In your case, the blue vectors are any vectors like (1,0), (2,0), (77,0), (-1,0) and so on.

3. Dec 9, 2011

### Kaldanis

So for the second eigenvalue λ = -1, I would have $$2x + 0y = -x$$$$0x - y = -y$$for the second equation it's obvious that -y=-y so that makes me think the vector would be $\begin{bmatrix}0\\1\end{bmatrix}$, but I really don't know how to interpret the first equation of 2x=-x and how this affects my final answer?

4. Dec 9, 2011

### Quinzio

It just an algebraic equation.
2x=-x
3x=0
x=0

In the end your eigenvectors will be 2.
(a,0) and (0,b)

5. Dec 9, 2011

### Kaldanis

Haha, well that was disgustingly obvious thank you for the help, I'll try and work through a few more problems and I'll post back here if I get stuck again.

6. Dec 9, 2011

### kru_

-y = -y is valid for any y in your field

What values of x satisfy the equation 2x = -x?

2x = -x => 2x + x = 0 => 3x = 0 => x = 0.

So, while any value of y will solve the system of equations, the only valid value for x is 0. Hence, the eigenvector for Lambda = -1 is [0, 1].

Basically, when you know the eigenvalues of a matrix or linear operator, you just need to solve the system of resulting equations to find the eigenvectors for each eigenvalue.

7. Dec 9, 2011

### HallsofIvy

Staff Emeritus
You need to understand that there do not exist just one or two eigenvectors. The fact that the determinant is 0 means that the equations will have an infinite number of vectors. In fact the set of all eigenvectors corresponding to one eigenvalue is a subspace.