# Formula for unit eigenvector

1. May 25, 2014

### Jhenrique

Last edited: May 25, 2014
2. May 25, 2014

### HallsofIvy

Staff Emeritus
?? Take any eigenvector and divide by its length.

3. May 25, 2014

### Jhenrique

And which is the formula for the eigenvector?

4. May 26, 2014

### DeIdeal

^ What do you mean by that? A (right) eigenvector of A, x, is a (nonzero) solution to Ax=λx, and λ is the corresponding eigenvalue. Any vector fulfilling the condition can be divided by its length ||x||, so that the resulting vector is still an eigenvector, since the eigenvector is not the zero vector.

(EDIT: Ok, technically talking about ||x|| obviously means that we have to be able to define a norm, but I don't think that was the issue?)

Last edited: May 26, 2014
5. May 26, 2014

### Jhenrique

Again: which is the formula for the eigenvectors?

6. May 26, 2014

### DeIdeal

What's wrong with Ax=λx? Given A and λ, it can be used to (numerically or analytically) solve x.

7. May 27, 2014

### HallsofIvy

Staff Emeritus
Your original question was about unit eigenvectors and that is what I responded to. There are a number of ways of finding eigenvectors but there is no "formula" you can just plug numbers into. Finding eigenvalues and eigenvectors is one of the harder problems in Linear Algebra.

8. May 27, 2014

### Jhenrique

If I want to express an eigenvector like (cos(Θ), sin(Θ)), is this form good way of constraint the expression, so that Θ is function of eigenvalues?

9. May 29, 2014

### Jhenrique

By defition:

$A\vec{v} =\lambda \vec{v}$
$(A - \lambda I)\vec{v} = \vec{0}$

So any eigenvector $\vec{v}$ is:
$\vec{v} = (A - \lambda I)^{-1}\vec{0} = \vec{0}$
$\vec{v} = \vec{0}$

What is wrong?

10. May 29, 2014

### DeIdeal

A square matrix M has an inverse iff $|M|\neq0$. To obtain the eigenvalues $\lambda$, you solve the equation $|A-\lambda I|=0$. In your post, you use the expression $(A-\lambda I)^{-1}$, which is meaningless, because the eigenvalues are exactly the values for which the inverse doesn't exist.

11. May 29, 2014

### Matterwave

Also, although the math used was wrong, the 0 vector really is technically an eigenvector of all matrices...it's the trivial eigenvector, with an ill-defined eigenvalue. A*0=lambda*0 for all A and all lambda.

12. May 29, 2014

### AlephZero

No, eigenvectors are defined to be non-zero vectors.

http://www.math.harvard.edu/archive/20_spring_05/handouts/ch05_notes.pdf

http://mathworld.wolfram.com/Eigenvector.html

http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors

The notion of the zero vector as a "trivial eigenvector with an ill-defined eigenvalue" doesn't have any practical (or even theoretical) value.

13. May 29, 2014

### Matterwave

Alright. Well, if it's right there in the definition, then looks like I was wrong. :)