# Eigenvalue of 0

1. Aug 10, 2007

### wakko101

This is just a general question:

If, when you are calculating the eigenvalues for a matrix, you get a root of 0 (eg. x^3 - x) --> x(x-1)(x+1), what does that mean for the eigenvectors?

thanks,
w.

2. Aug 10, 2007

### olgranpappy

Nothing. It just means that one of the eigenvalues is zero, it doesn't mean anything special about eigenvectors... When diagonalized the matrix of the operator looks like
$$\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0& 0 & 1 \end{array} \right)\;.$$
In this basis, the eigenvector with eigenvalue -1 is (1,0,0) and the eigenvector with eigenvalue 0 is (0,1,0) and the eigenvector with eigenvalue 1 is (0,0,1).

3. Aug 11, 2007

### CompuChip

There is nothing wrong with an eigenvalue being zero, and it is not more special than an eigenvalue being -1, i or $\pi$.

Only an eigenvector cannot be zero. Which makes sense, because the zero vector trivially satisfies A 0 = $\lambda$ 0 for any number $\lambda$.

4. Aug 11, 2007

### D H

Staff Emeritus
A zero eigenvalue means the matrix in question is singular. The eigenvectors corresponding to the zero eigenvalues form the basis for the null space of the matrix.

5. Aug 11, 2007

### uiulic

According to definition: Ax=cx,x is nonzero vector,then
we have Ax=0,
which means it has nonzero solutions,
also means A is signular,
also means the corresponding eigenvector is not uniquely determined.

6. Aug 11, 2007

### Hurkyl

Staff Emeritus
Eigenvectors are never1 uniquely determined; at the very least, any nonzero scalar multiple of an eigenvector is an eigenvector.

1: Unless your scalar field is GF(2).

7. Aug 11, 2007

### quasar987

And who, might I ask, is GF(2) ?

8. Aug 11, 2007

### uiulic

Thank Hurkyl for pointing out my misunderstanding (I was thinking about sth else.)

9. Aug 11, 2007

### Hurkyl

Staff Emeritus
GF(2) is the finite field with two elements. It's isomorphic to the integers modulo 2.