# Finding the eigenvectors of a 2nd multiplicity engenvalue

1. Feb 17, 2013

### xdrgnh

1. The problem statement, all variables and given/known data
I'm given this matrice 2 1 0
1 2 0
0 0 3
and I need to find it's eigenvectors

2. Relevant equations

3. The attempt at a solution

So I get the eigenvalues to be 1,3,3 with 3 being the one with multiplicity of 2. For the eigenvector for 1 I get 1,-1,0 and for 3 I get 1,1,1 but here is the problem. For the other eigen vector for 3 the answer can be anything that satisfies x1=x2
x2=x1
x3=x3

so can something like 1,1,2 be the answer?

2. Feb 17, 2013

### Dick

Sure it could. Try it out if you are unsure. There are a lot of choices for specifying the eigenvectors. Any linear combination of 1,1,1 and 1,1,2 will also be an eigenvector with eigenvalue 3.

3. Feb 17, 2013

### xdrgnh

But wouldn't that mean there is no one definitive eigenbasis?

4. Feb 17, 2013

### Dick

Sure it would. There never is. You have a two dimensional space of eigenvectors with eigenvalue 3. There are lots of ways to choose a basis. I wouldn't call any of them 'definitive'.