# Too many Eigenvectors?

1. Apr 11, 2012

### :Buddy:

Too many Eigenvectors!?

1. The problem statement, all variables and given/known data
I have to find the eigenvalues and eigenvectors of:

-1 2 -2
1 2 1
-1 -1 0

and I can find four eigenvectors I'm not sure how to tell which of my eigenvectors is
wrong as they all seem to satisfy Av=λv
(i also checked that they arent simply multiples of each other)

3. The attempt at a solution
i used
det(A-λI)=0
to get λ=-1,1,1

then i used the definition Av=λv

to get the eigenvectors
[0,1,-1], [1,-1,0], [1,0,-1], [1,-2,1]

im not sure which of these is wrong and why

2. Apr 11, 2012

### Dick

Re: Too many Eigenvectors!?

The eigenvalues of the matrix you showed are not -1,1,1. How did you conclude that? Is there a typo?

3. Apr 11, 2012

### :Buddy:

Re: Too many Eigenvectors!?

yes it was a typo sorry
the matrix was meant to be

-1 -2 -2
1 2 1
-1 -1 0

4. Apr 12, 2012

### Dick

Re: Too many Eigenvectors!?

Writing v1=[1,-1,0] and v2=[0,1,-1], v1 and v2 are both independent eigenvectors corresponding to the eigenvalue 1. And you are allowed to have two of those since the eigenvalue 1 has multiplicity 2. The other two vectors you wrote are v1+v2 and v1-v2. There are lots more eigenvectors corresponding the eigenvalue 1 as well, any linear combination of v1 and v2 will do. What you are missing is the eigenvector corresponding to the eigenvalue -1.

Last edited: Apr 12, 2012
5. Apr 19, 2012

### :Buddy:

Re: Too many Eigenvectors!?

thanks! I hadn't realised that linear combinations would be solutions. I now have a correct ( I think) set of vectors

v1=[-1 1 0]
v2=[-1 0 1]
v3=[2 -1 1]

:)

6. Apr 19, 2012

### HallsofIvy

Staff Emeritus
Re: Too many Eigenvectors!?

One of the very first things you should have learned about eigenvectors is there is NOT a single unique eigenvector corresponding to a given eigenvalue. In fact, the set of all eigenvectors corresponding to a given eigenvalue form a subspace which necessarily contains linear combinations.