Which Eigenvector is Wrong for Finding Eigenvalues of a Matrix?

[:Buddy:]

Too many Eigenvectors!?

Homework Statement

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)

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

 

[Dick]

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

 

[:Buddy:]

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

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

 

[Dick]

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

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

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.

 

[:Buddy:]

thanks! I hadn't realized 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]

:)

 

[HallsofIvy]

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.