How to find eigenvectors of a 3 by 3 matrix

[gboff21]

Question: find the eigenvectors of {{2,-1,-1},{-1,2,-1},{-1,-1,2}}.

Equations: none

Attempted answer:

I have the characteristic equation as x^3-6x^2+9 which gives eigenvalues as 0 3 and 3. This is correct says wolfram alpha

for x=0 {{2,-1,-1},{-1,2,-1},{-1,-1,2}}•{x,y,z}=0
I get 3 different equations! How does this give an answer?

For x=3 I get -{{1,1,1},{1,1,1},{1,1,1}•{x,y,z}=0
which gives v=anything
So what the hell??

Please help!

 

[Sethric]

For x=0 you get three equations for three unknowns. That's exactly what you need. Use whatever method you prefer to solve (substitute, gaussian elimination, etc.) for x,y, and z.

For x=3 you have repeated roots. Since this multiplicity is two, all you need to do is find two eigenvectors that work that are linearly independent. As you demonstrated, anything works, so just pick something simple for the first and then use that one to determine a second.

 

[gboff21]

For x=0. How do you get the right answer cos I constantly get {3,1,1} which is wrong.
And for x=3 how do you do do it. I don't understand how you use one randomly picked one, say {1,1,1} (which is one of the answers), to work out another!

 

[gboff21]

Never mind I've done it. Thanks