Finding eigenvectors of a matrix that has 2 equal eigenvalues

[aija]

Matrix A=
2 1 2
1 2 -2
2 -2 -1

It's known that it has eigenvalues d1=-3, d2=d3=3Because it has 3 eigenvalues, it should have 3 linearly independent eigenvectors, right?

I tried to solve it on paper and got only 1 linearly independent vector from d1=-3 and 1 from d2=d3=3.

The method I used was:
[A-dI]v=0
and from this equation I used Gaussian elimination to find v1, v2 and v3

Even wolfram alpha finds only 1 solution from this:
http://www.wolframalpha.com/input/?i=-x+++y+++2z+=+0,+x+-+5y+-+2z+=+0,+2x+-+2y+-+4z+=+0
^
this is the system of equations from [A-3I]v=0 (3 is the eigenvalue d2=d3)

I don't see any way to get 2 linearly independent vectors from this solution
y=0, z=x/2

all i get is vectors
t*[2 0 1]T, t is a member of ℝ

here's matrix A in wolfram alpha: http://www.wolframalpha.com/input/?i={{2,+1,+2},+{1,+2,+-2},+{2,+-2,+-1}}

It shows that there is an eigenvector v3 = [1 1 0]T, but i don't see how to get it. Obviously my way to solve this problem doesn't work, so what did I forget to do in my solution or what did I do wrong and why doesn't it work this way?

PS. I'm not sure if this should be in the homework section, because this is more like a general problem and I don't understand why doesn't it work the way i tried to solve it. Matrix A could be any matrix with two equal eigenvalues.

 

[mathwonk]

when a matrix has r equal eigenvalues, the number of eigenvectors (using complex numbers) can be anywhere from one to r.

e.g. a square 2by2 matrix which has a "1" in the upper right hand corner, and all other entries zero, has only one eigenvector.

a square r by r matrix which has s ones just above the diagonal and all other entries zero, should have r-s eigenvectors.

 

[aija]

when a matrix has r equal eigenvalues, the number of eigenvectors (using complex numbers) can be anywhere from one to r.

e.g. a square 2by2 matrix which has a "1" in the upper right hand corner, and all other entries zero, has only one eigenvector.

a square r by r matrix which has s ones just above the diagonal and all other entries zero, should have r-s eigenvectors.

Ok, but according to wolfram alpha this matrix still has 3 eigenvectors, and I'm wondering why can i only find the first two eigenvectors using the method i used?

 

[Erland]

When I solve (A-3I)X=0, I find two linearly independent solutions, i.e. eigenvectors to the eigenvalue 3:
[1 1 0]^T and [2 0 1]^T.

 

[aija]

When I solve (A-3I)X=0, I find two linearly independent solutions, i.e. eigenvectors to the eigenvalue 3:
[1 1 0]^T and [2 0 1]^T.

Thanks,

I tried it again and now I get it. I just made a little mistake calculating 2-3 (not -5)

It's weird because I counted this twice (did the same mistake twice) and checked that I had counted everything totally right but didn't notice this.

 

[mathwonk]

my point was that your question:

"Because it has 3 eigenvalues {-3,3,3}, it should have 3 linearly independent eigenvectors, right?"

has answer
"no, not right."

and misunderstanding this general principle is more harmful in the long run than adding 2 and -3 and getting -5.

 

[aija]

my point was that your question:

"Because it has 3 eigenvalues {-3,3,3}, it should have 3 linearly independent eigenvectors, right?"

has answer
"no, not right."

and misunderstanding this general principle is more harmful in the long run than adding 2 and -3 and getting -5.

yes, i understood that as well, thanks