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Finding eigenvectors of a matrix that has 2 equal eigenvalues

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aija
#1
Nov21-12, 07:01 AM
P: 15
Matrix A=
2 1 2
1 2 -2
2 -2 -1

It's known that it has eigenvalues d1=-3, d2=d3=3


Because 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...+2y+-+4z+%3D+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%2C+-1%7D%7D

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.
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mathwonk
#2
Nov21-12, 01:25 PM
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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
#3
Nov21-12, 01:34 PM
P: 15
Quote Quote by mathwonk View Post
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
#4
Nov21-12, 06:51 PM
P: 345
Finding eigenvectors of a matrix that has 2 equal eigenvalues

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
#5
Nov21-12, 07:20 PM
P: 15
Quote Quote by Erland View Post
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
#6
Nov22-12, 10:37 AM
Sci Advisor
HW Helper
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P: 9,470
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
#7
Nov22-12, 02:16 PM
P: 15
Quote Quote by mathwonk View Post
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


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