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Finding eigenvectors of a matrix that has 2 equal eigenvalues 
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#1
Nov2112, 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: [AdI]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 [A3I]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. 


#2
Nov2112, 01:25 PM

Sci Advisor
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P: 9,470

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 rs eigenvectors. 


#3
Nov2112, 01:34 PM

P: 15




#4
Nov2112, 06:51 PM

P: 345

Finding eigenvectors of a matrix that has 2 equal eigenvalues
When I solve (A3I)X=0, I find two linearly independent solutions, i.e. eigenvectors to the eigenvalue 3:
[1 1 0]^{T} and [2 0 1]^{T}. 


#5
Nov2112, 07:20 PM

P: 15

I tried it again and now I get it. I just made a little mistake calculating 23 (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. 


#6
Nov2212, 10:37 AM

Sci Advisor
HW Helper
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. 


#7
Nov2212, 02:16 PM

P: 15




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