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Eigenvectors of a 3x3 matrix

  1. Jul 20, 2011 #1
    1. The problem statement, all variables and given/known data
    So the 3x3 matrix involved is [3 -1 -1:-4 6 4:-1 1 1], The eigenvalues are L=6 and L=2.


    2. Relevant equations

    (A-LI)e=0

    3. The attempt at a solution

    I stuck the eigenvalues into the matrix and got (-1 1 1)(not sure if its right) for L=2 but when I use L=6 in I cant seem to get the answer, I cant bring it down to echelon form that easily either. Thanks for any help
     
  2. jcsd
  3. Jul 20, 2011 #2

    Dick

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    No, (-1,1,1) isn't an eigenvector corresponding to the eigenvalue 2, and 6 isn't even a eigenvalue of that matrix. How you would expect anyone to be able to guess what you are doing wrong given the amount of work you've shown beats me.
     
  4. Jul 20, 2011 #3
    Sorry the matrix is [3 -1 -1:-4 6 4:1 -1 1] and one of the eigenvalue is 6 and the other is 2
     
  5. Jul 20, 2011 #4

    Dick

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    Ok, that's better. But (-1,1,1) is still not an eigenvector corresponding to the eigenvalue 2. There are two linearly independent eigenvectors corresponding the eigenvalue 2 and one eigenvector corresponding to the eigenvalue 6. Now can you show how you are trying to find them so someone can figure out what you are doing wrong?
     
  6. Jul 20, 2011 #5
    Okay, So for L=2 im getting [1 -1 -1:-4 4 4:1 -1 -1] and when I try using G.E I get [1 -1 -1:0 0 0:0 0 0].

    From there x - y -z=0
    x=y+z, thats why I thought it was [-1 1 1], i've never done this before so im not sure if what I am doing is right. Im trying to learn off youtube but im not understanding how they are getting the eigenvectors.
     
  7. Jul 20, 2011 #6

    Dick

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    x-y-z=0 is correct. Good work. But [-1,1,1] doesn't solve x-y-z=0? Does it??
     
  8. Jul 20, 2011 #7
    So [1 -1 -1] ? And you said there were 2 corresponding eigenvectors for L=2, how would I find the other?
     
  9. Jul 20, 2011 #8

    Dick

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    [1,-1,-1] means x=1, y=(-1) and z=(-1). That doesn't solve x-y-z=0 either. Just put the numbers in and try it. There are an infinite number of solutions of x-y-z=0. Try and find ONE of them first.
     
  10. Jul 20, 2011 #9
    [2 -1 -1]?
     
  11. Jul 20, 2011 #10

    Dick

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    x=2, y=(-1), z=(-1) so x-y-z=2-(-1)-(-1)=2+1+1=4! Not 0. Can you try just once more, please? Find me ANY correct solution of x-y-z=0.
     
  12. Jul 20, 2011 #11
    Ha, [2 1 1], okay so it could be [4 2 2] aswell and [6 3 3] etc?
     
  13. Jul 20, 2011 #12

    Dick

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    There you go. Much better. But [4,2,2]=2*[2,1,1] and [6,3,3]=3*[2,1,1]. So they aren't really 'different'. They are linearly dependent, they all lie along the same line. The eigenvectors of 2 form a plane. To get a full set of eigenvectors you need to find one off that line. Any ideas?
     
  14. Jul 20, 2011 #13
    [4 3 1]?
     
  15. Jul 20, 2011 #14

    Dick

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    That works. So you could give the eigenvectors as the space spanned [2,1,1] and [4,3,1]. It's pretty likely your book will express it as something like the span of [1,1,0] and [1,0,1]. That also works and looks simpler. But it's equivalent to the span of your two vectors.
     
  16. Jul 20, 2011 #15
    Yes I was getting confused with the 1's and 0's everywhere in the notes. So for L=6 will it be easier to use Gaussian Elimination or simultaneous? Im trying to use G.E but then the matrix goes into fractions and makes it harder. What I got so far is [1 -4 1].
     
  17. Jul 20, 2011 #16

    Dick

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    You can do it any way that feels easier to you. You can test your answer of [1,-4,1]. If it's an eigenvector with eigenvalue 6 then if you multiply it by the original matrix, then you should get 6*[1,-4,1], right? It's easier to check them than to solve for them. Try it, you'll find you are correct.
     
  18. Jul 20, 2011 #17
    Oh yes I see that it is actually easier to check them, thanks alot for that man! Much appreciated. I just didn't understand the x-y-z=0 properly but now you've clearly explained it to me.
     
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