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Eigenvector solution problem

  1. Dec 13, 2005 #1
    Hi, I've got these two matrices (V & T) and omega square, which is what I have found to be the eigenvalues. Could anyone tell me if this is the way to find the eigenvectors for these matrices and if they are correct?
  2. jcsd
  3. Dec 13, 2005 #2
    I'm a little confused with your methodology...I believe it's the T matrix that is throwing me off. Why is there an 'm' in the vector? Are you multiplying an identity matrix by 'm' just to get rid of the 1/m in the omega value?

    If my understanding is correct it seems what you have done is fine except that I believe that your subtraction is backwards. I havn't done it myself to see if it makes a differce. But when finding eigenvectors I believe that you must have Vij - (wn)^2*Tij. Like I said, I havn't run through this on pencil and paper so I don't know how much of an effect this will make.

    Otherwise, if what I think you are doing is correct, everything looks right to me.
  4. Dec 13, 2005 #3
    yes I think that's why I multiplied m. One other question is x2

    [1 ]


    [1 ]

    how can you know which one it should be? or maybe those two are the same?

    thanks again...
  5. Dec 13, 2005 #4
    Eigenvectors are determined up to a multiplicative constant, so [1 -1] is the same eigenvector as [-1 1] as far as the eigenvalue goes. Now, if they form a degenerate subspace (i.e. they both have eigenvalue 2) things get slightly more interesting, but not really.
  6. Dec 13, 2005 #5
    exactly. Even more cool is the fact that they don't even have to be [-1,1]/[1,-1]! They could be anything you want as long as a1=-a2 or -a1=a2! It could be [4,523,231,-4,523,231] or [-87,87], the possibilities are endless! Math is cool!
  7. Dec 13, 2005 #6


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    Staff Emeritus
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    Gold Member

    We can say this more precisely: either of these eigenvectors constitute a basis for the eigenspace associated with this eigenvalue.

    Or, if you prefer, [1 -1] and [-1 1] both span the same subspace.
  8. Dec 14, 2005 #7
    I see, so how do you normalize these eigenvectors?
  9. Dec 14, 2005 #8
    Is this near close to the normalized form?

    Should I multiply the scalars into the vectors or leave them as is?

    Thanks again...
    Last edited: Dec 14, 2005
  10. Dec 14, 2005 #9

    Doc Al

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    Staff: Mentor

    Your normalized vectors look good to me. I'd multiply the scalars into the vectors (though it doesn't really matter).
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