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Homework Help: 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?
    Thanks...
    http://img305.imageshack.us/img305/6937/ok4zd.jpg [Broken]
     
    Last edited by a moderator: May 2, 2017
  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]

    or

    [-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

    Hurkyl

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    Staff Emeritus
    Science Advisor
    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?
    http://img124.imageshack.us/img124/4827/scan39nn.jpg [Broken]

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

    Thanks again...
     
    Last edited by a moderator: May 2, 2017
  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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