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Linear Algebra

  1. Nov 16, 2003 #1
    I am having trouble with the following question. (Just hoping to get some guidance, recommended texts etc.):

    "Consider an eigenvalue problem Ax = λx, where A is a real symmetric n*n matrix, the transpose of the matrix coincides with the matrix, (A)^T = A. Find all the eigenvalues and all the eigenvectors. Assume that n is a large number."

    Any help would be fantastic!
  2. jcsd
  3. Nov 25, 2003 #2
    Don't know how much help I can be, but since I am studying the same material at the moment, I will help with what I can.

    The eigenvalues of A are equal to the eigenvalues of A^T because det(A-λI)=det(A^T-λI). The diagonal/trace stays the same here.

    (I am guessing on the next part, as our book does not cover this)
    Normally, for non-symmetric matrices the eigenvectors are not the same. However, in your case, since A^T=A then (and I am guessing here) I would tend to believe that Laplace expansions would end up yielding equal eigenvectors as well.
  4. Nov 25, 2003 #3


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    Re: Re: Linear Algebra

    If A^T = A then A is symmetric, no? Does this help?
  5. Nov 25, 2003 #4
    Yes, A is symmetric when A^T=A.

    First of all, I was wrong about the eigenvectors of A and A^T being the same. They are not. However, I cannot help as to why, as our text offers only two sentences in this matter. Further, I do not believe I am at a level of knowledge upon which speculation would prove fruitful. Hmm.. let's see. I can give you some links that will hopefully be of some help.

    We are using a book by Gilbert Strang from MIT. He has quite a bit of information on his/the books website, as well as fully recorded lectures. Here is his site.


    I am sorry I can be of little help with this, but I hope this can help you shed some light on the problem.
  6. Nov 28, 2003 #5
    Thanx bro,

    I know, its a tough subject, cheers for the link. Now that all other coursework is out of the way I will crack on with this and post my findings when I find something.

    Good luck with your course as well,


  7. Nov 28, 2003 #6
    Please do, as I am quite curious now.
    Good luck as well with your course!
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