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Eigenvalues of a matrix

  1. Dec 16, 2008 #1

    JJ6

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    1. The problem statement, all variables and given/known data

    Hey guys, for my linear algebra class I need to find the signs of the eigenvalues (I just need to know how many are positive and how many are negative) of an nxn matrix with zeros everywhere except for the two diagonals directly above and directly below the main diagonal, which both have all entries 1/2. Here's a picture with n=10:

    [​IMG]
    [​IMG]

    2. Relevant equations



    3. The attempt at a solution

    I know that I can take teh determinant and solve the characteristic equation, but I don't know how I would do that for an nxn matrix. Also, it seems like it shouldn't require so much work, since I need to know their signs but not their values. Can anybody point me in the right direction?
     
  2. jcsd
  3. Dec 16, 2008 #2
    i would try to preform some elementary row/column operations on it.
    try for the n=4,5,6
     
  4. Dec 17, 2008 #3

    JJ6

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    If I apply row/column operations on the matrix, will the eigenvalues be the same for the original matrix as they are for the echelon form?
     
  5. Dec 17, 2008 #4

    Hurkyl

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    :confused: Surely you know an algorithm for computing the determinant of any matrix....

    edit: bleh, the closed form solution is ugly. :frown:
    edit: but you can still find the roots of the closed form solution
    edit: and there's a much easier trick if you already know the eigenvalues are real
     
    Last edited: Dec 17, 2008
  6. Dec 17, 2008 #5

    Defennder

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    Well I don't think so, since it's eigenvalues and not determinant. You can try it out for n=3,4.
     
  7. Dec 17, 2008 #6

    HallsofIvy

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    Unfortunately, just row- reducing a matrix will NOT tell you its eigenvalues. If it did, finding eigenvalues would be a lot easier!
     
  8. Dec 17, 2008 #7

    JJ6

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    I've been trying to come up with a "clever" solution for a while, but to no avail. In my book, we have the theorem:

    Let A be a matrix, and let its eigenvalues all be real numbers. Then A is diagonalizable if and only if the geometric multiplicity of each eigenvalue equals its algebraic multiplicity.

    This is the only theorem I could find in our book that deals with real eigenvalues. Is there a clever way of applying this that I'm missing?
     
  9. Dec 17, 2008 #8

    Hurkyl

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    However, row reducing [itex]A - \lambda I[/itex] does help you calculate the determinant, which gives you the characteristic polynomial of A.

    But given that the matrix is so sparse, I wonder why not just apply the normal, general purpose algorithm that works to compute the determinant of any matrix.
     
  10. Dec 17, 2008 #9

    Avodyne

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    If you knew that [itex]\det(A-\lambda I)[/itex] was an odd or even polynomial in [itex]\lambda[/itex], the answer would be immediately obvious.
     
  11. Dec 17, 2008 #10

    Hurkyl

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    Only if you also knew that the polynomial was totally real. Odd and even polynomials can still have complex roots, and so their sign would be undefined.
     
  12. Dec 17, 2008 #11

    Avodyne

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    I was assuming the OP would have read the 3rd edit in your post #4. :smile:
     
  13. Dec 17, 2008 #12

    Dick

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    Oh, I think we all know it has real eigenvalues, don't we?
     
  14. Dec 17, 2008 #13

    JJ6

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    I have a quick question. If we have n = 2, then the matrix (after subtracting λ*I) has top row [-λ 1/2] and bottom row [1/2 -λ]. Then the characteristic equation is:

    λ^2 - 1/4 = 0

    But in this case, λ could be either positive or negative, right? If these λ that can be either + or - pop up, how do I tell what the matrix's eigenvalues are?

    Edit: Wait, that was a really stupid question. We could factor to (λ - 1/2)(λ + 1/2), so our two eigenvalues would be 1/2 and -1/2, correct?
     
    Last edited: Dec 17, 2008
  15. Dec 17, 2008 #14

    JJ6

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    OK, I'm guessing from what you guys are saying that I need to use Decartes' rule of signs, but I'm not quite sure how I need to apply it. Assuming that I have a polynomial of degree n with all real roots, I should have number of positive roots + number of negative roots + multiplicity of 0 = n. But I don't know what 0's multiplicity will be, and I don't know how many sign changes there will be to calculate the number of positive roots.
     
  16. Dec 17, 2008 #15

    Avodyne

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    Zero is a root only if your original matrix has zero determinant. Does it?

    You don't need to solve [itex]\lambda^2-{1\over 4}=0[/itex] to see that if [itex]\lambda[/itex] is a solution, so is [itex]-\lambda[/itex].
     
  17. Dec 17, 2008 #16

    JJ6

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    OK, the original matrix will have determinant zero when n is odd. If n is even, det(A) = (-1/2)^n. So if n is even, we'll have n roots that are either odd or even, and if n is odd, we'll have n-1 roots that are either even or odd. Then can I say that if λ is a solution, -λ is a solution also, so we'll have the same number of even and odd roots?
     
  18. Dec 17, 2008 #17

    Dick

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    Avodyne is saying that for even n det(A-lambda*I) has only terms of even power in lambda and for odd, only odd powers. I don't think this follows from what you are saying. I know he's right, but it seems like it should have a lot easier proof than anything I've come up with. And it's not Descartes rule of signs. It's just that if lambda and -lambda both have to be eigenvalues and you know you have n real eigenvalues, it's pretty easy to count the positive ones. I hope Avodyne is a he. Correct me if I'm wrong.
     
    Last edited: Dec 17, 2008
  19. Dec 17, 2008 #18

    Avodyne

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    The method I had in mind was to find a recursion relation for [itex]d_n = \det(A-\lambda I)[/itex]. Using cofactors, it's not too hard to express [itex]d_n[/itex] in terms of [itex]d_{n-1}[/itex] and [itex]d_{n-2}[/itex]. Then, using [itex]d_1=-\lambda[/itex] and [itex]d_0=1[/itex], it's pretty easy to show that [itex]d_n[/itex] is an even (odd) function of [itex]\lambda[/itex] if [itex]n[/itex] is even (odd).

    And yes, I'm a he!
     
  20. Dec 18, 2008 #19

    Dick

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    That sounds promising. I've just started to realize I haven't been paying enough attention to showing that there aren't multiple zero roots which something like that could give you. And actually Descartes rule of signs does come in pretty handy. I forgot to look up exactly what it was before dismissing it. Sorry JJ6.
     
    Last edited: Dec 18, 2008
  21. Dec 18, 2008 #20

    JJ6

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    Oh, I finally understand this. I got the recursion to work out fine, and I managed to find the signs of the eigenvalues. Thank you very much, everyone.
     
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