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Eigenvalue homework problem

  1. Mar 30, 2008 #1
    For which value of k does the matrix

    |4 k|
    |-7 -5|

    have one real eigenvalue of multiplicity 2?

    3. The attempt at a solution

    - I tried by setting this problem up with det(A-(lambda)I) and trying to solve like that, but I can't seem to get it that way either. I am getting (lambda)^2+(lambda)-20+7k=0, but I don't know what to do from there. Please help.
  2. jcsd
  3. Mar 30, 2008 #2
    Well, what you have there is quadratic equation lambda .. do you know how to solve such an equation (hint; there is a formula ...:smile:)

    This formula will give you (in general) two different solutions (which might be complex). However, when you've figured out the formula for the eigenvalues (depending on k) you can think about what k has to be, such that these two eigenvalues are actually the same.

    So first you need to find the lambda's which fulfill your equation. And don't worry about k. It's just some constant number.
  4. Mar 30, 2008 #3
    look at the discriminant
  5. Mar 30, 2008 #4
    so the discriminant is negative, so that means there are no real roots and k would have to be something that would have the discriminant come out to 1, right? is it even possible to get the discriminant to equal 1 in this situation with b=1? am I not seeing what you all are pointing out?
  6. Mar 30, 2008 #5


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    The discriminant is b^2-4ac. You want it to be zero to get a double root.
  7. Mar 30, 2008 #6
    in order to get zero you would need the discriminant to come out to 1-1, but I don't know how you would get the 4ac to be equal to 1 with this problem. If you ignore the 7k, the discriminant is 81, but in order for the discriminant to be zero, wouldn't -20+7k have to equal 1/4?
  8. Mar 30, 2008 #7


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    Yes. 7k-20 should equal 1/4. Why is that a problem?
  9. Mar 30, 2008 #8
    I just made a simple mistake in my math. Thanks for the help.
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