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Are my eigenvalues right?

  1. Mar 4, 2004 #1
    Find the eigenvalues and eigenvectors of the general real symmetric 2 x 2 matrix A= a b
    b c

    The two eigenvalues that I got are a-b and c-b. I got these values from this:

    (a-eigenvalue)(c-eigenvalue) = b^2
    (a-eigenvalue)= b = a-b
    (c-eigenvalue)= b = c-b

    Will there be 4 eigen values instead of the two that I have? Like, a-b and a+b and c-b and c+b?
    I haven't gotten to the eigenvectors yet. I'll post what I have in a minute.

    Thanks for your help
  2. jcsd
  3. Mar 4, 2004 #2

    Are these the eigenvectors?

    For a-b,


    For c-b,

  4. Mar 4, 2004 #3
    For the eigenvalues, you can't quite perform one of the steps you did. It's an incorrect assumption that

    a - r = b = c - r
    where r is an eigenvalue.

    You've got the right equation up until that point, i.e.,

    (a-r)(c-r) - b^2 = 0

    but this is a quadratic equation in the r variable. Solve it the way you would normally solve a quadratic equation! Nothing too fancy.

  5. Mar 5, 2004 #4


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    Staff Emeritus
    Science Advisor

    If you are working with eigenvalues and eigenvectors it would be a really, really good idea not to mess up basic algebra!

    You have:
    (a-eigenvalue)(c-eigenvalue) = b^2
    (a-eigenvalue)= b = a-b
    (c-eigenvalue)= b = c-b

    You appear to be thinking that if xy= b^2, then x= b and y= b.

    That is not at all true!! (9*1= 32;5*(9/5)= 32, etc.)

    Once you have the equation (a-eigenvalue)(c-eigenvalue) = b^2

    (I'm going to use λ for the eigenvalue)

    (a- &lamda;)(c-&lamba;)- b2= 0
    λ2-(a+c)&lambda+ (ac- b2)= 0

    Now you can use the quadratic formula to solve for the two values of λ
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