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Math Help - Eigen value determination

  1. #1
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    Eigen value determination

    I need to determine eigen values for the following matrix.



    - 3 2 -1

    2 2 -6

    -1 -6 5


    I get Det of A - lambda I =
    I will use x instead of lambda

    (-3 - x) (2 - x) (5 - x) - 36(-3 - x) - 4 (2 - x) + 10 + x = 0

    I know my factoring skills need work, but is there another way besides trial and error to determine my eigenvalues??? I get that lambda (or x as I have used) = -2 is one of my eigen values. I also understand, I think, that I should have two more. Is there a method I am not aware of or a site on the internet that has a tutorial for this process. So far all the ones I have found just use 2 x 2 matrices which are so simple even I can do them!!! Thanks so much for looking at this. Frostking
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  2. #2
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    You already know that one eigenvalue is -2.

    Now, instead of factoring the polynomial, you can use these two facts to find the remaining two:

    det A = 60 = \prod_{i=1}^{3}\lambda_i = -2\lambda_2\lambda_3

    trace(A) = 4 = \sum_{j=1}^{3}\lambda_j= \lambda_2 + \lambda_3 - 2

    Spoiler:
    -4, 10
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  3. #3
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    doublepost
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  4. #4
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    Quote Originally Posted by Frostking View Post
    I need to determine eigen values for the following matrix.



    - 3 2 -1

    2 2 -6

    -1 -6 5


    I get Det of A - lambda I =
    I will use x instead of lambda

    (-3 - x) (2 - x) (5 - x) - 36(-3 - x) - 4 (2 - x) + 10 + x = 0

    I know my factoring skills need work, but is there another way besides trial and error to determine my eigenvalues??? I get that lambda (or x as I have used) = -2 is one of my eigen values. I also understand, I think, that I should have two more. Is there a method I am not aware of or a site on the internet that has a tutorial for this process. So far all the ones I have found just use 2 x 2 matrices which are so simple even I can do them!!! Thanks so much for looking at this. Frostking

    Long division: -2 is a root of the pol. iff x+2 divides the pol., so divide your pol. by x+2 and get as quotient x^2-6x-40=(x-10)(x+4) and the other two eigenvalus are 10 and -4

    Tonio
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