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Math Help - Finding eigenvalues

  1. #1
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    Finding eigenvalues

    I've worked out my characteristic equation to be  \lambda^{3} - 4\lambda^{2} - 4\lambda + 16 = 0

    I have been letting  p(\lambda) = \lambda^{3} - 4\lambda^{2} - 4\lambda + 16 and filling in values for  \lambda which are the factors of 16 i.e.  \lambda = 1,-1,2,-2,4,-4,8,-8,16,-16 .

    However this method cannot tell me if an the same eigenvalue occurs twice... which i've a feeling in this case occurs...

    Any thoughts?
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  2. #2
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    Hi!

    Quote Originally Posted by Tekken View Post
    I've worked out my characteristic equation to be  \lambda^{3} - 4\lambda^{2} - 4\lambda + 16 = 0

    I have been letting  p(\lambda) = \lambda^{3} - 4\lambda^{2} - 4\lambda + 16 and filling in values for  \lambda which are the factors of 16 i.e.  \lambda = 1,-1,2,-2,4,-4,8,-8,16,-16 .

    However this method cannot tell me if an the same eigenvalue occurs twice...
    I agree!

    Do you know that a polynomial long division is? After finding a zero/eigenvalue polynomial long division leads to a quadratic equation including the two remaining zeros.

    Quote Originally Posted by Tekken View Post
    Any thoughts?
    I think the eigenvalues are
    +4
    +2
    -2

    so there are 3 different zeros - in this case no zero occurs twice, because the polynom

     p(\lambda)= \lambda^{3} - 4\lambda^{2} - 4\lambda + 16 = 0

    has the degree 3, so there are 3 solutions in  \mathbb{R} max
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  3. #3
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    Quote Originally Posted by Rapha View Post
    Hi!



    I agree!

    Do you know that a polynomial long division is? After finding a zero/eigenvalue polynomial long division leads to a quadratic equation including the two remaining zeros.



    I think the eigenvalues are
    +4
    +2
    -2

    so there are 3 different zeros - in this case no zero occurs twice, because the polynom

     p(\lambda)= \lambda^{3} - 4\lambda^{2} - 4\lambda + 16 = 0

    has the degree 3, so there are 3 solutions in  \mathbb{R} max
    Yes i know polynomial long division. So what your saying is if i found the first eigenvalue to be  \lambda = 4 i could divide the characteristic equation by  \lambda - 4 which will leave me a quadratic equation to which the solutions would be  \lambda = 2 &  \lambda = -2

    Correct?
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  4. #4
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    Quote Originally Posted by Tekken View Post
    Yes i know polynomial long division. So what your saying is if i found the first eigenvalue to be  \lambda = 4 i could divide the characteristic equation by  \lambda - 4 which will leave me a quadratic equation to which the solutions would be  \lambda = 2 &  \lambda = -2

    Correct?
    That's correct.
    Your remaining quadratic equation should be \lambda^2-4
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  5. #5
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    Quote Originally Posted by Rapha View Post
    That's correct.
    Your remaining quadratic equation should be \lambda^2-4
    Many thanks for the help, you've just given me a bit more confidence coming up to exam time
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  6. #6
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    Quote Originally Posted by Tekken View Post
    Many thanks for the help, you've just given me a bit more confidence coming up to exam time
    You're welcome.

    Good luck on your exams.
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