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Math Help - 3X3 Matrix, I can find the Characteristic polynomial but not the eigenvalues

  1. #1
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    3X3 Matrix, I can find the Characteristic polynomial but not the eigenvalues

    I need to learn to find eigenvalues and vectors.
    I can find the Characteristic polynomial (and if i have the eigenvalues i can find the vectors)

    However i cannot get from the Characteristic polynomial to the eigenvalues

    for example if i have a 3x3 matrix

    (1 0 4)
    (0 2 0)
    (3 1 -3)

    The Characteristic polynomial is x^3 - 19x + 30 = 0 (x=lambda)

    I know that the eigenvalues are Real eigenvalues:
    {-5, 2, 3}

    and then i can work out the eigenvectors.

    But how do you get from this step (x^3 - 19x + 30 = 0) to the eigenvalues.

    Thankyou!
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  2. #2
    Senior Member yeKciM's Avatar
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    u can use Horner's scheme (are you familiar with it) ? sorry I forgot do they acquainted you with that on your level of education?

    or u can :

     x^3-19x+30=0

     x^3-19x+30=x^3-9x -10x +30 = x(x^2-9) -10(x-3)= x(x^2-3^2) -10(x-3) =

    = x(x-3)(x+3)-10(x-3) = (x-3) [x(x+3) -10]

     = (x-3)(x^2+3x-10) = (x-3)(x-2)(x+5) =0

    so u have

     x-3=0 \Rightarrow x_1=3

     x-2=0 \Rightarrow x_2=2

     x+5=0 \Rightarrow x_3=-5
    Last edited by yeKciM; August 10th 2010 at 01:17 PM.
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  3. #3
    MHF Contributor

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    Another thing you can do is use the "rational roots theorem": if a_nx^2+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0 is a polynomial equation with integer coefficients and \frac{m}{n} is a rational root of the equation, then the denominator, n, must divide the leading coefficient, a_n, and the numerator, m must divide the constant term, a_0.

    Here, your equation is x^3- 19x+ 30= 0, a polynomial equation with integer coefficients. The leading coefficient is 1 and the only integers that divide 1 are 1 and -1. Since the denominator of any rational root must be 1 or -1, the only rational roots must be integers. The constant term is 30 so any rational root must be a factor of 30. Since 30= 2(3)(5), the only possible rational roots are 1, -1, 2, -2, 3, -3, 5, -5, 6, -6, 10, -10, 15, -15, 30, and -30. Just put each of those into the equation to see whether or not they satisfy the equation and you will see that 2, 3, and -5 satisfy it.

    Note- this does not guarentee that there are any rational roots to a polynomial equation but if there are, it will find them.

    We should point out that, in general, finding eigenvalues is not at all an easy task!
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