How to solve eigenvalues from a characteristic equation.

Discussion in 'Homework Help' started by u-will-neva-no, Nov 5, 2011.

  1. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    Hey! I have worked out the characteristic equation using det(A-λI) = 0,
    where
    A = [\frac{a  b}{c   d}] (ignore the line, it is a matrix but don't know how to do it in LaTex so used the fraction function)

    So my characteristic equation is:
    \lambda^2 -tr(A)\lambda + det(A) = 0

    where tr(A) = a + d and  det(A) = ad - bc

    How would I solve the eigenvalues for the above function? I understand that it is the values of \lambda = 0 but not sure how to make the function to to zero in this general case... Thanks!
     
  2. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
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    Your characteristic equation in \lambda is a quadratic -- can't you just write down the two roots (eigenvalues) using the quadratic formula? I get \frac{a+b \pm \sqrt{(a+d)^2 - 4(ad-bc)}}{2}. As always, check the geezer's algebra. :p
     
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  3. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
    2
    My quadratic equation came to:
     \frac{a+d \pm \sqrt(a^2 +d^2 -2(ad -2bc))}{2}

    My complete question is the following. I need to obtain a relationship between the trace and the determinant of A. The question also talks about a formula
    A = M  \wedge (M^-1) Would it be possible if anyone could explain/ give a solid example on how to use this formula?
     
  4. Georacer

    Moderator

    Nov 25, 2009
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    u-will-neva-no likes this.
  5. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
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    @someonesdad, I think I was wrong to expand out and your form is alot better to deal with. Has anyone come across a problem like this before?
     
  6. Georacer

    Moderator

    Nov 25, 2009
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    I agree with someonesdad's answer too. It's not that unusual of a problem. The methodology to set up your equation is standard. The fact that in the end you get a quadratic equation shouldn't trouble you, it's pretty common.
     
  7. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
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    What I don't understand Is how I can find the eigenvectors when my two eigenvalues are messy. Can the above eigenvalues be expressed in a matrix form?
     
  8. Georacer

    Moderator

    Nov 25, 2009
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    Yes, practically your eigenvalues are:

    \left[ \begin{array}{c} \frac{a+b + \sqrt{(a+d)^2 - 4(ad-bc)}}{2} \\ \frac{a+b - \sqrt{(a+d)^2 - 4(ad-bc)}}{2} \end{array} \right]

    Which makes your eigenvectors the solutions of the equation:
    A \cdot x_i=\lambda_i \cdot x_i

    I know it's fuzzy, but remember that in the end λi is a constant for your use.
     
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  9. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
    2
    Cool, thanks Georacer!
     
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