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Homework Help: Please help, eigenvalues

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data

    find the eigenvalues of 3x3 matrix:

    I have to learn how to find eigenvalues of 3x3 matrix

    and this is the link, am I not supposed to do lamda-1 instead of 1-lamda like here?
    (the chapter name is "Eigenvalues of 3×3 matrices")

    is there any simpler rule???
  2. jcsd
  3. Mar 19, 2009 #2


    Staff: Mentor

    Sadly, no simple rules.

    For eigenvalues, you're solving the following equation for [itex]\lambda[/itex].
    A[itex]\lambda[/itex] = [itex]\lambda[/itex]x
    or equivalently,
    (A - [itex]\lambda[/itex]I)x = 0

    For the equation above to be true for arbitrary x (which turn out to be the eigenvectors),
    (A - [itex]\lambda[/itex]I) has to be singular, which means that its determinant must be zero.
  4. Mar 19, 2009 #3
    A= ( -1 -2 -2;
    1 2 1;
    -1 -1 0)

    this is the matrix

    I don´t understand how to get the eigenvalues which are supposed to be -1 1 1
    because when I calculate it with the wikipedia rule I get -lamda^3+lamda^2-3lamda+1 and that does not make sense to me.
  5. Mar 19, 2009 #4


    Staff: Mentor

    You're taking the determinant of A - [itex]\lambda[/itex]I and setting it to zero, so you'll have a cubic polynomial in [itex]\lambda[/itex] that is equal to zero.
    You need to solve for [itex]\lambda[/itex] in that polynomial.
  6. Mar 19, 2009 #5
    What do you mean? :confused:
  7. Mar 19, 2009 #6


    Staff: Mentor

    Which part of what I said don't you understand?
  8. Mar 19, 2009 #7
    so you'll have a cubic polynomial in LaTeX Code: \\lambda that is equal to zero.
    You need to solve for LaTeX Code: \\lambda in that polynomial.

    can you show me?
  9. Mar 19, 2009 #8


    Staff: Mentor

    Here's the equation you need to solve:
    [tex]-\lambda^3+\lambda^2-3\lambda+1 = 0[/tex]

    I haven't checked your work, so it's possible that this isn't the right equation.

    If ([itex]\lambda - a[/itex]) is a factor, a has to be a divisor of 1, which severely limits the possibilities for a. By divisor, I mean a has to go into 1 a whole number of times.

    Use polynomial long division to work this out. Once you get one factor, you'll be left with a quadratic in lambda, which should be pretty easy to factor or use the quadratic formula on.
  10. Mar 19, 2009 #9
    thanks for your help
    i just don´t know how to solve it :( I´be been trying but is there any rule for solving equations like this or....?
  11. Mar 19, 2009 #10


    Staff: Mentor

    Let me say it again.
    Didn't you learn about polynomial long division back when you took elementary or intermediate algebra?
  12. Mar 19, 2009 #11


    User Avatar
    Homework Helper

    It may be worth looking at your determinant calculation again as well, very easy to make a miscalulation in those steps.

    if you know the eigenvalues are -1, 1, 1 the you should be able to factorise the determinant in term of these... so try subsitituting 1 or -1 into
    [tex] -\lambda^3+\lambda^2-3\lambda+1=0[/tex]
    [tex] -(1)^3+(1)^2-3.(1)+1=-2\neq 0[/tex]
    which makes me think something may have gone wrong in the det calc

    and then
    [tex] -\lambda^3+\lambda^2-3\lambda+1 \neq (\lambda-1)^2(\lambda+1) [/tex]
  13. Mar 20, 2009 #12


    User Avatar
    Science Advisor

    Your matrix is
    [tex]A= \begin{bmatrix} -1 & -2 & -2 \\ 1 & 2 & 1 \\ -1 & -1 & 0\end{bmatrix}[/tex]

    So the characteristic equation is
    [tex]|A- \lambda I|= \left|\begin{array}{ccc}-1- \lambda & -2 & -2 \\ 1 & 2- \lambda & 1 \\ -1 & -1 & -\lambda\end{array}\right|= 0[/tex]
    expanding, on, say, the first row, that is
    [tex](-1-\lambda)\left|\begin{array}{cc}2-\lambda 1 \\ -1 & -\lambda\end{array}\right|+ 2\left|\begin{array}{cc}1 & 1 \\ -1 & -\lambda\end{array}\right|+ 2\left|\begin{array}{cc}1 & 2-\lambda \\ -1 & -1\end{array}\right|[/tex]
    [tex]= (-1-\lambda)(\lambda^2- 2\lambda+ 1)+ 2(-\lambda+ 1)-2(1-\lambda)[/tex]
    [tex]= -(1+\lambda)(\lambda- 1)^2- 2(\lambda-1)+ 2(\lambda-1)[/tex]
    and, in that form, it is obvious that the last two terms cancel leaving the already factored form
    [tex]-(\lambda+ 1)(\lambda- 1)^2[/tex]
  14. Mar 20, 2009 #13
    When you are learning all this keep in mind that no-one actually solves the characteristic polynomial to find eigenvalues in real-world problems. This is mostly useful for small toy problems, or to gain understanding of the theory.

    For real problems you would use something like the QR algorithm or the Arnoldi iteration.
  15. Mar 25, 2009 #14
  16. Mar 25, 2009 #15
    Thanks a lot everybody!!!!
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