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Eigenvectors of a 3x3 Matrix

  1. Aug 17, 2007 #1
    1. The problem statement, all variables and given/known data
    Given Matrix B:

    [ 1 2 1]
    [-1 2 -1]
    [ 2 -2 3]

    and knowing that one of the Eigenvalues is 4, find one other value and its corresponding eigenvector

    2. Relevant equations
    Bx=Lx (The basic idea behind eigenvectors)

    3. The attempt at a solution
    Ive set up the above determinant

    [ 1-L 2 1]
    [-1 2-L -1]
    [ 2 -2 3-L]

    Equal to zero, the only way I could figure how to do this question was using long division after getting the characteristic equation, but I keep getting a remainder which I shouldn't get If im not making a huge mistake.

    For the determinant I get either -L^3 + L^2 -9L +3 or -L^3 + 6L^2 -11L + 6
    but I checked both using a calculator and long division and both of them give me a remainder.

    I dont want the answer flat out maybe point out where im going wrong, this problem has been bugging me for ages and I really want to know what the hell is wrong.:grumpy:

  2. jcsd
  3. Aug 17, 2007 #2


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  4. Aug 17, 2007 #3


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    I don't get either of your polynomials. Maybe you should show your steps in calculating the determinant.
  5. Aug 18, 2007 #4
    He this is the determinant as I set it up







    Completely different but still wrong.
  6. Aug 18, 2007 #5


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    {[(-1)(-2)]-[(2)(2-X)]} = 2 -4 -(-2x) = -2 + 2x
  7. Aug 18, 2007 #6

    I just tried your problem twice. The first time, I did it the way you did, by calculating the determinant, and like you, I ended up with a factor left after long division. But then, I calculated det(LI-B)=0. It'd do it this way, if I were you, because then you don't get all the -L's in there, which just means there's less chance of numerical errors when you calculate the determinants.
  8. Aug 18, 2007 #7
    Oh I hate when that happens!!

    SO then

    The final solution is

  9. Aug 18, 2007 #8


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    Well, no, the "final solution" is not a polynomial! You still need to find the eigenvalues and corresponding eigenvectors.
  10. Aug 18, 2007 #9
    Yes, sorry I meant the polynomials solution.

    So all I have to do now is put the value for lamba back into Matrix B, multiply by the column matrix, make that equal to zero and work out the values of X, Y and Z.
  11. Aug 18, 2007 #10


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    Yes, finding the eigenvectors should be straightforward. Notice, however, that you have x=1 as a double root. There may be two independent eigenvectors corresponding to that.
  12. Aug 18, 2007 #11
    I would use row expansion.

    I get:

    [tex](1 - r)[(2 - r)·(3 - r) - 2] - 2[(-1)·(3 - r) + 2] + [2 - 2·(2 - r)] = 0[/tex]

    After simplifying you get:

    [tex](4 - r)·(r - 1)^2[/tex]

    So one eigenvalue is 4, like you said, and the others would be 1 and 1 (repeated).

    Finding the eigenvectors is straight forward from here. If you need anymore help I would be glad to go further :smile:

    Good luck!
  13. Aug 19, 2007 #12
    Thanks everyone, I managed to get the other eigenvector, there is only one from the repeated L=1 so it was straight forward.

    Thanks again.
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