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How to find the eigenvectors

  1. Jan 25, 2006 #1
    find the eigenvectors for this matrix
    [tex] \left( \begin{array}{ccc}3&0&6\\0&-3&0\\5&0&2 \end{array} \right) [/tex]

    easy to find the eigenvectors which are -3,-3 and 8
    now how to find the eigenvectors
    am i supposed to do
    [tex] \left| \lambda I - A \right| X = 3 X [/tex]?
    then
    [tex] \left( \begin{array}{ccc} -6&0&6\\0&0&0\\5&0&-5 \end{array} \right) \left( \begin{array}{c} X_{1}\\X_{2}\\X_{3} \end{array} \right) = 0 [/tex] ? an find the solution for X1 x2 and X3?
    it comes 1,0,1 then, which is not correct...

    pleas help!
     
    Last edited: Jan 25, 2006
  2. jcsd
  3. Jan 25, 2006 #2

    0rthodontist

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    Well, you set

    [tex] \left( \begin{array}{ccc} -6&0&6\\0&0&0\\5&0&-5 \end{array} \right) \left( \begin{array}{c} X_{1}\\X_{2}\\X_{3} \end{array} \right) = 0 [/tex]

    up right but

    [tex] \left| \lambda I - A \right| X = 3 X [/tex] is wrong. The equation you need to solve is
    [tex] ( \lambda I - A ) X = 0 [/tex] for [tex] \lambda [/tex] = -3, and then 8. You only take the determinant if you're trying to find the eigenvalues. You can derive it because you're trying to solve

    [tex] A X = \lambda X = \lambda I X[/tex]
    [tex]A X - \lambda I X = 0[/tex]
    [tex](A - \lambda I) X = 0[/tex]
    or
    [tex](\lambda I - A) X = 0
    [/tex]

    But the final matrix equation you gave is not solved by (1, 0, 1). Probably just an arithmetic error.
     
    Last edited: Jan 25, 2006
  4. Jan 25, 2006 #3
    this is the matrix for the characteristic polynomai lright ...
    [tex] \left( \begin{array}{ccc}\lambda -3&0&6\\0&\lambda+3&0\\5&0&\lambda -2 \end{array} \right) [/tex]

    when u substitute -3 in there you get
    [tex] \left( \begin{array}{ccc} -6&0&6\\0&0&0\\5&0&-5 \end{array} \right) [/tex]
    isnt htat correct?
    Please point out my math error here...
     
  5. Jan 25, 2006 #4

    0rthodontist

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    D'oh, I was confused. The error is the 6 and the 5 in the top right and bottom left corners. They should be negative.
     
  6. Jan 25, 2006 #5
    OF COURSE! I was completely disregarding that everything lese in the matrix A will be negated

    thank you
     
  7. Jan 25, 2006 #6
    so from the -3 one of the eigenvectors is [-1 0 1] but why not [1 0 -1]??

    since there are only 2 eigenvalues how would you get the third eigenvalue?
     
  8. Jan 25, 2006 #7

    0rthodontist

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    I don't know what your question is. (-1, 0, 1) is an eigenvector for -3 and (1, 0, -1) is another eigenvector for -3, though they are not independent. Use row reduction to find the general form that an eigenvector for -3 must take, and you can get 2 independent eigenvectors.
     
  9. Jan 25, 2006 #8
    the real question is to diagonalize this matrix
    and to diagonalizei need to find P where P = [X1 X2 X3 ... Xn]
    thus i need to find THREE eigenvectors
     
  10. Jan 25, 2006 #9

    0rthodontist

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    You don't need just any three eigenvectors, you need three independent eigenvectors. The third comes from 8. Can you give the general form that an eigenvector for -3 must take?
     
  11. Jan 25, 2006 #10
    however the text (and mathematica) say that hte eigenvectors are
    [-1 0 1]
    [0 1 0]
    [6 0 5]
    how how?
     
  12. Jan 25, 2006 #11

    0rthodontist

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    Can you give the general form that an eigenvector for -3 must take?
     
  13. Jan 25, 2006 #12
    nevermind...
    i found the last eigenvector when i used the eigenvalue 8 and it yielded 2 independant eigenvectors
     
  14. Jan 25, 2006 #13

    0rthodontist

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    8 yields only 1 independent eigenvector. It has algebraic multiplicity 1 so it can't yield more than that.
     
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