Find the eigenvectors given the eigenvalues

  1. Sep 27, 2011 #1

    sharks

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    The problem statement, all variables and given/known data
    This is the matrix A, which i need to find the eigenvalues and eigenvectors.
    3x3 matrix
    5 6 12
    0 2 0
    -1 -2 -2

    The attempt at a solution
    I have found the eigenvalues to be: 1, 2, 2.
    So, the final eigenvalues are : 1 and 2.

    Now, i found the eigenvector for eigenvalue = 1, which is:
    3x1 column matrix:
    [-3 0 1]^T

    But for the eigenvalue = 2, i am stuck, as these are the system equations that i have before me:

    3x1 + 6x2 + 12x3 = 0
    -x1 - 2x2 - 4x3 = 0

    I made x1 the subject of formula: -2x2 - 4x3
    And then i'm not sure how to proceed. But i'm going out on a limb here, so please correct me.

    Let x2 = 1 and x3 = 0

    Then i get this 3x1 column matrix:
    x2[-2 1 0]^T

    Let x3 = 1 and x2 = 0
    I get another 3x1 column matrix:
    x3[-4 0 1]^T

    So, all the eigenvectors in a 3x3 matrix P, are:
    -3 -2 -4
    0 1 0
    1 0 1

    Is this correct?? Most importantly, is my method correct? Is there a better method?
     
    Last edited: Sep 27, 2011
  2. jcsd
  3. Sep 27, 2011 #2

    micromass

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    Your equations reduce to one equation:
    [tex]x_1=-2x_2-4x_3[/tex]

    Try to set [itex]x_2,x_3[/itex] equal to something and see what you get. For example, set [itex](x_2,x_3)=(1,0)[/itex] and [itex](x_2,x_3)=(0,1)[/itex]. This will give rise to two linear independent eigenvectors which span the eigenspace.
     
  4. Sep 27, 2011 #3

    sharks

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    Thanks for your help, micromass.
    Could you please check on my final solution which i edited at the end of my first post above.
    BTW, you really have the best degree in the world. :)
     
  5. Sep 27, 2011 #4

    micromass

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    Looks good!!
     
  6. Sep 27, 2011 #5

    sharks

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    Thanks again.
    Now, i might be pushing into some daring territory here, but might you (or someone else) be familiar with matlab? Anyway, here goes the eigenvector solution from matlab:

    v =

    0.9701 -0.9487 -0.6963
    0 0 0.6963
    -0.2425 0.3162 -0.1741

    Notice that the last column should correspond to: [-4 0 1]^T (the ratio is what matters here) and since there are no middle zero from the matlab solution, i'm a bit uneasy that i might have made a mistake somewhere in my manual calculations, although i have doubled checked everything.
     
  7. Sep 27, 2011 #6

    Mark44

    Staff: Mentor

    It looks to me like the first column, not the third column, represents a vector that is a multiple of <-4, 0, 1>^T.
     
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