Find the eigenvectors given the eigenvalues

  1. sharks

    sharks 837
    Gold Member

    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. micromass

    micromass 20,039
    Staff Emeritus
    Science Advisor
    Education Advisor

    Your equations reduce to one equation:

    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. sharks

    sharks 837
    Gold Member

    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. micromass

    micromass 20,039
    Staff Emeritus
    Science Advisor
    Education Advisor

    Looks good!!
  6. sharks

    sharks 837
    Gold Member

    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. 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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