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Finding eigenvectors of a matrix

  1. Feb 11, 2007 #1
    1. The problem statement, all variables and given/known data
    I need to find the eigenvectors for the matrix shown below.


    2. Relevant equations
    Nothing relevant.


    3. The attempt at a solution
    I have the matrix
    2 1 -1 0
    0 4 -2 0
    0 3 -1 0
    0 3 -2 1

    The eigenvectors are 1,1,2,2.

    To get the eigenvectors for 1,1, I use the matrix
    -1 1 -1 0
    0 -3 -2 0
    0 3 2 0
    0 3 -2 0

    When I bring this to RREF, I have
    1 0 0 0
    0 1 0 0
    0 0 1 0
    0 0 0 0

    So I know that the vector
    0
    0
    0
    1
    is an eigenvector.

    But the book also says that
    1
    2
    3
    0
    is an eigenvector for the eigenvalue 1.

    If the matrix only has one solution, how do I get this other eigenvector?
     
  2. jcsd
  3. Feb 11, 2007 #2
    See if this equations help:

    [tex]det(A-{\lambda}I) = 0[/tex]

    [tex]A{\cdot}x = {\lambda}x[/tex]

    where [tex]A[/tex] is the matrix, [tex]x[/tex] is an eigenvector and [tex]\lambda[/tex] is it's corresponding eigenvalue
     
  4. Feb 11, 2007 #3

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    When an eigenvalue has a multiplicity > 1 it means that there multiple linearly independent eigenvectors corresponding to that eigenvalue. In particular, the eigenvectors span an N-dimensional subspace, where N is the multiplicity of the eigenvalue.

    As a trivial example, the n-dimensional identity matrix [itex]I_n[/itex] has eigenvalue 1 with multiplicity n. The eigenvalues span the original space. (In other words, any vector in [itex]\mathbb R^n[/itex] is an eigenvector of [itex]I_n[/itex]).
     
  5. Feb 11, 2007 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is, I take it, your original matrix with 1 subtracted from the diagonal values? Or, at least it's supposed to be: check the second row: 4-1= 3, not -3.

    1 1 -1 0
    0 3 -2 0
    0 3 -2 0
    0 3 -2 0

    row reduces to
    1 0 -1/3 0
    0 1 -2/2 0
    0 0 0 0
    0 0 0 0

    That gives both (1, 2, 3, 0) as eigen value.
     
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