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Linear Algebra - Eigenvectors

  1. Aug 1, 2010 #1
    1. The problem statement, all variables and given/known data
    A = [2,1,2;2,1,2;2,1,2]

    Find the Eigenvectors of A



    2. Relevant equations



    3. The attempt at a solution

    First I found the eigenvalues of A
    [tex]
    det(A - \lambda I) = 0
    [/tex]


    [tex]
    \lambda = 0,5
    [/tex]

    __________________________________________
    A-5I

    [-3,1,2;2,-4,2;2,1,-3] --(rref)--> [1,0,-1;0,1,1;0,0,0]

    x1 = x3
    x2 = -x3

    v1 = [1;-1;1]

    Is this the right eigenvector for A-5I?
    _______________________________________________

    A-0I

    [2,1,2;2,1,2;2,1,2] --(rref)--> [1,1/2,1;0,0,0;0,0,0]

    x1 + x2/2 + x3 = 0


    I'm not sure what to do in order to find the eigenvector for A-0I
     
  2. jcsd
  3. Aug 1, 2010 #2

    lanedance

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    Homework Helper

    the characteristic equation gives you a bit more help than that
    [tex] A = \begin{pmatrix}
    2 & 1 & 2 \\
    2 & 1 & 2 \\
    2 & 1 & 2
    \end{pmatrix}
    [/tex]

    [tex] A - \lambda I = \begin{pmatrix}
    2- \lambda & 1 & 2 \\
    2 & 1- \lambda & 2 \\
    2 & 1 & 2- \lambda
    \end{pmatrix}
    [/tex]

    taking the determinant simplifies to
    [tex] \lambda^2(5 -\lambda) [/tex]

    giving eigenvalues:
    5 with algebraic multiplicity of 1
    0 with algebraic multiplicity of 2

    the goemtric multiplicity (number of linearly independent eignevectors) is always less than or equal to the algebraic multiplicity

    this means there will be one eignevector corresponding to [itex] \lambda = 5 [\itex] and upto 2 corresponding to [itex] \lambda = 0 [\itex]
     
    Last edited: Aug 1, 2010
  4. Aug 1, 2010 #3

    lanedance

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    to check if its right try multiplying the vector by the matrix

    [tex] A.v_1 =
    \begin{pmatrix} 2 & 1 & 2 \\2 & 1 & 2 \\2 & 1 & 2 \end{pmatrix}
    \begin{pmatrix} 1 \\-1 \\1 \end{pmatrix}
    = \begin{pmatrix}3 \\3 \\3 \end{pmatrix}
    \neq 5 v_1
    [/tex]

    so this is not correct
     
  5. Aug 1, 2010 #4
    For A-5I
    x1 = x3
    x2 = x3

    so [1;1;1] is the eigenvector for [tex] \lambda = 5 [/tex]

    [2,1,2;2,1,2;2,1,2] * [1;1;1] = [5;5;5]


    But I'm still not sure how to find the eigenvectors for A-0I
     
  6. Aug 1, 2010 #5

    vela

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    You have x1 + x2/2 + x3 = 0. You can solve for any one of the variables in terms of the other two, so you have two degrees of freedom. So let x2=s and x3=t. Express the solutions in the form (x1, x2, x3) = s(some vector) + t(some vector). Those two vectors will be the eigenvectors.
     
  7. Aug 1, 2010 #6

    lanedance

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    ollowing on from Vela
    x1 + x2/2 + x3 = 0
    is the equaton of a plane through the origin, which can be spanned by any two linearly independent vectors lying in that plane, solving for x2 in term so of x1 & x3, gives
    x2 = -2(x1 + x3)

    note vela suggested solving for x1 first, which is fine, but i like the symmetry soving for x2

    then let x1=s, x3=t and follow on, note this is equivalent to solving the 2 cases below
    a) x1=1, x3=0
    b) x1=0, x3=1
    whih may be easier to see first time round
     
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