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Eigenvalue and Eigenvector problem

  1. May 8, 2006 #1
    Hi

    Given a 3x3 matrix


    [tex]A = \[ \left[ \begin{array}{ccc} 0 & 0 & 1+2i \\ 0 & 5 & 0 \\ 1-2i & 0 & 4 \end{array} \right][/tex]

    I need to a another 3x3 which satisfacies

    D = U^-1 A U

    Step 1.

    Finding the eigenvalues

    [tex]0 = det(A- \lambda I ) = (0- \lambda)(\lambda - 5) (\lambda -4 ), \lambda = 5,4,0[/tex]

    step 2.

    Finding the eigenvectors.

    A vector which satisfies (A-\lambda I) v = 0

    For \lambda = 5


    p(\lambda = 5) = [tex] \[ \left[ \begin{array}{ccc} -5 & 0 & 1+2i \\ 0 & 0 & 0 \\ 1-2i & 0 & -1 \end{array} \right][/tex] ~ [tex]\[ \left[ \begin{array}{ccc} 1 & 0 & -1/5-2/5i \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right][/tex]

    How do I read the eigenvector from the reduced matrix ???

    Sincerely Fred
     
    Last edited: May 8, 2006
  2. jcsd
  3. May 8, 2006 #2
    Your eigenvalues are wrong. Find the characteristic equation by working it out completely, and you'll find eigenvalues of -1, 5, and 5.
     
  4. May 8, 2006 #3
    Is 5 then a what is called a double root ?

    /Fred

     
  5. May 8, 2006 #4
    Kind of. The eigenvalue with the value of five is said to have a multiplicity of two.
     
  6. May 8, 2006 #5
    I found the first eigenvector for lambda = -1 to be [1+2i, 0, -1]^T

    To find for lambda = 5, do I row reduce the matrix A-5I ?

    If I put that matrix into reduced echelon form, I get [0,0,0], but that can't be right?

    /Fred

     
    Last edited: May 8, 2006
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