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Botched Eigen Vectors!

  1. May 6, 2009 #1
    1. The problem statement, all variables and given/known data

    (a) Find the characteristic equation and
    (b) the eigenvalues and corresponding eigen-vectors

    [-5 0 0
    3 7 0
    4 -2 3]

    2. Relevant equations

    det([tex]\lambda[/tex]*I - A)

    3. The attempt at a solution
    Finding the characteristic equation wasn't that challenging, I took lambda, [tex]\lambda[/tex] , and subtracted the given matrix values from it for every diagonal entry. The determiant of a 3x3 matrix is just the product of these diagonals.....I think so my characteristic equation is ([tex]\lambda[/tex] + 5)([tex]\lambda[/tex]-7)([tex]\lambda[/tex] - 3)
    Next the eigenvalues are the values for [tex]\lambda[/tex] that when plugged in would make the characteristic equation equal zero.....so the eigenvalues are -5, 7, 3
    This is where I'm confused.
    To find the eigen-vectors I'm supposed to multiply each eigenvalue by the identity matrix and subtract the original given matrix: so for the eigenvalue of -5....

    [-5 0 0
    0 -5 0
    0 0 -5]
    MINUS
    [-5 0 0
    3 7 0
    4 -2 3]
    which equals
    [0 0 0
    -3 -12 0
    -4 2 -8]
    I took the reduced row echelon form of this matrix and got...
    [1 0 1.777
    0 1 -.4444
    0 0 0]
    Is my eigen vector in the matrix above? I am so confused....I also tried to solve the system generated by multiplying a 3x1 vector of variable components x, y, z times
    [0 0 0
    -3 -12 0
    -4 2 -8]
    I appear to have failed miserably at this....because the vector I got did not work. Can someone help walk me through this?
     
  2. jcsd
  3. May 6, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, in an "upper triangular" or "lower triangular" matrix like this, the eigenvalues are just the numbers on the main diagonal.

     
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