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Quick Eigen Vector Question

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Hi, for the matrix A =
    0 1 0

    1 0 0

    0 0 2

    I have calculated the eigen values, and have successfully calculated the eigen vectors for lamda = -1 and 1. However for lamda = 2 the solutions says the answer is: Eigen vector: A( 0, 0, 1) where A is arbitrary and the vector is a column vector. The problem I am having is seeing where this answer came from.
    2. Relevant equations

    (A - lamda*I3)*(X, Y, Z) (column vector) = 0

    3. The attempt at a solution
    As I said above I already know the correct Eigen vector for this particular Eigen value, I just cannot see where the answer has come from. From using the equation above I got the three following equations:
    -2X + Y = 0
    -2Y = 0
    0 = 0

    From these equations I cannot see where you get an eigen vector including a value for Z. Any help would be very much appreciated, thanks in advance, Kamran.
     
  2. jcsd
  3. Mar 23, 2010 #2

    vela

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    The second equation tells you y=0, and then the first equation tells you x=0. There's no condition on z, so z is arbitrary. In other words, the eigenvector is of the form (0,0,t)=t(0,0,1).
     
  4. Mar 23, 2010 #3
    Ok I understand now, thank you for your help. :)
     
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