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How do you find this eigenvector?

  1. Oct 18, 2012 #1
    [0 1]
    [-2 -2] This is the 2x2 matrix.

    [λ -1]
    [2 λ+2] This is the matrix that equals λI - A.

    Here are the eigenvalues I found:
    λ = -1 + i, -1 - i

    I am really confused at what to do next to find the eigenvectors. I keep looking online for help but I still can't figure it out...

    Thanks!
     
  2. jcsd
  3. Oct 18, 2012 #2

    Dick

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    Now put one of your eigenvalues into λI - A and try to find a vector v such that (λI - A)v=0. Then v would be an eigenvector of λ. It's just a set of linear equations to solve with a free parameter.
     
  4. Oct 18, 2012 #3

    LCKurtz

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    For ##\lambda## equal an eigenvalue, your two equations are dependent, so you only need to solve one and you have one free variable. The first one is the same as ##\lambda x - y =0## or ##y = \lambda x##. So let ##x = 1## and what do you get for ##y##? Do that for each value of ##\lambda## and you will have two [x,y] eigenvectors.
     
  5. Oct 18, 2012 #4
    I get what I need to do in theory, I just don't actually know how to do it. How do you find v such that (λI-A)v=0?
     
  6. Oct 18, 2012 #5

    Dick

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    Just try it! What is the matrix (λI-A) when λ=-1+i? Write that matrix times a vector (x,y), set it equal to zero and try to find a solution for x and y.
     
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