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Math Help - Finding eigenvectors

  1. #1
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    Finding eigenvectors

    Just out of my exam there and I it went fine except for one question. Getting the eigenvalues and eigenvectors of

    1 0 8
    4 3 12
    0 0 7

    I got eigenvalues of 1,3 and 7. However when I used the equation (A-λI)v = 0 for these values I couldnt get a correct answer. For example when λ = 1 I get

    (0 0 8 ) x = 0
    (4 2 12) y = 0
    (0 0 6 ) z = 0

    That gives -
    8z = 0
    4x + 2y + 12z = 0
    6z = 0

    So I get no solution....


    When I take λ = 3 I get

    (-2 0 8 ) x = 0
    (4 0 12) y = 0
    (0 0 4 ) z = 0

    and that gives
    -2x + 8z = 0
    4x + 12z = 0
    4z = 0

    ...


    So how am I supposed to find the eigenvectors?
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  2. #2
    A Plied Mathematician
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    Quote Originally Posted by nukenuts View Post
    Just out of my exam there and I it went fine except for one question. Getting the eigenvalues and eigenvectors of

    1 0 8
    4 3 12
    0 0 7

    I got eigenvalues of 1,3 and 7. However when I used the equation (A-λI)v = 0 for these values I couldnt get a correct answer. For example when λ = 1 I get

    (0 0 8 ) x = 0
    (4 2 12) y = 0
    (0 0 6 ) z = 0

    That gives -
    8z = 0
    4x + 2y + 12z = 0
    6z = 0

    So I get no solution....
    You don't? How about z = 0, 2x + y = 0, for x not equal to 0?



    When I take λ = 3 I get

    (-2 0 8 ) x = 0
    (4 0 12) y = 0
    (0 0 4 ) z = 0

    and that gives
    -2x + 8z = 0
    4x + 12z = 0
    4z = 0

    ...


    So how am I supposed to find the eigenvectors?
    Here z = 0, and therefore x = 0, but y is allowed to be anything except zero. This is all assuming you've done your row reduction correctly. I haven't checked that.
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  3. #3
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    How did I miss that? Thanks for pointing it out mate.
    Last edited by Ackbeet; April 15th 2011 at 01:07 PM.
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  4. #4
    A Plied Mathematician
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    You're welcome!
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