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Math Help - Defect of an eigenvalue

  1. #1
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    Defect of an eigenvalue

    Hi,

    A = \left( \begin{array}{ccc}<br />
13 & 5 & 2 \\<br />
2 & 7 & -8 \\<br />
5 & 4 & 7 \end{array} \right)

    The characteristic polynomial is -(9-\lambda)^3=0, giving \lambda_{1,2,3}=9

    So the algebraic multiplicity M_{9}=3. I seem to be getting an incorrect answer for the geometric multiplicity m_{9}.

    By Gaussian Elimination: 4a_{11} + 5a_{12} + 2a_{13}=0 => a_{11} = -0.5[5a_{12} + 2a_{13}]

    It seems to me that we can obtain three linearly independent (LI) eigenvectors from this by setting a_{11}, a_{12}, and a_{13} to zero in turn. That would give m_{9} = 0. My textbook however gives m_{9}=2 (suggesting that we can only obtain one LI vector)

    Could you please clarify where I am going wrong in the LI part?

    Thanks.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Defect of an eigenvalue

    \dim V_9=3-\mbox{rank}(A-9I)=3-\mbox{rank}\begin{bmatrix}{4}&{\;\;5}&{\;\;2}\\{2}  &{-2}&{-8}\\{5}&{\;\;4}&{-2}\end{bmatrix}=\\3-\mbox{rank}\begin{bmatrix}{4}&{5}&{2}\\{0}&{9}&{18  }\\{0}&{9}&{18}\end{bmatrix}=3-2=1

    A basis of V_9 is \{(2,-2,1)\}.
    Thanks from algorithm
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  3. #3
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    Re: Defect of an eigenvalue

    Thanks.
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