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Thread: Defect of an eigenvalue

  1. #1
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    Defect of an eigenvalue

    Hi,

    $\displaystyle A = \left( \begin{array}{ccc}
    13 & 5 & 2 \\
    2 & 7 & -8 \\
    5 & 4 & 7 \end{array} \right)$

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

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

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

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

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

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

    Thanks.
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