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Math Help - Eigenvector problem

  1. #1
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    Eigenvector problem

    Never mind guys, I found the mistake. 0 - \lambda \neq 0

    Find the eigenvalues and eigenvectors of \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 1 & 1 & 0 \end{array} \right)

    Here's what I did.

    A - \lambda I = \left( \begin{array}{ccc} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & 4 \\ 1 & 1 & 0 \end{array} \right)

    det(A - \lambda I) = (2 - \lambda)(-4) + 1 = 0

    -8 + 4 \lambda + 1 = 0

    4\lambda = 7

    \lambda = \frac{4}{7}

    A - \lambda I = \left( \begin{array}{ccc} \frac{1}{4} & 0 & 0 \\ 0 & \frac{5}{7} & 4 \\ 1 & 1 & 0 \end{array} \right)

    \frac{1}{4}x_1 = 0

    \frac{5}{4}x_2 + 4x_3 = 0

    Eigenvector = \left( \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array} \right)

    I'm pretty sure this is wrong.
    Last edited by geft; October 26th 2009 at 12:14 AM.
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  2. #2
    Super Member
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    Quote Originally Posted by geft View Post
    Find the eigenvalues and eigenvectors of \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 1 & 1 & 0 \end{array} \right)

    Here's what I did.

    A - \lambda I = \left( \begin{array}{ccc} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & 4 \\ 1 & 1 & 0 \end{array} \right)

    det(A - \lambda I) = (2 - \lambda)(-4) + 1 = 0

    -8 + 4 \lambda + 1 = 0

    4\lambda = 7

    \lambda = \frac{4}{7}

    A - \lambda I = \left( \begin{array}{ccc} \frac{1}{4} & 0 & 0 \\ 0 & \frac{5}{7} & 4 \\ 1 & 1 & 0 \end{array} \right)

    \frac{1}{4}x_1 = 0

    \frac{5}{4}x_2 + 4x_3 = 0

    Eigenvector = \left( \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array} \right)

    I'm pretty sure this is wrong.
    First, the eigenvector can never be \left( \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array} \right), so this is wrong (can you see why?).

    Second, det(A-\lambda I) = (2-\lambda)((3-\lambda)(-\lambda) -4) = (2-\lambda)(\lambda(\lambda-3)-4) = (2-\lambda)(\lambda^2-3\lambda-4) = (2-\lambda)(\lambda+1)(\lambda-4)
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