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? (need help as soon as possible, thx)

  1. Dec 13, 2004 #1

    quasar987

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    ?!? (need help as soon as possible, thx)

    Ok, what am I doing wrong with that? It's the second problem like this with which I arrive at an absurdity.. Essentially I must show that A is not diagonalizable. So I want to find the eigenvalues and eigenvectors and show that there are not enough eigenvectors to diagonalize A. To ease the calculations, I am told that -1 is an eigenvalue.

    [tex]A=\left( \begin{array}{ccc} 2 & 1 & 0 \\
    2 & 1 & -2 \\ -1 & 0 & -2 \end{array} \right)[/tex]

    I must find the roots of the 3rd degree polynomial equation expressed by [itex]det(A-\lambda I)=0[/itex]. Let us develop that determinant according the 2nd column. We have

    [tex]det(A-\lambda I) = \sum_{i=1}^3 (-1)^{i+2}a'_{i2}det((A-\lambda I)_{i2}) = - \left| \begin{array}{cc} 2 & -2 \\ -1 & -2-\lambda \end{array} \right| + (1-\lambda)\left| \begin{array}{cc} 2-\lambda & 0 \\ -1 & -2-\lambda \end{array} \right|[/tex]
    [tex]= -[(2)(-2-\lambda)-(-2)(-1)]+(1-\lambda)[(2-\lambda)(-2-\lambda)] = (2\lambda + 6) + (1-\lambda)(\lambda^2 -4)[/tex]
    [tex] = -\lambda^3 + \lambda^2 -2\lambda + 2 [/tex]

    For which of course -1 is not a root!


    The final exam is tomorrow, so thanks for your quick replies! :smile:
     
    Last edited: Dec 13, 2004
  2. jcsd
  3. Dec 13, 2004 #2

    shmoe

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    Hi, it looks like you've made an error in finding the characteristic polynomial. In your second to last equality, you replaced [tex](2-\lambda)(-2-\lambda)[/tex] with [tex](-2\lambda -4)[/tex]. This still won't make -1 a root though.

    Is it possible there's a typo in the question?
     
  4. Dec 13, 2004 #3
    According to Derive, A doesn't have an eigenvalue of -1...
     
  5. Dec 13, 2004 #4

    quasar987

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    grr.. well alright, thanks guys.
     
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