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Math Help - Eigenvalues

  1. #1
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    Eigenvalues

    I am trying to determine the eigenvalues of A

    A =
    (6 0 2)
    (0 7 0)
    (2 0 3)
    Which is a 3x3 matrix.

    So far I have.

    The characteristic equation is det (A-lambda I)=0;

    |6-lam 0 2|
    |0 7-Lam 0| = 0
    |2 0 3-Lam|

    I expand the determinant, and obtain


    (6-Lam)
    |7-lam 0|
    |0 3-lam|
    -0+
    |0 6-Lam| = 0
    |2 0 |

    Not sure how to simplyfy and factorise this expession.
    Can anyone help?
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  2. #2
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    Re: Eigenvalues

    Quote Originally Posted by Arron View Post
    I am trying to determine the eigenvalues of A

    A =
    (6 0 2)
    (0 7 0)
    (2 0 3)
    Which is a 3x3 matrix.

    So far I have.

    The characteristic equation is det (A-lambda I)=0;

    |6-lam 0 2|
    |0 7-Lam 0| = 0
    |2 0 3-Lam|

    I expand the determinant, and obtain


    (6-Lam)
    |7-lam 0|
    |0 3-lam|
    -0+
    |0 6-Lam| = 0
    |2 0 |

    Can anyone help?
    You need to solve \displaystyle |\mathbf{A} - \lambda \mathbf{I}| = 0 for \displaystyle \lambda.

    \displaystyle \begin{align*} \mathbf{A} - \lambda \mathbf{I} &= \left[\begin{matrix}6 & 0 & 2 \\ 0 & 7 & 0 \\ 2 & 0 & 3\end{matrix}\right] - \left[\begin{matrix}\lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix}\right] \\ &= \left[\begin{matrix} 6 - \lambda & 0 & 2 \\ 0 & 7 - \lambda & 0 \\ 2 & 0 & 3 - \lambda \end{matrix}\right] \\ |\mathbf{A} - \lambda \mathbf{I}| &= \left|\begin{matrix}6 - \lambda & 0 & 2 \\ 0 & 7 - \lambda & 0 \\ 2 & 0 & 3 - \lambda\end{matrix}\right| \\ &= (6 - \lambda)\left|\begin{matrix} 7 - \lambda & 0 \\ 0 & 3 - \lambda \end{matrix}\right| - 0 \left|\begin{matrix} 0 & 0 \\ 2 & 3-\lambda \end{matrix}\right| + 2\left|\begin{matrix}0 & 7-\lambda \\ 2 & 0 \end{matrix}\right| \\ &= (6 - \lambda)(7 - \lambda)(3 - \lambda) - 2\cdot 2(7 - \lambda) \\ &= (7-\lambda)\left[(6-\lambda)(3-\lambda) - 4\right] \\ &= (7-\lambda)\left(18 -6\lambda -3\lambda + \lambda ^2 - 4\right) \\ &= (7-\lambda)(\lambda ^2 - 9\lambda + 14)\end{align*}

    So now solve \displaystyle (7-\lambda)(\lambda - 7)(\lambda - 2) = 0 for \displaystyle \lambda
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  3. #3
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    Re: Eigenvalues

    Quote Originally Posted by Arron View Post
    I am trying to determine the eigenvalues of A

    A =
    (6 0 2)
    (0 7 0)
    (2 0 3)
    Which is a 3x3 matrix.

    So far I have.

    The characteristic equation is det (A-lambda I)=0;

    |6-lam 0 2|
    |0 7-Lam 0| = 0
    |2 0 3-Lam|

    I expand the determinant, and obtain


    (6-Lam)
    |7-lam 0|
    |0 3-lam|
    -0+
    |0 6-Lam| = 0
    |2 0 |

    Not sure how to simplyfy and factorise this expession.
    Can anyone help?
    Expanding on the middle column (or middle row), because of those "0"s, give you
    (7- \lambda)\left|\begin{array}{cc}6-\lambda & 2 \\ 2 & 3- \lambda\end{array}\right|
    = (7- \lambda)(\lambda^2- 9\lambda+ 18- 4)= (7- \lambda)(\lambda^2- 9\lambda+ 14)
    and that last quadratic factors easily.
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  4. #4
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    Jan 2011
    Posts
    110

    Re: Eigenvalues

    Thanks, so the eigenvalues are;

    Lambda=7, lambda=7 and lambda=2.
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