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Math Help - eigenvalues

  1. #1
    Junior Member
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    eigenvalues

    Can someone please help me find the eigenvalues because i keep getting the wrong values..

    The matrix is:
    <br />
\left[\begin{array}{rrr}-4&0&3\\1&1&-1\\-6&0&5\end{array}\right]<br />


    but then I keep getting a weird answer sighs (what i did was subtract lamda from the diagonal: a11, a22, a33)

    the eigenvalues are supposed to be 2, -1 and 1.


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  2. #2
    Senior Member
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    Finding the det of the following:

    \begin{vmatrix} -4-\lambda&0&3\\1&1-\lambda&-1\\-6&0&5-\lambda\end{vmatrix} = (-4-\lambda )(1-\lambda )(5-\lambda ) + 3(-1+\lambda )\cdot (-6) = 0

    \implies - (\lambda - 2) (\lambda - 1) (\lambda + 1) = 0

    Hence \lambda_1 = 2, \lambda_2 = 1, \lambda_3 = -1

    QED.
    Last edited by scorpion007; June 11th 2009 at 04:42 AM.
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Thanks
    1
    \begin{vmatrix}\lambda+4&0&-3\\[3mm]<br />
-1&\lambda-1&1\\[3mm]<br />
6&0&\lambda-5\end{vmatrix}


    =\ (\lambda+4)\begin{vmatrix}\lambda-1&1\\[1mm]<br />
0&\lambda-5\end{vmatrix}-3\begin{vmatrix}-1&\lambda-1\\[1mm]<br />
6&0\end{vmatrix}

    =\ (\lambda+4)(\lambda-1)(\lambda-5)+18(\lambda-1)

    =\ (\lambda-1)\left[(\lambda+4)(\lambda-5)+18\right]

    =\ (\lambda-1)(\lambda^2-\lambda-20+18)

    =\ (\lambda-1)(\lambda^2-\lambda-2)

    =\ (\lambda-1)(\lambda+1)(\lambda-2)

    Looks pretty straightforward to me.
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  4. #4
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    i dunno, i think that i must have messed up my algebra along the way.. thanks guys
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