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Thread: find characteristic equation

  1. #1
    Super Member bigwave's Avatar
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    find characteristic equation

    Attachment 38634
    ok im doing #1

    ok I go a reduction of
    $$\frac{1}{3}
    \begin{bmatrix}
    1&1/3&0\\
    0&1&0\\
    0&0&1
    \end{bmatrix}$$
    but what next?

    the bk answer is
    $-\lambda^3 +18\lambda^2
    -95-\lambda+150$
    Attached Thumbnails Attached Thumbnails find characteristic equation-capture-_2018-04-01-11-15-54.png  
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  2. #2
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    Cool Re: find characteristic equation

    Beware Bigwave: your number of posts reached the beastly "666" on Easter day
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  3. #3
    Super Member bigwave's Avatar
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    Re: find characteristic equation

    God Bless...
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  4. #4
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    Re: find characteristic equation

    You say, in your post, that you are doing #1 but there is no "#1" in the attachment. Instead, #13 is highlighted so I assume that is what you mean.

    Your problem is that while "reducing" a matrix to upper triangular form is useful for finding an inverse matrix or solving matrix equations, it has nothing to do with finding eigenvalues. The eigenvalues of the reduced matrix are not those of the original matrix.

    Instead we need to solve the determinant equation $\displaystyle \left|\begin{array}{ccc}6-\lambda & -2 & 0 \\ -2 & 9-\lambda & 0 \\ 5 & 9 & 3- \lambda \end{array}\right|= 0$. I notice that the last column has '0's every where except at the bottom so we can "expand" on the final column to get
    $\displaystyle (3- \lambda)\left|\begin{array}{cc} 6-\lambda & -2 \\ -2 & 9- \lambda\end{array}\right|= (3- \lambda)((6- \lambda)(9- \lambda)- 4)= (3- \lambda)(54- 15\lambda+ \lambda^2- 4)= 150- 95\lambda+ 18\lambda^2- \lambda^3= 0$.
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  5. #5
    Super Member bigwave's Avatar
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    Re: find characteristic equation

    yeah I saw the #1 was actually #13 but the edit on the OP was not available

    So really no need to reduce then?
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  6. #6
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    Re: find characteristic equation

    Not for finding the characteristic equation. The characteristic equation for your reduced matrix is $\displaystyle \left|\begin{array}{ccc}1- \lambda & \frac{1}{3} & 0 \\ 0 & 1- \lambda & 0 \\ 0 & 0 & 1- \lambda \end{array}\right|= (1- \lambda)^3$ but that is NOT the characteristic equation of the original matrix.
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