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Math Help - Eigenvalues; help with calculating characteristic polynomial roots

  1. #1
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    Eigenvalues; help with calculating characteristic polynomial roots

    I need to find the eigenvalues of the linear transformation that the following matrix represent:

    A=\begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ 4 & -17 & 8\end{bmatrix}

    so I calculated the characteristic polynomial of A:

    f_A(\lambda)=det(\lambda I-A)=\lambda \begin{vmatrix}\lambda & 1\\ -17 & \lambda-8\end{vmatrix}-1\begin{vmatrix}0 & \lambda -8\\ 4 & 1\end{vmatrix}+0\begin{vmatrix}0 & \lambda\\ 4 & -17 \end{vmatrix}

    f_A(\lambda)=\lambda(\lambda(\lambda-8)+17)-1(-4)

    f_A(\lambda)=\lambda^3-8\lambda^2+17\lambda+4

    but I have no idea how to find its roots...

    any ideas would be greatly appreciated.
    TIA!


    **edit
    Is there another way to find the eigenvalues of A?
    Last edited by Stormey; April 7th 2013 at 02:54 PM.
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  2. #2
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    Re: Eigenvalues; help with calculating characteristic polynomial roots

    There is some methods that work on a 3rd degree polynomial. You could use a factoration like
    f_A(\lambda)=(\lambda - a_0)(\lambda - a_1)(\lambda - a_2) where a_0, a_1, a_2 are polynomial roots.
    There is another method that is called Ruffini's method (or Briot-Ruffini's method)
    Ruffini's rule - Wikipedia, the free encyclopedia, take a look!
    Last edited by bosabarbosa; April 7th 2013 at 03:08 PM.
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    Re: Eigenvalues; help with calculating characteristic polynomial roots

    Hi bosabarbosa, thanks for the help.

    I'm familiar with polynomial division, but it is not useful here, since i don't know what are the divisors to be begin with (to use the example from the wiki-link - I don't know what 'r' is).
    Division by (\lambda -r) when 'r' isn't a root will produce remainder.

    I'm looking for a way to find the roots without dealing with the polynomial's standard form ( a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0).
    (maybe calculating the determinant in some other way will produce an easier-to-handle polynomial?)
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    Re: Eigenvalues; help with calculating characteristic polynomial roots

    Quote Originally Posted by Stormey View Post
    I need to find the eigenvalues of the linear transformation that the following matrix represent:

    A=\begin{bmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ 4 & -17 & 8\end{bmatrix}

    so I calculated the characteristic polynomial of A:

    f_A(\lambda)=det(\lambda I-A)=\lambda \begin{vmatrix}\lambda & 1\\ -17 & \lambda-8\end{vmatrix}-1\begin{vmatrix}0 & \lambda -8\\ 4 & 1\end{vmatrix}+0\begin{vmatrix}0 & \lambda\\ 4 & -17 \end{vmatrix}
    You made a mistake calculating the determinant.

    f_A(\lambda)=det(\lambda I-A)=\lambda \begin{vmatrix}\lambda & -1\\ 17 & \lambda-8\end{vmatrix}-1\begin{vmatrix}0 & -1\\ -4 & \lambda-8\end{vmatrix}+0\begin{vmatrix}0 & \lambda\\ -4 & 17 \end{vmatrix}
    Last edited by Gusbob; April 8th 2013 at 01:56 AM.
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    Re: Eigenvalues; help with calculating characteristic polynomial roots

    I tried is some manipulations that seems to work fine:

    \lambda^3-8\lambda^2+17\lambda-4=0

    \lambda(\lambda^2-8\lambda)+17\lambda-4=0

    doing again, we get:

    \lambda[(\lambda^2-8\lambda)+17]-4=0

    Now, if you try to complete the square:

    \lambda[(\lambda-4)^2-16+17]-4=0

    \lambda[(\lambda-4)^2+1]-4=0

    \lambda(\lambda-4)^2+(\lambda-4)=0

    Factoring \lambda-4:

    (\lambda-4)[\lambda(\lambda-4)+1]=0

    (\lambda-4)[\lambda^2-4\lambda+1]=0

    Now, you have a root, 4! Now, the rest is up to you.
    Last edited by bosabarbosa; April 8th 2013 at 06:13 AM.
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  6. #6
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    Re: Eigenvalues; help with calculating characteristic polynomial roots

    Thats brilliant man.
    Thank you so much!
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