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Finding eigenvalues to use in Cayley-Hamilton theorem problem

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Let C =

    2,0,-2
    1,1,2
    -1,-1,-1

    Use the Cayley-Hamilton theorem to compute C^3.


    2. Relevant equations

    Cayley-Hamilton theorem says that every square matrix satisfies its own characteristic equation.

    [itex]C^3=PD^3P^{-1}[/itex]

    where P is the matrix formed from linearly independant eigenvectors of C and D is the diagonal matrix formed from the eigenvalues of C.

    3. The attempt at a solution

    I get the characteristic equation of C is

    [itex]\lambda^3 - 2\lambda^2 - \lambda - 2 = 0 [/itex]

    I get stuck because I can't factorise this and get the eigenvalues to proceed. Is there some trick to factorising cubics like this?
     
  2. jcsd
  3. Oct 28, 2012 #2
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  4. Oct 28, 2012 #3
    Thanks, I was just going back over the lecture notes and realized that I was absurdly confused in that section (I'm embarassed I even asked this question!)...anyway, I get it now, thanks for that :)
     
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