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Quadratic forms

  1. Dec 15, 2007 #1

    T-7

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    1. The problem statement, all variables and given/known data

    [tex]x^{2} + 2y^{2} + z^{2} + 2xy + 4xz + 6yz[/tex]

    Write down the symmetric matrix A for which the form is expressible as [tex]x^{t}Ax[/tex] where t denotes transpose. Diagonalise each of the forms and in each case find a real non-singular matrix P for which the matrix [tex]P^{t}AP[/tex] is diagonal with entries in {1,-1,0}.

    3. The attempt at a solution

    I first tried this by completing the square.

    [tex]
    x^{2} + 2y^{2} + z^{2} + 2xy + 4xz + 6yz
    = (x + y + 2z)^{2} + y^{2} - 3z^{2} + 2yz
    = (x + y + 2z)^{2} + (y + z)^2 - 4z^2
    = x_{1}^{2} + x_{2}^{2} - x_{3}^{2}[/tex]

    where

    [tex]
    x_{1} = x + y + 2z,
    x_{2} = y + z,
    x_{3} = 2z,
    [/tex]

    However, I just can't seem to find the eigenvalues for this form.

    The symmetric matrix A for this quadratic form is

    [tex]
    \[ \left( \begin{array}{ccc}
    1 & 1 & 2 \\
    1 & 2 & 3 \\
    2 & 3 & 1 \end{array} \right)\]
    [/tex]

    and the characteristic polynomial is given by

    [tex]
    \[ \chi(\lambda) = \left| \begin{array}{ccc}
    1-\lambda & 1 & 2 \\
    1 & 2-\lambda & 3\\
    2 & 3 & 1-\lambda \end{array} \right|.\]
    [/tex]

    I find this comes to

    [tex]f(\lambda) = \lambda^{3} - 4(\lambda^2) + 9(\lambda) - 4[/tex]

    which does not factorise -- so I can't get the eigenvalues, and can't form a matrix P. However, I have shown that a diagonal form is possible by completing the square. So surely I ought to be able to find three eigenvalues? Can someone point out where I've gone wrong?

    Cheers!
     
    Last edited: Dec 15, 2007
  2. jcsd
  3. Dec 15, 2007 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You've misplaced a couple of signs! The coefficient of [itex]\lambda[/itex] is +4+ 1- 4= 1. The correct equation is [itex]\lambda^3- 4\lambda^2- \lambda+ 4= 0[/itex] which obviously has [itex]\lambda- 1[/itex] as a factor.
     
  4. Dec 15, 2007 #3

    T-7

    User Avatar

    I am obviously brain-dead today. I'm still getting what I got before:

    [tex]
    \[ \chi(\lambda) = \left| \begin{array}{ccc}
    1-\lambda & 1 & 2 \\
    1 & 2-\lambda & 3\\
    2 & 3 & 1-\lambda \end{array} \right|\]
    [/tex]

    [tex]= (1-\lambda)((2-\lambda)(1-\lambda)-9)-(1-\lambda-6)+2(3-2(2-\lambda))[/tex]
    [tex]= (1-\lambda)(2-3\lambda+\lambda^{2}-9)-(-\lambda-5)+2(2\lambda-1)[/tex]
    [tex]=(1-\lambda)(\lambda^{2}-3\lambda-7)+\lambda+5+4\lambda-2[/tex]
    [tex]=(1-\lambda)(\lambda^{2}-3\lambda-7)+5\lambda+3[/tex]
    [tex]=\lambda^{2}-3\lambda-7-\lambda^{3}+3\lambda^{2}+7\lambda+5\lambda+3[/tex]
    [tex]=-\lambda^{3}+4\lambda^{2}+9\lambda-4[/tex]
     
  5. Dec 15, 2007 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well, that's not exactly what you had before but I messed up also.

    The part about "diagonal with entries in {1,-1,0}" implies that the eigenvalues are 0, 1, and -1. But that can't possibly be correct.
     
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