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Reduction of quadratic form (principal axis)

  1. Mar 18, 2006 #1
    i keep getting nonzero off diagonal elements when i try to reduce to simple sum of squares, of the equation
    [tex]2 x_{1}^{2}+2x_{2}^{2}+x_{3}^{2}+2x_{1}x_{3}+2x_{2}x_{3} [/tex]
    what i have is
    [tex] \left(\begin{array}{ccc} x_{1} & x_{2} & x_{3} \end{array}\right)
    2 & 1 & 0 \cr
    1 & 2 & 1 \cr
    0 & 1 & 1
    \end{array} \right)
    \left(\begin{array}{c} x_{1} \cr x_{2} \cr x_{3} \end{array} \right)
    so my thought was to calculate the eigenvalues of the coefficient matrix above, which yield complex solutions from the characteristic equation
    [tex] 1-6 \lambda+5 \lambda^{2}-\lambda^{3}=0 [/tex]
    From the complex eigenvalues I obtain complex eigenvectors, which i'll post if necessary, but are rather lengthy. From the eigenvectors I choose to use Gram-Schmidt orthogonalization to form an orthonormal basis set. From which I construct a matrix with the corresponding basis set, and use diagonalize the system I have the diagonalization matrix
    [tex] D = \left(\mid n \rangle \langle m \mid \right)^{T} A \left( \mid n \rangle \langle m \mid \right) [/tex]
    where the matrix
    [tex] \left(\mid n \rangle \langle m \mid \right) [/tex]
    is the orthonormal eigenvector matrix. When I'm done with all of this I'm not getting a diagonalized matrix. I was wondering if I am making a mistake in my approach, or if anyone else does get a diagonalized matrix equation.
  2. jcsd
  3. Mar 18, 2006 #2


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    The equation 1-6x+5x^2-x^3=0 has all real solutions. You could have known this because your matrix is symmetric and symmetric matrices in the reals are orthogonally diagonalizable in the reals.
  4. Mar 18, 2006 #3
    That's interesting, I was bad and usinig Maxima to calculate the roots of the equation and getting imaginary components, interesting, when I plot and find the roots, you're correct the roots are all real. Thanks, for the tip, I didn't know that about symmetric matrices, either. Thanks again.
  5. Mar 18, 2006 #4


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    :biggrin: I was using my TI-89.
  6. Mar 19, 2006 #5
    Perhaps the -89 is superior to my cas, even mathematica, maple given imaginary components, though on order of [tex]10^{-16}[/tex] or so. I wonder why that is. Not to mention I'm glad I'm not the only one whose cheats on algebra parts of problems...well that may get you into trouble as I found out today. Hey thanks again.
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