1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)
    \left(\begin{array}{ccc}
    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)
    [/tex]
    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

    0rthodontist

    User Avatar
    Science Advisor

    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

    0rthodontist

    User Avatar
    Science Advisor

    :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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Reduction of quadratic form (principal axis)
  1. Quadratic Form (Replies: 3)

  2. Quadratic Forms (Replies: 0)

  3. Quadratic forms (Replies: 3)

Loading...