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Problem reducing quadratic to diagonal form

  1. Nov 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Reduce ##xy+zy## to diagonal form.

    2. Relevant equations
    The desired diagonal form is ##Q(\vec{x})=(\alpha_1(\vec{x}))^2+...+(\alpha_k(\vec{x}))^2-(\alpha_{k+1}(\vec{x}))^2-...-(\alpha_{k+l}(\vec{x}))^2,## where ##\alpha_i## are linearly independent linear functions. Also known as 'changing the variables in quadratic form'.

    3. The attempt at a solution
    ##xy+zy=y^2+yx+yz+(\frac{x}{2})^2+\frac{xz}{2}+(\frac{z}{2})^2-((\frac{x}{2})^2+\frac{xz}{2}+(\frac{z}{2})^2)-y^2=(y+\frac{x}{2}+\frac{z}{2})^2+(\frac{x}{2}+\frac{z}{2})^2-y^2##, but now
    ##\alpha_1=y+\frac{x}{2}+\frac{z}{2}##
    ##\alpha_2=\frac{x}{2}+\frac{z}{2}##
    ##\alpha_3=y##
    are linearly dependent. I have tried several different substitutions without success. There's also the matrix method which I am not familiar with. This rather simple problem is giving me headache.
     
    Last edited: Nov 10, 2016
  2. jcsd
  3. Nov 10, 2016 #2

    lurflurf

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    hint
    $$xy=\left(\frac{x}{2}+\frac{y}{2}\right)^2-\left(\frac{x}{2}-\frac{y}{2}\right)^2$$
     
  4. Nov 10, 2016 #3
    Using that lead to situation where the ##\alpha##'s weren't linearly independent.
     
  5. Nov 10, 2016 #4

    lurflurf

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    sorry try
    $$\left( \frac{x}{2} \pm \frac{y}{\sqrt{2}}+\frac{z}{2} \right)^2$$
    As the two
     
  6. Nov 10, 2016 #5

    pasmith

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    What happens if you set [itex]u= x + z[/itex] so that [itex]Q = uy[/itex]? Doesn't the identity in post #2 then work?
     
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