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Proposition: Let

$$\tag 1 f(x,y) = (a_1x+b_1y+c_1)^{2} + (a_2x+b_2y+c_2)^{2} + (a_3x+b_3y+c_3)^{2}$$

Then there exist numbers $\hat a_1,\hat b_1, \hat c_1,\hat a_2,\hat b_2, \hat c_2$ such that

$$\tag 2 f(x,y) = (\hat a_1x+\hat b_1y+\hat c_1)^{2} + (\hat a_2x+\hat b_2y+\hat c_2)^{2} $$


My work

I was trying to prove this after studying

Minimize this real function on $\mathbb{R}^{2}$ without calculus?

I tried to develop some theory but decided to seek help.

In the same spirit as the motivating question, please show me an algebra-precalculus proof.

I've been left with 'egg on my face' before, so the proposition might not even be true. If it isn't, any answers would have to impose further assumptions/constraints.

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    $\begingroup$ I do not believe it. By Sylvester's Law of Inertia, if there were a way to write your $f$ as the sum of two squares, it would be necessary that my "hessian_matrix" have determinant zero, independently of the nine coefficients a,b,c,d,e,f,g,h,i Indeed, whenever that determinant is zero, a two-square sum is possible. Oh: notice that I put in a variable z. To get back to your version, set z to 1 $\endgroup$ – Will Jagy Jun 6 at 0:26
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Easy example: set up the coefficients so that your positive polynomial is $$ x^2 + y^2 + 1 $$ This is not the sum of two squares, by Sylvester's Law of Inertia

I don't believe that to be possible in general. This would take forever to typeset:

                                          Copyright (C) 2000-2015 The PARI Group

PARI/GP is free software, covered by the GNU General Public License, and comes WITHOUT ANY WARRANTY WHATSOEVER.

Type ? for help, \q to quit.
Type ?12 for how to get moral (and possibly technical) support.

parisize = 4000000, primelimit = 500000

? p = ( a*x + b*y + c * z )^2 + ( d*x + e*y + f * z )^2 + ( g*x + h*y + i* z )^2
%2 = (a^2 + (d^2 + g^2))*x^2 + ((2*y*b + 2*z*c)*a + ((2*e*d + 2*h*g)*y + (2*z*f*d + 2*z*i*g)))*x + (y^2*b^2 + 2*z*c*y*b + ((e^2 + h^2)*y^2 + (2*z*f*e + 2*z*i*h)*y + (z^2*c^2 + (z^2*f^2 + z^2*i^2))))
? 
? 
? 
? px = deriv( p, x)
%3 = (2*a^2 + (2*d^2 + 2*g^2))*x + ((2*y*b + 2*z*c)*a + ((2*e*d + 2*h*g)*y + (2*z*f*d + 2*z*i*g)))
? 
? pxx = deriv( px, x)
%4 = 2*a^2 + (2*d^2 + 2*g^2)
? pxy = deriv( px, y)
%5 = 2*b*a + (2*e*d + 2*h*g)
? pxz = deriv( px, z)
%6 = 2*c*a + (2*f*d + 2*i*g)
? 
? py = deriv( p, y)
%7 = (2*b*a + (2*e*d + 2*h*g))*x + (2*y*b^2 + 2*z*c*b + ((2*e^2 + 2*h^2)*y + (2*z*f*e + 2*z*i*h)))
? 
? pyy = deriv( py, y)
%8 = 2*b^2 + (2*e^2 + 2*h^2)
? pyz = deriv( py, z)
%9 = 2*c*b + (2*f*e + 2*i*h)
? 
? 
? pz = deriv( p, z)
%10 = (2*c*a + (2*f*d + 2*i*g))*x + (2*c*y*b + ((2*f*e + 2*i*h)*y + (2*z*c^2 + (2*z*f^2 + 2*z*i^2))))
? 

? pzz = deriv( pz, z)
%12 = 2*c^2 + (2*f^2 + 2*i^2)

? hessian_matrix = [ pxx, pxy, pxz; pxy, pyy, pyz; pxz,pyz, pzz]
%13 = 
[2*a^2 + (2*d^2 + 2*g^2) 2*b*a + (2*e*d + 2*h*g) 2*c*a + (2*f*d + 2*i*g)]

[2*b*a + (2*e*d + 2*h*g) 2*b^2 + (2*e^2 + 2*h^2) 2*c*b + (2*f*e + 2*i*h)]

[2*c*a + (2*f*d + 2*i*g) 2*c*b + (2*f*e + 2*i*h) 2*c^2 + (2*f^2 + 2*i^2)]



? disc = matdet( hessian_matrix  )
%16 = (8*i^2*e^2 - 16*i*h*f*e + 8*h^2*f^2)*a^2 + (((-16*i^2*e + 16*i*h*f)*d + (16*i*g*f*e - 16*h*g*f^2))*b + ((16*i*h*e - 16*h^2*f)*d + (-16*i*g*e^2 + 16*h*g*f*e))*c)*a + ((8*i^2*d^2 - 16*i*g*f*d + 8*g^2*f^2)*b^2 + (-16*i*h*d^2 + (16*i*g*e + 16*h*g*f)*d - 16*g^2*f*e)*c*b + (8*h^2*d^2 - 16*h*g*e*d + 8*g^2*e^2)*c^2)
? 
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  • $\begingroup$ It appears you have to 'boot up' to some more serious algebra. The question could be the motivation for studying/developing the machinery you are using. $\endgroup$ – CopyPasteIt Jun 6 at 0:39
  • $\begingroup$ @CopyPasteIt Why did you think two squares would work? $\endgroup$ – Will Jagy Jun 6 at 0:41
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    $\begingroup$ @CopyPasteIt anyway, take a look at math.stackexchange.com/questions/1388421/… $\endgroup$ – Will Jagy Jun 6 at 0:42
  • $\begingroup$ I was wondering what was the theory underpinning CY Aries's answer math.stackexchange.com/a/3250828/432081 $\quad$ It was algebraic manipulation of a the polynomial and I wanted to figure out the underlying theory for that to work. I saw answers on quadratic forms but thought I could figure it another way. Also, I looked at this as perturbations of a function, and my intuition stuck out! $\endgroup$ – CopyPasteIt Jun 6 at 0:50
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    $\begingroup$ @CopyPasteIt I made that a separate answer. Evidently "my" method is taught in, say, Europe. $\endgroup$ – Will Jagy Jun 6 at 1:11

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