$-6xy-6xz+3y^2+6yz-2z^2$
I've already tried to factor out some variables, but I am always left with 3 variables again after my transformation. I've tried $(a+b)^2$ and $(a+b+c)^2$, I guess I need some methodology, I really rely on you, guys >.<
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i only found this here $$-2, \left( z-3/2,y+3/2,x \right) ^{2}+3, \left( y-x \right) ^{2}-3 ,{x}^{2}+1/8, \left( 6,y-6,x \right) ^{2} $$ – Dr. Sonnhard Graubner Oct 21 '17 at 20:37
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$$3(x-y)^2+3(x-z)^2+3(y+z)^2-6x^2-3y^2-8z^2$$ Is this the expression you want ? – Peter Oct 21 '17 at 20:41
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Or with only $3$ squares ; $$3(y+z-x)^2-3x^2-5z^2$$ – Peter Oct 21 '17 at 20:46
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@Alice do you want to solve an equation or to simplify this expression ? – Peter Oct 21 '17 at 20:48
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@Peter this is probably indefinite quadratic form, I presume? – Alice Oct 21 '17 at 21:07
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@Peter but I really can't understand how are you people are managed to get to this elegant solution – Alice Oct 21 '17 at 21:08
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Alice, read algorithm at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr and some of the links i put there. The matrix is the Hessian matrix of second partial derivatives of the form. In your case, you can take half the Hessian matrix and still have all integer entries. – Will Jagy Oct 21 '17 at 21:31
2 Answers
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Complete the square to deal with all the $z$'s \begin{eqnarray*} -2(z-\frac{3}{2}y +\frac{3}{2}x)^2 + ... \end{eqnarray*} Now complete the square for the $x$'s \begin{eqnarray*} -2(z-\frac{3}{2}y +\frac{3}{2}x)^2 + \frac{1}{2} (3 x -5y)^2 -5y^2. \end{eqnarray*}
Donald Splutterwit
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Thank you, but how did you "saw" what variable we should handle first? I'm really interested in the algorithm of solution, I'd be glad to get a piece of advice, if u have time – Alice Oct 21 '17 at 21:05
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The order is up to handle the variables is up to you ... indeed Peter's solution shows that there is a better/easier way than the one that I give. – Donald Splutterwit Oct 21 '17 at 21:12
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The result of a purely algorithmic approach is $Q^T D Q = H,$ where half the Hessian of your quadratic form is $H,$ then $D$ is diagonal, and $Q$ has determinant $\pm 1.$ I show the matrices first. The method, an algorithm, is discussed in detail at my http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
===============================================================
? qt
%17 =
[-3/2 3/5 -1]
[3/2 -1 0]
[-1 0 0]
? d
%18 =
[-2 0 0]
[0 15/2 0]
[0 0 9/5]
? q
%19 =
[-3/2 3/2 -1]
[3/5 -1 0]
[-1 0 0]
? qt * d * q
%20 =
[0 0 -3]
[0 3 3]
[-3 3 -2]
?
===============================================================
The whole thing:
parisize = 4000000, primelimit = 500509
? h = [ 0,0,-3; 0,3,3; -3,3,-2]
%1 =
[0 0 -3]
[0 3 3]
[-3 3 -2]
? ht = mattranspose(h)
%2 =
[0 0 -3]
[0 3 3]
[-3 3 -2]
? h - ht
%3 =
[0 0 0]
[0 0 0]
[0 0 0]
? p1 = [ 0,0,1; 0,1,0; 1,0,0]
%4 =
[0 0 1]
[0 1 0]
[1 0 0]
? h1 = p1 * h * p1
%5 =
[-2 3 -3]
[3 3 0]
[-3 0 0]
? p2 = [ 1, 3/2, -3/2; 0,1,0; 0,0,1]
%6 =
[1 3/2 -3/2]
[0 1 0]
[0 0 1]
? p2t = mattranspose(p2)
%7 =
[1 0 0]
[3/2 1 0]
[-3/2 0 1]
? h2 = p2t * h1 * p2
%8 =
[-2 0 0]
[0 15/2 -9/2]
[0 -9/2 9/2]
? p3 = [ 1,0,0; 0,1, 9/15; 0,0,1]
%9 =
[1 0 0]
[0 1 3/5]
[0 0 1]
? p3t = mattranspose(p3)
%10 =
[1 0 0]
[0 1 0]
[0 3/5 1]
? h3 = p3t * h2 * p3
%11 =
[-2 0 0]
[0 15/2 0]
[0 0 9/5]
? d = h3
%12 =
[-2 0 0]
[0 15/2 0]
[0 0 9/5]
? p = p1 * p2 * p3
%13 =
[0 0 1]
[0 1 3/5]
[1 3/2 -3/5]
? matdet(p)
%14 = -1
? q = matadjoint(p)
%15 =
[-3/2 3/2 -1]
[3/5 -1 0]
[-1 0 0]
? qt = mattranspose(q)
%16 =
[-3/2 3/5 -1]
[3/2 -1 0]
[-1 0 0]
? qt
%17 =
[-3/2 3/5 -1]
[3/2 -1 0]
[-1 0 0]
? d
%18 =
[-2 0 0]
[0 15/2 0]
[0 0 9/5]
? q
%19 =
[-3/2 3/2 -1]
[3/5 -1 0]
[-1 0 0]
? qt * d * q
%20 =
[0 0 -3]
[0 3 3]
[-3 3 -2]
?
===============================================================
Will Jagy
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