Get the following quadratic form:
$$Q(x)=x_1^{2}+x_3^2+4x_1x_2-4x_1x_3 $$
to obtain the canonical form I tried the following:
$$Q(x)=x_1^{2}+x_3^{2}+4x_1x_2-4x_1x_3=4(x_1^{2})+(x_3^{2})-4x_1x_3-3x_1x_1+4x_1x_2-(4/3)x_2x_2+(4/3)x_2x_2=(2x_1-x_3)^2-... $$
There I stopped because I rememebered that I shouldn't change the coefficient of $x_1^2$.
From here I don't know how to continue.
I thank you in anticipation for your understanding and I wait forward your answer!
Since $x_1^2$ appears in $Q(x)$, you should start by writing
$$ Q(x) = (ax_1 + bx_2 + cx_3)^2 + \star $$
for $a, b, c \in \mathbb{R}$ in such a way that $\star$ won't involve $x_1$ at all. Since the terms
$$ x_1^2, 4x_1x_2, -4x_1x_3 $$
appear in $Q$, we choose $a = 1, b = 2, c = -2$ and get
$$ (x_1 + 2x_2 - 2x_3)^2 = (x_1 + 2x_2)^2 - 4(x_1 + 2x_2)x_3 + 4x_3^2 \\ = \color{blue}{x_1^2} + \color{blue}{4x_1x_2} + 4x_2^2 - \color{blue}{4x_1x_3} - 8x_2x_3 + 4x_3^2 $$
and so
$$ Q(x) = (x_1 + 2x_2 - 2x_3)^2 - 4x_2^2 + 8x_2x_3 - 4x_3^2 + x_3^2 \\ = (x_1 + 2x_2 - 2x_3)^2 -4 (\color{green}{x_2^2} - \color{green}{2x_2x_3}) - 3x_3^2. $$
Now we can repeat the process for the $x_2$ term. We have
$$ (x_2 - x_3)^2 = \color{green}{x_2^2} - \color{green}{2x_2x_3} + x_3^2 $$
and so
$$ Q(x) = (x_1 + 2x_2 - 2x_3)^2 - 4(x_2 - x_3)^2 + 4x_3^2 - 3x_3^2 \\ = (x_1 + 2x_2 - 2x_3)^2 - 4(x_2 - x_3)^2 + x_3^2 $$
and we're done.
I am not at all sure this is what you are looking for but dont you complete the square:
$$x_1^{2}+x_3^2+4x_1x_2-4x_1x_3=x_1^{2}+(4x_2-4x_3)x_1 +x_3^2$$ $$=x_1^{2}+(4x_2-4x_3)x_1 +(2x_2-2x_3)^2 -(2x_2-2x_3)^2 + x_3^2$$
$$=(x_1+2x_2-2x_3)^2 -(2x_2-2x_3)^2 + x_3^2$$
$$=(x_1+2x_2-2x_3)^2 -4(x_2^2-2x_2x_3)-3x_3^2$$
$$=(x_1+2x_2-2x_3)^2 -4(x_2+x_3)^2 +x_3^2$$
If I have not made a mistake this is the diagonalized form. That is $x^2-4y^2+z^2$ which is unique.
There is a method I asked about at
reference for linear algebra books that teach reverse Hermite method for symmetric matrices
The main advantage is that it is a recipe with matrices, no need to carry variable names. The main disadvantage is the need to invert one matrix at the end; however, the matrix has determinant $\pm 1$ and may very well be upper triangular (it is this time).
It leads to $$ x^2 + z^2 + 4 xy - 4 zx = (x +2y-2z)^2 - 4 (y-z)^2 + z^2 $$
and comes from this matrix stuff; I did it first by hand, it did work, but I thought i would check everything. The Pari code is not quite as readable as Latex but is not too bad.
parisize = 4000000, primelimit = 500509
? m = [ 1,2,-2; 2,0,0; -2,0,1]
%2 =
[1 2 -2]
[2 0 0]
[-2 0 1]
? m - mattranspose(m)
%3 =
[0 0 0]
[0 0 0]
[0 0 0]
? p1 = [1,-2,2; 0,1,0; 0,0,1]
%4 =
[1 -2 2]
[0 1 0]
[0 0 1]
? m1 = mattranspose(p1) * m * p1
%5 =
[1 0 0]
[0 -4 4]
[0 4 -3]
? p2 = [ 1,0,0; 0,1,1; 0,0,1]
%6 =
[1 0 0]
[0 1 1]
[0 0 1]
? d = mattranspose(p2) * m1 * p2
%7 =
[1 0 0]
[0 -4 0]
[0 0 1]
? p = p1 * p2
%8 =
[1 -2 0]
[0 1 1]
[0 0 1]
? matdet(p)
%9 = 1
? q = matadjoint(p)
%10 =
[1 2 -2]
[0 1 -1]
[0 0 1]
? confirm = mattranspose(q) * d * q
%12 =
[1 2 -2]
[2 0 0]
[-2 0 1]
? m
%13 =
[1 2 -2]
[2 0 0]
[-2 0 1]
? m - confirm
%14 =
[0 0 0]
[0 0 0]
[0 0 0]
?
? ( x + 2 * y - 2 * z)^2 - 4 * (y - z)^2 + z^2
%1 = x^2 + (4*y - 4*z)*x + z^2
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Places on this site I put this, several typeset:
reference for linear algebra books that teach reverse Hermite method for symmetric matrices
Find the transitional matrix that would transform this form to a diagonal form.
Writing an expression as a sum of squares
