Quadratic form over real number field

Question

I have to turn the quadratic form over real number filed in normal form. I'm stuck trying to find similar problems on the internet but couldn't.

$F(x) = x_1^2-3x_3^2-2x_1x_2+2x_1x_3-6x_2x_3$

Quadratic form

Answer 1

First, you need to complete the square in $x_1$: $$ \mathbf{x_1^2} -3x_3^2\mathbf{-2x_1x_2}+\mathbf{2x_1x_3}-6x_2x_3=(x_1-x_2+x_3)^2+p(x_2,x_3),$$ with $$ p(x_2,x_3)=-x_2^2-3x_3^2-4x_2x_3.$$

Now, you iterate and complete the square in $x_2$ in $p(x_2,x_3)$: $$-\mathbf{x_2^2}-3x_3^2\mathbf{-4x_2x_3}=-(x_2+2x_3)^2+x_3^2.$$

From here you conclude that the standard form is

$$ y_1^2-y_2^2+y_3^2=0,$$

where the new coordinates are defined by $y_1=x_1-x_2+x_3$, $y_2=x_2+2x_3$, $y_3=x_3$.

Answer 2

let $\vec{x}=\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{pmatrix}$, then the form quadratic is:

$q(\vec{x})=\vec{x}^{t}A\vec{x}$, hence:

$q(x_{1},x_{2},x_{3})=(x_{1},x_{2},x_{3}) \begin{pmatrix} 1 & -1 & 1 & \\ -1 &0 & -3 & \\ 1 & -3 & -3 & \\ \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{pmatrix}$, then you can find new coordinates diagonalizing the matrix.