Decompose form on $\Bbb R^4$ as the sum of squares of independent linear forms

Question

I have the following task:

Decompose the following forms on $\mathbb{R}^4$ as the sum of squares of independent linear forms:

(1) $\phi(x) = x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1$

(2) $\phi(x) = 9x_1^2-8x_2^2+5x_4^2+6x_1x_2+18x_1x_4+6x_2x_3-6x_2x_4+6x_3x_4$

Now for (1) I tried to write this as a matrix form, like this:

$\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 &0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2\\x_3\\x_4 \end{pmatrix}$.

Is it possible to get the sum of squares of independent linear forms in this way or am I totally wrong in this way?

Thanks for your help!

Answer 1

Hint: Use Gauß' method:

1. Since you only have double products, factorise $\phi(x)$ first: $$\phi(x)=(x_1+x_3)(x_2+x_4)$$ and use the polarisation identity: $$ab=\tfrac14\bigl((a+b)^2-(a-b)^2\bigr).$$
2. Collect first all terms containing $x_1$ and complete the square: \begin{align} \phi(x) &= 9x_1^2+6x_1x_2+18x_1x_4-8x_2^2+5x_4^2+6x_2x_3-6x_2x_4+6x_3x_4 \\ &=\bigl(3x_1+x_2+3x_4)^2-(x_2+3x_4)^2-8x_2^2 +5x_4^2+6x_2x_3 -6x_2x_4 + 6x_3x_4\\ &= \bigl(3x_1+x_2+3x_4)^2- 9x_2^2 -4x_4^2+6x_2x_3 -12x_2x_4 + 6x_3x_4 \end{align}

and proceed in the same way for the form in three variables $$\psi(x)=- 9x_2^2 -4x_4^2+6x_2x_3 -12x_2x_4 + 6x_3x_4.$$ As there is one variable less at each step, this ensures the forms are linearly independent.