# Problem reducing quadratic to diagonal form

1. Nov 10, 2016

### lep11

1. The problem statement, all variables and given/known data
Reduce $xy+zy$ to diagonal form.

2. Relevant equations
The desired diagonal form is $Q(\vec{x})=(\alpha_1(\vec{x}))^2+...+(\alpha_k(\vec{x}))^2-(\alpha_{k+1}(\vec{x}))^2-...-(\alpha_{k+l}(\vec{x}))^2,$ where $\alpha_i$ are linearly independent linear functions. Also known as 'changing the variables in quadratic form'.

3. The attempt at a solution
$xy+zy=y^2+yx+yz+(\frac{x}{2})^2+\frac{xz}{2}+(\frac{z}{2})^2-((\frac{x}{2})^2+\frac{xz}{2}+(\frac{z}{2})^2)-y^2=(y+\frac{x}{2}+\frac{z}{2})^2+(\frac{x}{2}+\frac{z}{2})^2-y^2$, but now
$\alpha_1=y+\frac{x}{2}+\frac{z}{2}$
$\alpha_2=\frac{x}{2}+\frac{z}{2}$
$\alpha_3=y$
are linearly dependent. I have tried several different substitutions without success. There's also the matrix method which I am not familiar with. This rather simple problem is giving me headache.

Last edited: Nov 10, 2016
2. Nov 10, 2016

### lurflurf

hint
$$xy=\left(\frac{x}{2}+\frac{y}{2}\right)^2-\left(\frac{x}{2}-\frac{y}{2}\right)^2$$

3. Nov 10, 2016

### lep11

Using that lead to situation where the $\alpha$'s weren't linearly independent.

4. Nov 10, 2016

### lurflurf

sorry try
$$\left( \frac{x}{2} \pm \frac{y}{\sqrt{2}}+\frac{z}{2} \right)^2$$
As the two

5. Nov 10, 2016

### pasmith

What happens if you set $u= x + z$ so that $Q = uy$? Doesn't the identity in post #2 then work?