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May 11th, 2018, 10:54 AM   #1
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Question Basis for a degenerate quadratic form


I have this quadratic form q(x,y,z)= x²-z²-4xy-4zy

I have reduced this form using the Gauss algorithm into q(x',y',z') = x'² -y'²
with x' = x-2y and y' = 2y+z , z=?

The problem statement wants me to find the reduced form (which I already did) and to find THE basis on which q is reduced.

Can you guys help me find this Basis?

Last edited by skipjack; May 11th, 2018 at 07:23 PM.
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May 15th, 2018, 03:00 AM   #2
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Looks like you have already done that- you have x'= x- 2y and y'= 2y+ z. Since you don't need an additional formula for z, just take z'= z. And that is the basis you are looking for.
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May 15th, 2018, 03:35 AM   #3
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Good morning !

Let $\displaystyle (e_1;e_2;e_3)$ the canonical basis of $\displaystyle \mathbb R^3$. We denote $\displaystyle (e_1^*;e_2^*;e_3^*)$ the associated dual basis. (Then we get a basis of the dual space of $\displaystyle \mathbb R^3$, the set of all linear functionals).

So, you have reduced $\displaystyle q$ into :
$\displaystyle q(X)=\left(\ell_1(X)\right)^2-\left(\ell_2(X)\right)^2,$
where : $\displaystyle X=(x,y,z)\in\mathbb R^3$, $\displaystyle \ell_1(X)=x-2y$ and $\displaystyle \ell_2(x)=2y+z$.
So we have two linearly independent functionals $\displaystyle \ell_1$ and $\displaystyle \ell_2$. We complete by a third functional $\displaystyle \ell_3$ to obtain a basis of the dual space of $\displaystyle \mathbb R^3$. For example : $\displaystyle \ell_3(X)=z$.
We can write :
$\displaystyle \ell_1=e_1^*-2e_2^*$, $\displaystyle \ell_2=2e_2^*+e_3^*$ and $\displaystyle \ell_3=e_3^*$.
Then, the change of basis matrix $\displaystyle Q$ is :


Finally we calcule $\displaystyle \left(Q^T\right)^{-1}$, to obtain the basis dual associated to the basis $\displaystyle (\ell_1,\ell_2,\ell_3)$ in which $\displaystyle q$ is reduced.

I got :
$\displaystyle \left(Q^T\right)^{-1}=\begin{pmatrix}

Thus, the wanted basis is : $\displaystyle (u,v,w)$ where : $\displaystyle u=(1,0,0)$ , $\displaystyle v=(1;\frac12,0)$ and $\displaystyle w=(-1,-\frac12,1)$
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