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May 11th, 2018, 10:54 AM  #1 
Newbie Joined: May 2018 From: Africa Posts: 1 Thanks: 0  Basis for a degenerate quadratic form
Hello, I have this quadratic form q(x,y,z)= x²z²4xy4zy I have reduced this form using the Gauss algorithm into q(x',y',z') = x'² y'² with x' = x2y 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. 
May 15th, 2018, 03:00 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 
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.

May 15th, 2018, 03:35 AM  #3 
Member Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 85 Thanks: 14 Math Focus: Algebra, Cryptography 
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 : $\displaystyle X=(x,y,z)\in\mathbb R^3$, $\displaystyle \ell_1(X)=x2y$ 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 : $\displaystyle \begin{pmatrix} 1&0&0\\2&2&0\\0&1&1 \end{pmatrix} $ 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 $\displaystyle w=(1,\frac12,1)$
1&1&1\\0&\frac12&\frac12\\0&0&1 \end{pmatrix} $ 

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basis, degenerate, form, quadratic 
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