Isotropic vectors of quadratic form

Question

How can i find all isotropic vectors of quadratic form $x_{1}^2+4x_{2}^2+8x_{3}^2-4x_{1}x_{2}+8x_{1}x_{3}-14x_{2}x_{3}$

I don't understand how to approach.

Answer 1

Use Lagrange's Reduction (complete the square). I'll use $\;x,y,z\;$ for ease:

$$x^2+4y^2+8z^2-4xy+8xz-14yz=(x^2-2x(2y+4z))+4y^2+8z^2-14yz=$$

$$=\left(x-2y-4z\right)^2-\color{red}{4y^2}-16yz-16z^2+\color{red}{4y^2}+8z^2-14yz=$$

$$=\left(x-2y-4z\right)^2-8\left(z^2+\frac{15}2yz\right)=(x-2y-4z)^2-8\left(z+\frac{15}4y\right)^2+\frac{225}2y^2$$

Thus, your quadratic is equivalent to $\;\tilde x^2+\tilde y^2-\tilde z^2\;$ (signature $\,(2,1)\,$) , so it actually is a cone.

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Answer 2

Integer and primitive: $$ \pm \; \; \left(4 u^2 + 2uv-4v^2, \; 4 u^2 - 2 v^2, \; u^2 \right) $$ $$ \pm \; \; \left(8 u^2 - 2uv-2v^2, \; 8 u^2 - v^2, \; 2 u^2 \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 4 & 0 & 1 \\ \frac{ 3 }{ 2 } & 1 & \frac{ 1 }{ 8 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & 4 \\ - 2 & 4 & - 7 \\ 4 & - 7 & 8 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 4 & \frac{ 3 }{ 2 } \\ 0 & 0 & 1 \\ 0 & 1 & \frac{ 1 }{ 8 } \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 8 & 0 \\ 0 & 0 & \frac{ 1 }{ 8 } \\ \end{array} \right) $$ tells us that we can take any column vector $C=(u,v,w)^T$ with $8u^2 + w^2 = 64 v^2,$ multiply that vector to get $X=PC,$ then $X^THX = 0$

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$$ H = \left( \begin{array}{rrr} 1 & - 2 & 4 \\ - 2 & 4 & - 7 \\ 4 & - 7 & 8 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 1 & - 2 & 4 \\ - 2 & 4 & - 7 \\ 4 & - 7 & 8 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & - 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 1 & 0 & 4 \\ 0 & 0 & 1 \\ 4 & 1 & 8 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & 2 & - 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & - 2 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & - 8 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - 4 & 2 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & - 2 & 4 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 8 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) $$

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$$ E_{4} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 1 }{ 8 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrr} 1 & - 4 & \frac{ 3 }{ 2 } \\ 0 & 0 & 1 \\ 0 & 1 & \frac{ 1 }{ 8 } \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrr} 1 & - 2 & 4 \\ 0 & - \frac{ 1 }{ 8 } & 1 \\ 0 & 1 & 0 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 8 & 0 \\ 0 & 0 & \frac{ 1 }{ 8 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 4 & 0 & 1 \\ \frac{ 3 }{ 2 } & 1 & \frac{ 1 }{ 8 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & 4 \\ - 2 & 4 & - 7 \\ 4 & - 7 & 8 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 4 & \frac{ 3 }{ 2 } \\ 0 & 0 & 1 \\ 0 & 1 & \frac{ 1 }{ 8 } \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 8 & 0 \\ 0 & 0 & \frac{ 1 }{ 8 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 2 & - \frac{ 1 }{ 8 } & 1 \\ 4 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 8 & 0 \\ 0 & 0 & \frac{ 1 }{ 8 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & 4 \\ 0 & - \frac{ 1 }{ 8 } & 1 \\ 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & - 2 & 4 \\ - 2 & 4 & - 7 \\ 4 & - 7 & 8 \\ \end{array} \right) $$