对偶二次曲线

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矩阵$\left(\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & \frac{1}{2} \\ 0 & \frac{1}{2} & 1\end{array}\right)$对应的二次曲线
在$(1,0)$有参数化$\left[x_{0}: x_{1}: x_{2}\right]=\left[q^{2}-1:-q(2+q): q^{2}+q+1\right]$
在Example 54中如何得出对偶二次曲线的参数化?

then the dual conic has a rational parametrization\begin{aligned} {\left[\alpha_{0}: \alpha_{1}: \alpha_{2}\right] } & =\left[1-q^{2}: \frac{1}{2}\left(q^{2}+q+1\right)-q(2+q):-\frac{1}{2} q(2+q)+q^{2}+q+1\right] \\ & =\left[2-2 q^{2}: 1-3 q-q^{2}: 2+q^{2}\right]\end{aligned}

page 11

without proof

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\[\left(\begin{array}{ccc}-1 & 0 & 0 \\ 0 & 1 & \frac{1}{2} \\ 0 & \frac{1}{2} & 1\end{array}\right)^{-1}=\frac13\pmatrix{-3&0&0\\
0&4&-2\\
0&-2&4}\]

Finding dual conic
二次曲线 $\mathbf x^tA\mathbf x=0$ 的对偶是 $\mathbf x^t(\operatorname{adj}A)\mathbf x=0$.

WolframAlpha
$$\operatorname{adj}\pmatrix{a& b& d\\b& c& e\\ d& e& f}=\pmatrix{c f - e^2&d e - b f&b e - c d\\
d e - b f&a f - d^2&b d - a e\\
b e - c d&b d - a e&a c - b^2}$$

1. ResourceFunction["Adjugate"][{{a, b, d}, {b, c, e}, {d, e, f}}]

Copy the Code

same as the formula in page 11

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When $A$ is singular, $A^{-1}$ is not defined, but $\operatorname{adj}A$ is defined.

Degenerate conics dualize to double lines, but double lines do not have a well defined dual.

• $A=\pmatrix{1\\&0\\&&0}$, then $\mathbf x^tA\mathbf x=0$ is a double line.
  $\operatorname{adj}A$ is zero matrix, so the dual of a double line is undefined.
• $A=\pmatrix{&1\\1\\&&0}$, then $\mathbf x^tA\mathbf x=0$ is a pair of intersecting lines.
  $\operatorname{adj}A=\pmatrix{0\\&0\\&&-1}$, so the dual of a pair of intersecting lines is a double line. WolframAlpha

Since $(A^{-1})^{-1}=A$, the dual of a conic is itself, but we see this is false for degenerate conics.