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Example: Simplistic Classification of Quadrics


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Given the quadric

$\displaystyle Q:\quad x_1^2 +\lambda x_2^2 +2x_2 +\lambda =0
$

with $ \lambda\in\mathbb{R}$, or in terms of matrices

$\displaystyle Q:\quad x^{\operatorname t}A x + 2b^{\operatorname t}x + c =0
$

where

\begin{displaymath}
A=\left(
\begin{array}{cc}
1 & 0 \\
0 & \lambda\\
\end{arr...
...0 & 1\\
0 & 1 & 0\\
1 & 0 & \lambda\\
\end{array}\right)\,.
\end{displaymath}

For $ \lambda=0$ we have $ \operatorname{Rg}(A)=1$. Otherwise: $ \operatorname{Rg}(A)=2$. For $ \lambda=\pm 1$ we have $ \operatorname{Rg}(\tilde{A})=2$, otherwise: $ \operatorname{Rg}(\tilde{A})=3$.

Thus, $ Q$ is parabolic for $ \lambda=0$, conic for $ \lambda=\pm 1$ and a central quadric for all other $ \lambda$.


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  automatisch erstellt am 13.  7. 2018