Mathematics-Online course: Linear Algebra-Analytic Geometry-Quadrics-Simplistic Classification of Quadrics

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	Mathematics-Online course: Linear Algebra - Analytic Geometry - Quadrics
	Simplistic Classification of Quadrics

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Starting from the equation of a quadric

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

let

$\begin{displaymath} \tilde{A} = \left( \begin{array}{c\vert c} c & b^{\operatorname t}\\ \hline b & A \end{array}\right)\,. \end{displaymath}$

is called

cone quadric, if $\operatorname{Rg}(\tilde{A})=\operatorname{Rg}(A)$ ,
central quadric, if $\operatorname{Rg}(\tilde{A})=\operatorname{Rg}(A)+1$ ,
parabolic quadric, if $\operatorname{Rg}(\tilde{A})=\operatorname{Rg}(A)+2$ ,

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

automatically generated 4/21/2005