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Example: Euclidean Normal Forms of Three-Dimensional Quadrics

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z overview
The quadric

$\displaystyle Q: \frac{1}{2} x^{\operatorname t}\left(\begin{array}{rrr}5 & 4 & 0 \\ 4 & 3 & 4 \\ 0 & 4 & 1\end{array}\right)x+ \left(-2,1,2\right)x+1=0 $

is to put into normal form.

The characteristical polynomial of the corresponding matrix,

$\displaystyle p(\lambda)=\lambda^3-9\lambda^2-9\lambda+81\,, $

has the zeros

$\displaystyle \lambda_1=9\,,\quad \lambda_2=3\,,\quad \lambda_3=-3\,. $

A basis of eigenvectors is

$\displaystyle v_9=\left(\begin{array}{r}2\\ 2\\ 1\end{array}\right)\,,\quad v_3... ...y}\right)\,,\quad v_{-3}=\left(\begin{array}{r}1\\ -2\\ 2\end{array}\right)\,. $

We obtain the following orthogonal transformation

$\displaystyle U=\frac{1}{3}\left(\begin{array}{rrr} 2 & -2 & 1\\ 2 & 1 &-2\\ 1&2&2\end{array}\right)\,. $

Transforming the equation by $ x=Uy$ yields

$\displaystyle \frac{1}{2} y^{\operatorname t}\left(\begin{array}{rrr}9 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -3\end{array}\right)y+ \left(0,3,0\right)y+1=0\,. $

Completing squares gives

0 $\displaystyle =$ $\displaystyle 9y_1^2+3y_2^2-3y_3^2+6y_2+2$  
  $\displaystyle =$ $\displaystyle 9y_1^2+3(y_2+1)^2-6y_2-3-3y_3^2+6y_2+2$  
  $\displaystyle =$ $\displaystyle 9z_1^2+3z_2^2-3z_3^2-1$  

or

$\displaystyle \frac{z_3^2}{\frac{1}{3}}-\frac{z_1^2}{\frac{1}{9}}-\frac{z_2^2}{\frac{1}{3}}+1=0\,. $

Thus, the equation describes a hyperboloid of 1 sheet.

see also:

  automatisch erstellt am 13.  7. 2018