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Mathematics-Online course: Linear Algebra - Analytic Geometry - Quadrics

Principal Axis Transformation

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By coordinate transformation (rotation and translation) a quadric in $ \mathbb{R}^n$ can be brought to normal form:

$\displaystyle x^{\operatorname t}A x + 2 b^{\operatorname t}x + c = \sum_{i=1}^m \lambda_i w_i^2 + 2\beta w_{m+1} + \gamma \,, $

where $ \beta\gamma=0$.

\includegraphics[width=\moimagesize]{a_hauptachsentrafo}

The columns of the rotation matrix $ U$ contain the eigenvectors $ u_i$ of $ A$ the directions of which are called principal axes.

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The quadric

$\displaystyle Q:\quad 8x_1^2 + 17x_2^2 + 20x_3^2 + 20x_1x_2 + 8x_1x_3 + 28x_2x_3 - 48x_1 - 114x_2 - 78x_3 + 207 =0 $

is to be brought to normal form.

In terms of matrices we have

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

where

\begin{displaymath} A=\left( \begin{array}{ccc} 8 & 10 & 4\\ 10 & 17 & 14\\ 4 ... ...ht)\,,\quad b=(-24,-57,-39)^{\operatorname t}\,,\quad c=207\,. \end{displaymath}

Matrix $ A$ has the eigenvalues $ 36$, $ 9$ and 0, corresponding normalized eigenvectors are, for example,

$\displaystyle u_1=\frac{1}{3}(1,2,2)^{\operatorname t}\,,\quad u_2=\frac{1}{3}(2,1,-2)^{\operatorname t}\,,\quad u_3=\frac{1}{3}(2,-2,1)^{\operatorname t}\,. $

After transformation $ x=Uy$ with $ U=(u_1,u_2,u_3)$ we obtain

$\displaystyle Q:\quad 36y_1^2 + 9 y_2^2 - 144y_1 - 18y_2 + 18y_3 + 207 = 0\,. $

Completion of squares $ z_1=y_1-2$, $ z_2=y_2-1$, $ z_3=y_3$ yields

$\displaystyle Q:\quad 36z_1^2 + 9z_2^2 + 18z_3 + 54 = 0\,. $

The translation $ w_1=z_1$, $ w_2=z_2$, $ w_3=z_3+3$ yields the normal form

$\displaystyle Q:\quad 36w_1^2 + 9w_2^2 + 18w_3 =0\,, $

or after division by $ 9$

$\displaystyle Q:\quad 4w_1^2 + w_2^2 + 2w_3 =0\,. $

  automatically generated 4/21/2005