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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.

(Authors: App/Burkhardt/Höllig)
For the symmetric matrix $ A$ there exists an orthonormal basis of eigenvectors $ u_i$, where the first eigenvectors belong to eigenvalues $ \lambda_i \ne 0$. After the first substitution $ x=Uy=(u_1,\ldots,u_n)y$ we obtain the diagonal form

$\displaystyle y^{\operatorname t}\operatorname{diag}(\lambda_1, \ldots,\lambda_... ... 2(\underbrace{U^{\operatorname t}b}_{\tilde{b}})^{\operatorname t}y + c \,. $

For $ \lambda_i \ne 0$ we can eliminate the linear terms by the substitution (completion of square)

$\displaystyle y_i = z_i - \tilde{b}_i/\lambda_i\,. $

The constant changes as follows:

$\displaystyle c\to \gamma = c - \sum_{i=1} ^m \tilde{b}_i^2 / \lambda_i \,. $

Hence, we obtain the form

$\displaystyle \sum_{i=1}^m \lambda_i z_i^2 + \sum_{i=m+1}^n 2\tilde{b}_i z_i +\gamma\,. $

If there exists $ \tilde{b}_i\ne 0$ with $ i\ge m+1$, then a further rotation, given by the vectors

$\displaystyle v_i$ $\displaystyle = z_i\,,\quad i\le m$    
$\displaystyle v_{m+1}$ $\displaystyle =(1/\beta)\sum_{m+1}^n \tilde{b}_i z_i\,,\quad\beta = \pm\vert\tilde{b}\vert$    

which are orthogonally completed by $ v_{m+2},\ldots,v_n$, yields

$\displaystyle \sum_{i=1}^m \lambda_i v_i^2 + 2\beta v_{m+1}+\gamma\,. $

Here we choose the sign for $ \beta$ so that, after division by $ \beta$, there are at least as many positive terms as negative ones ( $ \lambda_i/\beta$).

The following translation

$\displaystyle w_{m+1}$ $\displaystyle = v_{m+1} +\gamma/(2\beta)$    
$\displaystyle w_i$ $\displaystyle = v_i\,,\quad i\ne m+1$    

eliminates the constant and yields

$\displaystyle \sum_{i=1}^m \lambda_i w_i^2 + 2\beta w_{m+1}\,. $

(Authors: App/Burkhardt/Höllig)

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