Principal axis transformation

The principal axis transformation (HAT) is a process in Euclidean geometry with which the equations of quadrics ( ellipse , hyperbola , ...; ellipsoid , hyperboloid , ...) are brought to the respective normal form by means of a suitable coordinate transformation and thus their type and geometric properties (Center, vertex, semi-axes) can determine. Orthogonal coordinate transformations (rotations, reflections, shifts) must be used (see below) so that lengths and angles are not changed during the transformation . The essential aid of this procedure is the diagonalization of a symmetrical matrix with the help of an orthogonal matrix.

Major axis transformation of an ellipse with the help of a rotation of the coordinate system

In addition to the purely mathematical-geometric meaning of the main axis transformation for determining the type of quadrics, it is used in numerous disciplines of theoretical physics as well as in computer science and the geosciences (see section Application ).

Simple examples and motivation of the procedure

example 1

That the equation

${\displaystyle x^{2}+y^{2}-4x+2y-4=0}$

Describing the circle with its center and radius can be recognized by adding the square to the form . ${\displaystyle (2,-1)}$ ${\displaystyle 3}$ ${\displaystyle (x-2)^{2}+(y+1)^{2}=9}$

Ellipse with main axes parallel to the axes

Example 2

The equation too

${\displaystyle 4x^{2}+9y^{2}-8x+36y+4=0}$

can be brought to the shape by adding a square and you can see that it is an ellipse with a center and the semi-axes . ${\displaystyle {\tfrac {(x-1)^{2}}{9}}+{\tfrac {(y+2)^{2}}{4}}=1}$ ${\displaystyle (1,-2)}$ ${\displaystyle a=3,b=2}$

Example 3

It is much more difficult to find the equation

${\displaystyle 5x^{2}-2{\sqrt {3}}{\color {red}xy}+7y^{2}-16=0}$

to see that it is an ellipse with the semiaxes . The problem arises from the "mixed" term . It is a sign that the mutually perpendicular main axes in this case are not parallel to the coordinate axes. This can be changed by using a suitable rotation of the coordinate system (around the zero point). The angle of rotation results from the coefficients at (see conic sections ). However, this descriptive procedure becomes very confusing when examining quadrics in Euclidean space. Linear algebra provides a method that can be used in any dimension: the diagonalization of symmetric matrices. To do this, write the equation of the quadric with the help of a symmetrical matrix: the coefficients of are on the diagonal of the matrix, and half of the coefficients of are on the secondary diagonal . ${\displaystyle a={\sqrt {2}},b=2}$ ${\displaystyle \dots \color {red}xy}$ ${\displaystyle x^{2},y^{2},xy}$ ${\displaystyle x^{2},y^{2}}$ ${\displaystyle \color {red}xy}$

${\displaystyle 5x^{2}-2{\sqrt {3}}{\color {red}xy}+7y^{2}-16={\begin{pmatrix}x&y\end{pmatrix}}{\begin{pmatrix}5&{\color {red}-{\sqrt {3}}}\\{\color {red}-{\sqrt {3}}}&7\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}-16={\vec {x}}^{T}A{\vec {x}}-16=0}$

Now the matrix is ​​diagonalized by applying an orthogonal coordinate transformation (rotation or rotation mirroring im ). In the case of an orthogonal coordinate transformation, lengths are not changed, so that after the transformation and any necessary quadratic addition (see above), the lengths of the semi-axes and positions of the center and vertex can be read off. ${\displaystyle A}$ ${\displaystyle \mathbb {R} ^{2}}$

Diagonalization of a symmetric matrix (principal axis theorem)

A symmetric matrix always has an orthogonal matrix , so ${\displaystyle \left(n\times n\right)}$ ${\displaystyle A}$ ${\displaystyle S}$

${\displaystyle D_{A}=S^{T}AS}$

is a diagonal matrix. The main diagonal of the diagonal matrix consists of the eigenvalues ​​of the matrix . A symmetric matrix always has real eigenvalues ​​taking into account the respective multiplicity. For the matrix , orthonormal eigenvectors of the matrix are chosen as column vectors . (Eigenvectors of a symmetrical matrix with different eigenvalues ​​are always orthogonal. If a eigenspace has a dimension larger than 1, one has to determine an orthonormal basis of the eigenspace with the help of the Gram-Schmidt orthogonalization method.) The determinant of is . So that there is a rotation matrix in the plane case, one must choose the orientations of the eigenvectors used so that is. ${\displaystyle D_{A}}$ ${\displaystyle \lambda _{1},\ldots ,\lambda _{n}}$ ${\displaystyle A}$ ${\displaystyle \left(n\times n\right)}$ ${\displaystyle n}$ ${\displaystyle S}$ ${\displaystyle n}$ ${\displaystyle {\vec {e}}_{1},\ldots ,{\vec {e}}_{n}}$ ${\displaystyle S}$ ${\displaystyle \pm 1}$ ${\displaystyle S}$ ${\displaystyle \det S=1}$

If the matrix is ​​interpreted as a linear mapping , the matrix can be understood as a transformation on the new basis . The relationship exists between the old and new coordinates . The effect of the matrix in the new coordinate system is taken over by the diagonal matrix . An important property of the (orthogonal!) Matrix is . So can also be easily coordinate old into new convert: . ${\displaystyle A}$ ${\displaystyle \mathbb {R} ^{n}}$ ${\displaystyle S}$ ${\displaystyle {\vec {e}}_{1},\ldots ,{\vec {e}}_{n}}$ ${\displaystyle {\vec {x}}=S{\vec {\xi }}}$ ${\displaystyle A}$ ${\displaystyle D_{A}}$ ${\displaystyle S}$ ${\displaystyle S^{-1}=S^{T}}$ ${\displaystyle {\vec {\xi }}=S^{-1}{\vec {x}}=S^{T}{\vec {x}}}$

The equation of a quadric

${\displaystyle \sum _{i,j=1}^{n}a_{ij}x_{i}x_{j}+2\,\sum _{i=1}^{n}b_{i}x_{i}+c=0}$

has a square part that can be described by a ( oBdA symmetrical) matrix . With the main axis theorem, this quadratic part is transformed into the "diagonal shape " . After that, there are no more mixed terms with . ${\displaystyle \sum _{i,j=1}^{n}a_{ij}x_{i}x_{j}}$ ${\displaystyle \left(a_{ij}\right)}$ ${\displaystyle \sum _{i=1}^{n}\lambda _{i}\xi _{i}^{2}}$ ${\displaystyle \xi _{i}\xi _{j}}$ ${\displaystyle i\not =j}$

Major axis transformation of a conic section

Description of the method

A conic section suffices for one equation ${\displaystyle \mathbb {R} ^{2}}$

${\displaystyle {\color {blue}a}x^{2}+{\color {red}b}xy+{\color {blue}c}y^{2}+dx+ey+f=0}$.

This equation can be written in matrix form as follows:

${\displaystyle {\begin{pmatrix}x&y\end{pmatrix}}{\begin{pmatrix}a&b/2\\b/2&c\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}+dx+ey+f=0.}$

Step 1

Set . ${\displaystyle A={\begin{pmatrix}{\color {blue}a}&{\color {red}b}/2\\{\color {red}b}/2&{\color {blue}c}\end{pmatrix}}}$

2nd step

Find the eigenvalues ​​of the matrix as solutions to the eigenvalue equation ${\displaystyle A}$

${\displaystyle \det \left(A-\lambda E\right)=0}$. The eigenvalues ​​are . ${\displaystyle \lambda _{1},\lambda _{2}}$

3rd step

Determine normalized eigenvectors from the 2 systems of equations:

${\displaystyle \left(A-\lambda _{1}E\right)\cdot {\vec {x}}=0\quad \rightarrow \quad {\vec {e}}_{1}}$

${\displaystyle \left(A-\lambda _{2}E\right)\cdot {\vec {x}}=0\quad \rightarrow \quad {\vec {e}}_{2}}$

4th step

Set and replace by using . ${\displaystyle S=({\vec {e}}_{1}\;{\vec {e}}_{2})}$ ${\displaystyle x,y}$ ${\displaystyle \xi ,\eta }$ ${\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}=S{\begin{pmatrix}\xi \\\eta \end{pmatrix}}}$

5th step

The result is the equation of the conic section in the new coordinates:

${\displaystyle \lambda _{1}\xi ^{2}+\lambda _{2}\eta ^{2}+\varepsilon \xi +\delta \eta +\omega =0}$

Since the quadratic part in this equation is fixed by the eigenvalues ​​as coefficients and the disappearance of the mixed part, the old coordinates x and y only need to be replaced in the linear part . ${\displaystyle dx+ey}$

6th step

By adding a square, you get the center or apex shape of the conic section and can read off the center (for ellipse, hyperbola, ...) or apex (for parabola) and possibly semiaxes.

7th step

With the help of the relationship , the - -coordinates of the center point and vertex can finally be calculated. ${\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}=S{\begin{pmatrix}\xi \\\eta \end{pmatrix}}}$ ${\displaystyle {\color {red}x}}$ ${\displaystyle {\color {red}y}}$

Example 3 (continued)

Major axis transformation: Ellipse with major axes NOT parallel to the axes

${\displaystyle 5x^{2}-2{\sqrt {3}}{\color {red}xy}+7y^{2}-16=0}$

Step 1

${\displaystyle A={\begin{pmatrix}5&{\color {red}-{\sqrt {3}}}\\{\color {red}-{\sqrt {3}}}&7\end{pmatrix}}}$

2nd step

${\displaystyle \det \left(A-\lambda E\right)={\begin{vmatrix}5-\lambda &-{\sqrt {3}}\\-{\sqrt {3}}&7-\lambda \end{vmatrix}}=\lambda ^{2}-12\lambda +32=(\lambda -8)(\lambda -4)=0}$

${\displaystyle \lambda _{1}=8,\ \lambda _{2}=4}$

3rd step

A normalized eigenvector zu results from the linear system of equations ${\displaystyle \lambda _{1}=8}$

${\displaystyle {\begin{matrix}-3x-{\sqrt {3}}y&=&0\\-{\sqrt {3}}x-y&=&0\end{matrix}}}$to . ${\displaystyle {\vec {e}}_{1}={\begin{pmatrix}1/2\\-{\sqrt {3}}/2\end{pmatrix}}}$

A normalized eigenvector zu results from the linear system of equations ${\displaystyle \lambda _{2}=4}$

${\displaystyle {\begin{matrix}x-{\sqrt {3}}y&=&0\\-{\sqrt {3}}x+3y&=&0\end{matrix}}}$to . ${\displaystyle {\vec {e}}_{2}={\begin{pmatrix}{\sqrt {3}}/2\\1/2\end{pmatrix}}}$

4th step

${\displaystyle S={\begin{pmatrix}1/2&{\sqrt {3}}/2\\-{\sqrt {3}}/2&1/2\end{pmatrix}}}$and this results in: ${\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}=S{\begin{pmatrix}\xi \\\eta \end{pmatrix}}}$ ${\displaystyle {\begin{matrix}x&=&{\tfrac {1}{2}}\;\xi +{\tfrac {1}{2}}{\sqrt {3}}\;\eta \\y&=&-{\tfrac {1}{2}}{\sqrt {3}}\;\xi +{\tfrac {1}{2}}\;\eta \end{matrix}}}$

Because of this , the transformation is a rotation around the angle . The latter follows from (see rotation matrix ). ${\displaystyle \det S=1}$ ${\displaystyle \varphi =-60^{\circ }}$ ${\displaystyle \cos \varphi =1/2,\ \sin \varphi =-{\sqrt {3}}/2}$

5th step

${\displaystyle 8\xi ^{2}+4\eta ^{2}=16\quad \rightarrow \quad {\frac {\xi ^{2}}{2}}+{\frac {\eta ^{2}}{4}}=1}$

6th step

Since there is neither linear nor linear in the last equation , no quadratic addition is necessary. ${\displaystyle \xi }$ ${\displaystyle \eta }$

Result: The conic section is an ellipse with the center at the zero point and the semi-axes . The vertices are in - coordinates: ${\displaystyle a={\sqrt {2}},b=2}$ ${\displaystyle \xi }$ ${\displaystyle \eta }$ ${\displaystyle (\pm {\sqrt {2}},0)_{\xi \eta },\ (0,\pm 2)_{\xi \eta }}$

7th step

The - coordinates of the vertices are (see 4th step): ${\displaystyle x}$ ${\displaystyle y}$

${\displaystyle ({\tfrac {1}{\sqrt {2}}},-{\sqrt {\tfrac {3}{2}}}),\ (-{\tfrac {1}{\sqrt {2}}},{\sqrt {\tfrac {3}{2}}}),\ ({\sqrt {3}},1),\ (-{\sqrt {3}},-1)}$

Note: The new coordinate system and the matrix S are not clearly defined. Both depend on the order of the eigenvalues ​​and the orientation of the eigenvectors chosen. The position of the conic section (center point, vertex) in the xy coordinate system is clearly determined by the given conic section equation.

Example 4: hyperbola

Major axis transformation of a hyperbola

The conic section has the equation:

${\displaystyle 3x^{2}-6xy-5y^{2}+{\frac {48}{\sqrt {10}}}x+{\frac {64}{\sqrt {10}}}y-19=0}$

Step 1

${\displaystyle A={\begin{pmatrix}3&{\color {red}-3}\\{\color {red}-3}&-5\end{pmatrix}}}$

2nd step

${\displaystyle \det \left(A-\lambda E\right)={\begin{vmatrix}3-\lambda &-3\\-3&-5-\lambda \end{vmatrix}}=\lambda ^{2}+2\lambda -24=0}$

${\displaystyle \rightarrow \quad \lambda _{1}=-6,\ \lambda _{2}=4}$

3rd step

${\displaystyle {\vec {e}}_{1}={\frac {1}{\sqrt {10}}}{\begin{pmatrix}1\\3\end{pmatrix}},\quad {\vec {e}}_{2}={\frac {1}{\sqrt {10}}}{\begin{pmatrix}-3\\1\end{pmatrix}}}$

4th step

${\displaystyle S={\begin{pmatrix}{\tfrac {1}{\sqrt {10}}}&{\tfrac {-3}{\sqrt {10}}}\\{\tfrac {3}{\sqrt {10}}}&{\tfrac {1}{\sqrt {10}}}\end{pmatrix}}\quad \rightarrow \quad {\begin{matrix}x&=&{\tfrac {1}{\sqrt {10}}}\;\xi +{\tfrac {-3}{\sqrt {10}}}\;\eta \\y&=&{\tfrac {3}{\sqrt {10}}}\;\xi +{\tfrac {1}{\sqrt {10}}}\;\eta \end{matrix}}}$

5th step

${\displaystyle -6\xi ^{2}+4\eta ^{2}+24\xi -8\eta -19=0}$

6th step

${\displaystyle 6(\xi -2)^{2}-4(\eta -1)^{2}=1}$

The conic section is a hyperbola with the center point and the semiaxes . ${\displaystyle (2,1)_{\xi \eta }}$ ${\displaystyle a={\tfrac {1}{\sqrt {6}}},\ b={\tfrac {1}{2}}}$

7th step

The xy coordinates of the center are (see 4th step). ${\displaystyle ({\tfrac {-1}{\sqrt {10}}},{\tfrac {7}{\sqrt {10}}})}$

Example 5: parabola

Major axis transformation of a parabola

The conic section has the equation:

${\displaystyle x^{2}-2{\sqrt {3}}xy+3y^{2}-4{\sqrt {3}}x-4y+24=0}$

It is

${\displaystyle A={\begin{pmatrix}1&-{\sqrt {3}}\\-{\sqrt {3}}&3\end{pmatrix}}}$.

The associated eigenvalues ​​are and the transformation matrix is . In - -coordinates, the conic section satisfies the equation: ${\displaystyle \lambda _{1}={\color {red}0},\ \lambda _{2}=4}$ ${\displaystyle S={\begin{pmatrix}{\sqrt {3}}/2&-1/2\\1/2&{\sqrt {3}}/2\end{pmatrix}}}$ ${\displaystyle \xi }$ ${\displaystyle \eta }$

${\displaystyle 4\eta ^{2}-8\xi +24=0\quad \rightarrow \quad \xi ={\tfrac {1}{2}}\eta ^{2}+3}$

So the conic section is a parabola with the vertex or in xy coordinates . The matrix describes a rotation around the angle . ${\displaystyle (3,0)_{\xi \eta }}$ ${\displaystyle ({\tfrac {3}{2}}{\sqrt {3}},{\tfrac {3}{2}})}$ ${\displaystyle S}$ ${\displaystyle 30^{\circ }}$

Principal axis transformation of surfaces

The principal axis transformation for quadrics in space proceeds according to the same method as in the plane case for conic sections. However, the bills are much more extensive.

Example: hyperboloid

The principal axis transformation is used to determine which area is described by the following equation:

${\displaystyle -x^{2}+3y^{2}-z^{2}+6xz=1}$

Step 1

${\displaystyle A={\begin{pmatrix}-1&0&3\\0&3&0\\3&0&-1\end{pmatrix}}}$

2nd step

${\displaystyle \det \left(A-\lambda E\right)={\begin{vmatrix}-1-\lambda &0&3\\0&3-\lambda &0\\3&0&-1-\lambda \end{vmatrix}}=(3-\lambda )(\lambda ^{2}+2\lambda -8)=0}$

The eigenvalues ​​are: ${\displaystyle \lambda _{1}=3,\ \lambda _{2}=2,\ \lambda _{3}=-4}$

3rd step

Determination of the eigenvectors:

to ${\displaystyle \lambda _{1}=3\ :\qquad {\begin{matrix}-4x&&3z&=&0\\&&0&=&0\\3x&&-4z&=&0\end{matrix}}\quad \rightarrow \quad {\vec {e}}_{1}={\begin{pmatrix}0\\1\\0\end{pmatrix}}}$

to ${\displaystyle \lambda _{2}=2\ :\quad \rightarrow \quad {\vec {e}}_{2}={\tfrac {1}{\sqrt {2}}}{\begin{pmatrix}1\\0\\1\end{pmatrix}}}$

and to ${\displaystyle \lambda _{2}=-4\ :\quad \rightarrow \quad {\vec {e}}_{3}={\tfrac {1}{\sqrt {2}}}{\begin{pmatrix}1\\0\\-1\end{pmatrix}}}$

Major axis transformation: single-shell hyperboloid in - - coordinates with 2 vertices and a secondary vertex ${\displaystyle \xi }$ ${\displaystyle \eta }$ ${\displaystyle \zeta }$

4th step

${\displaystyle S={\begin{pmatrix}0&{\tfrac {1}{\sqrt {2}}}&{\tfrac {1}{\sqrt {2}}}\\1&0&0\\0&{\tfrac {1}{\sqrt {2}}}&-{\tfrac {1}{\sqrt {2}}}\end{pmatrix}}\quad \rightarrow \quad {\begin{matrix}x&=&{\tfrac {1}{\sqrt {2}}}\eta +{\tfrac {1}{\sqrt {2}}}\zeta \\y&=&\xi \\z&=&{\tfrac {1}{\sqrt {2}}}\eta -{\tfrac {1}{\sqrt {2}}}\zeta \end{matrix}}}$

5th step

${\displaystyle 3\xi ^{2}+2\eta ^{2}-4\zeta ^{2}=1}$

6th step

The quadric is a single-shell hyperboloid (see list of quadrics ) with the center in the zero point, the semiaxes , the vertices and the minor vertices . ${\displaystyle a={\tfrac {1}{\sqrt {3}}},\ b={\tfrac {1}{\sqrt {2}}},\ c={\tfrac {1}{2}}}$ ${\displaystyle (\pm a,0,0),(0,\pm b,0)}$ ${\displaystyle (0,0,\pm c)}$

7th step

With the relationships in step 4 you get the vertices or minor vertices in xyz coordinates: (the center is the zero point). ${\displaystyle (0,\pm a,0),\;({\tfrac {\pm b}{\sqrt {2}}},0,{\tfrac {\pm b}{\sqrt {2}}}),\;({\tfrac {\pm c}{\sqrt {2}}},0,-{\tfrac {\pm c}{\sqrt {2}}})}$

Example: cone

The equation is given

${\displaystyle x^{2}-2xy+y^{2}-2yz+z^{2}-2xz+2x+2y+2z-3=0.}$

This equation is to be converted into a normal form by means of a major axis transformation and the type of quadric represented by the equation is to be determined.

Step 1

${\displaystyle A={\begin{pmatrix}1&-1&-1\\-1&1&-1\\-1&-1&1\end{pmatrix}}}$

2nd step

The eigenvalues ​​of the matrix are . ${\displaystyle A}$ ${\displaystyle {\color {red}\lambda _{1}=\lambda _{2}}=2,\quad \lambda _{3}=-1}$

3rd step

The eigenspace zu is the solution of the (one!) Equation . Two mutually orthogonal solution vectors have to be determined. One solution is . A solution vector orthogonal to this must also satisfy the equation . Obviously the vector satisfies both equations. Now both vectors have to be normalized: ${\displaystyle \lambda _{1,2}=2}$ ${\displaystyle -x-y-z=0}$ ${\displaystyle {\vec {v}}_{1}=(-1,1,0)^{T}}$ ${\displaystyle -x+y=0}$ ${\displaystyle (1,1,-2)^{T}}$

${\displaystyle {\vec {e}}_{1}={\frac {1}{\sqrt {2}}}{\begin{pmatrix}-1\\1\\0\end{pmatrix}},\quad {\vec {e}}_{2}={\frac {1}{\sqrt {6}}}{\begin{pmatrix}-1\\-1\\2\end{pmatrix}}}$

A normalized eigenvector zu is ${\displaystyle \lambda _{3}=-1}$ ${\displaystyle :\quad {\vec {e}}_{3}={\frac {1}{\sqrt {3}}}{\begin{pmatrix}1\\1\\1\end{pmatrix}}.}$

4th step

${\displaystyle S={\begin{pmatrix}{\frac {-1}{\sqrt {2}}}&{\frac {-1}{\sqrt {6}}}&{\frac {1}{\sqrt {3}}}\\{\frac {1}{\sqrt {2}}}&{\frac {-1}{\sqrt {6}}}&{\frac {1}{\sqrt {3}}}\\0&{\frac {2}{\sqrt {6}}}&{\frac {1}{\sqrt {3}}}\end{pmatrix}}\quad \rightarrow \quad {\begin{matrix}x&=&-{\frac {1}{\sqrt {2}}}\xi &-&{\frac {1}{\sqrt {6}}}\eta &+&{\frac {1}{\sqrt {3}}}\zeta \\y&=&{\frac {1}{\sqrt {2}}}\xi &-&{\frac {1}{\sqrt {6}}}\eta &+&{\frac {1}{\sqrt {3}}}\zeta \\z&=&&&{\frac {2}{\sqrt {6}}}\eta &+&{\frac {1}{\sqrt {3}}}\zeta \end{matrix}}}$

Major axis transformation of a cone (the apex is the point (1,1,1), the center of the represented base circle is the zero point)

5th step

${\displaystyle 2\xi ^{2}+2\eta ^{2}-\zeta ^{2}+2{\sqrt {3}}\zeta -3=0}$

6th step

Square complement supplies

${\displaystyle \xi ^{2}+\eta ^{2}-{\frac {(\zeta -{\sqrt {3}})^{2}}{2}}=0}$.

The quadric is a vertical circular cone with the tip at the point and the axis as the axis of rotation. ${\displaystyle (0,0,{\sqrt {3}})}$ ${\displaystyle \zeta }$

7th step

The point is the point in xyz coordinates . The cone axis has the direction . ${\displaystyle (1,1,1)}$ ${\displaystyle (1,1,1)^{T}}$

Major axis transformation in any dimension

A quadric im is (analogous to n = 2) the solution set of a general quadratic equation (see quadric ): ${\displaystyle Q}$ ${\displaystyle \mathbb {R} ^{n}}$

${\displaystyle Q=\left\{x\in \mathbb {R} ^{n}\mid x^{T}Ax+2b^{T}x+c=0\right\}}$

where a symmetric matrix and and are column vectors . ${\displaystyle A=(a_{ij})\in \mathbb {R} ^{n\times n}}$ ${\displaystyle b=(b_{i})\in \mathbb {R} ^{n}}$ ${\displaystyle x=(x_{i})\in \mathbb {R} ^{n}}$

The main axis transformation in this general case proceeds according to the same scheme as for conic sections (see above). After the diagonalization, however, the zero point is often shifted to the center or vertex of the quadric, so that the normal shape of the quadric arises, from which the type and properties of the quadric can be read.

application

In theoretical physics , the main axis transformation is used in classical mechanics to describe the kinematics of rigid bodies : Here, using a main axis transformation of the inertia tensor , which specifies the inertia of the body with respect to rotations around different axes, any moments of deviation that may be present - for example in the case of a top  - for Disappear.

A moment of deviation is a measure of the tendency of a rigid body to change its axis of rotation. Moments of deviation are combined with the moments of inertia in inertia tensors, with the moments of inertia being on the main diagonal of the tensor and the moments of deviation on the secondary diagonals . As shown above, the symmetric inertia tensor can be made into a diagonal shape. The axes of the new, adjusted coordinate system determined by the main axis transformation are called the main axes of inertia , the new coordinate system is called the main axis system. The diagonal elements of the transformed tensor are consequently called the main moments of inertia .

The main axis transformation is also used in other sub-areas of classical mechanics, for example in strength theory to calculate the main stresses that act on a body. Major axis transformations are still frequently used in relativistic mechanics for the basic representation of spacetime in four-dimensional Minkowski space or, for example, in electrostatics for the quadrupole moment and other higher multipole moments.

In addition, the principal axis transformation in multivariate statistics is part of the principal component analysis , which is also known as the Karhunen-Loève transformation , especially in image processing . Sometimes the terms are used synonymously, but the two transformations are not identical.

In practice, the principal axis transformation is used as part of principal component analysis to reduce the size of large data sets without significant data loss. Existing relationships between individual statistical variables are reduced as much as possible by transferring them to a new, linearly independent, problem-adapted coordinate system. For example, the number of required signal channels can be reduced by arranging them according to variance and removing the channels with the lowest variance from the data set without any relevant loss of data. This can improve the efficiency and result of a later analysis of the data.

In electronic image processing, the reduction of the data set size through principal component analyzes is used, especially in remote sensing through satellite images and the associated scientific disciplines of geodesy , geography , cartography and climatology . Here, the quality of the satellite images can be significantly improved by suppressing the noise using principal component analysis.

In computer science, the main axis transformation is mainly used in pattern recognition , to create artificial neural networks , a sub-area of artificial intelligence , for data reduction (see main component analysis ).

literature

• Burg & Haf & Wille: Higher Mathematics for Engineers. Volume II (Linear Algebra), Teubner-Verlag, Stuttgart 1992, ISBN 3 519 22956 0 , pp. 214, 335.
• Meyberg & Vachenauer: Höhere Mathematik 1. Springer-Verlag, 1995, ISBN 3 540 59188 5 , p. 341.
• W. Nolting: Basic course Theoretical Physics 1: Classical Mechanics. 7th edition. Springer, Berlin 2004, ISBN 3-540-21474-7 .
• T. Fließbach: Mechanics. 2nd Edition. Spectrum, Heidelberg 1996, ISBN 3-86025-686-6 .
• W. Greiner: Theoretical Physics. Volume 2. Mechanics Part 2. 5th edition. Harri Deutsch, Thun / Frankfurt am Main, 1989, ISBN 3-8171-1136-3 .

Individual evidence

1. ↑ See pattern recognition script . Cape. 5.2: Karhunen-Loeve transformation. (PDF) Laboratory for Message Processing, University of Hanover.
2. ^ Society for data analysis and remote sensing Hanover. ( Memento of July 10, 2009 in the Internet Archive ).
3. ↑ ^{a b} Probabilistic major axis transformation for generic object recognition. ( Memento of July 18, 2007 in the Internet Archive ) ( Postscript ). Diploma thesis in computer science, Friedrich-Alexander-Universitat Erlangen-Nürnberg.