Rotation operator (vector space)

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This article derives the main properties of rotations in 3-dimensional space.

The three Euler rotations is an obvious way to bring a rigid body into any desired orientation by sequentially making rotations about axis fixed relative the body. But it is a non-trivial fact is that this also can be achieved with one single rotation. Using the concepts of linear algebra it is shown how this single rotation can be found.

[edit] Mathematical formulation

Let

$\hat e_1\ ,\ \hat e_2\ ,\ \hat e_3$

be a coordinate system fixed in the body that through a change in orientation is brought to the new directions

$\mathbf{A}\hat e_1\ ,\ \mathbf{A}\hat e_2\ ,\ \mathbf{A}\hat e_3.$

Any vector

$\bar x\ =x_1\hat e_1+x_2\hat e_2+x_3\hat e_3$

of the body is then brought to the new direction

$\mathbf{A}\bar x\ =x_1\mathbf{A}\hat e_1+x_2\mathbf{A}\hat e_2+x_3\mathbf{A}\hat e_3$

i.e. this is a linear operator

The matrix of this operator relative the coordinate system

$\hat e_1\ ,\ \hat e_2\ ,\ \hat e_3$

is

$\begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} \langle\hat e_1 | \mathbf{A}\hat e_1 \rangle & \langle\hat e_1 | \mathbf{A}\hat e_2 \rangle & \langle\hat e_1 | \mathbf{A}\hat e_3 \rangle \\ \langle\hat e_2 | \mathbf{A}\hat e_1 \rangle & \langle\hat e_2 | \mathbf{A}\hat e_2 \rangle & \langle\hat e_2 | \mathbf{A}\hat e_3 \rangle \\ \langle\hat e_3 | \mathbf{A}\hat e_1 \rangle & \langle\hat e_3 | \mathbf{A}\hat e_2 \rangle & \langle\hat e_3 | \mathbf{A}\hat e_3 \rangle \end{bmatrix}$

As

$\sum_{k=1}^3 A_{ki}A_{kj}= \langle \mathbf{A}\hat e_i | \mathbf{A}\hat e_j \rangle = \begin{cases} 0 & i\neq j, \\ 1 & i = j, \end{cases}$

or equivalently in matrix notation

$\begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix}^T \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

the matrix is orthogonal and as a "right hand" base vector system is re-orientated into another "right hand" system the determinant of this matrix has the value 1.

[edit] Rotation around an axis

Let

$\hat e_1\ ,\ \hat e_2\ ,\ \hat e_3$

be an orthogonal positively oriented base vector system in $R 3$

The linear operator

"Rotation with the angle $θ$ around the axis defined by $\hat e_3$ "

has the matrix representation

$\begin{bmatrix} Y_1 \\ Y_2 \\ Y_3 \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix}$

relative this basevector system

This then means that a vector

$\bar x=\begin{bmatrix} \hat e_1 & \hat e_2 & \hat e_3 \end{bmatrix} \begin{bmatrix} X_1 \\ X_2 \\ X_3 \end{bmatrix}$

is rotated to the vector

$\bar y=\begin{bmatrix} \hat e_1 & \hat e_2 & \hat e_3 \end{bmatrix} \begin{bmatrix} Y_1 \\ Y_2 \\ Y_3 \end{bmatrix}$

by the linear operator

The determinant of this matrix is

$\det \begin{bmatrix} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{bmatrix}=1$

and the characteristic polynomial is

$\begin{align} \det\begin{bmatrix} \cos\theta -\lambda & -\sin\theta & 0 \\ \sin\theta & \cos\theta -\lambda & 0 \\ 0 & 0 & 1-\lambda \end{bmatrix} &=\big({(\cos\theta -\lambda)}^2 + {\sin\theta}^2 \big)(1-\lambda) \\ &=-\lambda^3+(2\ \cos\theta\ +\ 1)\ \lambda^2 - (2\ \cos\theta\ +\ 1)\ \lambda +1 \\ \end{align}$

The matrix is symmetric if and only if $sinθ = 0$ , i.e. for $θ = 0$ and for $θ = π$

The case $θ = 0$ is the trivial case of an identity operator

For the case $θ = π$ the characteristic polynomial is

- (λ - 1)(λ + 1) 2

i.e. the rotation operator has the eigenvalues

$\lambda=1 \quad \lambda=-1$

The eigenspace corresponding to $λ = 1$ is all vectors on the rotation axis, i.e. all vectors

$\bar x =\alpha \ \hat e_3 \quad -\infty <\alpha < \infty$

The eigenspace corresponding to $λ = - 1$ consists of all vectors orthogonal to the rotation axis, i.e. all vectors

$\bar x =\alpha \ \hat e_1 + \beta \ \hat e_2 \quad -\infty <\alpha < \infty \quad -\infty <\beta < \infty$

For all other values of $θ$ the matrix is un-symmetric and as $sinθ 2 > 0$ there is only the eigenvalue $λ = 1$ with the one-dimensional eigenspace of the vectors on the rotation axis:

$\bar x =\alpha \ \hat e_3 \quad -\infty <\alpha < \infty$

[edit] The general case

The operator

"Rotation with the angle $θ$ around a specified axis"

discussed above is an orthogonal mapping and its matrix relative any base vector system is therefore an orthogonal matrix . Further more its determinant has the value 1. A non-trivial fact is the opposite, i.e. that for any orthogonal linear mapping in $R 3$ having determinant = 1 there exist base vectors

$\hat e_1\ ,\ \hat e_2\ ,\ \hat e_3$

such that the matrix takes the "canonical form"

$\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{bmatrix}$

for some value of $θ$ .

In fact, if a linear operator has the orthogonal matrix

$\begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix}$

relative some base vector system

$\hat f_1\ ,\ \hat f_2\ ,\ \hat f_3$

and this matrix is symmetric the "Symmetric operator theorem" valid in $R n$ (any dimension) applies saying

that it has n orthogonal eigenvectors. This means for the 3-dimensional case that there exists a coordinate system

$\hat e_1\ ,\ \hat e_2\ ,\ \hat e_3$

such that the matrix takes the form

$\begin{bmatrix} B_{11} & 0 & 0 \\ 0 & B_{22} & 0 \\ 0 & 0 & B_{33} \end{bmatrix}$

As it is an orthogonal matrix these diagonal elements $B i i$ are either 1 or −1. As the determinant is 1 these elements are either all 1 or one of the elements is 1 and the other two are −1.

In the first case it is the trivial identity operator corresponding to $θ = 0$ .

In the second case it has the form

$\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

if the basevectors are numbered such that the one with eigenvalue 1 has index 3. This matrix is then of the desired form for $θ = π$ .

If the matrix is un-symmetric the vector

$\bar E = \alpha_1\ \hat f_1 + \alpha_2\ \hat f_2 + \alpha_3\ \hat f_3$

where

$\alpha_1=\frac{A_{32}-A_{23} }{2}$

$\alpha_2=\frac{A_{13}-A_{31}}{2}$

$\alpha_3=\frac{A_{21}-A_{12}}{2}$

is non-zero. This vector is an eigenvector with eigenvalue

λ = 1

Setting

$\hat e_3=\frac{\bar E}{|\bar E|}$

and selecting any two orthogonal unit vectors in the plane orthogonal to $\hat e_3$ :

$\hat e_1\ ,\ \hat e_2$

such that

$\hat e_1\ ,\ \hat e_2,\ \hat e_3$

form a positively oriented trippel the operator takes the desired form with

$\cos \theta=\frac{A_{11}+A_{22}+A_{33}-1}{2}$

$\sin \theta=|\bar{E}|$

The expressions above are in fact valid also for the case of a symmetric rotation operator corresponding to a rotation with $θ = 0$ or $θ = π$ . But the difference is that for $θ = π$ the vector

$\bar E = \alpha_1\ \hat f_1 + \alpha_2\ \hat f_2 + \alpha_3\ \hat f_3$

is zero and of no use for finding the eigenspace of eigenvalue 1, i.e. the rotation axis.

Defining $E 4$ as $cosθ$ the matrix for the rotation operator is

$\frac{1-E_4}{{E_1}^2+{E_2}^2+{E_3}^2} \begin{bmatrix} E_1 E_1 & E_1 E_2 & E_1 E_3 \\ E_2 E_1 & E_2 E_2 & E_2 E_3 \\ E_3 E_1 & E_3 E_2 & E_3 E_3 \end{bmatrix} + \begin{bmatrix} E_4 & -E_3 & E_2 \\ E_3 & E_4 & -E_1 \\ -E_2 & E_1 & E_4 \end{bmatrix}$

provided that

${E_1}^2+{E_2}^2+{E_3}^2 > 0$

i.e. except for the cases $θ = 0$ (the identity operator) and $θ = π$

[edit] Quaternions

Main article: Quaternions and spatial rotation

Quaternions are defined similar to $E_1\ ,\ E_2\ ,\ E_3\ ,\ E_4$ with the difference that the half angle $\frac{\theta}{2}$ is used in stead of the full angle $θ$ .

This means that the first 3 components $q_1\ ,\ q_2\ ,\ q_3\$ are components of a vector defined from

$q_1\ \hat{f_1}\ +\ q_2\ \hat{f_2}\ +\ \ q_3\ \hat{f_1}\ =\ \sin \frac{\theta}{2}\quad \hat{e_3}=\frac{\sin \frac{\theta}{2}}{\sin\theta}\quad \bar E$

and that the fourth component is the scalar

$q_4=\cos \frac{\theta}{2}$

As the angle $θ$ defined from the canonical form is in the interval

$0 \le \theta \le \pi$

one would normally have that $q_4 \ge 0$ . But a "dual" representation of a rotation with quaternions is used, i.e.

$q_1\ ,\ q_2\ ,\ q_3\ ,\ q_4\$

and

$-q_1\ ,\ -q_2\ ,\ -q_3\ ,\ -q_4\$

are two alternative representations of one and the same rotation.

The entities $E k$ are defined from the quaternions by

E 1 = 2 q 4 q 1

E 2 = 2 q 4 q 2

E 3 = 2 q 4 q 3

$E_4={q_4}^2 -({q_1}^2+{q_2}^2+{q_3}^2)$

Using quaternions the matrix of the rotation operator is

$\begin{bmatrix} 2({q_1}^2+{q_4}^2)-1 &2({q_1}{q_2}-{q_3}{q_4}) &2({q_1}{q_3}+{q_2}{q_4}) \\ 2({q_1}{q_2}+{q_3}{q_4}) &2({q_2}^2+{q_4}^2)-1 &2({q_2}{q_3}-{q_1}{q_4}) \\ 2({q_1}{q_3}-{q_2}{q_4}) &2({q_2}{q_3}+{q_1}{q_4}) &2({q_3}^2+{q_4}^2)-1 \\ \end{bmatrix}$

[edit] Numerical example

Consider the reorientation corresponding to the Euler angles $\alpha=10^\circ \quad \beta=20^\circ \quad \gamma=30^\circ \quad$ relative a given base vector system

$\hat f_1\ ,\ \hat f_2,\ \hat f_3$

Corresponding matrix relative this base vector system is (see Euler angles#Matrix notation)

$\begin{bmatrix} 0.771281 & -0.633718 & 0.059391 \\ 0.613092 & 0.714610 & -0.336824 \\ 0.171010 & 0.296198 & 0.939693 \end{bmatrix}$

and the quaternion is

$(0.171010,\ -0.030154,\ 0.336824,\ 0.925417)$

The canonical form of this operator

$\begin{bmatrix} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{bmatrix}$

with $\theta=44.537^\circ$ is obtained with

$\hat e_3=(0.451272,-0.079571,0.888832)$

The quaternion relative this new system is then

$(0,\ 0,\ 0.378951,\ 0.925417) = (0,\ 0,\ \sin\frac{\theta}{2},\ \cos\frac{\theta}{2})$

Instead of making the three Euler rotations

$10^\circ,20^\circ,30^\circ$

the same orientation can be reached with one single rotation of size $44.537^\circ$ around $\hat e_3$

[edit] References

Shilov, Georgi (1961), An Introduction to the Theory of Linear Spaces, Prentice-Hall, Library of Congress 61-13845 .

[edit] External links

The Mathematics behind rotating and moving observer, explanation on how matrix algebra is used to render 3D scenery viewed by a moving or rotating observer into 2D screen.

Rotation operator (vector space)

From Wikipedia, the free encyclopedia

Contents

[edit] Mathematical formulation

[edit] Rotation around an axis

[edit] The general case

[edit] Quaternions

[edit] Numerical example

[edit] References

[edit] External links

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