# Rotation operator (vector space)

The three Euler rotations is an obvious way to bring a rigid body into any desired orientation bysequentially 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.

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

Any vector:

of the body is then brought to the new direction:

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 :

As :

or equivalently in matrix notation

: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.

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 $heta$ around the axis defined by $hat e_3$"

has the matrix representation

:

relative this basevector system

This then means that a vector

:

is rotated to the vector

:

by the linear operator

The determinant of this matrix is

:

and the characteristic polynomial is

:

The matrix is symmetric if and only if $sin heta=0$, i.e. for $heta=0$and for $heta=pi$

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

For the case $heta=pi$ the characteristic polynomial is

:$-\left(lambda-1\right)\left\{\left(lambda +1\right)\right\}^2$

i.e. the rotation operator has the eigenvalues:$lambda=1 quad lambda=-1$

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

:

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

:

For all other values of $heta$ the matrix is un-symmetric and as $\left\{sin heta\right\}^2 > 0$ there isonly the eigenvalue $lambda=1$ with the one-dimensional eigenspace of the vectors on the rotation axis:

:

The general case

The operator

"Rotation with the angle $heta$ 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"

:

for some value of $heta$.

In fact, if a linear operator has the orthogonal matrix

: 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: As it is an orthogonal matrix these diagonal elements $B_\left\{ii\right\}$ 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 correspondingto $heta=0$.

In the second case it has the form

:

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

If the matrix is un-symmetric the vector:

where

:$alpha_1=frac\left\{A_\left\{32\right\}-A_\left\{23\right\} \right\}\left\{2\right\}$:$alpha_2=frac\left\{A_\left\{13\right\}-A_\left\{31\left\{2\right\}$:$alpha_3=frac\left\{A_\left\{21\right\}-A_\left\{12\left\{2\right\}$

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

:$lambda=1$

Setting :

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:$cos heta=frac\left\{A_\left\{11\right\}+A_\left\{22\right\}+A_\left\{33\right\}-1\right\}\left\{2\right\}$:

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

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

Defining $E_4$ as $cos heta$ the matrix for therotation operator is

:$frac\left\{1-E_4\right\}$E_1}^2+{E_2}^2+{E_3}^2}egin{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}+egin{bmatrix}E_4 & -E_3 & E_2 \ E_3 & E_4 & -E_1 \-E_2 & E_1 & E_4 end{bmatrix}

provided that

:$\left\{E_1\right\}^2+\left\{E_2\right\}^2+\left\{E_3\right\}^2 > 0$

i.e. except for the cases $heta=0$ (the identity operator) and $heta=pi$

Quaternions

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

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

:$q_1 hat\left\{f_1\right\} + q_2 hat\left\{f_2\right\} + q_3 hat\left\{f_1\right\} = sin frac\left\{ heta\right\}\left\{2\right\}quad hat\left\{e_3\right\}=frac\left\{sin frac\left\{ heta\right\}\left\{2${sin heta}quad ar Eand that the forth component is the scalar:$q_4=cos frac\left\{ heta\right\}\left\{2\right\}$

As the angle $heta$ defined from the canonical form is in the interval :$0 le heta le pi$

one would normally have that $q_4 ge 0$. But a "dual" representation of a rotation with quaternionsis 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:$E_1=2 q_4 q_1$:$E_2=2 q_4 q_2$:$E_3=2 q_4 q_3$:$E_4=\left\{q_4\right\}^2 -\left(\left\{q_1\right\}^2+\left\{q_2\right\}^2+\left\{q_3\right\}^2\right)$

Using quaternions the matrix of the rotation operatator is:

Numerical example

Consider the reorientation corresponding to the Euler anglesrelative 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)

:

and the quaternion is:$\left(0.171010, -0.030154, 0.336824, 0.925417\right)$

The canonical form of this operator:with $heta=44.537deg$ is obtained with:$hat e_3=\left(0.451272,-0.079571,0.888832\right)$

The quaternion relative this new system is then:$\left(0, 0, 0.378951, 0.925417\right) = \left(0, 0, sinfrac\left\{ heta\right\}\left\{2\right\}, cosfrac\left\{ heta\right\}\left\{2\right\}\right)$

Instead of making the three Euler rotations

:$10 deg + 20 deg + 30 deg$

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

Reference

*.

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