Definitions

# Rotation operator (vector 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.

## 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

$bar x =x_1hat e_1+x_2hat e_2+x_3hat e_3$

of the body is then brought to the new direction

$mathbf\left\{A\right\}bar x =x_1mathbf\left\{A\right\}hat e_1+x_2mathbf\left\{A\right\}hat e_2+x_3mathbf\left\{A\right\}hat e_3$

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} langlehat e_1 | mathbf{A}hat e_1 rangle & langlehat e_1 | mathbf{A}hat e_2 rangle & langlehat e_1 | mathbf{A}hat e_3 rangle langlehat e_2 | mathbf{A}hat e_1 rangle & langlehat e_2 | mathbf{A}hat e_2 rangle & langlehat e_2 | mathbf{A}hat e_3 rangle langlehat e_3 | mathbf{A}hat e_1 rangle & langlehat e_3 | mathbf{A}hat e_2 rangle & langlehat e_3 | mathbf{A}hat e_3 rangle end{bmatrix}

As

$sum_\left\{k=1\right\}^3 A_\left\{ki\right\}A_\left\{kj\right\}= langle mathbf\left\{A\right\}hat e_i | mathbf\left\{A\right\}hat e_j rangle$
= begin{cases} 0 & ineq 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.

### 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 $theta$ around the axis defined by $hat e_3$"

has the matrix representation


begin{bmatrix}
` Y_1  `
` Y_2  `
` Y_3`
end{bmatrix} = begin{bmatrix}
` costheta & -sintheta & 0 `
` sintheta &  costheta & 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}
` costheta  & -sintheta & 0`
` sintheta  &  costheta & 0`
`          0  &           0 & 1`
end{bmatrix}=1

and the characteristic polynomial is


begin{align} detbegin{bmatrix}
` costheta -lambda  & -sintheta          & 0        `
` sintheta           &  costheta -lambda & 0        `
`                    0 &                    0 & 1-lambda`
end{bmatrix} &=big({(costheta -lambda)}^2 + {sintheta}^2 big)(1-lambda) &=-lambda^3+(2 costheta + 1) lambda^2 - (2 costheta + 1) lambda +1 end{align}

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

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

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


-(lambda-1){(lambda +1)}^2

i.e. the rotation operator has the eigenvalues



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


bar x =alpha hat e_3 quad -infty

The eigenspace corresponding to $lambda=-1$ 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

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


bar x =alpha hat e_3 quad -infty

### The general case

The operator

"Rotation with the angle $theta$ 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} costheta & -sintheta & 0 sintheta & costheta & 0 0 & 0 & 1end{bmatrix}

for some value of $theta$.

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 $B_\left\{ii\right\}$ 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 corresponding to $theta=0$.

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 $theta=pi$.

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

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

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


lambda=1

Setting


hat e_3=frac{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

$cos theta=frac\left\{A_\left\{11\right\}+A_\left\{22\right\}+A_\left\{33\right\}-1\right\}\left\{2\right\}$
$sin theta=|bar\left\{E\right\}|$

The expressions above are in fact valid also for the case of a symmetric rotation operator corresponding to a rotation with $theta = 0$ or $theta = pi$. But the difference is that for $theta = pi$ 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 $cos theta$ 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 $theta=0$ (the identity operator) and $theta=pi$

## Quaternions

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

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}}{sintheta}quad bar E and that the forth component is the scalar

q_4=cos frac{theta}{2}

As the angle $theta$ 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

$E_1=2 q_4 q_1$
$E_2=2 q_4 q_2$
$E_3=2 q_4 q_3$
$E_4=\left\{q_4\right\}^2 -\left(\left\{q_1\right\}^2+\left\{q_2\right\}^2+\left\{q_3\right\}^2\right)$

Using quaternions the matrix of the rotation operatator is

$begin\left\{bmatrix\right\}$
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}

## Numerical example

Consider the reorientation corresponding to the Euler angles $alpha=10deg quad beta=20deg quad gamma=30deg quad$ 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}
` costheta  & -sintheta & 0`
` sintheta  &  costheta & 0`
`          0  &           0 & 1`
end{bmatrix} with $theta=44.537deg$ is obtained with
$hat e_3=\left(0.451272,-0.079571,0.888832\right)$

The quaternion relative this new system is then


(0, 0, 0.378951, 0.925417) = (0, 0, sinfrac{theta}{2}, cosfrac{theta}{2})

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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