Rotation operator (vector space)


Rotation operator (vector space)

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

Any vector:ar x =x_1hat e_1+x_2hat e_2+x_3hat e_3

of the body is then brought to the new direction:mathbf{A}ar x =x_1mathbf{A}hat e_1+x_2mathbf{A}hat e_2+x_3mathbf{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 :egin{bmatrix} A_{11} & A_{12} & A_{13} \ A_{21} & A_{22} & A_{23} \ A_{31} & A_{32} & A_{33} end{bmatrix} =egin{bmatrix} langlehat e_1 | mathbf{A}hat e_1 angle & langlehat e_1 | mathbf{A}hat e_2 angle & langlehat e_1 | mathbf{A}hat e_3 angle \ langlehat e_2 | mathbf{A}hat e_1 angle & langlehat e_2 | mathbf{A}hat e_2 angle & langlehat e_2 | mathbf{A}hat e_3 angle \ langlehat e_3 | mathbf{A}hat e_1 angle & langlehat e_3 | mathbf{A}hat e_2 angle & langlehat e_3 | mathbf{A}hat e_3 angle end{bmatrix}

As : sum_{k=1}^3 A_{ki}A_{kj}= langle mathbf{A}hat e_i | mathbf{A}hat e_j angle= egin{cases}0 & i eq j, \ 1 & i = j,end{cases}

or equivalently in matrix notation

:egin{bmatrix} A_{11} & A_{12} & A_{13} \ A_{21} & A_{22} & A_{23} \ A_{31} & A_{32} & A_{33} end{bmatrix}^Tegin{bmatrix} A_{11} & A_{12} & A_{13} \ A_{21} & A_{22} & A_{23} \ A_{31} & A_{32} & A_{33} end{bmatrix} =egin{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 heta around the axis defined by hat e_3"

has the matrix representation

:egin{bmatrix} Y_1 \ Y_2 \ Y_3end{bmatrix}=egin{bmatrix} cos heta & -sin heta & 0 \ sin heta & cos heta & 0 \ 0 & 0 & 1end{bmatrix}egin{bmatrix} X_1 \ X_2 \ X_3end{bmatrix}

relative this basevector system

This then means that a vector

:ar x=egin{bmatrix} hat e_1 & hat e_2 & hat e_3end{bmatrix}egin{bmatrix} X_1 \ X_2 \ X_3end{bmatrix}

is rotated to the vector

:ar y=egin{bmatrix} hat e_1 & hat e_2 & hat e_3end{bmatrix}egin{bmatrix} Y_1 \ Y_2 \ Y_3end{bmatrix}

by the linear operator

The determinant of this matrix is

:detegin{bmatrix} cos heta & -sin heta & 0\ sin heta & cos heta & 0\ 0 & 0 & 1end{bmatrix}=1

and the characteristic polynomial is

:egin{align}detegin{bmatrix} cos heta -lambda & -sin heta & 0 \ sin heta & cos heta -lambda & 0 \ 0 & 0 & 1-lambdaend{bmatrix}&=ig({(cos heta -lambda)}^2 + {sin heta}^2 ig)(1-lambda) \&=-lambda^3+(2 cos heta + 1) lambda^2 - (2 cos heta + 1) lambda +1 \end{align}

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

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

For the case heta=pi the characteristic polynomial is

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

:ar 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

:ar x =alpha hat e_1 + eta hat e_2 quad -infty

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

:ar x =alpha hat e_3 quad -infty

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"

:egin{bmatrix}cos heta & -sin heta & 0 \sin heta & cos heta & 0 \ 0 & 0 & 1end{bmatrix}

for some value of heta.

In fact, if a linear operator has the orthogonal matrix

:egin{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:egin{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_{ii} 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

:egin{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 heta=pi.

If the matrix is un-symmetric the vector:ar 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

:lambda=1

Setting :hat e_3=frac{ar 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 heta=frac{A_{11}+A_{22}+A_{33}-1}{2}:sin heta=|ar{E}|

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 = pithe vector :ar 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. therotation axis.

Defining E_4 as cos heta the matrix for therotation operator is

:frac{1-E_4}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

:{E_1}^2+{E_2}^2+{E_3}^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{ heta}{2} 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{f_1} + q_2 hat{f_2} + q_3 hat{f_1} = sin frac{ heta}{2}quad hat{e_3}=frac{sin frac{ heta}{2{sin heta}quad ar Eand that the forth component is the scalar:q_4=cos frac{ heta}{2}

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

Using quaternions the matrix of the rotation operatator is:egin{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}

Numerical example

Consider the reorientation corresponding to the Euler anglesalpha=10deg quad eta=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)

:egin{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:egin{bmatrix} cos heta & -sin heta & 0\ sin heta & cos heta & 0\ 0 & 0 & 1end{bmatrix}with heta=44.537deg 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, sinfrac{ heta}{2}, cosfrac{ heta}{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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