show that both representations are equally valid. They exhibit a lot of similarities, both
conceptually and algebraically, as the reader will appreciate in Table 1. Perhaps, the
most important difference is that the unit quaternion group constitutes a double cover of
SO(3) (thus technically not being SO(3) itself), something that is not critical in most of
our applications.
5
The table is inserted upfront for the sake of a rapid comparison and
evaluation. The rotation matrix and quaternion representations of SO(3) are explored in
the following sections.
2.3 The rotation group and the rotation matrix
The operator r() is linear, since it is defined from the scalar and vector products, which
are linear. It can therefore be represented by a matrix R ∈ R
3×3
, which produces rotations
to vectors v ∈ R
3
through the matrix product,
r(v) = R v . (58)
Injecting it in (55a), using the dot product ha, bi = a
>
b and developing we have that for
all v,
(Rv)
>
(Rv) = v
>
R
>
Rv = v
>
v , (59)
yielding the orthogonality condition on R,
R
>
R = I = R R
>
. (60)
The condition above is indeed a condition of orthogonality, since we can observe from it
that, by writing R = [r
1
, r
2
, r
3
] and substituting above, the column vectors r
i
of R, with
i ∈ {1, 2, 3}, are of unit length and orthogonal to each other,
hr
i
, r
i
i = r
>
i
r
i
= 1
hr
i
, r
j
i = r
>
i
r
j
= 0 , if i 6= j .
The set of transformations keeping vector norms and angles is for this reason called the
Orthogonal group, denoted O(3). The orthogonal group includes rotations (which are rigid
motions) and reflections (which are not rigid). The notion of group here means essentially
(and informally) that the product of two orthogonal matrices is always an orthogonal
matrix,
6
and that each orthogonal matrix admits an inverse. In effect, the orthogonality
condition (60) implies that the inverse rotation is achieved with the transposed matrix,
R
−1
= R
>
. (61)
5
The effect of the double cover needs to be considered when performing interpolation in the space of
rotations. This is however easy, as we will see in Section 2.7.
6
Let Q
1
and Q
2
be orthogonal, and build Q = Q
1
Q
2
. Then Q
>
Q = Q
>
2
Q
>
1
Q
1
Q
2
= Q
>
2
IQ
2
= I.
15
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