Cross product
19
DISCUSSION.
For a given
A
∈
SO(3), the representation
A
=
R
( ˆ
w, θ
), where ˆ
w
is a unit
vector and 0
≤
θ
≤
π
, is virtually unique. To see this we start with the result of Problem 7–16,
and compute the trace of each matrix:
trace
A
=
3 cos
θ
+ (1

cos
θ
)(
w
2
1
+
w
2
2
+
w
2
3
)
=
3 cos
θ
+ (1

cos
θ
)
=
2 cos
θ
+ 1
.
Thus
cos
θ
=
trace
A

1
2
.
Therefore, as 0
≤
θ
≤
π
, the angle is completely determined by the knowledge of trace
A
.
Next notice that Problem 7–16 implies
A

A
t
= 2 sin
θ
0

w
3
w
2
w
3
0

w
1

w
2
w
1
0
.
We conclude immediately that if 0
< θ < π
, the vector
w
is also uniquely determined.
The only cases of nonuniqueness are thus when sin
θ
= 0, or, equivalently,
A
=
A
t
.
If
θ
= 0, then
A
=
R
( ˆ
w,
0) =
I
and the “axis” ˆ
w
is irrelevant. If
θ
=
π
, then we have
A
=

I
+ 2(
w
i
w
j
)
.
It is easy to check in this case that ˆ
w
is uniquely determined up to sign. This is because ˆ
w
6
= 0
and so some coordinate
w
j
6
= 0 and we obtain from the above equation
2
w
j
ˆ
w
=
j
th
column of
(
A
+
I
)
.
This determines ˆ
w
up to a scalar factor, and the fact that
k
ˆ
w
k
= 1 shows that the only
possibilities are ˆ
w
and

ˆ
w
.
We now have a truly intuitive understanding of SO(3). In particular, the righthand rule
makes wonderful sense in this context, as we now explain. Suppose
{
ˆ
ϕ
1
,
ˆ
ϕ
2
,
ˆ
ϕ
3
}
is an arbitrary
orthonormal frame with standard orientation, and form the usual matrix Φ with those columns,
Φ = ( ˆ
ϕ
1
ˆ
ϕ
2
ˆ
ϕ
3
)
.
Then Φ
∈
SO(3) and Φ

1
maps the frame
{
ˆ
ϕ
1
,
ˆ
ϕ
2
,
ˆ
ϕ
3
}
to
{
ˆ
ı,
ˆ
,
ˆ
k
}
. Moreover, Φ is a rotation
R
( ˆ
w, θ
). Thus if you place your right hand to lie in the correct
ˆ
k
= ˆ
ı
×
ˆ
position, then rotating
your hand with
R
( ˆ
w, θ
) produces the ˆ
ϕ
3
= ˆ
ϕ
1
×
ˆ
ϕ
2
position.