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terclockwise motion. Therefore, we need an appropriate equivalence relation. The key
to defining this relation is the matrix relating two ordered bases.
Let (
v
1
,
v
2
) and (
w
1
,
w
2
) be two ordered bases. Suppose that
v
=
a
ww
+
a
2 2
,
i
i
11
i
for a
ij
Œ
R
. Define (
v
1
,
v
2
) to be equivalent to (
w
1
,
w
2
) if the determinant of the matrix
(a
ij
) is positive. Since we are dealing with bases, we know that the a
ij
exist and are
unique and that the matrix (a
ij
) is nonsingular. It is easy to see that our relation is an
equivalence relation and that we have precisely two equivalence classes because the
nonzero determinant is either positive or negative. We could define an orientation of
R
2
to be such an equivalence class. As a quick check to see that we are getting what
we want, note that if
w
1
=
v
2
and
w
2
=
v
1
, then
01
10
a
i
()
=
Ê
Ë
ˆ
¯
and the determinant of this matrix is -1, so that (
v
1
,
v
2
) and (
v
2
,
v
1
) determine differ-
ent equivalence classes.
Because we only used vector space concepts, it is easy to generalize what we just
did.
Definition.
Let B
1
= (
v
1
,
v
2
,...,
v
n
) and B
2
= (
w
1
,
w
2
,...,
w
n
) be ordered bases for a
vector space
V
and let
n
Â
1
w
=
a
v
,
where
a
Œ
R
.
i
ij
j
ij
j
=
We say that B
1
is
equivalent
to B
2
, and write B
1
~ B
2
if the determinant of the matrix
(a
ij
) is positive.
1.6.1. Lemma.
~ is an equivalence relation on the set of ordered bases for
V
with
precisely two equivalence classes.
Proof.
Exercise 1.6.1.
Definition.
An
orientation
of a vector space
V
is defined to be an equivalence
class of ordered bases of
V
with respect to the relation ~. Given one orientation of
V
,
then the other one is called the
opposite orientation
. The equivalence class of an ordered
basis (
v
1
,
v
2
,...,
v
n
) will be denoted by [
v
1
,
v
2
,...,
v
n
]. We shall say that the ordered basis
(
v
1
,
v
2
,...,
v
n
)
induces
or
determines
the orientation [
v
1
,
v
2
,...,
v
n
]. An
oriented vector
space
is a pair (
V
,s), where
V
is vector space and s is an orientation of it.
1.6.2. Example.
To show that the ordered bases ((1,3),(2,1)) and ((1,1),(2,0)) deter-
mine the same orientation of the plane.
Solution.
See Figure 1.13. Note that
