Graphics Reference
In-Depth Information
This also shows that ta
i
+ (1 - t)b
i
£ 1; hence the point
p
belongs to
S
, proving that
S
is convex since
p
is a typical point on the segment from
w
to
w
¢.
Next, we show that
S
belongs to every convex set
C
containing the points
v
0
,
v
1
,
..., and
v
k
. The case k = 0 is trivial. Assume that k ≥ 1 and that the statement has
been proved for all values smaller than k. Let
k
Â
a
ii
i
wv
=
=
0
belong to
S
. Since not all a
i
can be zero, we may assume without loss of generality
that a
0
π 0. The case a
0
= 1 is trivial, and so assume that a
0
< 1. Thus we can write
k
Ê
Á
ˆ
˜
a
Â
i
(
)
wv
=
a
+
1
-
a
v
.
00
0
i
1
-
a
0
i
=
1
But
k
k
a
1
Â
i
Â
=
a
i
1
-
a
1
-
a
0
0
i
=
1
i
=
1
1
(
)
=
1
-
a
0
1
-
a
0
=
1
and 0 £ a
i
/(1 - a
0
) £ 1. By our inductive hypothesis
k
a
Â
i
u
=
v
i
1
-
a
0
i
=
1
belongs to every convex set containing
v
1
,
v
2
,..., and
v
k
. In particular,
u
belongs to
C
. Since
v
0
belongs to
C
, it follows that
w
= a
0
v
0
+ (1 - a
0
)
u
belongs to
C
and we are
done. Therefore,
(
{
}
)
S
=
conv
v
,
v
,...,
k
v
01
and (2) is proved.
An interesting consequence of Lemma 1.7.3(1) is that it gives us a homogeneous
way of defining a plane. We could define a k-dimensional plane as a set defined by k
+ 1 linearly independent points
v
0
,
v
1
,...,
v
k
which satisfy equation (1.25) instead of
the definition we gave in Section 1.5 that involved a point and a basis.
Lemma 1.7.3(2) motivates the following definition.
Definition.
An expression of the form
k
k
Â
Â
Œ
[]
a
v
,
where
a
01
,
and
a
=
1
,
ii
i
i
i
=
0
i
=
0
