Graphics Reference
In-Depth Information
If
w
belongs to aff({
v
0
,
v
1
,...,
v
k
}), then we know from Theorem 1.4.4 that
w v
=+
t
vv
+ +
....
t
vv
for some
t
Œ
R
.
0
101
k
0
k
i
This equation can be rewritten in the form
(
)
w
=-- -
1
t
...
t
v
+++
t
v
...
t
v
,
1
k
0
1
1
k
k
which shows that
w
belongs to
S
. Conversely, if
w
belongs to
S
, then
k
k
Â
Â
wv
=
a
for some a such that
a
=
1
.
ii
i
i
i
=
0
i
=
0
This equation can be rewritten in the form
+
+
..
.
a
k
vv
.
wv
=+
a
vv
0
101
0
k
Part (1) is proved.
To prove (2), let
k
k
Ï
Ó
¸
˛
Â
a
Â
Œ
[]
S
=
v
a
01
,
and
a
=
1
.
ii
i
i
i
=
0
i
=
0
We need to show that
S
is the smallest convex set containing {
v
0
,
v
1
,...,
v
k
}. We show
that
S
is convex first. Consider two points
k
k
Â
Â
wv
=
a
and
w v
¢ =
b
ii
ii
i
=
0
i
=
0
in
S
and let t Π[0,1]. Then
k
k
Ê
Á
ˆ
˜
+-
)
Ê
ˆ
˜
Â
Â
(
)
(
pw
=+-
t
1
t
w
¢=
t
a
v
1
t
b
v
Á
ii
ii
i
=
0
i
=
0
k
Â
(
(
)
)
=
ta
+
1
-
t b
v
.
i
i
i
i
=
0
Clearly, 0 £ ta
i
+ (1 - t)b
i
. Furthermore,
k
k
k
Ê
Á
ˆ
˜
+-
)
Ê
ˆ
˜
Â
Â
Â
(
(
)
)
=
(
ta
+-
1
t b
t
a
1
t
b
Á
i
i
i
i
i
=
0
i
=
0
i
=
0
(
)
◊
=◊+ -
t
11
t
1
=
1
.
