1.146
We have previously shown
∙
that the set
𝑈
is linearly independent (Exercise 1.135).
∙
the space
𝒢
𝑁
has dimension 2
𝑛
−
1
(Exercise 1.141).
There are 2
𝑛
−
1
distinct Tunanimity games
𝑢
𝑇
in
𝑈
. Hence
𝑈
spans the 2
𝑛
−
1
space
𝒢
𝑁
. Alternatively, note that any game
𝑤
∈ 𝒢
𝑁
can be written as a linear combination
of Tunanimity games (Exercise 1.75).
1.147
Let
𝐵
=
{
x
1
,
x
2
, . . . ,
x
𝑚
}
be a basis for
𝑆
.
Since
𝐵
is linearly independent,
𝑚
≤
𝑛
(Exercise 1.143). There are two possibilities.
Case 1:
𝑚
=
𝑛
.
𝐵
is a set of
𝑛
linearly independent elements in an
𝑛
dimensional
space
𝑋
. Hence
𝐵
is a basis for
𝑋
and
𝑆
= lin
𝐵
=
𝑋
.
Case 2:
𝑚 < 𝑛
.
Since
𝐵
is linearly independent but cannot be a basis for the
𝑛

dimensional space
𝑋
, we must have
𝑆
= lin
𝐵
⊂
𝑋
.
Therefore, we conclude that if
𝑆
⊂
𝑋
is a proper subspace, it has a lower dimension
than
𝑋
.
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 Fall '10
 Dr.DuMond
 Macroeconomics, Linear Algebra, Vector Space, basis, Foundations of Mathematical Economics, space ????

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