Unformatted text preview: 164 3. PRODUCTS OF GROUPS Deﬁnition 3.3.15. A linear combination of a subset S of a vector space V
is any element of V of the form ˛1 v1 C ˛2 v2 C
C ˛s vs , where for all
i , ˛i 2 K and vi 2 S . The span of S is the set of all linear combinations
of S. We denote the span of S by span.S /.
The span of the empty set is the set containing only the zero vector
Deﬁnition 3.3.16. A subset S of vector space V is linearly independent if
for all natural numbers s , for all ˛ D 4 : 5 2 K s , and for all sequences
.v1 ; : : : vs / of distinct vectors in S , if ˛1 v1 C ˛2 v2 C C ˛s vs D 0, then
˛ D 0. Otherwise, S is linearly dependent.
Note that a linear independent set cannot contain the zero vector. The
empty set is linearly independent, since there are no sequences of its elements!
Example 3.3.17. Deﬁne en .x/ D e i nx for n an integer and x 2 R. Then
fen W n 2 Zg is a linearly independent subset of the (complex) vector
space of C –valued functions on R. To show this, we have to prove that
for all natural numbers s , any set consisting of s of the functions en is
linearly independent. We prove this statement by induction on s . For
s D 1, suppose ˛ 2 C , n1 2 Z, and ˛en1 D 0. Evaluating at x D 0
gives 0 D ˛e i n1 0 D ˛ . This shows that fen1 g is linearly independent.
Now ﬁx s > 1 and suppose that any set consisting of fewer than s of the
functions en is linearly independent. Let n1 ; : : : ; ns be distinct integers,
˛1 ; : : : ; ˛s 2 C , and suppose that
˛1 en1 C C ˛s ens D 0: Notice that en em D enCm and e0 D 1. Also, the en are differentiable, with
.en /0 D i nen . Multiplying our equation by e n1 and rearranging gives
˛1 D ˛2 en2 n1 C C ˛s ens n1 : (3.3.1) Now we can differentiate to get
0 D i.n2 n1 /˛2 en2 n1 C C i.ns n1 /˛s ens n1 : The integers nj n1 for 2 Ä j Ä s are all nonzero and distinct, so the
induction hypothesis entails ˛2 D
D ˛s D 0. But then Equation (3.3.1)
gives ˛1 D 0 as well. ...
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