Linear Algebra

108 Chapter 2. Vector Spaces
Proof. One implication is clear: if �v ∈ [S] then �v = c1�s1 + c2�s2 + ···+ cn�sn
where each �si ∈ S and ci ∈ R, andso�0 =c1�s1 + c2�s2 + ···+ cn�sn +(−1)�v is a
nontrivial linear relationship among elements of S ∪{�v}.
The other implication requires the assumption that S is linearly independent.
With S ∪{�v} linearly dependent, there is a nontrivial linear relationship c0�v +
c1�s1 + c2�s2 + ···+ cn�sn = �0 and independence of S then implies that c0 �= 0,or
else that would be a nontrivial relationship among members of S. Now rewriting
this equation as �v = −(c1/c0)�s1 −···−(cn/c0)�sn shows that �v ∈ [S]. QED
(Compare this result with Lemma 1.1. Note the additional hypothesis here of
linear independence.)
1.16 Corollary A subset S = {�s1,...,�sn} of a vector space is linearly dependent
if and only if some �si is a linear combination of the vectors �s1, ... , �si−1
listed before it.
Proof. Consider S0 = {}, S1 = { �s1}, S2 = {�s1,�s2}, etc. Some index i ≥ 1is
the first one with Si−1 ∪{�si} linearly dependent, and there �si ∈ [Si−1]. QED
Lemma 1.15 can be restated in terms of independence instead of dependence:
if S is linearly independent (and �v �∈ S) then the set S ∪{�v} is also linearly
independent if and only if �v �∈ [S]. Applying Lemma 1.1, we conclude that if
S is linearly independent and �v �∈ S then S ∪{�v} is also linearly independent
if and only if [S ∪{�v}] �= [S]. Briefly, to preserve linear independence through
superset we must expand the span.
Example 1.14 shows that some linearly independent sets are maximal —
have as many elements as possible — in that they have no linearly independent
supersets. By the prior paragraph, linearly independent sets are maximal if and
only if their span is the entire space, because then no vector exists that is not
already in the span.
This table summarizes the interaction between the properties of independence
and dependence and the relations of subset and superset.
K ⊂ S K ⊃ S
S independent K must be independent K may be either
S dependent K may be either K must be dependent
In developing this table we’ve uncovered an intimate relationship between linear
independence and span. Complementing the fact that a spanning set is minimal
if and only if it is linearly independent, a linearly independent set is maximal if
and only if it spans the space.
We close with the result promised earlier that recasts Example 1.12 as a
theorem.
1.17 Theorem In a vector space, any finite subset has a linearly independent
subset with the same span.