Linear Algebra

102 Chapter 2. Vector Spaces
2.II Linear Independence
The prior section shows that a vector space can be understood as an unrestricted
linear combination of some of its elements — that is, as a span. For example,
the space of linear polynomials {a + bx � � a, b ∈ R} is spanned by the set {1,x}.
The prior section also showed that a space can have many sets that span it.
The space of linear polynomials is also spanned by {1, 2x} and {1,x,2x}.
At the end of that section we described some spanning sets as ‘minimal’,
but we never precisely defined that word. We could take ‘minimal’ to mean one
of two things. We could mean that a spanning set is minimal if it contains the
smallest number of members of any set with the same span. With this meaning
{1,x,2x} is not minimal because it has one member more than the other two.
Or we could mean that a spanning set is minimal when it has no elements that
can be removed without changing the span. Under this meaning {1,x,2x} is not
minimal because removing the 2x and getting {1,x} leaves the span unchanged.
The first sense of minimality appears to be a global requirement, in that to
check if a spanning set is minimal we seemingly must look at all the spanning sets
of a subspace and find one with the least number of elements. The second sense
of minimality is local in that we need to look only at the set under discussion
and consider the span with and without various elements. For instance, using
the second sense, we could compare the span of {1,x,2x} with the span of {1,x}
and note that the 2x is a “repeat” in that its removal doesn’t shrink the span.
In this section we will use the second sense of ‘minimal spanning set’ because
of this technical convenience. However, the most important result of this book
is that the two senses coincide; we will prove that in the section after this one.
2.II.1 Definition and Examples
We first characterize when a vector can be removed from a set without
changing its span.
1.1 Lemma Where S is a subset of a vector space,
for any �v in that space.
[S] =[S ∪{�v}] if and only if �v ∈ [S]
Proof. The left to right implication is easy. If [S] = [S ∪{�v}] then, since
obviously �v ∈ [S ∪{�v}], the equality of the two sets gives that �v ∈ [S].
For the right to left implication assume that �v ∈ [S] toshowthat[S] =[S ∪
{�v}] by mutual inclusion. The inclusion [S] ⊆ [S ∪{�v}] is obvious. For the other
inclusion [S] ⊇ [S ∪{�v}], write an element of [S ∪{�v}] asd0�v+d1�s1 +···+dm�sm,
and substitute �v’s expansion as a linear combination of members of the same set
d0(c0 �t0 + ···+ ck �tk)+d1�s1 + ···+ dm�sm. This is a linear combination of linear
combinations, and so after distributing d0 we end with a linear combination of
vectors from S. Hence each member of [S ∪{�v}] isalsoamemberof[S]. QED