Linear Algebra, 2020a

100 Chapter Two. Vector Spaces
2.12 Example This is a subspace of the 2×2 matrices M 2×2 .
( )
a 0
L = { | a + b + c = 0}
b c
To parametrize, express the condition as a =−b − c.
( )
( ) ( )
−b − c 0
−1 0 −1 0
L = {
| b, c ∈ R} = {b + c
b c
1 0 0 1
| b, c ∈ R}
As above, we’ve described the subspace as a collection of unrestricted linear
combinations. To show it is a subspace, note that a linear combination of vectors
from L is a linear combination of linear combinations and so statement (2) is
true.
2.13 Definition The span (or linear closure) of a nonempty subset S of a vector
space is the set of all linear combinations of vectors from S.
[S] ={c 1 ⃗s 1 + ···+ c n ⃗s n | c 1 ,...,c n ∈ R and ⃗s 1 ,...,⃗s n ∈ S}
The span of the empty subset of a vector space is its trivial subspace.
No notation for the span is completely standard. The square brackets used here
are common but so are ‘span(S)’ and ‘sp(S)’.
2.14 Remark In Chapter One, after we showed that we can write the solution
set of a homogeneous linear system as {c 1
⃗β 1 + ···+ c k
⃗β k | c 1 ,...,c k ∈ R}, we
described that as the set ‘generated’ by the ⃗β’s. We now call that the span of
{⃗β 1 ,...,⃗β k }.
Recall also from that proof that the span of the empty set is defined to
be the set {⃗0} because of the convention that a trivial linear combination, a
combination of zero-many vectors, adds to ⃗0. Besides, defining the empty set’s
span to be the trivial subspace is convenient because it keeps results like the
next one from needing exceptions for the empty set.
2.15 Lemma In a vector space, the span of any subset is a subspace.
Proof If the subset S is empty then by definition its span is the trivial subspace.
If S is not empty then by Lemma 2.9 we need only check that the span [S] is
closed under linear combinations of pairs of elements. For a pair of vectors from
that span, ⃗v = c 1 ⃗s 1 + ···+ c n ⃗s n and ⃗w = c n+1 ⃗s n+1 + ···+ c m ⃗s m , a linear
combination
p · (c 1 ⃗s 1 + ···+ c n ⃗s n )+r · (c n+1 ⃗s n+1 + ···+ c m ⃗s m )
= pc 1 ⃗s 1 + ···+ pc n ⃗s n + rc n+1 ⃗s n+1 + ···+ rc m ⃗s m