Page 199 - 35Linear Algebra

9.2 Building Subspaces 199

of the form
     
x 0 x
0
y
1
  + y   =  
0 0 0
is in span(S). On the other hand, any vector in span(S) must have a zero in the
z-coordinate. (Why?) So span(S) is the xy-plane, which is a vector space. (Try
drawing a picture to verify this!)

Reading homework: problem 2

Lemma 9.2.1. For any subset S ⊂ V , span(S) is a subspace of V .

Proof. We need to show that span(S) is a vector space.
It suffices to show that span(S) is closed under linear combinations. Let
u, v ∈ span(S) and λ, µ be constants. By the definition of span(S), there are
i
i
constants c and d (some of which could be zero) such that:
1
2
u = c s 1 + c s 2 + · · ·
2
1
v = d s 1 + d s 2 + · · ·
1
2
1
2
⇒ λu + µv = λ(c s 1 + c s 2 + · · · ) + µ(d s 1 + d s 2 + · · · )
1
1
2
2
= (λc + µd )s 1 + (λc + µd )s 2 + · · ·
This last sum is a linear combination of elements of S, and is thus in span(S).
Then span(S) is closed under linear combinations, and is thus a subspace
of V .
Note that this proof, like many proofs, consisted of little more than just
writing out the definitions.

Example 111 For which values of a does

     
 1 1 a 
3
1
span    2    = R ?
,
0
,
 
a −3 0
 
x
1
3
3
2
Given an arbitrary vector   in R , we need to find constants r , r , r such that
y
z
199