linear-guest

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
constants c i and d i (some of which could be zero) such that:
u = c 1 s 1 + c 2 s 2 + · · ·
v = d 1 s 1 + d 2 s 2 + · · ·
⇒ λu + µv = λ(c 1 s 1 + c 2 s 2 + · · · ) + µ(d 1 s 1 + d 2 s 2 + · · · )
= (λc 1 + µd 1 )s 1 + (λc 2 + µd 2 )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 ⎬
span ⎝0⎠ , ⎝ 2⎠ , ⎝1⎠
⎩
⎭ = R3 ?
a −3 0
⎛ ⎞
x
Given an arbitrary vector ⎝y⎠ in R 3 , we need to find constants r 1 , r 2 , r 3 such that
z
199