Linear Algebra, 2020a

Section I. Definition of Vector Space 99
is satisfied because for any ⃗s ∈ S, closure under linear combinations of pairs of
vectors shows that 0 · ⃗0 +(−1) · ⃗s is an element of S, and it is obviously the
additive inverse of ⃗s under the inherited operations. The verifications for the
scalar multiplication conditions are similar; see Exercise 35.
QED
We will usually verify that a subset is a subspace by checking that it satisfies
statement (2).
2.10 Remark At the start of this chapter we introduced vector spaces as collections
in which linear combinations “make sense.” Lemma 2.9’s statements (1)-(3) say
that we can always make sense of an expression like r 1 ⃗s 1 + r 2 ⃗s 2 in that the
vector described is in the set S.
As a contrast, consider the set T of two-tall vectors whose entries add to
a number greater than or equal to zero. Here we cannot just write any linear
combination such as 2⃗t 1 − 3⃗t 2 and be confident the result is an element of T.
Lemma 2.9 suggests that a good way to think of a vector space is as a
collection of unrestricted linear combinations. The next two examples take some
spaces and recasts their descriptions to be in that form.
2.11 Example We can show that this plane through the origin subset of R 3
⎛ ⎞
x
⎜ ⎟
S = { ⎝y⎠ | x − 2y + z = 0}
z
is a subspace under the usual addition and scalar multiplication operations
of column vectors by checking that it is nonempty and closed under linear
combinations of two vectors. But there is another way. Think of x − 2y + z = 0
as a one-equation linear system and parametrize it by expressing the leading
variable in terms of the free variables x = 2y − z.
⎛ ⎞
⎛ ⎞ ⎛ ⎞
⎜
2y − z ⎟
⎜
2 ⎟ ⎜
−1
S = { ⎝ y ⎠ | y, z ∈ R} = {y ⎝1⎠ + z ⎝ 0
z
0 1
⎟
⎠ | y, z ∈ R}
Now, to show that this is a subspace consider r 1 ⃗s 1 + r 2 ⃗s 2 . Each ⃗s i is a linear
combination of the two vectors in (∗) so this is a linear combination of linear
combinations.
⎛
⎜
2
⎞ ⎛
⎟ ⎜
−1
⎞ ⎛
⎟ ⎜
2
⎞ ⎛
⎟ ⎜
−1
⎞
⎟
r 1 · (y 1 ⎝1⎠ + z 1 ⎝ ⎠)+r 2 · (y 2 ⎝1⎠ + z 2 ⎝ ⎠)
0
0
1
The Linear Combination Lemma, Lemma One.III.2.3, shows that the total is
a linear combination of the two vectors and so Lemma 2.9’s statement (2) is
satisfied.
0
0
1
(∗)