Linear Algebra, 2020a

98 Chapter Two. Vector Spaces
2.8 Example Being vector spaces themselves, subspaces must satisfy the closure
conditions. The set R + is not a subspace of the vector space R 1 because with
the inherited operations it is not closed under scalar multiplication: if ⃗v = 1
then −1 · ⃗v ∉ R + .
The next result says that Example 2.8 is prototypical. The only way that
a subset can fail to be a subspace, if it is nonempty and uses the inherited
operations, is if it isn’t closed.
2.9 Lemma For a nonempty subset S of a vector space, under the inherited
operations the following are equivalent statements. ∗
(1) S is a subspace of that vector space
(2) S is closed under linear combinations of pairs of vectors: for any vectors
⃗s 1 ,⃗s 2 ∈ S and scalars r 1 ,r 2 the vector r 1 ⃗s 1 + r 2 ⃗s 2 is in S
(3) S is closed under linear combinations of any number of vectors: for any
vectors ⃗s 1 ,...,⃗s n ∈ S and scalars r 1 ,...,r n the vector r 1 ⃗s 1 + ···+ r n ⃗s n is
an element of S.
Briefly, a subset is a subspace if and only if it is closed under linear combinations.
Proof ‘The following are equivalent’ means that each pair of statements are
equivalent.
(1) ⇐⇒ (2) (2) ⇐⇒ (3) (3) ⇐⇒ (1)
We will prove the equivalence by establishing that (1) =⇒ (3) =⇒ (2) =⇒ (1).
This strategy is suggested by the observation that the implications (1) =⇒ (3)
and (3) =⇒ (2) are easy and so we need only argue that (2) =⇒ (1).
Assume that S is a nonempty subset of a vector space V that is closed under
combinations of pairs of vectors. We will show that S is a vector space by
checking the conditions.
The vector space definition has five conditions on addition. First, for closure
under addition, if ⃗s 1 ,⃗s 2 ∈ S then ⃗s 1 + ⃗s 2 ∈ S, as it is a combination of a pair
of vectors and we are assuming that S is closed under those. Second, for any
⃗s 1 ,⃗s 2 ∈ S, because addition is inherited from V, the sum ⃗s 1 + ⃗s 2 in S equals the
sum ⃗s 1 + ⃗s 2 in V, and that equals the sum ⃗s 2 + ⃗s 1 in V (because V is a vector
space, its addition is commutative), and that in turn equals the sum ⃗s 2 + ⃗s 1 in
S. The argument for the third condition is similar to that for the second. For
the fourth, consider the zero vector of V and note that closure of S under linear
combinations of pairs of vectors gives that 0 · ⃗s + 0 · ⃗s = ⃗0 is an element of S
(where ⃗s is any member of the nonempty set S); checking that ⃗0 acts under the
inherited operations as the additive identity of S is easy. The fifth condition
∗ More information on equivalence of statements is in the appendix.