Linear Algebra

Section I. Definition of Vector Space 93
2.6 Example All kinds of vector spaces, not just R n ’s, have subspaces. The
vector space of cubic polynomials {a + bx + cx 2 + dx 3 � � a, b, c, d ∈ R} has a subspace
comprised of all linear polynomials {m + nx � � m, n ∈ R}.
2.7 Example Another example of a subspace not taken from an R n is one
from the examples following the definition of a vector space. The space of all
real-valued functions of one real variable f : R → R has a subspace of functions
satisfying the restriction (d 2 f/dx 2 )+f =0.
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 the inherited operations
are used) 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
�s1,�s2 ∈ S and scalars r1,r2 the vector r1�s1 + r2�s2 is in S
(3) S is closed under linear combinations of any number of vectors: for any
vectors �s1,... ,�sn ∈ S and scalars r1,... ,rn the vector r1�s1 + ···+ rn�sn is
in S.
Briefly, the way that a subset gets to be a subspace is by being 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 show this equivalence by establishing that (1) =⇒ (3) =⇒ (2) =⇒
(1). This strategy is suggested by noticing that (1) =⇒ (3) and (3) =⇒ (2)
are easy and so we need only argue the single implication (2) =⇒ (1).
For that argument, assume that S is a nonempty subset of a vector space V
and that S is closed under combinations of pairs of vectors. We will show that
S is a vector space by checking the conditions.
The first item in the vector space definition has five conditions. First, for
closure under addition, if �s1,�s2 ∈ S then �s1 + �s2 ∈ S, as�s1 + �s2 =1· �s1 +1· �s2.
Second, for any �s1,�s2 ∈ S, because addition is inherited from V , the sum �s1 +�s2
in S equals the sum �s1 +�s2 in V , and that equals the sum �s2 +�s1 in V (because
V is a vector space, its addition is commutative), and that in turn equals the
sum �s2 +�s1 in S. The argument for the third condition is similar to that for the
∗ More information on equivalence of statements is in the appendix.