Linear Algebra

Section I. Definition of Vector Space 932.9 Lemma For a nonempty subset S of a vector space, under the inheritedoperations, 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 vectors⃗s 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 anyvectors ⃗s 1 , . . . , ⃗s n ∈ S and scalars r 1 , . . . , r n the vector r 1 ⃗s 1 + · · · + r n ⃗s n isin S.Briefly, the way that a subset gets to be a subspace is by being closed underlinear combinations.Proof. ‘The following are equivalent’ means that each pair of statements areequivalent.(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 Vand that S is closed under combinations of pairs of vectors. We will show thatS is a vector space by checking the conditions.The first item in the vector space definition has five conditions. First, forclosure under addition, if ⃗s 1 , ⃗s 2 ∈ S then ⃗s 1 + ⃗s 2 ∈ S, as ⃗s 1 + ⃗s 2 = 1 · ⃗s 1 + 1 · ⃗s 2 .Second, for any ⃗s 1 , ⃗s 2 ∈ S, because addition is inherited from V , the sum ⃗s 1 +⃗s 2in S equals the sum ⃗s 1 +⃗s 2 in V , and that equals the sum ⃗s 2 +⃗s 1 in V (becauseV is a vector space, its addition is commutative), and that in turn equals thesum ⃗s 2 +⃗s 1 in S. The argument for the third condition is similar to that for thesecond. For the fourth, consider the zero vector of V and note that closure of Sunder linear combinations of pairs of vectors gives that (where ⃗s is any memberof the nonempty set S) 0 · ⃗s + 0 · ⃗s = ⃗0 is in S; showing that ⃗0 acts under theinherited operations as the additive identity of S is easy. The fifth condition issatisfied because for any ⃗s ∈ S, closure under linear combinations shows thatthe vector 0 · ⃗0 + (−1) · ⃗s is in S; showing that it is the additive inverse of ⃗sunder the inherited operations is routine.The checks for item (2) are similar and are saved for Exercise 32. QEDWe usually show that a subset is a subspace with (2) =⇒ (1).2.10 Remark At the start of this chapter we introduced vector spaces ascollections in which linear combinations are “sensible”. The above result speaksto this.The vector space definition has ten conditions but eight of them — the conditionsnot about closure — simply ensure that referring to the operations as an‘addition’ and a ‘scalar multiplication’ is sensible. The proof above checks thatthese eight are inherited from the surrounding vector space provided that the∗ More information on equivalence of statements is in the appendix.