A. MillerM340April 97Vector SpacesA vector space, V , is a set with two operations, vector addition (writtenu + v) and scalar multiplication (written av). The elements of V will bedenoted using u, v, w, etc. The formula ‘u ∈ V ’ is short hand for ‘u is anelement of V’ or ‘u in V’ or just ‘u is a vector’. Vector spaces will be writtenusing capital letters V, W , etc. Scalars are elements of some field, which inthis class is either the real numbers, R, or the complex numbers, C. Scalarswill be written using the letters a, b, c, etc.Closure axioms:1. If u ∈ V and v ∈ V , then u + v ∈ V .2. If u ∈ V and a a scalar, then au ∈ V .Associative, commutative, distributive axioms:1. For all u, v, w ∈ V (u + v) + w = u + (v + w).2. For all u, v ∈ V u + v = v + u.3. For all scalars a and b and vectors u ∈ V (ab)u = a(bu).4. For all scalars a and b and vectors u ∈ V (a + b)u = au + bu.5. For all scalars a and vectors u, v ∈ V a(u + v) = au + av.Zero vector, additive inverse, identity axioms:1. There exists a vector z ∈ V such that for all u ∈ V z+u = u+z = u.2. For every u ∈ V there exists a vector v ∈ V (for which we write v = −u)such that u + v = v + u = z.3. For every u ∈ V , 1u = u.1Any abstract set V with two operations, vector addition and scalar mul-tiplication which satisfy all the above axioms is a vector space.Most author’s use either 0 or~0 to denote the zero vector but studentspersistently confuse the zero vector with the zero scalar, so I decided to writethe zero vector as z.Exercise 1 Prove that 0u = z for any u ∈ V a vector space.Exercise 2 Prove that (−1)u = −u for any u ∈ V a vector space.Definition 3 For W a subset of a vector space V (written W ⊆ V )we say that W is a subspace of V iff1. for every u, v ∈ V if u ∈ W and v ∈ W , then u + v ∈ W , and2. for every u ∈ V and scalar a if u ∈ W , then au ∈ W .I usually add to the definition that W be a nonempty set, but this semesterI have decided not to worry ab out the empty set or should I say I have noworries? or should I say my set of worries is the empty set?Theorem 4 If W is a subspace of V , then W is itself a vector space underthe operations defined in V .proof:The closure axioms are easy since they are practically the same as thedefinition of subspace. The associative, commutative, distributive axiomsare true in W because they are true in V and W is a subset of V . The zerovector z is in W because 0u=z (exercise 1) so (assuming W is nonempty) ifanything is in W , then z is in W. Similarly (−1)u = −u (exercise 2), so ifu ∈ W , then also −u ∈ W .QEDTheorem 5 Suppose W is a subset of V (i.e., W ⊆ V ). Then1. W is a subspace of Viff2. for every u, v ∈ W and scalars a, b we have au + bv ∈ W .2proof:(1) implies (2):Assume W is a subspace of V . Suppose u, v ∈ W and a, b are scalars. Bythe second axiom of subspaces we have that au ∈ W and bv ∈ W . Lettingw1= au and w2= bv we have that w1∈ W and w2∈ W , therefore by thefirst axiom of subspaces we have that w1+ w2∈ W and so au + bv ∈ W.(2) implies (1):Assume (2): for every u, v ∈ W and scalars a, b we have au + bv ∈ W.We must show the two axioms of a subspace hold for W . Suppose u, v ∈ W.Then letting a = b = 1 we have that 1u + 1v ∈ W, so 1u + 1v = u + v ∈ W .For the second axiom, suppose u ∈ W and a any scalar, then we have thatau + 0u ∈ W by (2), but au = au + z = au + 0u so au ∈ W .QEDDefinition 6 For u1, . . . , unelements of a vector space V , define theirspan:span({u1, u2, . . . , un}) = {a1u1+ a2u2+ · · · + anun: a1, a2, ..., anscalars}Each of the vectors a1u1+a2u2+· · ·+anunis called a linear combination ofthe u’s so we could also say that the span is the set of all linear combinations.If W = span({u1, u2, . . . , un}), we say that ‘W is spanned by u1, u2, . . . , un’or ‘u1, u2, . . . , unspan W’. The closure axioms of a vector space V guaranteethat if u1, u2, . . . , un∈ V , then span({u1, u2, . . . , un}) ⊆ V . This is truebecause the se cond closure axiom says each aiuiis in V , while the first axiomguarantees that their sum is in V , e.g., if we write vi= aiui, v1, v2∈ V impliesv1+ v2∈ V and so v1+ v2, v3∈ V implies v1+ v2+ v3= (v1+ v2) + v3∈ V ,and so on.Theorem 7 Suppose u1, u2, . . . , unare elements of W which is a subspaceof V . Then span({u1, u2, . . . , un}) ⊆ W .proof:Suppose v ∈ span({u1, u2, . . . , un}). Then for some scalars, a1, . . . , anwehave thatv = a1u1+ · · · + anun.Since W is a subspace of V we have that aiui∈ W for each i. Now letvi= aiuito simplify our writing. Since v1∈ W and v2∈ W we have by the3first axiom of subspaces that v1+v2∈ W . Thus we have that the two vectorsv1+ v2and v3are elements of W . This means there sum (v1+ v2) + v3is inW . Continuing on like this we see that v1+ v2+ · · · + vk∈ W for each k andsov = a1u1+ · · · + anun= v1+ v2+ · · · + vn∈ Was we needed to show.QEDTheorem 8 Suppose u1, u2, . . . , unare elements of a vector space V . Thenspan({u1, u2, . . . , un}) is a subspace of V .proof:We verify each of the axioms of a subspace. LetW = span({u1, u2, . . . , un}).Suppose v, w are elements of W . Then since W is the span of the u0sthere exists scalars c1, . . . , cnand d1, . . . , dnsuch thatv = c1u1+ · · · + cnunand w = d1u1+ · · · + dnun.But thenv + w = (c1+ d1)u1+ (c2+ d2)u2+ · · · …