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Chapter 3

Abstract Vector Spaces

3.1 The Definition

Let’s look back carefully at what we have done. As mentioned inthebeginning,theonly algebraic or operations we have performed in Rn or Cn are vector and scalar . In fact, looking back over the proofsofallthetheorems,youmay notice that we have never explicitly performed any of these operations. In other words, a proof has never required writing out the terms of a vector andthenmultiplyingthem all by some . Rather it has been sufficient to refer to a vector as v and multiply it by a to form av.Thesamegoesforvectoraddition.Certainlywe’vedonemanyexplicit examples where we computed things like (1, 2, 3)+3(2, 1, 0),butintheproofsthese expressions always looked like v + aw. What this means is that it has never really mattered what v really was. It didn’t matter if it had entries of entries, for instance. So now we pose an important question: did it even matter that it was a list of at all? Could v have stood for something entirely unrelated? Would the theorems and their proofs still work? The short answer is yes, but with some caveats. While we didn’treallyuseany information about what v actually was, we certainly did use some information about the operations. For example, in our proofs about linear independence, we used some arithmetic properties like the fact that vector addition is commutative and associative, and the fact that scalar multiplication commutes over vectoraddition.Butthesearefacts shared by lots of operations with lots of different mathematical. Would the definitions and proofs work for them? Let’s look at one specific example. Suppose we let denote the of all continuous functions f : → R.Weknowwhatitmeanstomultiplysuchafunctionbyareal number, and we know what it means to add to such functions. Thus I can define the notion of a linear combination of functions in C:alinearcombinationofthefunctions { f1,..., fn} is a of the form a1 f1 + ···+ an fn. But now I can define what it means for a of C to be a subspace:acollectionof functions in C is a subspace of C if it is closed under lineaar combinations. And I can

21 22 CHAPTER 3. ABSTRACT VECTOR SPACES

say what it means for a set of functions to span asubspace:theset{ f1,..., fn} spans the subspace of C if every function in D is equal to a linear combination of the fj. This is fine for definitions, but what about theorems? These require proof. Look back at the proof of the theorem that the span of any set of vectors in V is a subspace of V. Now replace V with C,andreplacethev and w symbols with f and g.Everythingstill works. There is one real worry: scalars. When we talk about Rn,we’rethinkingofthescalars as being real numbers. When we talk about Cn,wethinkofscalarsasbeingcomplex numbers. This can lead to confusion when talking about linearmaps,forexample,as we need to multiply things in both the domain and the range by the same scalars. Thus an important part of the definition of a concerns precisely what numbers will play the role of scalars. To make the issue clear, we need to define the notion of a field. A field is a nonempty set K on which two operations, denoted + and ·, are defined so that the following properties hold:

(i) x + y = y + x for all x and y in K; (ii) (x + y)+z = x +(y + z) for all x, y

Your favorite examples of fields are R, C, and Q.Therearealsofinitesetsthatarefields, but we will not have time to consider these. Suppose V is a set and K is a field. We say that V is a vector space over K if there is an + between two elements of V and an operation · defined between an element of K and one of V, so that the following hold:

(i) for any u and v in V, we have that u + v is again an element of V

(ii) for any u and v in V, we have that u + v = v + u

(iii) for any u, v,andw in V,wehavethatu +(v + w)=(u + v)+w

(iv) there is some element 0 in V so that u + 0 = u

(v) for each V in V there is some element −v so that u +(−u)=0

(vi) for any u in V and a in K, we have that au is again an element of V 3.2. REVIEW 23

(vii) for any u in V and any a and b in K,wehavethata(bu)=(ab)u

(viii) for any u and v in V and any a in K we have that a(u + v)=au + av

(ix) for any u in V and any a and b in K,wehavethat(a + b)u = au + bu

(x) for any u in V we have that 1u = u

The elements of V are called vectors and the element 0 is called the zero vector.The operations are called vector addition and scalar multiplication.

Example. The set Rn of ordered n- of real numbers is a vector space over R,where vector addition is defined componentwise, as is scalar multiplication The set Cn is simi- larly a vector space over C.

Example. The set C(K) of continuous functions K → K is a vector space over K.Vector addition is function addition, and scalar multiplication istheusualscalarmultiplication of functions.

Example. The set P(K) of all in a single variable with coefficients in K is a vector space over K with the usual operations.

Example. The collection Mmn(K) of all m × n matrices with entries in K is a vector field over K, where addition and scalar multiplication are term-by-term.

Example. The set R+ of is a vector space over R,wherevector addition is multiplication and scalar multiplication is exponentiation. In other words, if u is the positive real number p and v is the positive real number q,wedefineu + v to be the positive real number pq.Ifa is any real number at all, we define au to be the positive real number pa.Youshouldcheckthatalltheconditionsabovearesatisfied.

3.2 Review

Let’s recall some of the main definitions and theorems from theprevioussectionand look at some examples in these other more general vector spaces. First we prove some basic results that we took for granted in Rn and Cn,butnow need proof.

Proposition 3.2.1. Suppose V is a vector space over a field K.

(i) for any scalar a ∈ Kwehavethata0 = 0; (ii) for any vector v ∈ V we have that 0v = 0; (iii) if av = 0,theneithera= 0 or v = 0; (iv) for any a ∈ Kandv ∈ V we have that (−a)v = a(−v)=−av. 24 CHAPTER 3. ABSTRACT VECTOR SPACES

Proof. We prove part (ii), leaving the others as exercises. From property (viii) we know that (0 + 0)v = 0u + 0v.Since0+ 0 = 0, this shows that 0v = 0v + 0v.Buproperty(v),wecanadd−(0v) to both sides and use the associativity granted by property (iii) to obtain 0 = 0v + 0.Property(iv)thenshowsthat0 = 0v.

A subspace of a vector space V is a subset W of V that is closed under vector addition and scalar multiplication. A linear combination of vectors in V is any expression of the form a1v1 + ···+ anvn. The span of a set of vectors is the set of all linear combinations of those vectors. Spans are subspaces. A set of vectors is linearly independent if none of them lies inthespanoftheothers. Otherwise the set is linearly dependent. Abasisforavectorspaceisasetoflinearlyindependentvectors that spans the space. Any two bases for the same space have the same number of vectors. Avectorspaceiscalledfinite-dimensionalifithasafinitebasis. Otherwise it is infinite dimensional. If it is finite dimensional, the dimension is defined to be the number of vectors in any basis. If a vector space has dimension n,anysetwithmorethann vectors cannot be linearly independent. Any set with fewer than n vectors cannot span. Any set of linearly independent vectors in a finite-dimensional vector space can be extended to a basis. Any finite set of vectors that span a space contains a basis for that space. Suppose V and W are two vector spaces over the same field K.Wesayafunction L: V → W is linear if L(v1 + v2)=L(v1)+L(v2) for all v1 and v2 in V,andL(av)= aL(v) for all v in V and all scalars a in K. The kernel of L is the set of all solutions to the equation L(v)=0.Itisasubspaceof V. The image of L is the set of all w in W so that the equation L(v)=w has a solution. It is a subspace of W. If V is finite-dimensional, then for any L: V →W we have that dim ker L + dim imL = dim V.

3.3 Examples

2 Example. In P(R),expressv = 3t + 5t − 5asalinearcombinationof{v1, v2, v3},where 2 2 2 v1 = t + 2t + 1, v2 = 2t + 5t + 4, and v3 = t + 3t + 6.

11 12 25 26 Example. Show that the vectors , , ,and in M (R) are lin- 11 32 64 85 2,2 ! " ! " ! " ! " early dependent, and find a basis for the subspace they span.

Example. Show that P(R) is a suspace of C(R). 3.3. EXAMPLES 25

Example. Show that the set Pn(R) of all polynomials in P(R) having degree no more than n forms a subspace of P(R).

Example. Let W denote the subspace of P(R) spanned by t3 − 2t2 + 5t − 3, 2t3 + 3t2 + t − 4, and 3t3 + 8t2 − 3t − 5. Find a basis for W,andthenextendthisbasistoonefor P3(R).

Example. Let D denote the set of all differentiable functions R → R.ShowthatD is a subspace of C(R).

Example. Show that the map D : D → C,definedbyD( f )=f &,isalinearmap.Findits kernel. Can you find a basis for its image?

Example. Verify the rank-nullity theorem for the map L: P2(R) → P2(R) defined by L(p(t)) = 2tp&(t)+p(0).

Example. Show that the set of twice-differentiable functions f such that f && + 4 f & − 3 f is the zero function forms a suspace of C.Thisisthesuperpositionprincipleforhomoge- neous linear differential equations.