Chapter 3. Abstract Vector Spaces. 3.1 The Definition

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1 Chapter 3 Abstract Vector Spaces 3.1 The Definition Let s look back carefully at what we have done. As mentioned in thebeginning,theonly algebraic or arithmetic operations we have performed in R n or C n are vector addition and scalar multiplication. In fact, looking back over the proofs ofallthetheorems, 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 and thenmultiplyingthem all by some number. 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 real number entries of complex number entries, for instance. So now we pose an important question: did it even matter that it was a list of numbers 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 basic arithmetic properties like the fact that vector addition is commutative and associative, and the fact that scalar multiplication commutes over vector addition.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 C denote the set of all continuous functions f : R R. We know what it means to multiply such a function by a real 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: a linear combination of the functions { f 1,..., f n } is a function of the form a 1 f a n f n. But now I can define what it means for a subset 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

2 22 CHAPTER 3. ABSTRACT VECTOR SPACES say what it means for a set of functions to span asubspace:theset{ f 1,..., f n } spans the subspace D of C if every function in D is equal to a linear combination of the f j. 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 R n,we rethinkingofthescalars as being real numbers. When we talk about C n,wethinkofscalarsasbeingcomplex numbers. This can lead to confusion when talking about linear maps,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 vector space 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< andz in K; (iii) x + 0 = x for all x in K; (iv) for every x in K, thereisanumber x in K so that x +( x) =0; (v) x y = y x for all x and y in K; (vi) (x y) z = x (y z) for all x, y, and z in K; (vii) x 1 = x for all x in K; (viii) for every x = 0 in K there is a number x 1 in K so that x x 1 = 1; (ix) x (y + z) =x y + x z for all x, y, andz in K. 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 operation + 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 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. operations are called vector addition and scalar multiplication. The Example. The set R n of ordered n-tuples of real numbers is a vector space over R, where vector addition is defined componentwise, as is scalar multiplication The set C n is similarly 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 is theusualscalarmultiplication of functions. Example. The set P(K) of all polynomials in a single variable with coefficients in K is a vector space over K with the usual operations. Example. The collection M mn (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 positive real numbers 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 p a.youshouldcheckthatalltheconditionsabovearesatisfied. 3.2 Review Let s recall some of the main definitions and theorems from the previoussectionand look at some examples in these other more general vector spaces. First we prove some basic results that we took for granted in R n and C n,butnow need proof. Proposition 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.

4 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 a 1 v a n v n. 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 in thespanoftheothers. 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. We say a function L : V W is linear if L(v 1 + v 2 )=L(v 1 )+L(v 2 ) for all v 1 and v 2 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 linear map L : V W we have that dim ker L + dim iml = dim V. 3.3 Examples Example. In P(R), expressv = 3t 2 + 5t 5asalinearcombinationof{v 1, v 2, v 3 },where v 1 = t 2 + 2t + 1, v 2 = 2t 2 + 5t + 4, and v 3 = t 2 + 3t + 6. [ ] [ ] [ ] [ ] Example. Show that the vectors,,,and in M ,2 (R) are linearly dependent, and find a basis for the subspace they span. Example. Show that P(R) is a suspace of C(R).

5 3.3. EXAMPLES 25 Example. Show that the set P n (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 t 3 2t 2 + 5t 3, 2t 3 + 3t 2 + t 4, and 3t 3 + 8t 2 3t 5. Find a basis for W, andthenextendthisbasistoonefor P 3 (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 : P 2 (R) P 2 (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. Thisisthesuperpositionprincipleforhomogeneous linear differential equations.

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