Vector Spaces C H A P T E R 2

Transcription

1 C H A P T E R 2 Vector Spaces We ow begi our treatmet of the pricipal subject matter of this text. We shall see that all of liear algebra is essetially a study of various trasformatio properties defied o a vector space, ad hece it is oly atural that we carefully defie vector spaces. This chapter therefore presets a fairly rigorous developmet of (fiite-dimesioal) vector spaces, ad a discussio of their most importat fudametal properties. Basically, the geeral defiitio of a vector space is simply a axiomatizatio of the elemetary properties of ordiary three-dimesioal Euclidea space. 2.1 DEFINITIONS A oempty set V is said to be a vector space over a field F if: (i) there exists a operatio called additio that associates to each pair x, y V a ew vector x + y V called the sum of x ad y; (ii) there exists a operatio called scalar multiplicatio that associates to each a F ad x V a ew vector ax V called the product of a ad x; (iii) these operatios satisfy the followig axioms: (V1) x + y = y + x for all x, y V. (V2) (x + y) + z = x + (y + z) for all x, y, z V. (V3) There exists a elemet 0 V such that 0 + x = x for all x V. 68

2 2.1 DEFINITIONS 69 (V4) For all x V there exists a elemet -x V such that x + (-x) = 0. (V5) a(x + y) = ax + ay for all x, y V ad all a F. (V6) (a + b)x = ax + bx for all x V ad all a, b F. (V7) a(bx) = (ab)x for all x V ad all a, b F. (V8) 1x = x for all x V where 1 is the (multiplicative) idetity i F. Note that (V1) - (V4) simply require that V be a additive abelia group. The members of V are called vectors, ad the members of F are called scalars. The vector 0 V is called the zero vector, ad the vector -x is called the egative of the vector x. We metio oly i passig that if we replace the field F by a arbitrary rig R, the we obtai what is called a R-module (or simply a module over R). If R is a rig with uit elemet, the the module is called a uital R- module. I fact, this is the oly kid of module that is usually cosidered i treatmets of liear algebra. We shall ot discuss modules i this text, although the iterested reader ca lear somethig about them from several of the books listed i the bibliography. Throughout this chapter V will always deote a vector space, ad the correspodig field F will be uderstood eve if it is ot explicitly metioed. If F is the real field, the we obtai a real vector space while if F is the complex field ç, the we obtai a complex vector space. It may be easiest for the reader to first thik i terms of these spaces rather tha the more abstract geeral case. Example 2.1 Probably the best kow example of a vector space is the set F = F ª ~ ~ ~ ª F of all -tuples (aè,..., añ) where each aá F. To make F ito a vector space, we defie the sum of two elemets (aè,..., añ) F ad (bè,..., bñ) F by (aè,..., añ) + (bè,..., bñ) = (aè + bè,..., añ + bñ) ad scalar multiplicatio for ay k F by k(aè,..., añ) = (kaè,..., kañ). If A = (aè,..., añ) ad B = (bè,..., bñ), the we say that A = B if ad oly if aá = bá for each i = 1,...,. Defiig 0 = (0,..., 0) ad -A = (-aè,..., -añ) as the idetity ad iverse elemets respectively of F, the reader should have o trouble verifyig properties (V1) - (V8). The most commo examples of the space F come from cosiderig the fields ad ç. For istace, the space 3 is (with the Pythagorea otio of

3 70 VECTOR SPACES distace defied o it) just the ordiary three-dimesioal Euclidea space (x, y, z) of elemetary physics ad geometry. We shall soo see that ay fiite-dimesioal vector space V over a field F is essetially the same as the space F. I particular, we will prove that V is isomorphic to F for some positive iteger. Example 2.2 Aother very useful vector space is the space F[x] of all polyomials i the idetermiate x over the field F (polyomials will be defied carefully i Chapter 6). I other words, every elemet i F[x] is a polyomial of the form aà + aèx + ~ ~ ~ + añx where each aá F ad is ay positive iteger (called the degree of the polyomial). Additio ad scalar multiplicatio are defied i the obvious way by a i x i + b i x i = a i + b i i=0 i=0 i=0 ( ) x i ad c a i x i = ( ca i ) x i. i=0 i=0 (If we wish to add together two polyomials Íi ˆ= 0 aáxi ad Íi = 0 báxi where m >, the we simply defie aá = 0 for i = + 1,..., m.) Sice we have ot yet defied the multiplicatio of vectors, we igore the fact that polyomials ca be multiplied together. It should be clear that F[x] does ideed form a vector space. Example 2.3 We ca also view the field ç as a vector space over. I fact, we may geerally cosider the set of -tuples (xè,..., xñ), where each xá ç, to be a vector space over by defiig additio ad scalar multiplicatio (by real umbers) as i Example 2.1. We thus obtai a real vector space that is quite distict from the space ç. We ow prove several useful properties of vector spaces that are aalogous to the properties give i Theorem 1.7 for rigs. Theorem 2.1 Let V be a vector space over F. The for all x, y, z V ad every a F we have (a) x + y = z + y implies x = z. (b) ax = 0 if ad oly if a = 0 or x = 0. (c) -(ax) = (-a)x = a(-x).

4 2.1 DEFINITIONS 71 Proof We first remark that there is a certai amout of sloppiess i our otatio sice the symbol 0 is used both as a elemet of V ad as a elemet of F. However, there should ever be ay cofusio as to which of these sets 0 lies i, ad we will cotiue with this commo practice. (a) If x + y = z + y, the implies (x + y) + (-y) = (z + y) + (-y) x + (y + (-y)) = z + (y + (-y)) which implies x + 0 = z + 0 ad hece x = z. This is frequetly called the (right) cacellatio law. It is also clear that x + y = x + z implies y = z (left cacellatio). (This is just a special case of the geeral result proved for groups i Sectio 1.1.) (b) If a = 0, the 0x = (0 + 0)x = 0x + 0x. But 0x = 0 + 0x so that 0 + 0x = 0x + 0x, ad hece (a) implies 0 = 0x. If x = 0, the a0 = a(0 + 0) = a0 + a0. But a0 = 0 + a0 so that 0 + a0 = a0 + a0, ad agai we have 0 = a0. Coversely, assume that ax = 0. If a 0 the aî exists, ad hece x = 1x = (aîa)x = aî(ax) = aî0 = 0 by the previous paragraph. (c) By (V4) we have ax + (-(ax)) = 0, whereas by (b) ad (V6), we have 0 = 0x = (a + (-a))x = ax + (-a)x. Hece ax + (-(ax)) = ax + (-a)x implies -(ax) = (-a)x by (a). Similarly, 0 = x + (-x) so that 0 = a0 = a(x + (-x)) = ax + a(-x). The 0 = ax + (-(ax)) = ax + a(-x) implies -(ax) = a(-x). I view of this theorem, it makes sese to defie subtractio i V by x - y = x + (-y). It should the be clear that a vector space will also have the properties we expect, such as a(x - y) = ax - ay, ad -(x - y) = -x + y.

5 72 VECTOR SPACES If we take a arbitrary subset of vectors i a vector space the, i geeral, this subset will ot be a vector space itself. The reaso for this is that i geeral, eve the additio of two vectors i the subset will ot result i a vector that is agai a member of the subset. Because of this, we make the followig defiitio. Suppose V is a vector space over F ad W V. The if x, y W ad c F implies x + y W ad cx W, we say that W is a subspace of V. Ideed, if c = 0 the 0 = 0x W so that 0 W, ad similarly -x = (-1)x W so that -x W also. It is ow easy to see that W obeys (V1) - (V8) if V does. It should also be clear that a equivalet way to defie a subspace is to require that cx + y W for all x, y W ad all c F. If W is a subspace of V ad W V, the W is called a proper subspace of V. I particular, W = {0} is a subspace of V, but it is ot very iterestig, ad hece from ow o we assume that ay proper subspace cotais more tha simply the zero vector. (Oe sometimes refers to {0} ad V as trivial subspaces of V.) Example 2.4 Cosider the elemetary Euclidea space 3 cosistig of all triples (x, y, z) of scalars. If we restrict our cosideratio to those vectors of the form (x, y, 0), the we obtai a subspace of 3. I fact, this subspace is essetially just the space 2 which we thik of as the usual xy-plae. We leave it as a simple exercise for the reader to show that this does ideed defie a subspace of 3. Note that ay other plae parallel to the xy-plae is ot a subspace. Example 2.5 Let V be a vector space over F, ad let S = {xè,..., xñ} be ay vectors i V. Give ay set of scalars {aè,..., añ}, the vector a i x i = a 1 x 1 ++ a x is called a liear combiatio of the vectors xá S, ad the set S of all such liear combiatios of elemets i S is called the subspace spaed (or geerated) by S. Ideed, if A = Í i ˆ=1 aáxá ad B = Í i ˆ=1 báxá are vectors i S ad c F, the both ad A + B = ca = ( a i + b i ) x i (ca i )x i are vectors i S. Hece S is a subspace of V. S is sometimes called the liear spa of S, ad we say that S spas S.

6 2.1 DEFINITIONS 73 I view of this example, we might ask whether or ot every vector space is i fact the liear spa of some set of vectors i the space. I the ext sectio we shall show that this leads aturally to the cocept of the dimesio of a vector space. Exercises 1. Verify axioms (V1) - (V8) for the space F. 2. Let S be ay set, ad cosider the collectio V of all mappigs f of S ito a field F. For ay f, g V ad å F, we defie (f + g)(x) = f(x) + g(x) ad (åf)(x) = åf(x) for every x S. Show that V together with these operatios defies a vector space over F. 3. Cosider the two elemet set {x, y} with additio ad scalar multiplicatio by c F defied by x + x = x x + y = y + x = y y + y = x cx = x cy = x. Does this defie a vector space over F? 4. Let V be a vector space over F. Show that if x V ad a, b F with a b, the ax = bx implies that x = Let (V, +, Â) be a real vector space with the additio operatio deoted by + ad the scalar multiplicatio operatio deoted by Â. Let và V be fixed. We defie a ew additio operatio by x y = x + y + và, ad a ew scalar multiplicatio operatio by å x = åâx + (å - 1)Âvà. Show that (V,, ) defies a real vector space. 6. Let F[ ] deote the space of all real-valued fuctios defied o with additio ad scalar multiplicatio defied as i Exercise 1.2. I other words, f F[ ] meas f:. (a) Let C[ ] deote the set of all cotiuous real-valued fuctios defied o. Show that C[ ] is a subspace of F[ ]. (b) Repeat part (a) with the set D[ ] of all such differetiable fuctios. 7. Referrig to the previous exercise, let D[ ] deote the set of all -times differetiable fuctios from to. Cosider the subset V of D[ ] give by the set of all fuctios that satisfy the differetial equatio

7 74 VECTOR SPACES f () (x) + a 1 f (1) (x) + a 2 f (2) (x) ++ a 1 f (1) (x) + a 0 f (x) = 0 where f(i)(x) deotes the ith derivative of f(x) ad aá is a fixed real costat. Show that V is a vector space. 8. Let V = 3. I each of the followig cases, determie whether or ot the subset W is a subspace of V: (a) W = {(x, y, 0): x, y } (see Example 2.4). (b) W = {(x, y, z) 3: z 0}. (c) W = {(x, y, z) 3: x2 + y2 + z2 1}. (d) W = {(x, y, z) 3: x + y + z = 0}. (e) W = {(x, y, z) 3: x, y, z Œ}. (f) W = {(x, y, z) 3 - {0, 0, 0}}. 9. Let S be a oempty subset of a vector space V. I Example 2.5 we showed that the liear spa S of S is a subspace of V. Show that if W is ay other subspace of V cotaiig S, the S W. 10. (a) Determie whether or ot the itersectio iˆ= 1 Wá of a fiite umber of subspaces Wá of a vector space V is a subspace of V. (b) Determie whether or ot the uio i ˆ=1 Wá of a fiite umber of subspaces Wá of a space V is a subspace of V. 11. Let Wè ad Wì be subspaces of a space V such that Wè Wì is also a subspace of V. Show that oe of the Wá is subset of the other. 12. Let Wè ad Wì be subspaces of a vector space V. If for every v V we have v = w1 + w2 where wá Wá, the we write V = W1 + W2 ad say that V is the sum of the subspaces Wá. If V = W1 + W2 ad W1 W2 = {0}, show that every v V has a uique represetatio v = w1 + w2 with wá Wá. 13. Let V be the set of all (ifiite) real sequeces. I other words, ay v V is of the form (xè, xì, x3,... ) where each xá. If we defie the additio ad scalar multiplicatio of distict sequeces compoetwise exactly as i Example 2.1, the it should be clear that V is a vector space over. Determie whether or ot each of the followig subsets of V i fact forms a subspace of V: (a) All sequeces cotaiig oly a fiite umber of ozero terms.

8 2.1 DEFINITIONS 75 (b) All sequeces of the form {x1, x2,..., xn, 0, 0,... } where N is fixed. (c) All decreasig sequeces, i.e., sequeces where xk+1 xk for each k = 1, 2,.... (d) All coverget sequeces, i.e., sequeces for which lim k Ÿ xk exists. 14. For which value of k will the vector v = (1, -2, k) 3 be a liear combiatio of the vectors xè = (3, 0, -2) ad xì = (2, -1, -5)? 15. Write the vector v = (1, -2, 5) as a liear combiatio of the vectors x1 = (1, 1, 1), x2 = (1, 2, 3) ad x3 = (2, -1, 1). 2.2 LINEAR INDEPENDENCE AND BASES Let xè,..., xñ be vectors i a vector space V. We say that these vectors are liearly depedet if there exist scalars a 1,..., a F, ot all equal to 0, such that a 1 x 1 + a 2 x 2 ++ a x = a i x i = 0. The vectors xá are said to be liearly idepedet if they are ot liearly depedet. From these defiitios, it follows that ay set cotaiig a liearly depedet subset must be liearly depedet, ad ay subset of a liearly idepedet set is ecessarily liearly idepedet. It is importat to realize that a set of vectors may be liearly depedet with respect to oe field, but idepedet with respect to aother. For example, the set ç of all complex umbers is itself a vector space over either the field of real umbers or over the field of complex umbers. However, the set {xè = 1, xì = i} is liearly idepedet if F =, but liearly depedet if F = ç sice ixè + (-1)xì = 0. We will always assume that a liear combiatio is take with respect to the same field that V is defied over. As a meas of simplifyig our otatio, we will frequetly leave off the limits of a sum whe there is o possibility of ambiguity. Thus, if we are cosiderig the set {xè,..., xñ}, the a liear combiatio of the xá will ofte be writte as Íaáxá rather tha Í iˆ=1 aáxá. I additio, we will ofte deote a collectio {x1,..., x} of vectors simply by {xi}. Example 2.6 Cosider the three vectors i 3 give by

9 76 VECTOR SPACES e 1 = (1,0,0) e 2 = (0,1,0) e 3 = (0,0,1). Usig the defiitios of additio ad scalar multiplicatio give i Example 2.1, it is easy to see that these three vectors are liearly idepedet. This is because the zero vector i 3 is give by (0, 0, 0), ad hece a1 e1 + a2 e2 + a3 e3 = (a1, a2, a3) = (0, 0, 0) implies that a1 = a2 = a3 = 0. O the other had, the vectors x 1 = (1,0,0) x 2 = (0,1,2) x 3 = (1,3,6) are liearly depedet sice x3 = x1 + 3x2. Theorem 2.2 A fiite set S of vectors i a space V is liearly depedet if ad oly if oe vector i the set is a liear combiatio of the others. I other words, S is liearly depedet if oe vector i S is i the subspace spaed by the remaiig vectors i S. Proof If S = {x1,..., x} is a liearly depedet subset of V, the a1x1 + a2x2 + ~ ~ ~ + ax = 0 for some set of scalars a1,..., a F ot all equal to 0. Suppose, to be specific, that a1 0. The we may write x1 = -(a2/a1)x2 - ~ ~ ~ - (a /a1)x which shows that xè is a liear combiatio of xì,..., xñ. Coversely, if x 1 = i"1 a i x i the xè + (-aì)xì + ~ ~ ~ + (-añ)xñ = 0 which shows that the collectio {xè,..., xñ} is liearly depedet.

10 2.2 LINEAR INDEPENDENCE AND BASES 77 It is importat to realize that o liearly idepedet set of vectors ca cotai the zero vector. To see this, ote that if S = {xè,..., xñ} ad xè = 0, the axè + 0xì + ~ ~ ~ + 0xñ = 0 for all a F, ad hece by defiitio, S is a liearly depedet set. Theorem 2.3 Let S = {xè,..., xñ} V be a liearly idepedet set, ad let S be the liear spa of S. The every v S has a uique represetatio where each aá F. v = Proof By defiitio of S, we ca always write v = Ía áxá. As to uiqueess, it must be show that if we also have v = Íbáxá, the it follows that bá = a á for every i = 1,...,. But this is easy sice Íaáxá = Íbáxá implies Í(aá - bá)xá = 0, ad hece aá - bá = 0 (sice {xá} is liearly idepedet). Therefore aá = bá for each i = 1,...,. If S is a fiite subset of a vector space V such that V = S (the liear spa of S), the we say that V is fiite-dimesioal. However, we must defie what is meat i geeral by the dimesio of V. If S V is a liearly idepedet set of vectors with the property that V = S, the we say that S is a basis for V. I other words, a basis for V is a liearly idepedet set that spas V. We shall see that the umber of elemets i a basis is what is meat by the dimesio of V. But before we ca state this precisely, we must be sure that such a umber is well-defied. I other words, we must show that ay basis has the same umber of elemets. We prove this (see the corollary to Theorem 2.6) i several steps. Theorem 2.4 Let S be the liear spa of S = {xè,..., xñ} V. If k ad {xè,..., xé} is liearly idepedet, the there exists a liearly idepedet subset of S of the form {xè,..., xé, xiè,..., xià} whose liear spa also equals S. Proof If k = there is othig left to prove, so we assume that k <. Sice xè,..., xé are liearly idepedet, we let xé (where j > k) be the first vector i S that is a liear combiatio of the precedig xè,..., xj-1. If o such j exists, the take (i1,..., iå) = (k + 1,..., ). The the set of - 1 vectors xè,..., xj-1, xj+1,..., xñ has a liear spa that must be cotaied i S (sice this set is just a subset of S). However, if v is ay vector i S, we ca write v = Í i ˆ=1 aá xá where xé is just a liear combiatio of the first j - 1 vectors. I a i x i

11 78 VECTOR SPACES other words, v is a liear combiatio of xè,..., xj-1, xj+1,..., xñ ad hece these - 1 vectors also spa S. We ow cotiue this process by pickig out the first vector i this set of - 1 vectors that is a liear combiatio of the precedig vectors. A idetical argumet shows that the liear spa of this set of - 2 vectors must also be S. It is clear that we will evetually obtai a set {xè,..., xé, xiè,..., xià} whose liear spa is still S, but i which o vector is a liear combiatio of the precedig oes. This meas that the set must be liearly idepedet (Theorem 2.2). Corollary 1 If V is a fiite-dimesioal vector space such that the set S = {xè,..., xm} V spas V, the some subset of S is a basis for V. Proof By Theorem 2.4, S cotais a liearly idepedet subset that also spas V. But this is precisely the requiremet that S cotai a basis for V. Corollary 2 Let V be a fiite-dimesioal vector space ad let {xè,..., xñ} be a basis for V. The ay elemet v V has a uique represetatio of the form where each aá F. v = Proof Sice {x á} is liearly idepedet ad spas V, Theorem 2.3 shows us that ay v V may be writte i the form v = Íi ˆ=1 aáxá where each aá F is uique (for this particular basis). It is importat to realize that Corollary 1 asserts the existece of a fiite basis i ay fiite-dimesioal vector space, but says othig about the uiqueess of this basis. I fact, there are a ifiite umber of possible bases for ay such space. However, by Corollary 2, oce a particular basis has bee chose, the ay vector has a uique expasio i terms of this basis. Example 2.7 Returig to the space F, we see that ay (aè,..., añ) F ca be writte as the liear combiatio a i x i aè(1, 0,..., 0) + aì(0, 1, 0,..., 0) + ~ ~ ~ + añ(0,..., 0, 1). This meas that the vectors

12 2.2 LINEAR INDEPENDENCE AND BASES 79 e 1 = (1,0,0,,0) e 2 = (0,1,0,,0) e = (0,0,0,,1) spa F. They are also liearly idepedet sice Íaáeá = (aè,..., añ) = 0 if ad oly if aá = 0 for all i = 1,...,. The set {e á} is extremely useful, ad will be referred to as the stadard basis for F. This example leads us to make the followig geeralizatio. By a ordered basis for a fiite-dimesioal space V, we mea a fiite sequece of vectors that is liearly idepedet ad spas V. If the sequece xè,..., xñ is a ordered basis for V, the the set {xè,..., xñ} is a basis for V. I other words, the set {xè,..., xñ} gives rise to differet ordered bases. Sice there is usually othig lost i assumig that a basis is ordered, we shall cotiue to assume that {xè,..., xñ} deotes a ordered basis uless otherwise oted. Give ay (ordered) basis {xè,..., xñ} for V, we kow that ay v V has a uique represetatio v = Í iˆ=1 aáxá. We call the scalars aè,..., añ the coordiates of v relative to the (ordered) basis {xè,..., xñ}. I particular, we call a á the ith coordiate of v. Moreover, we ow proceed to show that these coordiates defie a isomorphism betwee V ad F. Sice a vector space is also a (additive abelia) group, it is reasoable that we make the followig defiitio. Let V ad W be vector spaces over F. We say that a mappig ƒ: V W is a vector space homomorphism (or, as we shall call it later, a liear trasformatio) if ad ƒ(x + y) = ƒ(x) + ƒ(y) ƒ(ax) = aƒ(x) for all x, y V ad a F. This agrees with our previous defiitio for groups, except that ow we must take ito accout the multiplicatio by scalars. If ƒ is ijective, the we say that ƒ is a isomorphism, ad if ƒ is bijective, that V ad W are isomorphic. As before, we defie the kerel of ƒ to be the set Ker ƒ = {x V: ƒ(x) = 0 W}. If x, y Ker ƒ ad c we have

13 80 VECTOR SPACES ad ƒ(x + y) = ƒ(x) + ƒ(y) = 0 ƒ(cx) = cƒ(x) = c0 = 0. This shows that both x + y ad cx are i Ker ƒ, ad hece Ker ƒ is a subspace of V. Note also that if a = 0 ad x V the Alteratively, we could also ote that ƒ(0) = ƒ(ax) = aƒ(x) = 0. ƒ(x) = ƒ(x + 0) = ƒ(x) + ƒ(0) ad hece ƒ(0) = 0. Fially, we see that ad therefore 0 = ƒ(0) = ƒ(x + (-x)) = ƒ(x) + ƒ(-x) ƒ(-x) = -ƒ(x). Our ext result is essetially the cotet of Theorem 1.6 ad its corollary. Theorem 2.5 Let ƒ: V W be a vector space homomorphism. The ƒ is a isomorphism if ad oly if Ker ƒ = {0}. Proof If ƒ is ijective, the the fact that ƒ(0) = 0 implies that we must have Ker ƒ = {0}. Coversely, if Ker ƒ = {0} ad ƒ(x) = ƒ(y), the implies that x - y = 0, or x = y. 0 = ƒ(x) - ƒ(y) = ƒ(x - y) Now let us retur to the above otio of a ordered basis. For ay fiitedimesioal vector space V over F ad ay (ordered) basis {xè,..., xñ}, we defie a mappig ƒ: V F by # (v) = % $ " & a i x i ( = (a 1,,a ) ' for each v = a i x i " V.

14 2.2 LINEAR INDEPENDENCE AND BASES 81 Sice ad ("a i x i + "b i x i ) = ("(a i + b i )x i ) = (a 1 + b 1,,a + b ) = (a 1,,a ) + (b 1,,b ) = ("a i x i ) + ("b i x i ) (kv) = (k"a i x i ) = ("(ka i )x i ) = (ka 1,,ka ) = k(a 1,,a ) = k(v) we see that ƒ is a vector space homomorphism. Because the coordiates of ay vector are uique for a fixed basis, we see that this mappig is ideed well-defied ad oe-to-oe. (Alteratively, the idetity elemet i the space F is (0,..., 0), ad the oly vector that maps ito this is the zero vector i V. Hece Ker ƒ = {0} ad ƒ is a isomorphism.) It is clear that ƒ is surjective sice, give ay ordered set of scalars aè,..., añ F, we ca defie the vector v = Íaáxá V. Therefore we have show that V ad F are isomorphic for some, where is the umber of vectors i a ordered basis for V. If V has a basis cosistig of elemets, is it possible to fid aother basis cosistig of m elemets? Ituitively we guess ot, for if this were true the V would be isomorphic to Fm as well as to F, which implies that Fm is isomorphic to F for m. That this is ot possible should be obvious by simply cosiderig the projectio of a poit i 3 dow oto the plae 2. Ay poit i 2 is thus the image of a etire vertical lie i 3, ad hece this projectio ca ot possibly be a isomorphism. Nevertheless, we proceed to prove this i detail begiig with our ext theorem. Theorem 2.6 Let {xè,..., xñ} be a basis for V, ad let {yè,..., ym} be liearly idepedet vectors i V. The m. Proof Sice {xè,..., xñ} spas V, we may write each yá as a liear combiatio of the xé. I particular, choosig ym, it follows that the set {ym, xè,..., xñ} is liearly depedet (Theorem 2.2) ad spas V (sice the xé already do so). Hece there must be a proper subset {ym, xiè,..., xi } with r - 1 that forms a basis for V (Theorem 2.4). Now this set spas V so that ym-1 is a liear combiatio of this set, ad hece

15 82 VECTOR SPACES {ym-1, ym, xiè,..., xi } is liearly depedet ad spas V. By Theorem 2.4 agai, we ca fid a set {ym-1, ym, xjè,..., xj } with s - 2 that is also a basis for V. Cotiuig our process, we evetually obtai the set {yì,..., ym, xå, x,... } which spas V ad must cotai at least oe of the xé (sice yè is ot a liear combiatio of the set {yì,..., ym} by hypothesis). This set was costructed by addig m - 1 vectors yá to the origial set of vectors xé, ad deletig at least m - 1 of the xé alog the way. However, we still have at least oe of the xé i our set, ad hece it follows that m or m. Corollary Ay two bases for a fiite-dimesioal vector space must cosist of the same umber of elemets. Proof Let {x1,..., x} ad {y1,..., ym} be bases for V. Sice the yá are liearly idepedet, Theorem 2.6 says that m. O the other had, the xé are liearly idepedet so that m. Therefore we must have = m. We ow retur to the proof that Fm is isomorphic to F if ad oly if m =. Let us first show that a isomorphism maps a basis to a basis. Theorem 2.7 Let ƒ: V W be a isomorphism of fiite-dimesioal vector spaces. The a set of vectors {ƒ(vè),..., ƒ(vñ)} is liearly depedet i W if ad oly if the set {vè,..., vñ} is liearly depedet i V. Proof If the set {vè,..., vñ} is liearly depedet, the for some set of scalars {aè,..., añ} ot all equal to 0 we have Íi ˆ= 1 aává = 0. Applyig ƒ to both sides of this equatio yields 0 = ƒ(0) = ƒ(íaává) = ̓(aává) = Íaáƒ(vá). But sice ot all of the aá are 0, this meas that {ƒ(vá)} must be liearly depedet. Coversely, if ƒ(vè),..., ƒ(vñ) are liearly depedet, the there exists a set of scalars bè,..., bñ ot all 0 such that Íbáƒ(vá) = 0. But this meas 0 = Íbáƒ(vá) = ̓(bává) = ƒ(íbává)

16 2.2 LINEAR INDEPENDENCE AND BASES 83 which implies that Íbává = 0 (sice Ker ƒ = {0}). This shows that the set {vá} is liearly depedet. Corollary If ƒ: V W is a isomorphism of fiite-dimesioal vector spaces, the {ƒ(xá)} = {ƒ(xè),..., ƒ(xñ)} is a basis for W if ad oly if {xá} = {xè,..., xñ} is a basis for V. Proof Sice ƒ is a isomorphism, for ay vector w W there exists a uique v V such that ƒ(v) = w. If {x á} is a basis for V, the v = Íi ˆ=1 aáxá ad w = ƒ(v) = ƒ(íaáxá) = Íaáƒ(xá). Hece the ƒ(xá) spa W, ad they are liearly idepedet by Theorem 2.7. O the other had, if {ƒ(xá)} is a basis for W, the there exist scalars {bá} such that for ay v V we have ƒ(v) = w = Íbáƒ(xá) = ƒ(íbáxá). Sice ƒ is a isomorphism, this implies that v = Íbáxá, ad hece {xá} spas V. The fact that it is liearly idepedet follows from Theorem 2.7. This shows that {xá} is a basis for V. Theorem 2.8 F is isomorphic to Fm if ad oly if = m. Proof If = m the result is obvious. Now assume that F ad Fm are isomorphic. We have see i Example 2.7 that the stadard basis of F cosists of vectors. Sice a isomorphism carries oe basis oto aother (corollary to Theorem 2.7), ay space isomorphic to F must have a basis cosistig of vectors. Hece by the corollary to Theorem 2.6 we must have m =. Corollary If V is a fiite-dimesioal vector space over F, the V is isomorphic to F for a uique iteger. Proof It was show followig Theorem 2.5 that V is isomorphic to F for some iteger, ad Theorem 2.8 shows that must be uique. The corollary to Theorem 2.6 shows us that the umber of elemets i ay basis for a fiite-dimesioal vector space is fixed. We call this uique umber the dimesio of V over F, ad we write dim V =. Our ext result agrees with our ituitio, ad is quite useful i provig other theorems.

17 84 VECTOR SPACES Theorem 2.9 Every subspace W of a fiite-dimesioal vector space V is fiite-dimesioal, ad dim W dim V. Proof We must show that W has a basis, ad that this basis cotais at most = dim V elemets. If W = {0}, the dim W = 0 ad we are doe. If W cotais some xè 0, the let Wè W be the subspace spaed by xè. If W = Wè, the dim W = 1 ad we are doe. If W Wè, the there exists some xì W with xì Wè, ad we let Wì be the subspace spaed by {xè, xì}. Agai, if W = Wì, the dim W = 2. If W Wì, the choose some x3 W with x3 Wì ad cotiue this procedure. However, by Theorem 2.6, there ca be at most liearly idepedet vectors i V, ad hece dim W. Note that the zero subspace is spaed by the vector 0, but {0} is ot liearly idepedet so it ca ot form a basis. Therefore the zero subspace is defied to have dimesio zero. Fially, let us show that ay set of liearly idepedet vectors may be exteded to form a complete basis. Theorem 2.10 Let V be fiite-dimesioal ad S = {xè,..., xm} ay set of m liearly idepedet vectors i V. The there exists a set {xm+1,..., xm+r} of vectors i V such that {xè,..., xm+r} is a basis for V. Proof Sice V is fiite-dimesioal, it has a basis {vè,..., vñ}. The the set {xè,..., xm, vè,..., vñ} spas V so, by Theorem 2.4, we ca choose a subset {xè,..., xm, viè,..., vi } of liearly idepedet vectors that spa V. Lettig viè = xm+1,..., vi = xm+r proves the theorem. Exercises 1. Determie whether or ot the three vectors xè = (2, -1, 0), xì = (1, -1, 1) ad x3 = (0, 2, 3) form a basis for I each of the followig, show that the give set of vectors is liearly idepedet, ad decide whether or ot it forms a basis for the idicated space: (a) {(1, 1), (1, -1)} i 2. (b) {(2, 0, 1), (1, 2, 0), (0, 1, 0)} i 3. (c) {(1, 0, 0, 0), (1, 1, 0, 0), (1, 1, 1, 0), (1, 1, 1, 1)} i 4.

18 2.2 LINEAR INDEPENDENCE AND BASES Exted each of the followig sets to a basis for the give space: (a) {(1, 1, 0), (2, -2, 0)} i 3. (b) {(1, 0, 0, 0), (1, 0, 1, 0), (1, 0, 0, 1)} i 4. (c) {(1, 1, 0, 0), (1, -1, 0, 0), (1, 0, 1, 0)} i Show that the vectors u = (1 + i, 2i), v = (1, 1 + i) ç2 are liearly depedet over ç, but liearly idepedet over. 5. Fid the coordiates of the vector (3, 1, -4) 3 relative to the basis xè = (1, 1, 1), xì = (0, 1, 1) ad x3 = (0, 0, 1). 6. Let 3[x] be the space of all real polyomials of degree 3. Determie whether or ot each of the followig sets of polyomials is liearly idepedet: (a) {x3-3x2 + 5x + 1, x3 - x2 + 8x + 2, 2x3-4x2 + 9x + 5}. (b) {x3 + 4x2-2x + 3, x3 + 6x2 - x + 4, 3x3 + 8x2-8x + 7}. 7. Let V be a fiite-dimesioal space, ad let W be ay subspace of V. Show that there exists a subspace Wæ of V such that W Wæ = {0} ad V = W + Wæ (see Exercise 1.12 for the defiitio of W + Wæ). 8. Let ƒ: V W be a homomorphism of two vector spaces V ad W. (a) Show that ƒ maps ay subspace of V oto a subspace of W. (b) Let Sæ be a subspace of W, ad defie the set S = {x V: ƒ(x) Sæ}. Show that S is a subspace of V. 9. Let V be fiite-dimesioal, ad assume that ƒ: V V is a surjective homomorphism. Prove that ƒ is i fact a isomorphism of V oto V. 10. Let V have basis xè, xì,..., xñ, ad let vè, vì,..., vñ be ay elemets i V. Defie a mappig ƒ: V V by # & %" a i x i ( = " a i v i $ ' where each aá F. (a) Show that ƒ is a surjective homomorphism. (b) Whe is ƒ a isomorphism?

19 86 VECTOR SPACES 2.3 DIRECT SUMS We ow preset some useful ways of costructig a ew vector space from several give spaces. The reader is advised to thik carefully about these cocepts, as they will become quite importat later i this book. We also repeat our earlier remark that all of the vector spaces that we are discussig are cosidered to be defied over the same field F. Let A ad B be subspaces of a fiite-dimesioal vector space V. The we may defie the sum of A ad B to be the set A + B give by A + B = {a + b: a A ad b B}. It is importat to ote that A ad B must both be subspaces of the same space V, or else the additio of a A to b B is ot defied. I fact, sice A ad B are subspaces of V, it is easy to show that A + B is also subspace of V. Ideed, give ay aè + bè ad aì + bì i A + B ad ay k F we see that ad (aè + bè) + (aì + bì) = (aè + aì) + (bè + bì) A + B k(aè + bè) = kaè + kbè A + B as required. This defiitio ca clearly be exteded by iductio to ay fiite collectio {Aá} of subspaces. I additio to the sum of the subspaces A ad B, we may defie their itersectio A B by A B = {x V: x A ad x B}. Sice A ad B are subspaces, we see that for ay x, y A B we have both x + y A ad x + y B so that x + y A B, ad if x A B the kx A ad kx B so that kx A B. Sice 0 A B, we the see that A B is a oempty subspace of V. This ca also be exteded to ay fiite collectio of subspaces of V. Our ext theorem shows that the dimesio of the sum of A ad B is just the sum of the dimesios of A ad B mius the dimesio of their itersectio. Theorem 2.11 If A ad B are subspaces of a fiite-dimesioal space V, the dim(a + B) = dim A + dim B - dim(a B).

20 2.3 DIRECT SUMS 87 Proof Sice A + B ad A B are subspaces of V, it follows that both A + B ad A B are fiite-dimesioal (Theorem 2.9). We thus let dim A = m, dim B = ad dim A B = r. Let {uè,..., ur} be a basis for A B. By Theorem 2.10 there exists a set {vè,..., vm-r} of liearly idepedet vectors i V such that {uè,..., ur, vè,..., vm-r} is a basis for A. Similarly, we have a basis for B. It is clear that the set {uè,..., ur, wè,..., w-r} {uè,..., ur, vè,..., vm-r, wè,..., w-r} spas A + B sice ay a + b A + B (with a A ad b B) ca be writte as a liear combiatio of these r + (m - r) + ( - r) = m + - r vectors. To prove that they form a basis for A + B, we eed oly show that these m + - r vectors are liearly idepedet. Suppose we have sets of scalars {aá}, {bé} ad {cé} such that The r a i u i + b j v j + c k w k = 0 r m"r j=1 "r k=1 a i u i + b j v j = " c k w k. m"r j=1 Sice the left side of this equatio is a elemet of A while the right side is a elemet of B, their equality implies that they both belog to A B, ad hece r "r k=1 " c k w k = " d i u i k=1 for some set of scalars {dá}. But {uè,..., ur, wè,..., w-r} forms a basis for B ad hece they are liearly idepedet. Therefore, writig the above equatio as implies that We are ow left with r r d i u i + c k w k = 0 "r k=1 dè = ~ ~ ~ = dr = cè = ~ ~ ~ = c-r = 0.

21 88 VECTOR SPACES r a i u i + b j v j = 0. But {uè,..., ur, vè,..., vm-r} is also liearly idepedet so that m"r j=1 aè = ~ ~ ~ = ar = bè = ~ ~ ~ = bm-r = 0. This proves that {uè,..., ur, vè,..., vm-r, wè,..., w-r} is liearly idepedet as claimed. The proof is completed by simply otig that we have show dim(a + B) = m + - r = dim A + dim B - dim(a B). We ow cosider a particularly importat special case of the sum. If A ad B are subspaces of V such that A B = {0} ad V = A + B, the we say that V is the iteral direct sum of A ad B. A completely equivalet way of defiig the iteral direct sum is give i the followig theorem. Theorem 2.12 Let A ad B be subspaces of a fiite-dimesioal vector space V. The V is the iteral direct sum of A ad B if ad oly if every v V ca be uiquely writte i the form v = a + b where a A ad b B. Proof Let us first assume that V is the iteral direct sum of A ad B. I other words, V = A + B ad A B = {0}. The by defiitio, for ay v V we have v = a + b for some a A ad b B. Suppose we also have v = aæ + bæ where aæ A ad bæ B. The a + b = aæ + bæ so that a - aæ = bæ - b. But ote that a - aæ A ad bæ - b B, ad hece the fact that A B = {0} implies that a - aæ = bæ - b = 0. Therefore a = aæ ad b = bæ so that the expressio for v is uique. Coversely, suppose that every v V may be writte uiquely i the form v = a + b with a A ad b B. This meas that V = A + B, ad we must still show that A B = {0}. I particular, if v A B we may write v = v + 0 with v A ad 0 B, or alteratively, we may write v = 0 + v with 0 A ad v B. But we are assumig that the expressio for v is uique, ad hece we must have v = 0 (sice the cotributios from A ad B must be the same i both cases). Thus A B = {0} ad the sum is direct. We emphasize that the iteral direct sum is defied for two subspaces A ad B of a give space V. As we stated above, this is because the additio of two vectors from distict spaces is ot defied. I spite of this, we ow proceed to show that it is evertheless possible to defie the sum of two distict vector spaces.

22 2.3 DIRECT SUMS 89 Let A ad B be distict vector spaces (over the same field F, of course). While the sum of a vector i A ad a vector i B makes o sese, we may relate these two spaces by cosiderig the Cartesia product A ª B defied as (see Sectio 0.1) A ª B = {(a, b): a A ad b B}. Usig the ordered pairs (a, b), it is ow easy to tur A ª B ito a vector space by makig the followig defiitios (see Example 2.1). First, we say that two elemets (a, b) ad (aæ, bæ) of A ª B are equal if ad oly if a = aæ ad b = bæ. Next, we defie additio ad scalar multiplicatio i the obvious maer by ad (a, b) + (aæ, bæ) = (a + aæ, b + bæ) k(a, b) = (ka, kb). We leave it as a exercise for the reader to show that with these defiitios, the set A ª B defies a vector space V over F. This vector space is called the exteral direct sum of the spaces A ad B, ad is deoted by A B. While the exteral direct sum was defied for arbitrary spaces A ad B, there is o reaso why this defiitio ca ot be applied to two subspaces of a larger space V. We ow show that i such a case, the iteral ad exteral direct sums are isomorphic. Theorem 2.13 If V is the iteral direct sum of A ad B, the V is isomorphic to the exteral direct sum A B. Proof If V is the iteral direct sum of A ad B, the ay v V may be writte uiquely i the form v = a + b. This uiqueess allows us to defie the mappig ƒ: V A B by ƒ(v) = ƒ(a + b) = (a, b). Sice for ay v = a + b ad væ = aæ + bæ, ad for ay scalar k we have ad ƒ(v + væ) = (a + aæ, b + bæ) = (a, b) + (aæ, bæ) = ƒ(v) + ƒ(væ) ƒ(kv) = (ka, kb) = k(a, b) = kƒ(v) it follows that ƒ is a vector space homomorphism. It is clear that ƒ is surjective, sice for ay (a, b) A B we have ƒ(v) = (a, b) where v = a + b V. Fially, if ƒ(v) = (0, 0) the we must have a = b = 0 = v ad hece Ker ƒ =

23 90 VECTOR SPACES {0}. This shows that ƒ is also ijective (Theorem 2.5). I other words, we have show that V is isomorphic to A B. Because of this theorem, we shall heceforth refer oly to the direct sum of A ad B, ad deote this sum by A B. It follows trivially from Theorem 2.11 that dim(a B) = dim A + dim B. Example 2.8 Cosider the ordiary Euclidea three-space V = 3. Note that ay v 3 may be writte as (vè, vì, v3) = (vè, vì, 0) + (0, 0, v3) which is just the sum of a vector i the xy-plae ad a vector o the z-axis. It should also be clear that the oly vector i the itersectio of the xy-plae with the z-axis is the zero vector. I other words, defiig the space A to be the xy-plae 2 ad the space B to be the z-axis 1, we see that V = A B or 3 = 2 1. O the other had, if we try to write 3 as the direct sum of the xy-plae A with say, the yz-plae B, the the itersectio coditio is violated sice A B is just the etire y-axis. I this case, ay vector lyig o the y-axis ca be specified i terms of its compoets i either the xy-plae or i the yzplae. I may of our later applicatios we shall eed to take the direct sum of several vector spaces. While it should be obvious that this follows simply by iductio from the above case, we go through the details evertheless. We say that a vector space V is the direct sum of the subspaces Wè,..., Wr if the followig properties are true: (a) Wá {0} for each i = 1,..., r; (b) Wá (Wè + ~ ~ ~ + Wi-1 + Wi+1 + ~ ~ ~ + Wr) = {0} for i = 1,..., r; (c) V = Wè + ~ ~ ~ + Wr. If V is the direct sum of the Wá, the we write V = Wè ~ ~ ~ Wr. The geeralizatio of Theorem 2.12 is the followig. Theorem 2.14 If Wè,..., Wr are subspaces of V, the V = Wè ~ ~ ~ Wr if ad oly if every v V has a uique represetatio of the form

24 2.3 DIRECT SUMS 91 where vá Wá for each i = 1,..., r. v = vè + ~ ~ ~ + vr Proof First assume that V is the direct sum of Wè,..., Wr. Give ay v V, part (c) i the defiitio of direct sum tells us that we have v = vè + ~ ~ ~ + vr where vá Wá for each i = 1,..., r. If we also have aother represetatio with væá Wá, the v = væè + ~ ~ ~ + vær vè + ~ ~ ~ + vr = væè + ~ ~ ~ + vær so that for ay i = 1,..., r we have væi - vi = (v1 - væ1) + ~ ~ ~ + (vi-1 - væi-1) + (vi+1 - væi+1) + ~ ~ ~ + (vr - vær). Sice væá - vá Wá ad the right had side of this equatio is a elemet of Wè + ~ ~ ~ + Wi-1 + Wi+1 + ~ ~ ~ + Wr, we see that part (b) of the defiitio requires that væá - vá = 0, ad hece væá = vá. This proves the uiqueess of the represetatio. Coversely, assume that each v V has a uique represetatio of the form v = vè + ~ ~ ~ + vr where vá Wá for each i = 1,..., r. Sice part (c) of the defiitio of direct sum is automatically satisfied, we must show that part (b) is also satisfied. Suppose Sice we must also have v1 Wè (Wì + ~ ~ ~ + Wr). v1 Wì + ~ ~ ~ + Wr vè = vì + ~ ~ ~ + vr for some vì Wì,..., vr Wr. But the ad 0 = -vè + vì + ~ ~ ~ + vr 0 = 0 + ~ ~ ~ + 0

25 92 VECTOR SPACES are two represetatios of the vector 0, ad hece the uiqueess of the represetatios implies that vá = 0 for each i = 1,..., r. I particular, the case i = 1 meas that Wè (Wì + ~ ~ ~ + Wr) = {0}. A similar argumet applies to Wá (Wì + ~ ~ ~ + Wi-1 + Wi+1 + ~ ~ ~ + Wr) for ay i = 1,..., r. This proves part (b) i the defiitio of direct sum. If V = Wè ~ ~ ~ Wr, the it seems reasoable that we should be able to form a basis for V by addig up the bases of the subspaces Wá. This is ideed the case as we ow show. Theorem 2.15 Let Wè,..., Wr be subspaces of V, ad for each i = 1,..., r let Wá have basis Bá = {wáè,..., wáá}. The V is the direct sum of the Wá if ad oly if the uio of bases is a basis for V. B = i Â=1 Bá = {w11,..., w1è,..., wr 1,..., wr } Proof Suppose that B is a basis for V. The for ay v V we may write v = (a 11 w a 11 w 11 ) ++ (a r1 w r1 ++ a rr w rr ) where = w 1 ++ w r w i = a i1 w i1 ++ a ii w ii W i ad ai j F. Now let v = wæè + ~ ~ ~ + wær be ay other expasio of v, where each wæá Wá. Usig the fact that Bá is a basis for Wá we have wæá = b i 1 wi 1 + ~ ~ ~ + b iáwiá for some set of scalars bi j. This meas that we may also write v = (b11wèè + ~ ~ ~ + b1è w1è) + ~ ~ ~ + (br 1wr 1 + ~ ~ ~ + br wr ). However, sice B is a basis for V, we may equate the coefficiets of wáé i these two expressios for v to obtai ai j = bi j for all i, j. We have thus proved

26 2.3 DIRECT SUMS 93 that the represetatio of v is uique, ad hece Theorem 2.14 tells us that V is the direct sum of the Wá. Now suppose that V is the direct sum of the Wá. This meas that ay v V may be expressed i the uique form v = wè + ~ ~ ~ + wr where wá Wá for each I = 1,..., r. Give that Bá = {wi 1,..., wi á} is a basis for Wá, we must show that B = Bá is a basis for V. We first ote that each wá Wá may be expaded i terms of the members of Bá, ad therefore Bá clearly spas V. It remais to show that the elemets of B are liearly idepedet. We first write (c11w11 + ~ ~ ~ + c1èw1è) + ~ ~ ~ + (c r 1 wr 1 + ~ ~ ~ + c r wr ) = 0 ad ote that ci 1 wi 1 + ~ ~ ~ + ciáwiá Wá. Usig the fact that 0 + ~ ~ ~ + 0 = 0 (where each 0 Wá) alog with the uiqueess of the represetatio i ay direct sum, we see that for each i = 1,..., r we must have ci1 wi1 + ~ ~ ~ + ciá wiá = 0. However, sice Bá is a basis for Wá, this meas that ci j = 0 for every i ad j, ad hece the elemets of B = Bá are liearly idepedet. Corollary If V = Wè ~ ~ ~ Wr, the dim V = dim W i. r i = 1 Proof Obvious from Theorem This also follows by iductio from Theorem Exercises 1. Let Wè ad Wì be subspaces of 3 defied by Wè = {(x, y, z): x = y = z} ad Wì = {(x, y, z): x = 0}. Show that 3 = Wè Wì. 2. Let Wè be ay subspace of a fiite-dimesioal space V. Prove that there exists a subspace Wì of V such that V = Wè Wì. 3. Let W1, W2 ad W3 be subspaces of a vector space V. Show that (Wè Wì) + (Wè W3) Wè (Wì + W3).

27 94 VECTOR SPACES Give a example i V = 2 for which equality does ot hold. 4. Let V = F[ ] be as i Exercise Let W+ ad W- be the subsets of V defied by W+ = {f V: f(-x) = f(x)} ad W- = {f V: f(-x) = -f(x)}. I other words, W+ is the subset of all eve fuctios, ad W- is the subset of all odd fuctios. (a) Show that W+ ad W- are subspaces of V. (b) Show that V = W+ W-. 5. Let Wè ad Wì be subspaces of a vector space V. (a) Show that Wè Wè + Wì ad Wì Wè + Wì. (b) Prove that Wè + Wì is the smallest subspace of V that cotais both Wè ad Wì. I other words, if S(Wè, Wì) deotes the liear spa of Wè ad Wì, show that Wè + Wì = S(Wè, Wì). [Hit: Show that Wè + Wì S(Wè, Wì) ad S(Wè, Wì) Wè + Wì.] 6. Let V be a fiite-dimesioal vector space. For ay x V, we defie Fx = {ax: a F }. Prove that {xè, xì,..., xñ} is a basis for V if ad oly if V = Fxè Fxì ~ ~ ~ Fxñ. 7. If A ad B are vector spaces, show that A + B is the spa of A B. 2.4 INNER PRODUCT SPACES Before proceedig with the geeral theory of ier products, let us briefly review what the reader should already kow from more elemetary courses. It is assumed that the reader is familiar with vectors i 3, ad we show that for ay aï, b ë 3 the scalar product (also called the dot product ) aï  b ë may be writte as either a b = where {aá} ad {bá} are the coordiates of aï ad b ë relative to the stadard basis for 3 (see Example 2.7), or as where œ = (aï, b ë) ad 3 i = 1 a i b i a b = a b cos

28 2.4 INNER PRODUCT SPACES 95 3 a 2 = a 2 i with a similar equatio for b ë. The symbol is just the vector space geeralizatio of the absolute value of umbers, ad will be defied carefully below (see Example 2.9). For ow, just thik of aï as meaig the legth of the vector aï i 3. Just for fu, for the sake of completeess, ad to show exactly what these equatios deped o, we prove this as a series of simple lemmas. Our first lemma is kow as the Pythagorea theorem. Lemma 2.1 Give a right triagle with sides a, b, ad c as show, i = 1 c a b we have c2 = a2 + b2. Proof Draw the lie PQ perpedicular to the hypoteuse c = AB. Note that we ca ow write c as the sum of the two parts cè ad cì. First observe that the triagle ABP is similar to triagle APQ because they are both right triagles ad they have the agle at A i commo (so they must have their third agle the same). If we let this third agle be θ =(ABP), the we also have θ =(APQ). cì A Q cè θ a B θ b P Note that the three triagles ABP, APQ ad PBQ are all similar, ad hece we have (remember c = cè + cì) c 1 b = b c ad c 2 a = a c.

29 96 VECTOR SPACES Therefore c = c 1 + c 2 = a2 + b 2 c from which the lemma follows immediately. Our ext lemma is kow as the law of cosies. This law, together with Lemma 2.1, shows that for ay triagle T with sides a b c, it is true that a2 + b2 = c2 if ad oly if T is a right triagle. Lemma 2.2 For ay triagle as show, c a b θ we have c2 = a2 + b2-2ab cos θ. Proof Draw a perpedicular to side b as show: c h b a θ By the Pythagorea theorem we have c 2 = h 2 + (b acos") 2 = (asi") 2 + (b acos") 2 = a 2 si 2 " + b 2 2abcos" + a 2 cos 2 " = a 2 + b 2 2abcos" where we used si2 θ + cos2 θ = 1 which follows directly from Lemma 2.1 with a = c(si θ) ad b = c(cos θ). We ow defie the scalar product aï Â bë for ay aï, bë 3 by a b = 3 a i b i = b a where aï = (aè, aì, a3) ad bë = (bè, bì, b3). It is easy to see that i = 1

30 2.4 INNER PRODUCT SPACES 97 3 a ( b + c) = a i (b i + c i ) = (a i b i + a i c i ) = a b + a c i = 1 i = 1 ad similarly, it is easy to show that 3 ad (aï + bë) Â c ï = aï Â cï + bë Â cï (kaï) Â bë = k(aï Â bë) where k. From the figure below, we see that the Pythagorea theorem also shows us that 3 a 2 = a i a i = a a. x3 i = 1 xè aì aï a3 aè xì This is the justificatio for writig aï to mea the legth of the vector aï 3. Notig that ay two vectors (with a commo origi) i 3 lie i a plae, we have the followig well-kow formula for the dot product. Lemma 2.3 For ay aï, bë 3 we have aï Â bë = ab cos θ where a = aï, b = bë ad θ = (aï, b ë). Proof Draw the vectors aï ad b ë alog with their differece cï = aï - bë:

31 98 VECTOR SPACES aï θ bë cï = aï - bë By the law of cosies we have c2 = a2 + b2-2ab cos θ, while o the other had c2 = aï - bë 2 = ( aï - bë)  ( aï - bë) = a2 + b2-2 aï  bë. Therefore we see that aï  bë = ab cos θ. The mai reaso that we wet through all of this is to motivate the geeralizatio to arbitrary vector spaces. For example, if u, v, the to say that u v = i = 1 makes sese, whereas to say that u  v = u v cos œ leaves oe woderig just what the agle œ meas i higher dimesios. I fact, this will be used to defie the agle œ. We ow proceed to defie a geeral scalar (or ier) product Óu, vô of vectors u, v V. Throughout this sectio, we let V be a vector space over either the real field or the complex field ç. By way of motivatio, we will wat the ier product Ó, Ô applied to a sigle vector v V to yield the legth (or orm) of v, so that v 2 = Óv, vô. But v must be a real umber eve if the field we are workig with is ç. Notig that for ay complex umber z ç we have \z\2 = zz*, we are led to make the followig defiitio. Let V be a vector space over F (where F is either or ç). By a ier product o V (sometimes called the Hermitia ier product), we mea a mappig Ó, Ô: V ª V F such that for all u, v, w V ad a, b F we have u i v i (IP1) Óau + bv, wô = a*óu, wô + b*óv, wô; (IP2) Óu, vô = Óv, uô*; (IP3) Óu, uô 0 ad Óu, uô = 0 if ad oly if u = 0. Usig these properties, we also see that u,av + bw = av + bw,u * = (a * v,u + b * w,u ) * = a u,v + b u,w

32 2.4 INNER PRODUCT SPACES 99 ad hece, for the sake of referece, we call this (IP1æ) Óu, av + bwô = aóu, vô + bóu, wô. (The reader should be aware that istead of Óau, vô = a*óu, vô, may authors defie Óau, vô = aóu, vô ad Óu, avô = a*óu, vô. This is particularly true i mathematics texts, whereas we have chose the covetio used by most physics texts. Of course, this has o effect o ay of our results.) A space V together with a ier product is called a ier product space. If V is a ier product space over the field ç, the V is called a complex ier product space, ad if the field is, the V is called a real ier product space. A complex ier product space is frequetly called a uitary space, ad a real ier product space is frequetly called a Euclidea space. Note that i the case of a real space, the complex cojugates i (IP1) ad (IP2) are superfluous. By (IP2) we have Óu, uô so that we may defie the legth (or orm) of u to be the oegative real umber u = Óu, uô1/2. If u = 1, the u is said to be a uit vector. If v 0, the we ca ormalize v by settig u = v/ v. Oe sometimes writes vä to mea the uit vector i the directio of v, i.e., v = v vä. Example 2.9 Let X = (xè,..., xñ) ad Y = (yè,..., yñ) be vectors i ç. We defie X,Y = x i * y i ad leave it to the reader to show that this satisfies (IP1) - (IP3). I the case of the space, we have ÓX, YÔ = XÂY = Íxáyá. This ier product is called the stadard ier product i ç (or ). We also see that if X, Y the X Y 2 = X Y,X Y = (x i y i ) 2. Thus X - Y is ideed just the distace betwee the poits X = (xè,..., xñ) ad Y = (yè,..., yñ) that we would expect by applyig the Pythagorea theorem to poits i. I particular, X is simply the legth of the vector X. "

33 100 VECTOR SPACES It is ow easy to see why we defied the ier product as we did. For example, cosider simply the space ç3. The with respect to the stadard ier product o ç3, the vector X = (1, i, 0) will have orm X 2 = ÓX, XÔ = = 2, while if we had used the expressio correspodig to the stadard ier product o 3, we would have foud X 2 = = 0 eve though X 0. Example 2.10 Let V be the vector space of cotiuous complex-valued fuctios defied o the real iterval [a, b]. We may defie a ier product o V by f,g = b a f *(x)g(x)dx for all f, g V. It should be obvious that this satisfies the three required properties of a ier product. We ow prove the geeralizatio of Theorem 0.7, a importat result kow as the Cauchy-Schwartz iequality. Theorem 2.16 Let V be a ier product space. The for ay u, v V we have \Óu, vô\ u v. Proof If either u or v is zero the theorem is trivially true. We therefore assume that u 0 ad v 0. The, for ay real umber c, we have (usig (IP2) ad the fact that \z\2 = zz*) 0 v " c u,v u 2 = v " c u,v u,v " c u,v u = v,v " c u,v v,u " c u,v * u,v + c 2 u,v * u,v u,u = v 2 " 2c u,v 2 + c 2 u,v 2 u 2. Now let c = 1/ u 2 to obtai or 0 v 2 - \Óu, vô\2/ u 2 \Óu, vô\2 u 2 v 2. Takig the square root proves the theorem.