A brief introduction to linear algebra
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- Arnold Harrell
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1 CHAPTER 6 A brief itroductio to liear algebra 1. Vector spaces ad liear maps I what follows, fix K 2{Q, R, C}. More geerally, K ca be ay field Vector spaces. Motivated by our ituitio of addig ad scalig vectors i the plae (see Figure 1.1), we make the followig defiitio: Defiitio AK-vector space cosists of a triple (V, +, ), where V is a set, ad + : V V! V ad : K V! V are maps, satisfyig the followig properties: (1) (Group laws) (a) (Additive idetity) There exists a elemet O 2 V such that for all v 2 V, v + O = v; (b) (Additive iverse) For each v 2 V there exists a elemet v 2 V such that v +( v) =O; (c) (Associativity of additio) For all v 1, v 2, v 3 2 V, (v 1 + v 2 )+v 3 = v 1 +(v 2 + v 3 ); (2) (Abelia property) (a) (Commutativity of additio) For all v 1, v 2 2 V, v 1 + v 2 = v 2 + v 1 ; (3) (Module coditios) (a) For all l 2 K ad all v 1, v 2 2 V, l (v 1 + v 2 )=(l v 1 )+(l v 2 ); (b) For all l 1, l 2 2 K, ad all v 2 V, (l 1 + l 2 ) v =(l 1 v)+(l 2 v); (c) For all l 1, l 2 2 K, ad all v 2 V, (d) For all v 2 V, (l 1 l 2 ) v = l 1 (l 2 v); 1 v = v. I the above, for all l 2 K ad all v, v 1, v 2 2 V we have deoted +(v 1, v 2 ) by v 1 + v 2 ad (l, v) by l v. I additio, for brevity, we will ofte write lv for l v. EXAMPLE (The vector space K ). By defiitio, K = {(x 1,...,x ) : x i 2 K, 1apple i apple }. The map + : K K! K is defied by the rule (x 1,...,x )+(y 1,...,y )=(x 1 + y 1,...,x + y ) 63
2 64 6. A BRIEF INTRODUCTION TO LINEAR ALGEBRA v 1 + v 2 v 1 v v 2 v 2 O FIGURE 1. Addig ad scalig vectors i the plae for all (x 1,...,x ), (y 1,...,y ) 2 K. The map : K K! K is defied by the rule l (x 1,...,x )=(lx 1,...,lx ) for all l 2 K ad (x 1,...,x ) 2 K. Exercise Show that (K, +, ), defied i the example above, is a K-vector space. Exercise (Cacelatio rule). Let (V, +, ) be a K-vector space. Show that if we have v 1, v 2, w 2 V, the v 1 + w = v 2 + w () v 1 = v 2. Exercise (Uique additive idetity). Let (V, +, ) be a K-vector space. Fix a elemet O 2 V such that for all v 2 V, we have v + O = v. Show that if w 2 V satisfies v 0 + w = v 0 for all v 0 2 V, the w = O. Exercise (Uique additive iverse). Let (V, +, ) be a K-vector space. Let v 2 V. Fix a elemet v 2 V such that v +( v) =O. Suppose that there is w 2 V such that v + w = O. Show that w = v. Exercise Let (V, +, ) be a K-vector space. Show the followig properties hold for all v, v 1, v 2 2 V ad all l, l 1, l 2 2 K. (1) 0v = O. (2) lo = O. (3) ( l)v = (lv) =l( v). (4) If lv = O, the either l = 0 or v = O. (5) If lv 1 = lv 2, the either l = 0 or v 1 = v 2. (6) If l 1 v = l 2 v, the either l 1 = l 2 or v = O. (7) (v 1 + v 2 )=( v 1 )+( v 2 ). (8) v + v = 2v, v + v + v = 3v, ad i geeral i=1 v = v. Exercise Cosider the set of maps from a set S to K. Let us deote this set by Map(S, K). Defie additio ad multiplicatio maps ad + : Map(S, K) Map(S, K)! Map(S, K) : K Map(S, K)! Map(S, K) i the followig way. For all f, g 2 Map(S, K), set f + g to be the fuctio defied by ( f + g)(x) = f (x)+g(x) for all x 2 S. For all l 2 K ad all f 2 Map(S, K), set l f to be the fuctio defied by (l f )(x) =l f (x) for all x 2 S. Show that if S 6= the (Map(S, K), +, ) is a K-vector space.
3 3. LINEAR MAPS Sub-vector spaces Defiitio (sub-k-vector space). Let (V, +, ) be a K-vector space. A sub-kvector space of (V, +, ) is a K-vector space (V 0, + 0, 0) such that V 0 V ad such that for all v 0, v 0 1, v0 2 2 V0 ad all l 2 K, We will write (V 0, + 0, 0) (V, +, ). v v 0 2 = v0 1 + v0 2 ad l 0 v 0 = l v 0. Defiitio If (V, +, ) is a K-vector space, ad V 0 V is a subset, we say that V 0 is closed uder + (resp. closed uder ) if for all v 0 1, v0 2 2 V0 (resp. for all l 2 K ad all v 0 2 V 0 ) we have v v0 2 2 V0 (resp. l v 0 2 V 0 ). I this case, we defie + V 0 : V 0 V 0! V 0 (resp. V 0 : K V 0! V 0 ) to be the map give by v V 0v0 2 = v0 1 + v0 2 (resp. l V 0v0 = l v 0 ), for all v 0 1, v0 2 2 V0 (resp. for all l 2 K ad all v 0 2 V 0 ). REMARK Note that if (V 0, + 0, 0) is a sub-k-vector space of (V, +, ), the V 0 is closed uder + ad. Exercise Show that if a o-empty subset V 0 V is closed uder + ad, the (V 0, + V 0, V 0) is a sub-k-vector space of (V, +, ). Exercise Show that if (V 0, + 0, 0) is a sub-k-vector space of a K-vector space (V, +, ), the the additive idetity elemet O 0 2 V 0 is equal to the additive idetity elemet O 2 V. Exercise Recall the R-vector space (Map(R, R), +, ) from Exercise I this exercise, show that the subsets of Map(R, R) listed below are closed uder + ad, ad so defie sub-r-vector spaces of (Map(R, R), +, ). (1) The set of all polyomial fuctios. (2) The set of all polyomial fuctios of degree less tha. (3) The set of all fuctios that are cotiuos o a iterval (a, b) R. (4) The set of all fuctios differetiable at a poit a 2 R. (5) The set of all fuctios differetiable o a iterval (a, b) R. (6) The set of all fuctios with f (1) =0. (7) The set of all solutios to the differetial equatio f 00 + af 0 + bf = 0 for some a, b 2 R. Exercise I this exercise, show that the subsets of Map(R, R) listed below are NOT closed uder + ad, ad so do ot defie sub-r-vector spaces of (Map(R, R), +, ). (1) Fix a 2 R with a 6= 0. The set of all fuctios with f (1) =a. (2) The set of all solutios to the differetial equatio f 00 + af 0 + bf = c for some a, b, c 2 R with c 6= Liear maps Defiitio (Liear map). Let (V, +, ) ad (V 0, + 0, 0) be K-vector spaces. A liear map F : (V, +, )! (V 0, + 0, 0) is a map of sets f : V! V 0
4 66 6. A BRIEF INTRODUCTION TO LINEAR ALGEBRA such that for all l 2 K ad v, v 1, v 2 2 V, f (v 1 + v 2 )= f (v 1 )+ 0 f (v 2 ) ad f (l v) =l 0 f (v). Note that we will frequetly use the same letter for the liear map ad the map of sets. The K-vector space (V, +, ) is called the source (or domai) of the liear map ad the K-vector space (V 0, + 0, 0) is called the target (or codomai) of the liear map. The set f (V) V 0 is called the image (or rage) of f. Exercise Let f : (V, +, )! (V 0, + 0, 0) be a liear map of K-vector spaces. Show that the image of f is closed uder + 0, 0, ad so defies a sub-k-vector space of the target (V 0, + 0, 0). Exercise Let f : (V, +, )! (V 0, + 0, 0) be a liear map of K-vector spaces. Show that f (O) =O 0. Exercise Show that the followig maps of sets defie liear maps of the K-vector spaces. (1) Let (V, +, ) be a K-vector space. Show that the idetity map f : V! V, give by f (v) =v for all v 2 V, is a liear map. This liear map will frequetly be deoted by Id V. (2) Let (V, +, ) ad (V 0, + 0, 0) be K-vector spaces. Show that the zero map f : V! V 0, give by f (v) =O 0 for all v 2 V, is a liear map. (3) Let (V, +, ) be a K-vector space ad let a 2 K. Show that the multiplicatio map f : V! V give by f (v) =a v for all v 2 V is a liear map. This liear map will frequetly be deoted by a Id V. (4) Let a ij 2 K for 1 apple i apple m ad 1 apple j apple. Show that the map f : K! K m give by! f (x 1,...,x )= a 1j x j,..., a ij x j,..., a mj x j is a liear map. (5) Let (V, +, ) be the R-vector space of all differetiable real fuctios g : R! R. Let (V 0, + 0, 0) be the R-vector space of all real fuctios g : R! R. Show that the map f : (V, +, )! (V 0, + 0, 0) that seds a differetiable fuctio g to its derivative g 0 is a liear map. (6) Let (V, +, ) be the R-vector space of all cotiuous real fuctios f : R! R. Show that the map f : (V, +, )! (V, +, ) that seds a fuctio g 2 V to the fuctio f (g) 2 V determied by f (g)(x) := Z x a g(t)dt for all x 2 R is a liear map. Make sure to show that f (g) 2 V for all g 2 V. Defiitio (Kerel). Let f : (V, +, )! (V 0, + 0, 0) be a liear map of K-vector spaces. The kerel of f (or Null space of f ), deoted ker( f ) (or Null( f )), is the set ker( f ) := f 1 (O 0 )={v 2 V : f (v) =O 0 }. Exercise Let f : (V, +, )! (V 0, + 0, 0) be a liear map of K-vector spaces. Show that ker( f ) is a sub-k-vector space of (V, +, ).
5 4. BASES AND DIMENSION 67 Exercise Fid the kerel of each of the liear maps listed below (see Problem ). (1) The liear map Id V. (2) The zero map V! V 0. (3) The liear map a Id V. (4) Let a ij 2 K for 1 apple i apple m ad 1 apple j apple. The liear map f : K! K m defied by! f (x 1,...,x )= a 1j x j,..., a ij x j,..., a mj x j. (5) Let (V, +, ) be the R-vector space of all differetiable real fuctios g : R! R. Let (V 0, + 0, 0) be the R-vector space of all real fuctios g : R! R. The liear map f : (V, +, )! (V 0, + 0, 0) that seds a differetiable fuctio g to its derivative g 0. (6) Let (V, +, ) be the R-vector space of all cotious real fuctios g : R! R. Let a 2 R. The liear map f : (V, +, )! (V, +, ) that seds a fuctio g 2 V to the fuctio f (g) 2 V determied by f (g)(x) := Z x a g(t)dt for all x 2 R. Exercise Show that the compositio of liear maps is a liear map. Defiitio (Isomorphism). Let f : (V, +, )! (V 0, + 0, 0) be a liear map of K-vector spaces. We say that f is a isomorphism of K-vector spaces if there is a liear map g : (V 0, + 0, 0)! (V, +, ) of K-vector spaces such that g f = Id (V,+, ) ad f g = Id (V 0,+ 0, 0). Exercise Show that a liear map is a isomorphism if ad oly if it is bijective. 4. Bases ad dimesio 4.1. Liear maps determied by elemets of a vector space. The basic example we are iterested i is the followig. Let V be a K-vector space. We fix From this we obtai a map v =(v 1,...,v ) 2 V. L v : K! V (a 1,...,a ) 7! a i v i. i=1 Exercise Show that L v is a liear map Spa, liear idepedece, ad bases. For every permutatio s 2 S, the symmetric group o -letters, we set v s :=(v s(1),...,v s() ). Defiitio Let V be a K-vector space, ad let v 1,...,v 2 V. Set v =(v 1,...,v ). We say: (1) The elemets v 1,...,v spa V (or geerate V) if for every s 2 S, the liear map L v s is surjective.
6 68 6. A BRIEF INTRODUCTION TO LINEAR ALGEBRA (2) The elemets v 1,...,v are liearly idepedet if for every s 2 S, the liear map L v s is ijective. (3) The elemets v 1,...,v are a basis for V if for every s 2 S, the liear map L v s is a isomorphism. Exercise Let V be a K-vector space, ad let v 1,...,v 2 V. Set v =(v 1,...,v ). (1) The elemets v 1,...,v spa V (or geerate V) if for ay s 2 S, the liear map L v s is surjective. (2) The elemets v 1,...,v are liearly idepedet if for ay s 2 S, the liear map L v s is ijective. (3) The elemets v 1,...,v are a basis for V if for ay s 2 S, the liear map L v s is a isomorphism. Exercise Let V be a K-vector space, ad let v 1,...,v 2 V. (1) The elemets v 1,...,v spa V (or geerate V) if for ay v 2 V, there exists (a 1,...,a ) 2 K such that i=1 a iv i = v. (2) The elemets v 1,...,v are liearly idepedet if wheever (a 1,...,a ) 2 K ad i=1 a iv i = 0, we have (a 1,...,a )=0. (3) The elemets v 1,...,v are a basis for V if they spa V ad are liearly idepedet Dimesio. We start with the followig motivatioal exercise: Exercise If K = K m, the = m. Defiitio A K-vector space V is said to be of dimesio if there is a isomorphism V = K. Exercise Show that a K-vector space V has dimesio if ad oly if it has a basis cosistig of elemets. 5. Direct products of vector spaces EXAMPLE Suppose that (V 1, + 1, 1) ad (V 2, + 2, 2) are K-vector spaces. There is a K-vector space (V 1, + 1, 1) (V 2, + 2, 2) := (V 1 V 2, +, ) where V 1 V 2 is the product of the sets V 1 ad V 2, where is defied by ad is defied by + : (V 1 V 2 ) (V 1 V 2 )! V 1 V 2 (v 1, v 2 )+(v1 0, v0 2 )=(v v1 0, v v2 0 ) + : K (V 1 V 2 )! V 1 V 2 l (v 1, v 2 )=(l 1 v 1, l 2 v 2 ). Exercise Show that the triple (V 1, + 1, 1) (V 2, + 2, 2) := (V 1 V 2, +, ) i the example above is a K-vector space. Defiitio (Direct product). Suppose that (V 1, + 1, 1) ad (V 2, + 2, 2) are K- vector spaces. We defie the direct product of (V 1, + 1, 1) ad (V 2, + 2, 2), writte (V 1, + 1, 1) (V 2, + 2, 2), to be the K-vector space (V 1 V 2, +, ) defied above.
7 7. FURTHER EXERCISES 69 Exercise Let V 1 ad V 2 be K-vector spaces. Show the followig: (1) There is a ijective liear map i 1 : V 1! V 1 V 2 give by v 1 7! (v 1, O V2 ), ad a surjective liear map p 1 : V 1 V 2! V 1 give by (v 1, v 2 ) 7! v 1. (2) There is a ijective liear map i 2 : V 1! V 1 V 2 give by v 2 7! (O V1, v 2 ), ad a surjective liear map p 2 : V 1 V 2! V 2 give by (v 1, v 2 ) 7! v Quotiet vector spaces Suppose that (V, +, ) is a K-vector space, ad W V is a sub-k-vector space. Defie a equivalece relatio o V by the rule v 1 v 2 () v 1 v 2 2 W. Exercise Show that this defies a equivalece relatio o V. Let V/W be the set of equivalece classes, ad let p : V! V/W be the quotiet map of sets. For ay elemet 2 V/W, there is a elemet v 2 V such that =[v], where [v] is the equivalece class of v. Exercise Let V be a K-vector space ad suppose that W V is a sub-k-vector space. (1) Suppose that [v 1 ], [v 2 ] 2 V/W. Show that the rule defies a map [v 1 ]+[v 2 ]=[v 1 + v 2 ] + : V/W V/W! V/W. (2) Suppose that l 2 K ad [v] 2 V/W. Show that the rule defies a map l [v] =[l v] : K V/W! V/W. (3) Show that V/W is a K-vector space with + ad defied as above. (4) Show that p : V! V/W is a surjective liear map with kerel W. Defiitio (Quotiet K-vector space). Let V be a K-vector space ad let W V be a sub-k-vector space. The quotiet (K-vector space) of V by W is the K-vector space V/W costructed above. Exercise Suppose that f : V V 0 is a surjective liear map of K-vector spaces. (1) Show that V 0 = V/ ker f. (2) If V 0 is fiite dimesioal, show that V = (ker f) V 0. (3) If V ad V 0 are fiite dimesioal, show that dim V = dim V 0 + dim(ker f). 7. Further exercises Exercise Fid a example of a triple (V, +, ) satisfyig all of the coditios of the defiitio of a K-vector space, except for coditio (3)(d).
8 70 6. A BRIEF INTRODUCTION TO LINEAR ALGEBRA Exercise Suppose that L : K! K m is a liear map. For j = 1,..., defie e j =(0,..., 1,..., 0) 2 K to be the elemet with all etries 0 except for the j-th place, which is 1. Similarly, for i = 1,..., m defie f i _ : K m! K to be the liear map defied by (y 1,...,y m ) 7! y i. Show that L is the same as the liear map defied i Example (4) with a ij = f i _ (L(e j )).
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