A brief introduction to linear algebra

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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. 1.1. Vector spaces. Motivated by our ituitio of addig ad scalig vectors i the plae (see Figure 1.1), we make the followig defiitio: Defiitio 6.1.1. 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 6.1.2 (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

64 6. A BRIEF INTRODUCTION TO LINEAR ALGEBRA v 1 + v 2 v 1 v 2 3 2 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 6.1.3. Show that (K, +, ), defied i the example above, is a K-vector space. Exercise 6.1.4 (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 6.1.5 (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 6.1.6 (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 6.1.7. 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 6.1.8. 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. LINEAR MAPS 65 2. Sub-vector spaces Defiitio 6.2.9 (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 0 1 +0 v 0 2 = v0 1 + v0 2 ad l 0 v 0 = l v 0. Defiitio 6.2.10. 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 0 1 + 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 0 1 + 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 6.2.11. Note that if (V 0, + 0, 0) is a sub-k-vector space of (V, +, ), the V 0 is closed uder + ad. Exercise 6.2.12. 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 6.2.13. 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 6.2.14. Recall the R-vector space (Map(R, R), +, ) from Exercise 6.1.8. 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 6.2.15. 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= 0. 3. Liear maps Defiitio 6.3.16 (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

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 6.3.17. 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 6.3.18. Let f : (V, +, )! (V 0, + 0, 0) be a liear map of K-vector spaces. Show that f (O) =O 0. Exercise 6.3.19. 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 6.3.20 (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 6.3.21. 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, +, ).

4. BASES AND DIMENSION 67 Exercise 6.3.22. Fid the kerel of each of the liear maps listed below (see Problem 6.3.19). (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 6.3.23. Show that the compositio of liear maps is a liear map. Defiitio 6.3.24 (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 6.3.25. 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 6.4.26. Show that L v is a liear map. 4.2. 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 6.4.27. 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.

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 6.4.28. 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 6.4.29. 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. 4.3. Dimesio. We start with the followig motivatioal exercise: Exercise 6.4.30. If K = K m, the = m. Defiitio 6.4.31. A K-vector space V is said to be of dimesio if there is a isomorphism V = K. Exercise 6.4.32. 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 6.5.33. 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 1 + 1 v1 0, v 2 + 2 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 6.5.34. 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 6.5.35 (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. FURTHER EXERCISES 69 Exercise 6.5.36. 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 2. 6. 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 6.6.37. 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 6.6.38. 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 6.6.39 (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 6.6.40. 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 6.7.41. 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).

70 6. A BRIEF INTRODUCTION TO LINEAR ALGEBRA Exercise 6.7.42. 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 6.3.19(4) with a ij = f i _ (L(e j )).

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