42 Dependence and Bases

Transcription

1 42 Depedece ad Bases The spa s(a) of a subset A i vector space V is a subspace of V. This spa ay be the whole vector space V (we say the A spas V). I this paragraph we study subsets A of V which spa V ad which are ost ecooical i the sese that ay proper subset of A spas a proper subspace of V. We begi with a defiitio that will be iportat for everythig i the sequel Defiitio: Let V be a vector space over a field K.. A fiite uber of vectors i V are called liearly depedet over K if there are scalars i K ot all of the beig zero such that = 0 (here 0 is the zero vector). If are ot liearly depedet over K the are said to be liearly idepedet over K. A fiite subset A of V is called liearly depedet (resp. liearly idepedet) over K if the fiitely ay vectors i A are liearly depedet (resp. liearly idepedet) over K. A ifiite subset A of V is called liearly depedet over K if there is a fiite subset of A which is liearly depedet over K. A ifiite subset A of V is called liearly idepedet over K if A is ot liearly depedet over K i.e. A is called liearly idepedet over K if every fiite subset of A is liearly idepedet over K. I place of the phrase "liearly (i)depedet over K" we shall also use the expressio "K-liearly (i)depedet". Whe the field of scalars is clear fro the cotext we drop the phrase "over K" or the prefix "K-". Accordig to our defiitio the vectors 1 K are liearly idepedet over K provided of a vector space over 500

2 K = 0 1 = 2 =... = = 0. That is to say are K-liearly idepedet if the vector zero ca be writte as a liear cobiatio of oly i the trivial way where the scalars are zero Exaples: (a) Let V be a vector space over a field K ad let be a ozero vector i V. The = 0 iplies = 0 (Lea 39.4(10)). Hece (ad { }) is liearly idepedet over K. O the other had {0} is liearly depedet over K because 1.0 = 0 ad 1 0. (b) Cosider the vector space 3 over. The vectors = (100) = (010) = (001) of 3 are liearly idepedet over for if K ad = 0 the (100) (010) (001) = (000) ( 00) (0 0) (00 ) = (000) ( ) = (000) = = = 0. (c) More geerally the vectors 1 = ( ) 2 = ( )... = ( ) i the vector space K over a field K are liearly idepedet over K: if K ad = 0 the 1 ( ) 2 ( )... ( ) = ( ) ) ( )... (00... ) = ( ) 2... ) = ( ) 1 = 2 =... = = 0. The reader is probably acquaited with the vectors i the vector space 3 over uder the aes i j. (d) The vectors (10) ad 0) i the -vector space 2 are liearly depedet over because 1 0 i ad 1(10) 10) = (00) = zero vector i 2. (e) Let V be a vector space over a field K ad let be vectors i V which are liearly idepedet over K.. The ay oepty subset of { } is liearly idepedet over K. I fact if say are liearly depedet over K ( ) the there are scalars i K ot all equal to zero such that = 0; 501

3 hece whe we put (i case ) 1 =... = = 0 we obtai = 0 where ot all of are equal to zero cotradictig the assuptio that are liearly idepedet over K. Thus ay oepty subset of a liearly idepedet fiite set of vectors is liearly idepedet. But this stateet is true also for ifiite liearly idepedet sets. Ideed let A be a ifiite liearly idepedet subset of V ad let B be a oepty subset of A. If B is fiite the B is liearly idepedet by defiitio. If B is ifiite the ay fiite subset of B beig a fiite subset of A is liearly idepedet over K ad hece B itself is liearly idepedet over K. Thus we have show that every oepty subset of a liearly idepedet set of vectors is liearly idepedet. Equivaletly ay set of vectors cotaiig a liearly depedet subset is liearly depedet. (f) Let V be a vector space over a field K ad let A be a subset of V cotaiig 0 V. The A is liearly depedet over K by Exaple 42.2(a) ad Exaple 42.2(e). Alteratively just choose a fiite uber of vectors fro A icludig 0 say. = 0 ad observe that = 0 so that are liearly depedet over K ad cosequetly A too is liearly depedet over K. (g) Let V be the vector space liearly depedet over because. 2 over. The vectors (10) ( i0) i V are i(10) 1( i0) = (00) = zero vector i V. However whe V is regarded as a are ot liearly depedet: if -vector space these two vectors ad (10) ( i0) = (00) the ( i0) = (00) hece the coplex uber i is equal to 0 so = = 0. Thus (10) ( i0) are liearly depedet over but liearly idepedet over. This exaple shows that the field of scalars ust be specified (uless it is clear fro the cotext) wheever oe discusses liear (i)depedece of vectors. (h) Let V be a vector space over a field K ad let 1... be ifiitely 2 ay vectors i V. The liear depedece of 1... does ot ea that there are scalars ot all equal to zero such that =1 =

4 This equatio is eaigless for its left had side is ot defied. What is defied (Defiitio 8.4) is a su =1 vectors i V. The defiitio of =1 of a fiite uber of would ivolve soe liitig process ad this is ot possible i a arbitrary vector space. (i) Cosider the vector space C 1 ([01]) over (Exaple 40.6(j)). The fuctios f: [01] ad g: [01] where f(x) = e x ad g(x) = e 2x for all x i [01] are vectors i C 1 ([01]).We clai that f ad g are liearly idepedet over. To prove this let us assue ad f g = zero vector i C 1 ([01]). The zero vector i C 1 ([01]) is the fuctio z: [01] such that z(x) = 0 for all x i [01]. Hece Differetiatig we obtai ( f g)(x) = 0 for all x [01] f(x) g(x) = 0 for all x [01] e x e 2x = 0 for all x [01]. e x 2 e 2x = 0 for all x [01]. We have thus e 2x = e x = 2 e 2x for all x [01] hece = 0 so = 0. Therefore f ad g are liearly idepedet over. (j) Let V be a vector space over a field K ad let 1 2 be vectors i V which are liearly depedet over K. The there are scalars K ot both zero such that 1 2 = 0. If say 0 the has a iverse 1 i K ad we obtai 1 ( 1 ) 2 = 1 ( 1 2 ) = 1 0 = 0 so 1 = 2 if we put = 1. So 1 is a scalar ultiple of 2. Coversely if 1 ad 2 are vectors i V ad if oe of the is a scalar ultiple of the other. for istace if 1 = 2 with soe K the 1 1 ( ) 2 = 0 ad 1 2 are liearly depedet over K.. Thus the liear depedece of two vectors eas that oe of the is a scalar ultiple of the other.. We geeralize the last exaple Lea: Let V be a vector space over a field K ad let be vectors i V where 2. These vectors are liearly depedet over K if ad oly if oe of the is a K-liear cobiatio of the other vectors. 503

5 Proof: We first assue that are liearly depedet over K. The there are scalars ot all of the zero such that = 0. To fix the ideas let us suppose 1 0. The 1 has a iverse 1 1 i K ad we obtai 1 ( ) = 1 0 = = 0 where we put j = 1 1 the vectors j 1 = K (j = 2... ). So 1 is a K-liear cobiatio of Coversely let us suppose that oe of the vectors for exaple 1 is a liear cobiatio of the rest so that there are scalars 2... K such that 1 = The we get = 0 whe we write 1 = 1. Sice 1 = 1 0 we see that are liearly depedet over K. A vector space over a field K ca be spaed by ay subsets of V. Aog the subsets of V which spa V we wat to fid the oes with the least uber of eleets. The ext two theores which are coverses of each other tell us that liearly depedet subsets are ot useful for this purpose Theore: Let V be a vector space over a field K ad let A be a oepty subset of V. If A is liearly depedet over K the there is a proper subset B of A such that (A) = (B).. Proof: Suppose A i-liearly depedet. If A is ifiite the by defiitio there is a fiite liearly depedet subset A 0 of A. If A is fiite let us put A 0 = A. Hece i both cases A 0 is a fiite liearly depedet subset of A. Let A 0 = { }. We first dispose of the trivial case A 0 = 1 V = A 0. I this case we have = 0 ad A 0 = { 0 } so 0 = 0 by Exaple 42.2(a) so A 0 = {0}. Thus {0} = A 0 A (A) V = A 0 = {0} ad (A) is equal to the K-spa of the proper subset B = of A. 504

6 Suppose ow A 0 1 or A 0 = 1 but V A 0. The we ay ad do joi ozero vectors to A without disturbig the liear 0 depedece ad fiiteess of A 0. Oe of the vectors which we ay assue to be 0 without loss of geerality is a K-liear cobiatio of the others (Lea 42.3). So there are scalars such that 0 =. We will show that 0 is redutat. We put B = A\{ 0 }. The B is a proper subset of A ad (B) (A). We prove (A) (B). Let (A). The there are vectors i A ad scalars i K such that =. Here we ay suppose that are pairwise distict (if 1 = 2 we write 2 ) 1 istead of etc.). If oe of the vectors 1 cobiatio of the vectors 2... is equal to the is a K-liear i B so (B). If oe of the vectors is equal to 0 for istace if 1 = 0 the we have = = ) = so is a K-liear cobiatio of the vectors i B = A\{ 0 } (soe i ight equal a but this does ot atter) so (B). j I both cases (B). Thus (A) (B) ad (A) = (B) as was to be proved Theore: Let V be a vector space over a field K ad let A be a oepty subset of V. If there is a proper subset B of A such that (B) = (A) the A is liearly depedet over K. Proof: We first dispose of the trivial case B =. If B = the A (A) = (B) = ( ) = {0}. gives A = {0} ad A i-liearly depedet by Exaple 42.2(a). Suppose ow B. Sice B A there is a vector i A\B. Fro 505

7 A (A) = (B) we coclude that there are vectors i B ad scalars i K with = So the vector i A is a K-liear cobiatio of the vectors ad the subset { } of A i-liearly depedet by Lea Fro Exaple 42.2(e) it follows that A i-liearly depedet. The last two theores lead us to cosider liearly idepedet subsets of V spaig V. Whether a arbitrary vector space does have such a subset will be discussed later. We give a ae to the subsets i questio Defiitio: Let V be a vector space over a field K. A oepty subset B of V is called a basis of V over K or a K-basis of V if B is liearly idepedet over K ad spas V over K (i.e. (B) = V).. By covetio the epty set will be called a K-basis of the vector space {0} Exaples: (a) Cosider the vector space K over a field K. The vectors 1 = ( ) 2 = ( )... = ( ) are liearly idepedet over K (Exaple 42.2(c)). Moreover { } spa over K because ay vector 2... ) i K is a K-liear cobiatio of the vectors Hece { } is a basis of K over K. (b) Let V = {h C 2 ([01]): h (x) 3h (x) 2h(x) = 0 for all x [01]}. The V is a -subspace of C 2 ([01]) as ca be verified directly ad also follows fro Exaple 40.6(). Fro the theory of ordiary differetial equatios it is ow that every fuctio i V (that is every solutio of y 3y 2 = 0) ca be writte i the for c 1 f c 2 g where c 1 c 2 ad f(x) = e x g(x) = e 2x for all x [01]. Thus {fg} spas V over. Also {fg} is liearly idepedet over basis of V. by Exaple 42.2(i). Hece {fg} is a 506

8 42.8 Theore: Let V be a vector space over a field K ad let B. = { } be a oepty subset of V. The B is a K-basis of V if ad oly if every eleet of V ca be writte i the for i a uique way (i.e. with uique scalars ) Proof: Assue first that B is a K-basis of V. The V = (B) ad every eleet of V ca be writte as with suitable scalars We are to show the uiqueess of this represetatio. I other words we ust prove that if = (1) the 1 = 1 2 = 2... =. This is easy: if (1) holds the 1 ) 1 ( 2 2 ) 2... ( ) = 0 ad we obtaisice B = { = 2 2 =... =. K) 2... } i-liearly idepedet that = 0. This proves uiqueess. Coversely let us suppose that every vector i V ca be writte i the for with uique scalars The V = 2... ) = (B). Moreover B is liearly idepedet over K for if K are scalars such that. = 0 the = ad the uiqueess of the scalars. i the represetatio of 0 V as a K- liear cobiatio of iplies that 1 = 2 =... = = 0. Thus B i-liearly idepedet ad cosequetly B is a K-basis of V.. We prove ext that ay fiitely spaed vector space has a basis Theore:. Let V be a vector space over a field K ad assue T is a fiite subset of V spaig V so that (T) = V. The V has a fiite K- basis. I fact a suitable subset of T is a K-basis of V.. 507

9 Proof: If V happes to be the vector space {0}. the V has a K-basis aely the epty set (Defiitio 42.6). ad T. Havig disposed of this degeerate case let us assue V 0. Now (T) = V. Sice V 0 we have T.. If T is liearly idepedet over K the T is a K-basis of V. Otherwise there is a proper subset T 1 of T with (T 1 ) = (T) = V (Theore 42.4). Here T 1 because (T 1 ) = V {0}. If T 1 is liearly idepedet over K the T 1 is a K-basis of V. Otherwise there is a proper subset T 2 of T 1 with (T 2 ) = (T 1 ) = V. Here T 2 because (T 2 ) = V {0}. If T 2 is liearly idepedet over K the T 2 is a K-basis of V. Otherwise there is a proper subset T 3 of T 2 with (T 3 ) = (T 2 ) = V. Here T 3 because (T 3 ) = V {0}. We cotiue i this way. Each tie we get a oepty subset T i1 of T i such that (T i1 ) = V ad T i1 has less eleets tha T i. Sice T is a fiite set this process caot go o idefiitely.. Sooer or later we will eet a K-liearly idepedet subset T of T with (T ) = V. This T is therefore a K-basis of V ad of course T is fiite. Havig coviced ourselves of the existece of bases i soe vector spaces we tur our attetio to the uber of vectors i a fiite basis. We show that the uber of liearly idepedet vectors i a subspace caot exceed the uber of vectors spaig the subspace. This theore due to E. Steiitz ( ) is the source of ay deep results cocerig the diesio of a vector space. The idea is to replace soe vectors i the spaig set by the vectors i the liearly idepedet set without chagig the spa Theore (Steiitz' replaceet theore):. Let V be a vector space over a field K ad be fiitely ay vectors i V. Let be liearly idepedet vectors i the K-spa 2... ) of The. Moreover there are vectors aog which we ay assue to be such that ) = 2... ). Proof: For 1 h let A h be the assertio "there are h vectors aog say 1... such that h 508

10 ... h h1... ) =... h h1... )". We show that (1) A 1 is true (2) if 2 h ad A h 1 is true the A h is true. This will establish A 1 A 2... A 1 A. The secod clai A i the theore will be proved i this way.. (1) A 1 is true. We have ) so 1 = with soe scalars K. Sice are liearly idepedet over K 0 (Exaple 42.2(f)) hece ot all of are equal to 0 K. So oe of the is distict fro 0. Reaig if ecessary we ay suppose 1 0. The 1 has a iverse 1 1 i K ad we get 1 ( = 1 ) ) { } 2... ). (i) Sice ) we also have { } 2... ). (ii) Usig (i) ad (ii) ad applyig Lea (with A = { } ad B = { } we obtai 2... ) = 2... ). This proves A 1. (2) Suppose 2 h ad A h 1 is true. The A h is true. The truth of A h 1 eas... h 1 h... ) = 2... ) provided the 's are idexed suitably. We have i h... h 1 h... ) h... h 1 h... ) so h = h 1 h 1 h h... for soe appropriate 1... h 1 h... K. Here ot all of h... equal to 0 K for the h would be a K-liear cobiatio are h 1 h 1 of the vectors 1... h 1 ad the vectors 1... h 1 h would ot be liearly idepedet over K (Lea 42.3) so would ot be liearly idepedet over K (Exaple 42.2(e)). cotrary to the hypothesis. So oe of h... is distict fro 0. Reaig h... if ecessary we ay suppose 0. The h h has a iverse 1 h i K ad we get h h = h 1 h 1 h h1 h

11 h = 1 ( h h 1 h 1 h h1 h1... ) s h K... h 1 h h1... ). Now each oe of the vectors 1... beig a eleet of the spa 2... ) =... h 1 h... ) ca be writte i the for h 1 h 1 h h h1 h1... with scalars 1... h 1 h h1... K. Thus each oe of 1... h 1 ca be writte as h h 1 h 1 h ( h h 1 h 1 h h1 h1... h1 h1... )) ad so { 1... Therefore { 1... h 1 }... h 1 h h1... ). h 1 h h1... }... h 1 h h1... ). (i ) Sice we also have { h 1 h... h 1 h h1... ) h 1 h h1... }... h 1 h h1... ). (ii ) Usig (i ) ad (ii ) ad applyig Lea with A = { 1... h 1 h h1... } B = { 1... h 1 h h1... } we obtai... h 1 h h1... ) = 2... ). Thus A h is true. As reared earlier this establishes the truth of A 1 A 2... A 1 A. I particular A is true ad the secod stateet i the euciatio is proved. Now it reais to establish.. If we had the A would be true ad we would get 2... ) = 2... ). The 2... ) would give K 2... ) cotrary to the hypothesis that are liearly idepedet over K.. So is ipossible ad ecessarily. This copletes the proof.. 510

12 42.11 Theore: Let V be a vector space over a field K ad assue that V has a fiite K-basis. The ay two K-bases of V have the sae uber of eleets. Proof: There is a fiite K-basis of V by hypothesis say B.. Assue that B has exactly vectors ( 0). We prove that ay K-basis B 1 of V has also vectors i it. If = 0 the B = ad V = {0}. Thus ad {0} are the oly subsets of V ad B = is the oly K-basis of V. The ay K-basis of V has exactly 0 eleets.. Suppose ow 1 ad let B 1 be ay K-basis of V. First we show that B 1 caot be ifiite. Otherwise B 1 would be a ifiite K-liearly idepedet subset of V. Every fiite subset of B 1 would be K-liearly idepedet by defiitio. Let be 1 K-liearly idepedet vectors i B 1 1. These 1 vectors lie i the K-spa (B) of B ad B has eleets.. Steiitz' replaceet theore gives 1 absurd. Thus B 1 caot be ifiite. We put B 1 = 1. Here 1 0 because 1 = 0 would iply B 1 = ad B which is (B) = V = (B 1 ) = ( ) = {0} so B = {0} ad B would be liearly idepedet over K. cotrary to the hypothesis that B is a K-basis of V. So 1. B 1 is a K-liearly idepedet subset of V i (B) therefore 1 by Steiitz' replaceet theore.. Liewise B is a K-liearly idepedet subset of V i (B 1 ) so 1. Therefore = 1 as was to be proved Defiitio: Let V be a vector space over a field K. If V has a fiite K-basis the uber of eleets i ay K-basis of V which is the sae for all K-bases of V by Theore is called the diesio of V over K or the K-diesio of V.. It is deoted as di K V or as di V. If V has o fiite K-basis. the the K-diesio of V is defied to be ifiity ad we write i this case di K V =. 511

13 Thus di K K = (Exaple 42.7(a)) ad di V = 2 where V is the - vector space of Exaple 42.7(b). We frequetly say that V is -diesioal whe di V =. A vector space is said to be fiite diesioal if di V is a oegative iteger ad ifiite diesioal if di V =. Notice that the diesio of the vector space {0} is zero Lea: Let V be a vector space over a field K let di K V = ad let be vectors i V. (1) If are liearly idepedet over K the 2... ) = V. (2) If 2... ) = V the are liearly idepedet over K. Proof: (1) We are give di K V =. Let { } be a basis of V over K. If are K-liearly idepedet vectors i V = 2... ) the we obtai 2... ) = 2... ) by Steiitz' replaceet theore. (2) Suppose 2... ) = V. If are ot liearly idepedet over K the there is a proper subset T { } of { } with (T) = 2... ) = V (Theore 42.4) ad there is a K-basis B of V such that B T (Theore 42.9). Usig Theore we obtai the cotradictio = di K V = B T. Hece have to be liearly idepedet over K. Ay fiite set spaig a vector space ca be stripped off to a basis of that vector space (Theore 42.9). Siilarly ay liearly idepedet subset of a vector space ca be exteded to a basis as we show ow Theore: Let V be a -diesioal vector space over a field K with 1. Let 1 ad let be liearly idepedet vectors i V. The there is a K-basis B of V such that { } B. 512

14 Proof: Let { } be a K-basis of V. The are liearly idepedet vectors i V = 2... ) ad Steiitz' replaceet theore gives ) = V ad o idexig 's suitably. The the = di K V vectors are liearly idepedet over K by Lea 42.13(2). Hece B = { } is a K-basis of V cotaiig the vectors Lea: Let V be a fiite diesioal vector space over a field K ad let W be a subspace of V. (1) W is fiite diesioal; i fact di K W di K V. (2) di K W = di K V if ad oly if W = V. Proof: Let = di K V. (1) The assertio is trivial whe W = {0} so let us assue W {0}. The there is a ozero vector i V ad { } is a K-liearly idepedet subset with oe eleet. O the other had ay 1 vectors i V (i fact i V) are liearly depedet over K by Steiitz' replaceet theore. Therefore there exists a atural uber such that (a) 1 (b) there are liearly idepedet vectors i W (c) ay 1 vectors i W are liearly depedet. This is clearly uique i view of (b) ad (c). The atural uber havig bee defied i this way let be K-liearly idepedet vectors i W. We clai that { } is a K-basis of W. To show this we ust prove oly that these vectors spa W over K. Let be a arbitrary vector i W. The are liearly depedet over K by (c). Hece = 0 with soe scalars i K. Here 0 for otherwise the equatio above would iply that are K-liearly depedet. Hece has a iverse 1 i K ad we get 1 = ( 1 ) 1 2 ) 2... ) 2... ). 513

15 This gives W 2... ). But belog to W so 2... ) W (Lea 40.5). The vectors therefore spa W over K so { } is a K-basis of W. Thus W is fiite diesioal ad i fact di K W = = di K V. (2) If W = V the of course di K W = di K V. Suppose coversely di K W = di K V = ad let A be a K-basis of W. The there is a K-basis B of V with A B: this follows fro Theore whe A ad is obvious whe A =. The = di K W = A B = di K V = iplies that A = B. Thus W = (A) = (B) = V Lea: Let VU be vector spaces over a field K.. Suppose V is fiite diesioal ad let. : V U be a vector space hooorphis. Let be vectors i V. (1) If is oe-to-oe ad { } is liearly idepedet over K the { } is liearly idepedet over K. (2) If is oto U ad { } spas V over K the { } spas U over K. (3) If is a vector space isoorphis ad { 1 the { } is a K-basis of V 2... } is a K-basis of U. I particular di K U = di K V. Proof: (1) Suppose are scalars such that 1 ) 2 ( 2 )... ( ) = 0. The ) = 0 so Ker ad = 0 sice Ker = 0 as is oe-to-oe. Sice are K-liearly idepedet we get 1 = 2 =... = = 0. Hece are liearly idepedet over K. (2) We ust show that ay eleet. of U ca be writte as a K-liear cobiatio of the vectors Let U. The there is a i V with = ad = where are suitable scalars i K. This yields = = ) = 1 ) 2 ( 2 )... ( ) 2... ) 514

16 as was to be proved. (3) This follows iediately fro (1) ad (2). Fro Lea it follows that di K U = wheever U K. The coverse of this stateet is also true Theore: Let V be a vector space over a field K. The di K V = if ad oly if V K (as vector spaces). Proof: If V K the di K V = di K K = by Lea 42.16(3). Suppose coversely that di K V = ad let { } be a K-basis of V. Every eleet of V ca be writte i a uique way as where K. We cosider the appig : V K 2... ). This is a K-liear trasforatio sice for ay K ad = = i V we have ( ) = ( ) ( = (( 1 1 ) 1 ( 2 2 ) 2... ( ) ) = ( ) = 2... ) 2... ) = ( ) ( ). )) Furtherore = V belogs to Ker if ad oly if 2... ) = ( ) thus if ad oly if = = 0. So Ker = {0} ad is oe-to-oe. Sice ay -tuple 2... ) i K is the iage uder of the vector i V we see that is oto. Hece is a vector space isoorphis ad V K. 515

17 42.18 Theore:. Let V ad U be fiite diesioal vector spaces over a field K. The V U if ad oly if di K V = di K U. Proof: The case whe di K V = 0 or di K U = 0 is trivial. Let us suppose di K V 1 ad di K U 1. If V U the di K V = di K U by Lea 42.16(3). If di K V = di K U the V K dikv = K di KU U by Theore hece V U Theore: Let V be a vector space over a field K ad let W be a subspace of V. If V is fiite diesioal the V/W is fiite diesioal. I fact di K V = di K W di K V/W. Proof: We eliiate the trivial cases. We ow that W is fiite diesioal (Lea 42.15(1)). If di K W = 0 the W = {0} so V V/{0} = V/W so di K V/W = di K V ad di K V = 0 di K V = di K W di K V/W. If di K W = di K V the W = V (Lea 42.15(2)) so V/W {0} ad di K V = di K V 0 = di K W di K V/W. Thus the theore is proved i case di K W = 0 or di K W = di K V (i particular i case di K V = 0). Let us assue ow 0 di K W di K V. Let di K W = ad let { } be a K-basis of W. There are vectors i V such that { } is a K-basis of V (Theore 42.14). Here 1 ad = di K V. We clai that { 1 W 2 W... W} is a K-basis of V/W. This will iply = di K V/W hece di K V = = di K W di K V/W. To establish our clai we ote first that 1 W 2 W... W are K-liearly idepedet vectors i V/W. Ideed if are scalars such that 1 W) 2 ( 2 W)... ( W) = 0 W the W = s K 2... ) = where are appropriate scalars i K. The = 0 ad liearly idepedece of iplies that 1 = 2 =... = = 0. Thus 1 W 2 W... W i V/W are liearly idepedet over K. 516

18 Secodly these vectors spa V/W. To see this let us tae a arbitrary vector W i V/W where V. The = where are scalars ad thus W = ) W = 1 W) 2 ( 2 W)... ( W) 1 W) 2 ( 2 W)... ( W) = 1 W) 2 ( 2 W)... ( W) W 2 W... W) V/W hece V/W = W 2 W... W). This proves that { 1 W 2 W... W} is a basis of V/W over K. As we reared above this gives di K V = di K W di K V/W. We deduce iportat corollaries fro Theore Theore: Let V be a vector space over a field K. Let WU be fiite diesioal subspaces of V. The W U is a fiite diesioal subspace of V ad i fact di K (W U) = di K W di K U di K (W U). Proof: If { } is a K-basis of W ad { } is a K-basis of U the W U = { V: W U}. is clearly spaed by the fiite set { } hece W U is fiite diesioal (Theore 42.9).. Fro W U/U W /W U (Theore 41.16) we obtai the di K (W U) di K U = di K (W U/U ) = di K (W /W U) = di K W di K (W U) Theore: Let V be a vector space over a field K ad let be a K- liear trasforatio fro V. If V is fiite diesioal the di K Ker di K I = di K V. 517

19 Proof: Theore tells us V/Ker I ad Theore gives di K V di K Ker = di K I Theore: Let VU be vector spaces over a field K ad let : V U be a K-liear appig. Suppose that V ad U have the sae fiite diesio. The the followig stateets are equivalet. (1) is oe-to-oe. (2) is oto. (3) is a vector space isoorphis. Proof: (1) (2) If is oe-to-oe the Ker = {0} so di K Ker = 0 ad di K I = di K Ker di K I = di K V = di K U. Thus I is a subspace of U with di K I = di K U ad Lea 42.15(2) gives the I = U. Hece is oto.. (2) (1) If is oto the I = U so di K I = di K U ad di K Ker = di K V di K I = di K U di K U = 0. Thuer = {0} ad is oe-to-oe. Hece ay oe of (1)(2) iplies the other ad these together iply (3). Coversely if is a isoorphis the of course is oe-to-oe ad oto. Thus (3) iplies both (1) ad (2).. We close this paragraph with a brief discussio of ifiite diesioal vector spaces. Do ifiite diesioal vector spaces have bases? Fro Theore 42.9 we ow that such a vector space caot be spaed by a fiite set. But if B is a spaig set ecessarily ifiite the arguet of Theore 42.9 does ot wor. To prove the existece of bases of ifiite diesioal vector spaces we have to resort to ore sophisticated eas. It is i fact true that every vector space has a basis ad a proof is give i the appedix. The proof of this stateet for ifiite diesioal vector spaces requires a fudaetal tool ow as Zor's lea. This lea ca be used i a variety of situatios to establish the existece of certai objects. 518

20 The existece of bases havig bee assured by Zor's lea we ight as whether ay two bases have the sae cardiality. The aswer tured out to be "yes" i the fiite diesioal case (Theore 42.11) ad this was proved by usig Steiitz' replaceet theore. The proof of Steiitz' replaceet theore does ot exted to the ifiite diesioal case. Nevertheless theores of set theory ca be eployed to show that two bases of a vector space have the sae cardial uber. Thus reders it possible to defie the diesio of a vector space as the cardiality of a basis. Hece it is possible to distiguish betwee various types of ifiities. This is uch fier tha Defiitio by which ifiite diesioality is erely a crude egatio of fiite diesioality. Theore which states that ay liearly idepedet subset ca be exteded to a basis is true i the ifiite diesioal case too. The proof aes use of Zor's lea. Lea 42.15(1) reais valid also i the ifiite diesioal case i the sese that a basis of a subspace has a cardial uber less tha or equal to the cardiality of a basis of the whole space. Lea 42.15(2) however is ot ecessarily true for ifiite diesioal vector spaces: a proper subspace ay have the sae diesio as the whole space (thi of ad as -vector spaces). Lea ad its proof wors i the ifiite diesioal case. Lea ad its proof wors i the ifiite diesioal case provided we refer to the geeralizatio of Theore at the appropriate place. Geerally speaig ifiite diesioal vector spaces are wild objects. To reder the ore aageable oe equips the with soe additioal structure perhaps with a topological or aalytic oe. Exercises 1. Let V be a vector space over a field K ad let W be a subspace of V. Show that there is a subspace U of V such that V = W U ad 519

21 W U = {0}. (U is called a direct copleet of W i V. We write the V = W U ad call V the direct su of W ad U.) 2. Is {(111) (110) (100)} a -basis of 3? 3. Is {(126) (001) (210) a 3 -basis of 3? 3 4. Fid a -basis of {f C 2 ([01]): f (x) 7f (x) 12f(x) = 0 for all x [01]}. 5. Fid all -liear appigs fro 4 oto Fid all 2 -bases of 3 ad 2 3 -bases of Show that the vectors (121) (020) (12 1) ad also the vectors (110) (101) (111) i 3 are liearly idepedet over. 8. Let f (x) = si x for x [01] ( = ). Prove that the fuctios {f 1 f 2 f 3... } i C ([01]) are liearly idepedet over. 520