g as the function in which for every element x be the set of polynomials of a degree less than or equal to n with , for each i from 0 to n )

Transcription

1 Vector Spce Observtio: I this book, we will distiguish vectors d sclr eleets. For sclr eleets we will use Greek letters,,,, etc. d for the vectors i V we will use letters x,y,u,v,w,s,, etc., fro our ow lphbet. Depedig o the cotext if there is cofusio we will use dsh o top, for exple with the cero vector we will lwys use the ottio 0, this wy we will distiguish it fro the sclr 0 i the field K. Cotext is iportt i deducig which type of opertio we re pplyig. For exple whe we write x, or siply x, we kow it is the exterl lw tht opertes sclr, i the field K, with vector i the group V. Whe we write or siply, we e the product of two sclrs ofk. For dditio we will lwys use to dd sclrs or vectors. VECTOR SPACE VECTOR SPACE EXAMPLES : 1 I the setf of rel fuctios with rel vribles, give y two fuctios f d g i F, we defie dditio f g s the fuctio i which for every eleet x we obti the uber f x g x i. 2 LetP x be the set of polyoils of degree less th or equl to with x. For the dditio of y two polyoils: Ad 2 p x 0 1x 2x... x (Where i, for ech i fro 0 to ) 2 q xb bx b x... b x (Whereb i, for ech i fro 0 to ) 0 We defie the polyoil: p x q x b b x b x... b x p x x x... x, i For every i d for every polyoil P x we defie, for the reds to prove tht P x,,,, is vector spce p x x x... x. Is give s exercise 1

2 3 Let the set of ll rel trices with trices with rows d colus: The su of two trices d the product of rel uber by trix is other exple. Additio is iterl opertio i trix d, s we hve see,. is beli group or couttive group. The product of rel uber by trix is exterl opertio which coplies with the properties (1). We hve the defied the vector spce,,,, of trices with rows d colus, over the rel ubers tht fro ow o we will siply cll. 4 The zero vector spce: Let V cosist of sigle object, which we deote by 0, d defie For ll sclrs k d k0 0 VECTOR SUBSPACE A subspce H (differet fro the epty set) of vector spce V,,, will lso be vector spce with respect to the iherited opertios fro the spce V (both iterl d exterl opertios o field ) if these re closed oh. It es, tht the iterl opertio of V is lso iterl i H ; d the exterl opertio of over V, is lso exterl i over H. We eed to verify tht: 1. H V, H 2. xy, H, we verify tht x y H 3. d x H, the x H CHARACTERIZATION OF A VECTOR SUBSPACE If V is vector spce over field, d H is o-epty subset of V, the: H is Vector Subspce of V x y H,, K ; x, y H Observtio: The proof of how this is equivlet to the three poits previously show is siple eough to be left to the reder LINEAR COMBINATION If V is vector spce over field, it is sid tht the vector x V, is lier cobitio of the vectors i the set v,v,...,v V, if there exist sclrs,,..., K, such tht 2

3 x v v... v 1 2 LINEAR DEPENDENCY If vector x V c be expressed s lier cobitio of set of vectors v,v,...,v of V, we sy tht this x vector is lierly depedet o the. The vectors of the setv,v,...,v V, c be sid to be lierly idepedet, if oe of the is lierly depedet o the rest. CHARACTERIZATION OF LINEARLY INDEPENDENT VECTORS The vectors of the subset v,v,...,v V, re lierly idepedet the oly lier cobitio where v v... v, which result is the cero 1 2 vector, is tht i which ll the i re equl to cero. Proof. If there exists y i 0, for which v... v... v 0, 1 1 i i 1 We could divide everythig by i, give tht v v... v, so i 1 i i tht v, could be expressed s lier cobitio of the rest so the the i vectors v,v,...,v, would ot be lierly idepedet. The reciprocl c be eqully show. SYSTEMS OF GENERATORS If V is vector spce over field S v,v,...,v V, the set:, the, give vector set L S x V x v v... v,,,.., K, / 1 2 (We e, the subset of ll possible lier cobitiosof the vectors, v 1,v 2,...,v ) is vector subspce ofv deoted s geerted vector spce (or sped) bys. The vectors v 1,v 2,...,v will be clled syste of geertors of sid subspce, d we sy thtlsis the geerted spce by vectorsv 1,v 2,...,v. We c lso esily prove tht LS is the iiu vector spce tht cotis the vectors v,v,...,v. 3

4 DIMENSION The vector spce V is sid to hve diesio if i it we c fid lierly idepedet vectors, but it is ipossible to fid ore th lierly idepedet vectors. The diesio of vector spce is the xiu uber of lierly idepedet vectors tht the vector spce cotis. We will deote the diesio of the 2 vector spce V with di V. The diesio of the vector spce is equl to 3 two; the diesio of the vector spce is equl to three. Vector spces tht hve fiite diesio re clled fiite-diesiol. A spce i which we c fid s y lierly idepedet vectors s we wt is clled ifiite-diesiol. Oe exple of ifiite-diesiol vector spce is the set P x of every polyoil with rel coefficiets. The set of ll cotiuous rel fuctios o give itervl (or cotiul log the uericl lie) is lso ifiite-diesiol. BASIS A whole collectio of lierly idepedet vectors i -diesiol vector spce V is clled bsis of this spce. Observtio: By vector spces ow properties it is obvious tht y lier cobitio of eleets of the bse re prt of the spce; dditiolly there cot be y vector i the spce tht cot be expressed s lier cobitio of the vectors i the bse, if this ws flse the the lierly idepedet vectors i the bse would lso be lierly idepedet with ew vector d we would hve 1 lierly idepedet vectors i spce with diesio which is ipossible. This es tht every bse of syste of geertors spce is fored of lierly idepedet vectors. Coo Bsis I the vector spce, the usul bse, clled the coicl bse is B 1,0,...,0, 0,1,...,0,..., 0,0,...,1. I the vector spce P x of the polyoils with degree equl to or less 2 B 1, x, x,..., x. th, with rel coefficiets the usul bse is: I the set of ll rel trices with trices with rows d colus, the usul bse is: B ,,...,,,..., ,...,....,...,

5 Propositio: Let B e,e,...,e be bsis for V. The every vector x of the spce c be represeted s lier cobitio of the vectors i the bsis, d, furtherore, this represettio is uique. Proof. B e 1,e 2,...,e be y bsis of -diesiol spce V, d let x. Give tht y 1 vectors re lierly depedet, i prticulr, the vectors e 1,e 2,...,e, x, re lso lierly depedet, we e by this tht there exist soe ubers,,...,,, of which ot every uber is equl to 0 such tht Let e e... e x Furtherore, we hve tht 0 sice, otherwise, we would hve 1 e e 2... e 0 with soe of the coefficiets 1, 2,..., differet fro cero d, cosequetly, the vectors e 1,e 2,...,e would result lierly depedet (flse sice they for bsis). The 0 d, so, we c write x 1 e 2 1 e 2... e Ad, settig i x i, we hve: x x 1 e 1 x 2 e 2... x e This represettio of x is uique becuse, if we lso hd x y 1 e 1 y 2 e 2... y e, subtrctig both expressios, we would hve x y e x y e... x y e Ad, due to the liel idepedece of the vectorse,e,...,e, we get x y, x y,..., x y 1 2 COORDINATES 5 B e,e,...,e be bsis for V. For every x V, the ubers Let x 1, x 2,..., x tht ke x x e x e... x e re clled the coordites of x o the bsis B. The previous propositio estblishes tht, give bsis of - diesiol vector spce V, every vector i V hs coordites i this bse. It is lso cler tht, if the coordites tch for two vectors x d y these vectors re equivlet, d, i this cse x x e x e... x e y 1 2

6 Becuse of this we hve tht oce we set bsis for V, to deterie vector x it is oly ecessry to idicte its coordites x, x,..., x, expressed by Give vector spce x x, x,..., x V,,, of fiite diesio, with of it; the followig isoorphis c be estblished: : V x ( x, x,..., x ) B e,e,...,e s bse The isoorphis will be defied i such wy tht for every vector x i V, there is ige which will be the -tuple x, x,..., x of, de up of the coordites of x i the bse B. It is left to the reder to prove tht is isoorphic, we e, to prove tht is bijective d tht: x y x y, x,y V x x, x V, R ROW AND COLUMN SPACES OF A MATRIX Give trix with rows d colus ( eleets of diesiol trix) of Ech ith row c be cosidered s vector i1, i2,...,i. The seta of ll the row vectors geertes vector spce LA of ll its possible lier cobitios. The uber r of vectors lierly idepedet i A (which xiu will be, becuse there re oly rows) will be the diesio of the spce LA which is clled the vector row spce of the trix. RANK OF A SET OF VECTORS 6 Give y vector spce V,,, with fiite diesio, d bse B e,e,...,e of the spce; the rger of the set S v 1,v 2,...,v V, is the highest uber of lierly idepedet vectors tht c be extrcted fros. If y vector v i S, is expressed i the bse B s:

7 v e e... e i i1 1 i2 2 The, workig i coordites, we could sy tht the spce LS geerted by the vectors fros, is the subspce, geerted by the row spce of the trix: A Which es tht LS is the row spce of the trixa d, i cosequece the rk r of the set of vectors S, is equivlet to the rk of A. To obti bse LS, we oly eed to obti the row echelo for of the trixa. for the spce i IMPLICIT AND PARAMETRIC EQUATIONS OF A SUBSPACE. Give fiite diesiol vector spce B e 1,e 2,...,e fro it, the iplicit equtios of the spce geerted by the set of vectors S v 1,v 2,...,v V, is the hoogeous syste which solutio spce tches the row spce of trixa defied previously. The, if r is the rk of S, the trix A, i row echelo for, would be writte s: V,,,, d bse b b b b b b b b b b b b b b r11 r b r The the row spce ofa, would be the spce geerted by the set b 11,b 12,...,b,b 1, 0,b 12,...,b ,b 1, 0,0,...,b 21,b 2,..., 0,0,...,0,b r the vector x, x,..., x, will be prt of the row spce if d oly if x, x,..., x b,b,...,b 0,b,...,b... 0,0,...,0,b r r Otherwise expressed s: 7

8 x b b... b b x b. b... b r11 r x b r1 r1 r r These re the pretric equtios of the spce geerted by the set of vectors S v 1,v 2,...,v V. Reducig the r preters, r hoogeous equtios re obtied: 11x 1 12 x x x x... x r1x 1 r2x 2... rx 0 Which re the iplicit equtios of the spce geerted by the set of vectors S v,v,...,v V EXTENDING TO A BASIS I fiite diesiol vector spce every set of lierly idepedet vectors c be icluded s prt of bsis. Proof: Let e 1,e 2,...,e, be vectors lierly idepedet i V. If ll the reiig k vectors i V re expressed lierly i ters of the vectors e 1,e 2,...,e k, these lredy costitute bsis. If there is vector e tht is ot expressed lierly k 1 i ters of e 1,e 2,...,e, the k 1 vectors k e 1,e 2,...,e k,e will be lierly k1 idepedet. If the followig equlity is true 8 1ee 2... kek k 1ek 10 with y i 0, the k 1 0 (due to the lier idepedece of the vectors e 1,e 2,...,e k ) d, i cosequece, the vector e c be expressed lierly i ters k 1 of e 1,e 2,...,e k. Add the vector e to k 1 e 1,e 2,...,e. If ll the vectors i V re lierly expressed k i ters of e 1,e 2,...,e k,e, these lredy costitute bsis. If there exists k1 vector e tht cot be expressed lierly i ters of k2 e 1,e 2,...,e k,e, by the k1 previous logic, it c be dded to the set of vectors to for ew set

9 e,e,...,e,e,e d the set will be lierly idepedet, etc. This process k k1 k2 cot be exteded idefiitely, due to the fct tht V is, by hypothesis, fiite diesiol, i cosequece the spce cot hold ifiite set of lierly idepedet vectors. Fiishig the process, lierly idepedet set of vectors fro which every other c be lierly expressed i V is obtied. This syste will be bsis for the spce V tht cotis the give vectors. THEOREM OF BASIS Every vector spce V 0, with fiite syste of geertors possesses, t lest, oe bse. co u siste de geerdores fiito posee, l eos, u bse. Proof: Let S v 1,v 2,...,v be syste of geertors for V. If S is lierly idepedet, the B S is bsis for V. Otherwise there will be vector, supposedly v, tht is lier cobitio of the rest, by which, if vector x is lier cobitio of the vectors i S, substitutig v i the lier cobitio with respect to the vectors i S 1 v 1,v 2,...,v1 x becoes lier cobitio of the vectors i S 1d, s such: V LS LS 1 If S 1 is lierly idepedet, the B S 1 is bsis forv. Otherwise, the previous resoig is repeted util obtiig soe Si v 1,v 2,...,v i tht is lierly idepedet d bsis. The ed of the process is esured, becuse i the worst cse scerio, fter de S v is reched with V 0 ) d this will be the 1steps bsis. i 1 v1 0 (becuse OPERATIONS ON SUBSPACES INTERSECTION Give fiite diesiol vector spce V, if S d T re two vector subspces of V, the the itersectio S T is lso subspce of V. Proof: The followig eeds to be proved: x,y S T x y S T, K 9

10 To prove it if x,y S T, the x,y S, d becuse S is vector spce d x y S ; logous resoig deostrtes the se for x y T. It is x y S T. Give fiite diesiol vector spce V,,, d bsis B e,e,...,e, if S d T re two vector subspces i V, to fid the equtios for S T, the best is to express S d T by their iplicit equtios, the equtios idicte the coditios to which vector s copoets hve to coply for it to belog to the spce S, d the coditios to which these se copoets hve to coply to i order to belog tot. Obviously if we joi every coditio, by this it is et, every equtio, the the vectors tht re subject to these coditios will belog to both S dt, d will defie the itersectio. Observtio: Eve if the itersectio of subspces is subspce, it is ot the se for the uio (see figure) Tkig the spce 3, d the subspces S (of equtio x 0 ) dt (of equtio z 0 ). The uio would be costituted by ll vectors tht belog to the uio of the gry ples i the figure.it is observed tht 1,0,0S T d 0,0,1S T but, however, 1,0,0 0,0,1 1,0,1S T.The S T is ot vector spce, becuse dditio is ot iterl opertio i S T. However, if the vectors tht re the result of ddig vectors i S with vectors i i T re selected, the set of vectors tht re obtied,will for vector spce. SUM Give vector spce V, if S d T re two vector subspces fro V, the su of S dt is defied s 10 S T x V / u S y v T / x u v. The result of ddig S d T is lso subspce i V.

11 Lookig t the previous figure, it is deducted esily tht S T, is equl to the whole of 3. Additiolly, s S is the spce defied by the equtio x 0 d T is defied by the equtio z 0, it is cler tht y x,y,z 3, c be expressed s the su of vector i S with other it, sice x,y,z 0,y,z x,0,0. It c lso be observed tht there is o uique wy to express x,y,zbecuse it is lso true tht x,y,z 0,y 1,z x,1,0 S 3 However, differet subspces could be used to obti. This wy, for exple, (see the figure bove), the subspces S of equtio x 0 d T of equtios y 0 z 0, lso verify tht 3 3 S T, sice y x,y,z c be expressed s: x,y,z 0,y,z x,0,0, but i this cse there is oly oe uique wy to express x,y,z s the result of ddig vector is with other it while i the first exple this ws ot the cse. This is wht will distiguishsufro the other deoiteddirect su. Additiolly, the fct tht every eleet i V c be uiquely expressed s the dditio of eleet i S with other it, is wht is goig to chrcterize the direct su. DIRECT SUM Give vector spce V, if S d T re two vector subspces i V, thethe direct su S T x V /! u S y! v T / x = u + v (where!, es there exists oe d oly oe), is lso subspce ofv. DIMENSION FORMULA Give fiite diesiol vector spce V, if S d T re two vector subspces iv, the 11

12 Proof: 12 di S T di S T di S di T Let di S, di T d di S T r, the if r S T, the by extedig to bsis, e,e,...,e bsis for such wy tht: d B e,e,...,e,e,...,e B e,e,...,e,v,...,v 1 r r1 2 r r1 r e,e,...,e is could be expded i Are bsis fors dt respectively. It will be proved tht B e 1,e 2,...,e r,e r1,...,e,v r1,...,v is bsis fors T. For this, i the first plce, it is ecessry to prove tht the vectors re lierly idepedet. I effect: if e e... e e..., e v... v r r1 1 r1 r1 r1 it is verified tht: v... v e e... e e..., e S S T (*) r1 r1 1 1 r r1 1 r1 d if r 1v r1... v, belogs to S T, it c be used s lier cobitio of the eleets of its bsis e,e,...,e, it is to sy there will exist 1, 2,..., r sclrs i such wy tht: wht is equivlet to syig v... v e e... e r1 r1 1 2 r r e e... e v... v r r r1 r1 B2 e 1,e 2,...,e r,v r1,...,v bsis of T, its vectors re lierly idepedet, the it is required to verify tht, r , d i the expressio (*), it would be hd tht d, for beig v... v e e... e e..., e 0 r1 r1 1 1 r r1 1 r1 which es tht e e... e e..., e 0 d, s 1 1 r r1 1 r1 B1 e 1,e 2,...,e r,e r1,...,e is bsis for T, its vectors re lierly idepedet, the it is required to verify tht, , d i cosequece: r1 r

13 B e,e,...,e,e,...,e,v,...,v, re lierly r r1 r1 d, the the vectors of idepedet. Additiolly, the vectors i B geerte the spce S T sice x S T s S d t T / x s t d, expressig vectors s d t i the bsis 1 r r1 B e,e,...,e,v,...,v fro the spces S dt respectively: 2 r r1 B e,e,...,e,e,...,e d x s t 1 e e 2... r er r 1 e r1... e e e... e v... v 1 2 r r r1 r1 1 e... e e... e v... v r r r r1 r1 r1 r1 Which es it is geerted by the vectors i the bsis B. Furtherore, y vector x geerted by the bsis B c be decoposed s the su of vector s is d othert it, such s x 1 ee 2... r e r... e r 1v r1... v 1ee 2... r e r... e0 e1 0 e er r 1v r1... v s t The the vectors i B geerte the spce S T d, pyig ttetio to the uber of vectors i the bsis S T, ofs T, ofs d oft, it is hs bee proved tht: di S T di S T di S di T r rrr 13