BU I XUA NHA IAI S0TUYFNTNHNANC CA0IUUHA NH N0 I B0CA N TH0200812Nuc lucChuongI. RU T C0 N N0 T T0A N TUTUYF N TNH 51. Ia t va n dc . . . . . . . . . . . . . . . . . . . . . . . . . b2. 1r} rc ng vavc c torc ng. . . . . . . . . . . . . . . . . . 7J. Ia thu c da c trung cu a mo t toa ntutuyc ntnh. . . . . . 94. 1oa n tu chc o ho a duo c . . . . . . . . . . . . . . . . . . . 10b. No t va u ng du ng cu a suchc o ho a . . . . . . . . . . . . 14b. 1am ga c ho a . . . . . . . . . . . . . . . . . . . . . . . . 227. Ia thu c trc t tc u. I}nh lyHamlton-Callcy . . . . . . . 278. odcca n ba n . . . . . . . . . . . . . . . . . . . . . . . J19. Ia thu c to tc u . . . . . . . . . . . . . . . . . . . . . . . Jb10. Ia ng tamga c kho . . . . . . . . . . . . . . . . . . . . J811. Ia ng chnh ta c Jordan . . . . . . . . . . . . . . . . . . 4Ja ta p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . bJJChuongII. KH0 NC CIAN FUCIID 571. 1ch vohuo ng . . . . . . . . . . . . . . . . . . . . . . . b72. Ia ng ru t go n Causs . . . . . . . . . . . . . . . . . . . . b9J. Chua n cu a vc c to. . . . . . . . . . . . . . . . . . . . . . bb4. Sutru cgao . . . . . . . . . . . . . . . . . . . . . . . . . b8b. Co sotru c gaovacosotru c chua n. . . . . . . . . . . . 72b. Na tra n cu a da ng song tuyc ntnh . . . . . . . . . . . . 7b7. Nho m tru c gao. . . . . . . . . . . . . . . . . . . . . . . 798. Nho m O(2, R) . . . . . . . . . . . . . . . . . . . . . . . 8J9. Nho m O(3, R) . . . . . . . . . . . . . . . . . . . . . . . 8b10. Chc oho a toa ntu tulc n ho p trong kho ng ganLucld. 90a ta p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9J4Chuong IRUT C0N N0TT0AN TUTUYFN TNH1. at van deCa su V la kho ngganvc ctonchc utrc ntruo ngKvaB=(e1, . . . , en) lamo t co soduo c sa p cu a V . Nc u flamo t toa n tutuyc ntnh trong kho ng ganVthi ma tra n bc u dc nftrong co so Bduo ckyhc u la A = [f]B. Casu B

= (e

1, . . . , e

n) lamo t cosokha c cu aV , A

= [f]B vaP= (B B

) lama tra n chuyc n co sotu B sangB

thiA va A

lc n hcvo nhaubo co ng thu cA

= P1AP.b1rongchuongna ychu ngtanghc ncu uva ndc timkc mtrongkho ng gan vc c toVmo t co so B sao cho trong doma tra n cu a toa ntu fco da ngdao}/a oo/a /co thc duo c. Cu thc hon, tasc timkc m Bsao cho trong doma tra n cu a fcoda ng chc o, hoa c cothclada ng tamga c cha ng ha n.|nh ngha 1.1. 1a no toa n tutuyc n tnhf EndK(V ) c/e c /c adaa c nc u to n ta co so B = (e1, . . . , en) sao cho [f]Blama tra n chc o.|nh ngha 1.2. 1ano toa ntutuyc ntnhf EndK(V ) /am }/a c/c a daa c nc uto n ta coso B = (e1, . . . , en) saocho [f]Blamatra ntamga c (trc nhoa c duo ).\a n dcru t go nmo t toa ntu tuyc n tnhdua vcca c buo csau:1) 1im dc u kc n dcmo t toa n tutuyc n tnh chc o ho a duo c (tuongu ng,tamga c ho a duo c).2)Nc uto antu fchc oho aduo c(haytamga cho aduo c), ha ytimmo tcoso saochotrongdo matra ncu afco da ng chc o(tuongu ng,da ng tamga c).Rora ng ta cu ng cothc tc p ca n va n dcna y tho ngqua ngo nnguma tra n. Cuthcnhusau:1) Io vo ma tra n A Mn(K), tim dc u kc n dcto n ta ma tra nkhangh}ch Psao choA

= P1APlama tra n chc o (tuong u ng, matra n tamga c).2) 1imPva A

.b2. Tr| riengvavecto rieng|nh ngha 2.1. Chof EndK(V ). 1ano vc ctov V la mo tvc c torc ng cu afnc u:()v ,= 0;() 1o n ta Ksaochof(v) = v.Kh dota no lamo t /r/ r/e o} //a /r/ da c /rao} cu a f, va vlace c /a r/e o} //a ce c /a da c /rao} u ng vo tr}rc ng.Nha n xe t.1) 1hcod}nhnghia, mo vc ctorc ngdc ukha c0. 1uynhc n,tr} rc ngco thc ba ng0. 1ha t va y, nc ukerf,=0thi mo vc cto0 ,= v Kerfdc u lavc c torc ng u ng vo tr} rc ng = 0.2) Nc u v lamo t vc c to rc ng u ng vo tr} rc ng thi 0 ,= K, vcu ng lamo t vc c torc ng u ng vo tr}rc ng.\a y, ca c vc c torc ng cu afchi cothcla :- Ca c vc c tokha c 0 cu aKerf;- Ca c vc c tokho ngdo phuong duo ta c do ng cu a f.Ngoa ra,do nha nxc t 2) nc nnc uDlamo t duo ng tha ng snh rabo mo t vc c torc ng cu afthi Dba t bc nduo ta c do ng cu af.V du1.ChoV la kho ngganca cvc ctotu dotrongkho nggan 3chc utho ngthuo ng, la mo tma tpha ng va Dla mo tduo ngtha ngca tvo vc ctochiphuongw. Co fla phc pchc uxuo ngma tphaa ngsongsongvo duo ngtha ngD. Khdo , v tadc uco f(v) =v,nghiala vla vc ctorc ngcu afu ng vo tr}rc ng = 1. No vc cto7kha c 0 cu a D lavc c to rc ng cu a fu ng vo tr} rc ng = 0. Ngo a ra,ba t kymo t vc c tokha c 0 na okha c cu aVcu ng dc u do phuongduo ta c do ng cu af. \a y, 0 va1 laca c tr} rc ng duy nha t cu af.V du2.1rongma tpha ng xc tfla mo tphc pquaymo tgo cquanhta mO. Nc u ,=k, k Zthimo vc ctokha c 0dc udo phuong, do ngtho cu ngkho ngco vc ctokha c0na obc ntha nhvc cto0. \a yfkho ng coca c vc c torc ng.V du3.Chok Kvaa nh xaV Vv kvlaphc p v} tuhcso k.I}nhlyduo da y chotha ytnhcha tquantro ngcu avc cnghc ncu u ca c vc c torc ng.|nh ly2.2.Tca o/a /ae o//o/f EndK(V )c/e c/c adaa c///cac// /// /c o /a / mc / ca saca a V}c m /ca o ca c ce c /a r/e o} ca a f.Chu ngminh. Nc u B = (e1, . . . , en)la mo tcoso go mtoa nca cvc ctorc ng cu afthif(v1) = 1v1, . . . , f(vn) = nvn.Kh do[f]B = diag(1, . . . , n).Nguo c la , gasu B lamo t co socu a Vtrong doma tra n cu a fcoda ng chc o [f]B = diag(1, . . . , n). Kh do , f(v1) = 1v1, . . . , f(vn) =nvn, nghiala v1, . . . , vndc u laca c vc c torc ng cu af.83. a thuc dactrung cua mot toan tutuyen tnhCasu f EndK(V ) va lamo t tr} rc ng cu a f. Kh do , to n ta 0 ,= v V saochof(v) = v, nghiala(f IdV)v = 0. Nc u Blamo t co soduo c sa p cu a Vthi[f IdV]B[v]B = 0.Iov ,=0nc n[v]B ,=0. 1u do suyradet[f IdV]B=0.Nc uda t A=[f]Bthi tu do suy ra tho a [A In[ =0. Ia tPf()= [A In[. Nc u B

la mo t coso duo csa pkha ccu aV vaA

=[f]B thi to nta matra nkha ngh}chP(Pchnhla matra nchuyc n cosotu Bsang B

) saochoA

= P1AP. 1u dotaco[A

In[ = [P1AP In[ = [P1(AIn)P[ = [A In[.\a y, Pf()chi phu thuo cva of ma kho ngphu thuo cva ovc ccho n co socu aV . 1a go Pf() lada //a c da c /rao} cu a to an tu f.1unhu ng pha n tch no trc n tanha n tha y ra ng, dctim tr} da c trungcu atoa ntu ftachivc ccho nmo tcosoBna odo cu aV , xa c d}nhmatra nA = [f]B,sau do ga phuong trinhda so[AIn[dctimta t caca c nghc mna m trongtruo ngKcu a no .|nh ngha 3.1. Ca su dathu cda ctrungPf()co ca cnghc m1, . . . , p K, vo kilabo cu ai. Kh dota vc tSpK(f) = 1, . . . , 1. .k1, . . . , p, . . . , p. .kp9vago no la p/ccu a toa ntu f. Nc uAlamatra n bc udc nftrongmo t coso duo csa p na odo cu aV thitaco thc vc tSpK(A)thayvivc tSpK(f).Ca nthc t luuy ra ngpho cu atoa ntu tuyc ntnhphu thuo cva otruo ng co so K. 1ha t va y, nc u A =_2 15 2_thi PA() = 2+1.Kh do SpR(A)=, nhungSpC(A)= i. 1uynhc n, kh na okho ng thccogi ga y nha m la n thi ta scdu ng ca c kyhc u don ga n laSp(f), Sp(A) thay choSpK(f), SpK(A).V du4. Chof: R3R3f(x1, x2) = (x1 + 2x2, x1 + 4x2)Natra ncu aftrongcoso chnhta cla A =_1 21 4_, suyraPf() = ( 2)( 3). \a y,Sp(f) = 2, 3.4. Toan tucheohoa duoc|nh ngha 4.1. Chof EndK(V ). Nc u lamo t tr}rc ng cu afthiE() := v V [f(v) = v10la mo tkho ngganconcu aV va tago no la //c o}}/aoccor/e o}cu aVa o} ca / /r/ r/e o}.Ne nh de4.2. C/c1, . . . , p/a ca c/r/ r/e o}//a co/aaca afEndK(V ). K// dc E(1) +. . . + E(p) /amc / /c o} /ra c //e p.Chu ngminh. 1asc chu ngmnhba ngqu na pthcop. Kh p =1thi kho ngcogidcchu ng mnh. Casu p 1 vadc u kha ng d}nhladu ng do vo p. 1a ca n chu ng mnh(E(1) +. . . + E(p)) E(p+1) = 0.Casu v = v1 + . . . + vp E(p+1),trongdo vk E(k), k 1, k. Kh dof(v) = 1v1+. . .+pvp = p+1v = p+1(v1+. . .+vp).1udosuy ra(1p+1)v1 + . . . + (pp+1)vp = 0.1ugathc t quna p suy ra (k p+1)vk = 0, k 1, p. Nhungk ,=p+1, k 1, p, nc ntu do suyravk= 0, k 1, p, kc othcov = 0.Hequa4.3. Tca o /afc/e c /c a daa c oe a cac// oe a V/a/c o} /ra c//e pca aca c//c o}}/aoccor/e o}ca aoc . Nc / c//o/va c/ao, }/asa 1, . . . , p/a ca c /r/ r/e o} //a c o/aa ca af. K// dc . fc/e c /c adaa c oe a cac// oe aV= E(1) . . . E(p).Chu ng minh. Casu V= E(1). . .E(p). Kh do , nc u B1, . . . , Bptuong u ng laca c co socu a E(1), . . . , E(p) thi B = (B1, . . . , Bp) laco socu aV . 1udo , a p du ng I}nh ly2.2 suy rafchc o ho a duo c.Nguo c la , gasufchc o ho a duo c. Kh do , cu ng thco I}nh ly2.2,to nta mo t cosoBcu aVgo mtoa nnhu ng vc ctorc ng cu af. CasuB = v1, . . . , vn1. .E(1), . . . , w1, . . . , wnp. .E(p).11Kh do , rora ng dimE(1) +. . . +dimE(p) = dimV. 1udosuyraV= E(1) . . . E(p).0uaHc qua no trc ntatha yra ngso chc ucu aca ckho ngganconrc ng do ngmo t vatro quantro ngtrongba toa nchc oho amo ttoa ntu tuyc ntnh. Iodokha o sa t ca c sochc u na y lamo t vc c ca nthc t.Ne nh de4.4. C/c f EndK(V ). Ne a /amc / /r/ r/e o} bc / m ca af///dimE() m.Chu ngminh. Ca su dimE() >mva v1, . . . , vm, vm+1la ca cvc cto do c la p tuyc n tnh cu aE(). otu c hoca c vc c tona y tha nh mo tco so Bcu aV :B = (v1, . . . , vm, vm+1, wm+2, . . . , wn).Kh do[f]B = O...O AO B.1udosuy raPf(t) = det t O...O tAO B tInm112= ( t)m+1det(B tInm1).Suy ra latr} rc ng bo kho ng nhohonm+ 1 vata como t ma uthua n.a y gota dacodudc u kc n dcchu ng mnh d}nh lychnh vcsuchc o ho a mo t toa ntutuyc ntnh.|nh ly4.5.Tca o/a /ae o//o/f EndK(V )c/e c/c adaa c///cac// /// ca c d/e a //e o daa / da daa c //c a ma o.// Pf() p/a o ra/re o K, o}//a /aPf() cc//ep/a o //c/ //a o/da o}Pf() = (1)n( 1)m1. . . ( p)mp, (1)ca /1, . . . , p Kca m1 +. . . +mp = n./// i 1, p, dimE(i) = mi.Chu ngminh. Nc uca cdc ukc n()va ()duo ctho ama nthi V=E(1) . . . E(p). Iodo , thcoHc qua4.J,fchc o ho a duo c.Nguo c la , gasufchc o ho a duo c. Nc u Pf() kho ng pha n ratrc nKthi noco da ngPf() = Q()( 1)m1. . . ( s)ms,vo m1 + . . . +ms< n. 1hcoNc nh dc4.4, tacodimE(1) +. . . + dimE(s) m1 +. . . + ms< n,kc othcofkho ngchc oho aduo c. \a yPf()pha pha nra trc nK,nghia lano coda ng (1).Nc u to n ta jsao chodimE(j) < mjthidimE(1) + . . . +dimE(p) m1 +. . . + mp< n,cu ngma uthua nvo tnhchc oho aduo ccu af. \a ydc ukc n()cu ng pha tho a ma n.1JHequa4.6. Ne afcc n/r/ r/e o} //a c o/aa ///fc/e c /c a daa c.5. Not vaiung dung cua sucheohoa5.1. Tnh lu y thu a cu a ma tra nChoA Mn(K). Ca su Achc oho aduo ctrc nK. Kh do to nta mo t matra nkha ngh}chP Mn(K)saochoA

=P1APlamo t matra n chc o. Casu A

= diag(1, . . . , n) vataca n tnhAk.1a co A = PA

P1. 1udosuy raAk= (PA

P1)k= PAkP1= Pdiag(k1, . . . , kn)P1.V du5. Cho A =_1 12 4_. 1a tnh duo c da thu c da c trung cu aA la PA() = ( 2)( 3). Ca c kho nggan conrc ng cu aA la :E(2) = u = (1, 1)) va E(3) = v = (1, 2)).\a y P=_ 1 11 2_ lama tra n la m chc o A vamo t da ng chc ocu aA laA

= P1AP=_2 00 3_.\i A = PA

P1nc n vo mo sotunhc n n ta coAn= PAnP1.14IoA

la matra nchc onc ndc da ngtnhduo cAn=_2n00 3n_.1c p thco,tnhduo cP1=_ 2 11 1_. IodoAn= PAnP1=_2n+13n2n3n2n+1+ 2.3n2n+ 2.3n_.5.2. Tmmo t heda y sotho a co ngthu c truy ho i1a mnh ho a ytuo ng tho ng quamo t v dusau da y:V du6. Ca su ca cda yso thu c(un)nZ+va (vn)nZ+tho aca cco ng thu c truy ho _un+1 = un vnvn+1 = 2un + 4vnvo _u0 = 2v0 = 1.Ha y timco ng thu c tnhca c soha ng to ng qua t (phuthuo cn)unva vn.Ia tXn =_unvn_va A =_1 12 4_.1bCo ng thu ctrc n duo c vc t la nhu sau:Xn+1 = AXnvo X0 =_21_.1udotnh duo cXn = AnX0.\o Andaduo c tnh ba ng co ng thu c pha trc nta co_unvn_=_2n+13n2n3n2n+1+ 2.3n2n+ 2.3n__21_=_2n+22.3n+ 2n3n2n+2+ 4.3n2n+ 2.3n_.\a y_un = 5.2n3n+1vn = 5.2n+ 6.3n.5.3. Cia i he phuongtrnh vi pha n tuye n tnh hesoha ngHcphuong trinh v pha n tuyc n tnh hcsoha ng coda ng sau da y:dx1dt= a11x1 + a12x2 +. . . + a1nxn;dx2dt= a21x1 + a22x2 +. . . + a2nxn;.......................................................dxndt= an1x1 +an2x2 + . . . +annxn,1btrongdo mo aijdc ula so thu cva mo xidc ula ha mthu ckha vtrc n R. Hc no trc ncothcduo c vc t la duo da ng ma tra n nhusau:dXdt= AX, vo A = (aij), X =x1x2...xn.1a go A lama tra ncu a hcphuong trinhv pha n dacho. CasuAchc oho aduo c, nghiala to nta matra nchc oA

va matra nkhangh}chPsao choA

= P1AP.Xc tAnhumatra ncu atoa ntu tuyc ntnhf: Rn Rntrongcoso chnhta c B0. Kh do to nta mo t cosoB=(u1, u2, . . . , un)go mtoa nca cvc ctorc ngcu afsaochomatra ncu aftrong BlaA

. \o x = (x1, x2, . . . , xn) Rnthi to ado Xcu axtrongcosochnh ta c B0va X

trongco so Bcomo lc n hcsau:X

= P1X,trongdo P= (B0 B) lama tra nchuyc n co so . 1udota codX

dt= P1dXdt.Luuy ra ngdoAla matra nvo ca chc so ha ngnc nPcu nglama tra nvo ca c hcsoha ng. 1uca c co ng thu c trc n,nha n duo cdX

dt= A

X

.17\i A

lama tra n chc o nc n hctrc n duo c ga mo t ca ch dcda ng dctimranghc mX

. Cuo cu ngnghc mXcu ahc phuongtrinhbanda u duo c tnh thcoco ng thu c:X = PX

.1o mla , nc uAlama tra nchc oho a duo c thihcphuong trinhvpha n dacho cothc duo c ga qua ca c buo csau:1. Chc oho a matra nA,nghialatimmatra nkha ngh}chPsaochoA

= P1APlama tra nchc o.2. Ca hcdX

dt= A

X

.J. 1imXbo co ng thu cX = PX

.a y gotascmnh ho a ba ng mo t v dusau:V du7. Ca hcphuong trinhv pha ndxdt= x y;dydt= 2x + 4y.Na tra ncu ahcla A =_1 12 4_. Ia y lamatra ndaduo cxc ttrong \ du1.b. 1a dabc t A chc o ho a duo c, P=_ 1 11 2_ la m18chc oAvaA

= P1AP=_2 00 3_.\c t la hcdX

dt= A

X

tha nhhcdx

dt= 2x

;dy

dt= 3y

.Nghc mcu ahc na yla_x

= C1e2ty

= C2e3t, trongdo C1va C2laca cha ng so . 1udoX =_xy_ = PX

=_ 1 11 2__x

y

_=_x

y

x

+ 2y

_.Suy ra_x = C1e2t C2e3t;y = C1e2t+ 2C2e3t.5.4. Da y Fihonacii11Ibonac(1170-12b0) conduocgola Lconardo dcPza, motthuongga nguoYno tcng boncm dam mctoanhoc.19Ia y sosauda y duo c go lada l/bcoac//:0, 1, 1, 2, 3, 5, 8, 13, 21, . . .No soha ng trong da y Ibonac (kctusoha ng thuba) ba ng to ngcu a ha soha ng du ng ngaytruo cno :Fk+2 = Fk+1 + Fk, k 0, F0 = 0, F1 = 1.1a cothcda t ca u ho lala m thcna o dctnh soha ng dulo n (thu1000 cha ng ha n) trong da y Ibonac makho ng pha tnh la n luo t tuca c so F0 = 0, F1 = 1:Ia tuk :=_Fk+1Fk_ va A =_1 11 0_. Kh douk+1 = Auk.1udosuy rauk = Aku0, vo u0 =_10_.\a n dcda n dc n vc c tnh Ak. Ia thu c da c trung fA() = 21coca c nghc m kha cnhau la1 =1 +52, 2 =1 52.20Iodo Achc o ho a duo c vamo t da ng chc ocu aA laD = P1AP=_100 2_, vo P=_121 1_.1a coP1=112_1 21 1_.1uca c co ng thu c trc nta tnhduo c_Fk+1Fk_= uk = Aku0 =11 2_k+11k+22k1k2_.1udosuy raFk =15__1 +52_k_1 52_k_.Co ng thu c cuo cu ng rora ng co thcta oramo t ba t ngo thu v} vica c soIbonac vo n laca c songuyc n nhung chu ng la duo c bc u dc nqua ca c pha n sovaca c ca n ba c ha. 1a t nhc n chu ng pha duo c ga nuo c thco mo t ca ch na o dodccuo cu ng nha n duo c nhu ng songuyc n.Nha n xc t ra ng15_1 52_k j vaA lama /ra o /am }/a c daa /nc uaij = 0, i < j.Ne nh de6.2. Hc / ma /ra o /am }/a c /re o de a dc o} da o} ca / mc / ma/ra o /am }/a c daa /.Chu ngminh. Ca su Ala matra ntamga ctrc nva f End(Kn)sao choma tra n cu aftrong cosochnh ta c B0 = (e1, . . . , en) la A.Xc t cosoB = (en, . . . , e1). 1atha y matra nA

cu aftrongcoso Blama tra n tamga c duo va A do ng da ng vo A

.a toa n da t ra lakh na o thi ma tra n A Mn(K) do ng da ng vo mo tmatra ntamga c: IoNc nhdc b.2nc ntachica nxc tkh na oma tra n A do ng da ng vo mo t ma tra n tam ga c trc n. 1hco ngo n ngucu aca ctu do ngca utuyc ntnhthi va ndc da trala kh na omo ttudo ng ca u tuyc n tnh duo c bc u dc n ba ng mo t ma tra n tam ga c trc ntrongmo t cosona o do .|nh ly6.3. f End(Kn) /am }/a c /c a daa c /// cac// /// da //a cda c /rao} ca a fp/a o ra/re oK.Chu ng minh. Casuftam ga c ho a duo c va B = (e1, . . . , en) lamo tco socu aV= Knsaocho [f]B = ...0 .1udota coPf() = deta11 ...0 ann = (a11 ) . . . (ann),2Jnghia la Pf() pha n ratrc nK.Nguo c la , gasuPf() pha n ratrc n K. 1a chu ng mnh ba ng quna p ra ng toa ntu ftam ga c ho a duo c.Nc un = 1 thi kho ng cogidcchu ng mnh. \a y, gasu n > 1 vakha ng d}nh du ng vo n = 1. Co 1 Klamo t nghc mna odocu aPf() va u1lamo t vc c torc ng u ng vo tr} rc ng1. otu c (u1) dccomo t co so c = (u1, u2, . . . , un) cu aV . 1acoA = [f]C =1b2. . . bn0... B0,vo Bla ma tra nvuo ng ca p n 1. Xc t kho ng ganconW=u2, . . . , en)va g: WWsaochomatra ncu agtrongcoso(u2, . . . , un) la B. 1a coPf() = det(AIn) = (1)det(B In1) = (1)Pg().\i Pf()pha nra trc nKnc nPg()cu ngpha nra trc nK, dodo thcogathc tquna pmatra nBtamga cho aduo c. \a y to nta mo t co so(e2, . . . , en) cu aWsao cho matra n cu agtrong dolamatra n tam ga c trc n. Kh doma tra n cu a ftrong co so(u1, e2, . . . , en)cu ng coda ng tamga c trc n.Hequa6.4. Hc / ma /ra o A Mn(C) de a /am }/a c /c a daa c.Nha n xe t.1)Nc umatra nAdo ngda ngvo matra ntamga cA

thi trc nduo ng chc o chnh cu aA

chi toa n laca c tr} rc ng cu aA.242) No matra nA Mn(R)dc utamga cho aduo ctrc ntrc ntruo ng sophu c C.Hequa6.5.C/cA Mn(R) ca SpA = 1, . . . , n C. K//dc/a ccTr(A) = 1 +. . . +nca detA = 1. . . n.Chu ngminh. Ioca cmatra ndo ngda ngdc uco cu ngvc tva cu ngd}nh thu c nc nnhu ng dc u ca n chu ng mnh lahc n nhc n.V du8. Na tra nA=4 0 20 1 05 1 3co dathu c da c trungPA() = ( +2)(1 )2nc n thco I}nh lyb.J,A tam ga c ho a duo ctrc n R. Xcm A nhu ma tra n bc u dc n tudo ng ca u tuyc n tnh ftrongcoso chnhta c. Kh do to nta mo tcosoB= (u1, u2, u3)saochoma tra ncu aftrong Bcoda ng tamga c trc n[f]B =1 a b0 1 c0 0 2.1a sctnh ca c vc c tou1, u2 vau3. Nha n xc t ra ng u1 chnh lavc ctorc ng u ng vo tr}rc ng1 = 1. 1acoA I3 =5 0 20 0 05 1 25 0 20 0 00 1 0.2bChox3 = 0 suy rax1 = 2. \a y co thcla yu1 = (2, 0, 5).1nhu2:1a co f(u2) = au1 + u2 =(f Id)(u2) = au1. Iodo5 0 20 0 05 1 2x1x2x3= a205.Ca hcphuong trinh trc n:5 0 2 2a0 0 0 05 1 2 5a5 0 2 2a0 1 0 3a0 0 0 0.Choa = 1, x3= 4 =x1= 2, x2= 3. Co thc la yu2=(2, 3, 4).1nhu3: 1abc t ra ngto nta vc ctorc ngvu ngvo tr} rc ng2 = 2,nghiala f(v) = 2v. 1aco thc cho nu3=v, b =c = 0.1a coA+2I3 =2 0 20 3 05 1 51 0 10 1 05 1 51 0 10 1 00 0 0.Io docothcla y u3 = (1, 0, 1). Kc m tra dcda ng u1, u2, u3 do cla ptuyc ntnh, dodo chu ngta otha nhmo tcoso cu a R3. 1rongco2bso Bmatra n bc udc n cu aflaA

=1 1 00 1 00 0 2.Na tra n chuyc n tu co sochnh ta c sang coso BlaP=2 2 10 3 05 4 1.Cuo cu ng taco A

= P1AP.7. a thuc triet tieu. |nhlyHamilton- CalleyChoVlamo t kho ng ganvc c totrc ntruo ngKva Q K[t]:Q(t) = amtm+am1tm1+. . . + a1t +a0.Nc u f EndK(V ) thi ta kyhc u Q(f) lamo t tudo ng ca u tuyc ntnh cu aVxa c d}nh bo Q(f) = amfm+am1fm1+ . . . +a1f + a0IdV.27Nha n xe t. Nc uP, Q K[t] thiP(f) Q(f) = Q(f) P(f), f EndK(V ).|nh ngha 7.1. Chof EndK(V )va Q(t) K[t]. 1ano Q(t)lada //a c /r/e / //e a toa ntu fnc uQ(f) = 0.Ne nh de7.2. C/asa Q(t) /ada //a c /r/e / //e a /ca o /afca /amc //r/ r/e o} ca a f. K// dc /ao}//e mca a Q(t).Chu ngminh. Co vla mo tvc ctorc ngcu afu ngvo tr} rc ng.Kh do fk(v) = kv, k N. CasuQ(t) = amtm+ am1tm1+. . . + a1t + a0lada thu c trc t tc uf. Kh dotacoamfm+am1fm1+ . . . +a1f + a0IdV= 0=(amfm+ am1fm1+. . . + a1f +a0IdV )v = 0=(amm+am1m1+ . . . +a1 +a0)v = 0.Iov ,= 0 nc n tu dosuy raamm+ am1m1+ . . . +a1 +a0 = 0 hayQ() = 0.A pdu ngmc nhdc vu achu ngmnhtatha yra ngnc utoa ntu ftho af2=fthi ca cga tr}rc ngcu afchico thc la 0hoa c 1. Nc uf3= fthi ca c tr} rc ng cu afchi co thcla0, 1 hoa c 1.1uy nhc n, cu ng ca n thc t luu yra ng kho ng pha ta t caca c nghc mcu aQ(t)dc ula tr} rc ngcu af. \ du , nc uf =IdVthi dathu cQ(t) = t2 t trc t tc ufnhung 0 kho ng pha latr} rc ng cu af.Ca uho da utc nma ta co thc da t ra la : Pha cha ngdo vo mo toa ntu tuyc ntnhf EndK(V ) dc uto nta mo t da thu c280 ,= Q(t) K[t] trc t tc u f:Ca u tralo lakha ng d}nh. 1ha t va y, nc udimK(V ) = nthiEndK(V )= Mn(K), suy radimK(EndK(V )) =n2. Io doca c pha n tuIdV, f, f2, . . . , fn2phuthuo c tuyc n tnh trongEndK(V ), suy ra to n ta ca c pha n tu a0, a1, a2, . . . , an2 K, kho ngpha ta t cadc u ba ng 0 saochoa0IdV+a1f +a2f2+ . . . +an2 fn2= 0.\a yQ(t) = a0 + a1t +a2t2+ . . . + an2tn2lada thu ctrc t tc uf.I}nh lyHamlton - Callcy mata scchu ng mnh duo da y cho tha yda thu c da c trung cu aflada thu c trc t tc uf.|nh ly7.3. /Hamlton- Callcy) Ne af EndK(V ) /// da //a c da c/rao}Pf() /r/e / //e af, o}//a /a Pf(f) = 0.Chu ng minh. Co Klabaodo ng da so cu aK. 1ruo chc t tachu ngmnhchotruo ngho pf EndK(V ). 1rongtruo ngho pna yftamga cho aduo c. Ca suB= (e 1, . . . , en)la mo tcoso cu aV saocho trongdoma tra n bc udc nfcoda ng tamga c trc n:[f]B =1...0 n.Kh dotacoPf() = (1) . . . (n ).1a ca n chu ng mnhPf(f) = (1IdV f) . . . (nIdV f) = 0.29i 1, n,da tgi = (1IdV f) . . . (iIdV f).1a sc chu ngmnhba ngqu na pthco i ra ng gi(e1) =. . . =gi(ei) = 0. Kh do , vo i = n tasccodc u ca n pha chu ng mnh.\o i = 1 ta cog1(e1) = (1IdV f)e1 = 1e1f(e1) = 0.Casu i > 1 vagi1(e1) = . . . = gi1(ei1) = 0.1a cogi = gi1(iIdV f) = (iIdV f)gi1.Iodogi(e1) = . . . = gi(ei1) = 0.Xc tgi(ei). 1a cof(ei) = a1e1 + . . . +ai1ei1 +iei,vo a1, . . . , ai1lanhu ng pha n tuna o dothuo cK. 1udosuy ragi(ei) = gi1(iIdV f)(ei)= gi1(iei (a1e1 + . . . +ai1ei1 +iei))= a1gi1(e1) . . . ai1gi1(ei1) = 0.1ada chu ngmnhPf(f)=0hayPA(A)=0, A Mn(K).a y go ,nc uA Mn(K) thi taxcmAnhumo t matra n trc nKvaa p du ng dc u vu a chu ng mnh ta sc co PA(A) = 0.J08. Bodecan hanBode8.1.C/cf EndK(V )ca Q(t) =Q1(t) . . . Qp(t), /rco}dcQ1, . . . , Qp/ao/a o} da//a c o}ae o /cca o} o/aa. K// dc .Ne aQ(t) /r/e / //e a f///V= KerQ1(f) . . . KerQp(f).Chu ng minh. 1ascchu ng mnh ba ng quna p thcop.Nc up = 1 thi Q(t) = Q1(t), do donc uQ(f) = 0 thiQ1(f) = 0va V= KerQ1(f).Traa o} /a p p = 2: Casu Q(t) = Q1(t)Q2(t), trong do Q1va Q2lanhu ng da thu c nguyc nto cu ng nhau. Kh do , to nta nhu ng da thu cU1va U2saochoU1Q1 +U2Q2 = 1.1udosuy raU1(f)Q1(f) + U2(f)Q2(f) = IdV.Iodo x V , tacox = U1(f)Q1(f)(x) + U2(f)Q2(f)(x) (1),kc othcoV= Im(U1(f)Q1(f)) + Im(U2(f)Q2(f)).\iQ2(f)Q1(f) = 0 nc nQ2(f)U1(f)Q1(f) = 0,suy raIm(U1(f)Q1(f)) KerQ2(f).J1Ho anto antuong tu ,tacoIm(U2(f)Q2(f)) KerQ1(f).Iodo V= KerQ1(f) +KerQ2(f).Casu x KerQ1(f) Q2(f). 1u(1) suy ra ngayx = 0. \a yV= KerQ1(f) KerQ2(f).Traa o} /a pp > 2: 1a coQ(t) = (Q1(t) . . . Qp1(t))Qp(t).Ia t Q(t) = Q1(t) . . . Qp1(t), ta co Q(t) vaQp(t) lanhu ng dathu c nguyc nto cu ng nhau. 1hcotruo ngho pp = 2 ta coV= KerQ(f) KerQp(f).Ia t W= KerQ(f) va f = f[W. 1a chu ng mnh f EndK(W).1ha t va y, x W, ta cof(x) = f(x). Io x Wnc n Q(f)(x) = 0,suy raQ(f)f(x) = fQ(f)(x) = 0,nghia la f(x) KerQ(f).Ngo a ra, do f= f[Wnc nQ(f) = 0. \a y, a p du ng gathc t quna p, nha n duo cW= KerQ1(f) . . . KerQp1(f).Nhung i 1, p 1 tacoKerQi(f) = x W[Qi(f)(x) = 0~x W[Qi(f)(x) = 0~KerQi(f).\a yV= KerQ1(f) . . . KerQp1(f) KerQp(f).J2Hequa8.2.C/cf/a mc / /ca o /a /ae o //o/ /rco}//c o}}/ao ce c/a n c//e a V/re o Kca}/asada //a c da c /rao} Pf() p/a o ra/re oK.Pf() = (1)n( 1)1. . . ( p)p, i,= j, i ,= j.K// dcV= Ker(f 1Id)1 . . . Ker(f pId)p.|nh ngha 8.3. Choflamo t toa ntutuyc ntnh trong kho ng ganvc c tonchc uVtrc nKvagasudathu c da c trungPf() pha n ratrc nK. 1ago N(i) := Ker(f iId)ila//c o} }/ao cco da c /rao},u ng vo tr}da c trungi.1hcoHc qua 8.2, nc udathu cda ctrungpha nra trc nKthi Vpha n tch tha nh to ng tru ctc p cu a ca c kho ng ganconda c trung.Nha n xe t. 1)Kho ng ganconrc ng luo nna m trongkho ng ganconda c trung(u ng vo cu ng mo t tr}da c trung):E() N().1ha t va y, nc u x E() thi (f Id)x = 0, suy ra (f Id)x =0 ( lasobo cu a) hayx N().2) Kho ng gan conda c trung laba t bc ndo vo f,nghia laf(N()) N().1ha t va y, gasu x N(). Kh doJJ(f Id)x = 0 =f (f Id)x = 0=(f Id) f(x) = 0 =f(x) N().Ickc t thu c mu c na y, duo da y ta scxc t mo t u ng du ng cu a odcca n ba n.Casu f EndK(V ) sao chof2= f. Kh do Q(t) = t(t 1) lada thu c trc t tc u f. \a y, thco odc8.1, V= Kerf Ker(f Id).1atha yVlato ng tru ctc p cu a ca c kho ngganconrc ng cu afnc nfchc oho a duo c.Iuo da y lamo t kc t qua to ng qua t honv duchu ng tavu a xc t.|nh ly8.4. Tca o /a/ae o //o/ fc/e c /c a daa c /// cac// /// /c o/a / mc / da //a c p/a o ra/re o K, cc/ca o o}//e m dao ca/r/e / //e a f.Chu ngminh. Ca su f chc oho aduo c. Kh do to nta mo t cosoB = (v1, . . . , vn) go mtoa nca c vc c torc ng cu af. Co 1, . . . , placa ctr} rc ngdo mo tkha cnhaucu af. Kh do , v Bto nta mo tj, 1 j p saocho (f jId)v = 0. 1udosuy ra(f 1Id) . . . (f pId)v = 0.\a y dathu cQ(t) = (t 1) . . . (t p)pha nratrc nK, chicotoa nnghc m donvatrc t tc uf.Nguo c la , gasuQ(t) = (t 1) . . . (t p), i,= j, i ,= jva Q(t) trc t tc uf. Kh dothcoodc8.1V= Ker(f 1Id) . . . Ker(f pId)= E(1) . . . E(p),suy rafchc o ho a duo c.J49. a thuc toi tie u|nh ngha 9.1. Iathu cdonkho ba cnho nha ttrc ttc utoa ntutuyc ntnhfduo c go lada//a c /c / //e a cu afvakyhc u la mf.Ne nh de9.2. a //a c Q(t) K[t] /r/e / //e a f/// cac// ///Q c//a/e / c/cmf/rco}K[t].Chu ng minh. Casu Q(f) = 0. ChaQ chomf:Q(t) = P(t).mf + R(t), deg(R) < deg(mf)(thco qu uo c, dathu c0co ba c ). \i Q(f) =0nc nsuyraR(f) = 0, kc othcoR(t) = 0dod}nhnghiadathu cto tc u. IodoQ(t) = P(t).mf.Nguo cla , nc uQ(t)=P(t).mfthi Q(f)=P(f)mf(f)=0,nghia la Q(t) trc t tc uf.Hequa9.3.mf/aaa c ca aPf.Chu ng minh. A p du ng I}nh lyHamlton-CallcyvaNc nh dc9.2.Hequa9.4. a //a c /c / //e a /adao/a /.Chu ng minh. Casu m1vam2laha da thu c to tc u cu a toa n tu f.Kh do , thcoNc nhdc 9.2, m1[m2va m2[m1. Iom1va m2dc ulaca c da thu c don kho nc ntu dosuy ram1 = m2.Ne nh de9.5. Ta po}//e mca amf/ra o}ca / /a po}//e mca aPf.Nc / ca c/ //a c, oe a Pfp/a o ra/re o bac dc o} da / scK ca a K//a o/Pf= (1)n(t 1)1. . . (t p)p, i,= j, i ,= jJb///mf= (t 1)1. . . (t p)p, 1 i i, i.Chu ngminh. 1u Hc qua 9.Jsuyramo nghc mcu a mfdc ulanghc mcu aPf. 1u Nc nhdc 7.2suyramo nghc mcu aPfdc ulanghc mcu amf.V du9.1)A =0 1 21 0 21 2 0.1a co PA(t) = (t + 1)(t + 2)(t 3). A p du ng Nc nh dc9.b suyramA(t) = (t + 1)(t + 2)(t 3).2)A =1 1 11 1 11 1 1.1a co PA(t) = (t 1)(t + 2)2. A p du ng Nc nh dc9.b suy ramA(t) =_(t 1)(t + 2)(t 1)(t + 2)2.1a co(A I3)(A+ 2I3) = 0. \a ymA(t) = (t 1)(t + 2).|nh ly9.6. Tca o /a/ae o //o/ fc/e c /c a daa c oe a cac// oe a daJb//a c/c / //e aca aoc p/a ora /re oKca /a / ca ca co}//e mca aocde a /ao}//e mdao.Chu ng minh. Ic u kc n duladu ng do I}nh ly8.4. Nguo c la , gasufchc oho a duo c. Kh doPf= (1)n(t 1)1. . . (t p)p, i,= j, i ,= j.1hco chu ng mnh I}nh ly8.4 da thu c Q(t) = (t 1) . . . (t p)trc t tc uf. A p du ng Nc nh dc9.b suy ramf= Q(t).V du10.1) Xc t ma tra nA =1 1 11 1 11 1 1. Iathu c to tc ucu aAla mA(t) = (t 1)(t 2). 1hcoI}nhly9.b,Achc o ho a duo c.2) Na tra nA =3 2 21 0 11 1 0 coda thu c da c trungPA(t) =(t 1)3nc nmA(t) =t 1(t 1)2(t 1)3.1hcoI}nhly 9.b, Achc oho aduo c mA(t)=t 1 A I3 = 0. IoA ,= I3nc ntu dosuy raAkho ng chc o ho a duo c.J7J) Na tra n A =3 1 12 0 11 1 2coda thu c da c trung laPA(t) =(t 1)(t 2)2,do domA(t) =_(t 1)(t 2)(t 1)(t 2)2.1hcoI}nh ly9.b, tacoA chc oho a duo c mA(t) = (t1)(t2) (AI3)(A2I3) = 0.Nhung ba ng ca ch tnh toa ntru ctc p ta tha y ra ng(A I3)(A2I3) ,= 0,do do A kho ng chc oho a duo c.10. Dang tamgiac khoi1rong mu c b ta dabc t nc u da thu c da c trung cu a toa n tuftrongkho ng ganvc c tohu u ha n chc u pha n ratrc ntruo ngcoso Kthi ftam ga c ho a duo c, nghia lato n ta mo t co socu a Vsao cho trong domatra nbc udc nfco da ng tamga c(trc nhoa cduo ). No chung,dc u na y cu ng dacho chu ng ta nhu ng u ng du ng khato t. 1rong mu cna y chu ng ta tc p tu c vc c ru t go n toa n tu fsao cho cothcto t honnu a. Chnh xa c hon, nc u da thu c da c trung cu a fpha n ratrc nKthitacothcduafvcda ng tamga c kho .J8Bode10.1. C/cV= V1 V2. . . Vp, /rco} dc Vi/aca c //c o}}/aoccoba / b/e odc / ca / f. K// dc , oe a B1, B2, . . . , Bp/aao}a o}/a ca ccasa ca aV1, V2, . . . , Vp/// ma/ra oca af/rco}casaB=(B1, B2, . . . , Bp) /a[f]B =M10M2...0 Mp=: diag(M1, M2, . . . , Mn),/rco} dc Mi/ama /ra o ca a /a o c/eca a f/e oVi.Chu ng minh. Casu B1 = (u1, . . . , un1), . . . , Bp = (v1, . . . , vnp). \if(Vi) Vi, i nc ntacof(u1) = a11u1 + . . . +an11un1;...............................................f(un1) = a1n1u1 + . . . +an1n1..................................................f(v1) = b11v1 +. . . + bnp1vnp;..............................................f(vnp) = b1npv1 + . . . +bnpnp.Iodo[f]B = diag(M1, M2, . . . , Mp).J9|nh ly10.2. /Ru t go n theo da ng tam gia c kho i) C/c f/amc / /ca o/a/ae o //o/ /re o //c o} }/ao ce c /a n c//e a /re o /raa o}K. C/asada //a c da c /rao} ca a fp/a o ra/re oK.Pf() = (1)n( 1)1. . . ( p)p, i,= j, i ,= j.K// dc/c o /a / mc / ca sa B = B1B2. . . Bpca a V , /rco} dcBi/amc / ca saca a N(i) sacc/c[f]B = diag(M1, M2, . . . , Mp),ca /Mi/ama /ra o b/e a d/e o ca a /a o c/eca a f/e o //c o} }/ao ccoda c /rao}N(i) /rco} ca sa Bica Miccda o} /am }/a c /re o.Chu ngminh. \i N(i)ba tbc ndo vo fnc nthcoHc qua 8.2vaodc10.1, to nta mo t co so B = B1 B2 . . . Bpcu aVsaocho[f]B = diag(M1, M2, . . . , Mp),trongdoBila mo tcoso cu aN(i), Mi= [fi]Bi, vo fila ha nchccu af lc nN(i). Iodo tachi co nca nchu ngmnhra ngMitamga c ho a duo c va SpMi = i, . . . , i ladu .1hco d}nh nghia, N(i) = Ker(f iID)inc n (f iId)ix =0, x N(i). \a y (t i)ila da thu ctrc ttc ufi, dododathu cto tc ucu aficoda ngmfi(t) = (t i)i, vo 1 i i.A p du ng Nc nh dc9.b,suy raPfi(t) = (1)i(t i)i, vo i i.\a yPfi(t)pha nra trc nKva Spfi= i, . . . , i, kc othcoMitamga cho aduo c. 1aco nca npha chu ngmnhi =i, i 1, p.1hu c va y, taco40Pf(t) = [M1tI[ . . . [Mp tI[= Pf1(t) . . . Pfp(t) = (1)n(t 1)1. . . (t p)p.= (1)n(t 1)1. . . (t p)p.1udosuy rai = i, i 1, p.V du11. Chomatra nA=1 1 2 20 0 1 11 1 1 01 1 1 0. XcmAnhumatra ncu atoa ntu tuyc ntnhf: R4 R4, trongcoso chnhta c B0=(e1, e2, e3, e4). \i Pf(t)=t2(t 1)2nc nto nta cosoB = (u1, u2, u3, u4)saochotrongcoso na ymatra ncu afco da ngtamga c kho [f]B =1 a 0 00 1 0 00 0 0 b0 0 0 0.Ic u na y conghiaf(u1) = u1f(u2) = au1 +u2f(u3) = 0f(u4) = bu3.\c c to u1 lavc c to rc ng cu a fu ng vo tr} rc ng = 1. Cothcla yu1 = (0, 0, 1, 1). 1hc u1va o co ng thu c (f Id)u2 = au1 dctnh u2.41a ngca chchoa = 1, co thc la yu2= (1, 0, 1, 1) (takho ngthc la ya = 0 vikhdoca cvc ctou1va u2sc phuthuo ctuyc ntnh). u3lavc ctorc ng u ngvo tr} rc ng = 0nc nco thc la yu3 = (1, 1, 0, 0).1hc u3va o co ng thu c f(u4) = bu3dctnhu4. \o b = 1 cothctnhduo cu4 = (0, 1, 1, 0). Luu yra ng ta cu ng kho ng thcla yb = 0 vi khdoca c vc c tou3va u4sc phuthuo ctuyc n tnh. \a y[f]B =1 1 0 00 1 0 00 0 0 10 0 0 0.Na tra n chuyc n cosotu B0sang BlaP=0 1 1 00 0 1 11 1 0 11 1 0 0.Nhuva ychu ngtada tha yra ngnc umo tmatra ntamga cho aduo cthi taco thc la mduo cnhc uhonthc nu aba ngca chduamatra nbc udc nno vc da ng tamga ckho . Ic una yco lo hontrongca c u ng du ng. 1hu c va y, nc u do vo mo t ma tra n tam ga c vc c na nglc nlu y thu ano chunglakho ngthc duo c. 1uynhc n, nc uchu ngtaduano vc da ngtamga ckho thi vc cna nglc nlu ythu asc duavctnhlu ythu acu aca ckho trc nduo ngchc ochnh. \c pha nminh,mo kho nhu va y dc u lamo t ma tra n tam ga c mata t caca c pha n tuna mtrc nduo ngchc ochnhdc uba ngnhau. \a y ba toa nthu ccha t42dua vcvc c na ng lc nlu y thu a cu a matra n tamga c da ngA = ...0 .Pha n tch A = I +N, trong doN=0 ...0 0. 1a chu ngmnhto nta p nsaochoNp=0. 1hu cva y, ta co PN(t) =(1)ntn,nc nmN(t) = tp,vo p n. IomN(N) = 0 nc nNp= 0.a ygo nha nxc tra ngIva Ngaohoa nvo nhaunc ntadc da ngtnh duo cAk= (I + N)kba ng ca ch kha trc nnh} thu c Ncwton.11. Dangchnhtac JordanIa ng tam ga c kho no chung dacothcdu ng khato t cho nhu ngu ngdu ng. 1uynhc n, taco thc tc nha nhru tgo ntrongtu ngkho chodc nkhnha nduo cmo t da ng ma thcomo tnghiana odo ladao}/a o o/a /. Io chnh la da o} c//o//a c Icrdaoma tasc dcca pduo da y.4J|nh ngha 11.1. 1a go ma tra n da ng sau da y lamo t //c / Icrdao:J() = 1 0......... 10 .Nc u ca p cu a kho ba ng 1 thi taqu uo cJ() = ().Bode11.2. C/asa J() /amc / //c / Icrdao ca pn. K// dc/a cc .//PJ(t) = (1)n(t )n./// mJ(t) = (t )n.//// dimE() = 1.Chu ng minh. a ng ca ch tnh toa n tru c tc p ta tha y ngay nhu ng dc uca n chu ng mnh.|nh ly11.3. /Jordan)C/cf EndK(V )sacc/cPf(t)p/a ora/re oK.1 C/asa fc// ccmc / /r/ da c /rao} caPf(t) = (1)n(t )n, mf(t) = (t ), dimE() = .K// dc , /c o /a / mc / ca sa Bca aVsacc/c[f]B =J1() 0J2()...0 J()=: J(),44/rco} dc .aJk() /a//c / Icrdao,b Ca p ca a //c / /a o o/a / /a ,c Scca c //c / Icrdao /a .? Ne afccca c /r/ r/e o} //a c o/aa1, . . . , pcaPf(t) = (1)n(t 1)1. . . (t p)p/// /c o /a / mc / ca sa Bca aVsacc/c[f]B =J(1) 0J(2)...0J(p).Chu ngminh. 1)Ca su Pf(t)= (1)n(t )n. \i Pf(t)pha nratrc n Knc n ftam ga c ho a duo c. Io doto n ta mo t co so B

sao cho[f]B = ...0 =: A.Pha ntchA =In + N,vo N=0 ...0 0 = [u]B . Luuy ra ngf=IdV+ u. 1atha yula toa ntu lu ylnh, nghiala to n4bta mo t so tu nhc npsaochoup= 0. \imatra nbc udc ntoa ntuIdVtrongmo coso dc ula Innc nba toa nduavc vc cru tgo ntoa n tulu y lnh. \a y, gasu u latoa n tulu y lnh va laba c lu y lnhcu au. Kh do ,da thu cto tc umu(t) = t. Iodo ,uchc oho a duo c=1 u= 0. \a y, trongnhu ngchu ngmnhsauda ytascga thc tu ,= 0. \c c chu ng mnhd}nh lysc duo c thu chc n tu ngbuo cqua mo t loa t ca c bo dc .Bode11.4. C/cu/a /ca o/a /a //o/ca /ba c/a //o/> 1. K//dc , ca c d/e a //e o daa / da /aao} daao}.// = n, o}//a/a Pu(t) = (1)nca mu(t) = tn./// Tc o /a / ce c /a 0 ,= x Vsac c/c(x, u(x), u2(x), . . . , un1(x))/a mc / casa ca aV /mc / casa o/aca se daa c}c / /a mc / cosoxyclc, cc o /c ao /a usedaa c }c / /atoa ntuxyclc.//// Tc o /a / mc / ca sa Bca a Vsacc/c [u]B = J(0).Chu ngminh. () =(). Ia t B= (un1(x), . . . , u2(x), u(x), x).1a co[u]B = J(0).() =(). Ca suB= (e1, . . . , en)saocho [u]B=J(0).Khdo , da t x = en ta co(x, u(x), u2(x), . . . , un1(x)) lamo t co soxyclccu aV .() =(). Nc u co() thi thco11.2, = n.() =(). Ca su co (). \i un1,= 0nc nto nta 0 ,=x Vsao choun1(x) ,= 0. Ia ten = x, en1 = u(x), . . . , e1 = un1(x),4bta scchu ng mnh (e1, . . . , en) lamo t co socu a V . Casu

nk=1kek =0, nghiala nk=1kunk(x)=0. La ya nhha vc quaun1, nha nduo cn = 0. 1uongtu nhuthc , ba ng ca ch na y tala nluo t thuduo cn1 = 0, . . . , e1 = 0. \a y B = (e1, . . . , en) lamo t co socu a V . Honnu a [u]B = J(0).Nhocoodc11.4 ba toa n cu a chu ng ta duo c dua vcvc c chu ngmnh dc u sau da y:Nc uula toa ntu lu ylnhthi V la to ngtru ctc pcu aca ckho ngganconba tbc ndo vo u, saochoha nchc cu aulc nmo kho nggancon na y lamo t toa ntu xyclc.Bode11.5. K//e a Kp := Kerup, /a ccda 0 = K0 K1 K2 . . . K1 K = V.Chu ng minh. \iup(x) = 0 =up+1(x) = 0, nc nKp Kp+1. Na tkha c, nc uto nta p 1, 1 sao choKp = Kp+1thi tacoKp = Kp+1 = . . . = K = V.Ic u na y da n dc n p laba c lu y lnh cu a u, ladc u ma u thua n vi p < .Bode11.6. Tc o/a / ca c//c o}}/aoccoM1, . . . , M//c o}/a m//aa o} //c a ma o ca c d/e a//e o.//Kp = Kp1 Mp, p 1, ;/// u(Mp) Mp1, p 2, p.Chu ngminh. \o p=, cho nMla mo t pha nbu na odo cu aK1trongK=V . Ca su da xa y du ngduo cca ckho ngganconM, M1, . . . , Mp tho a ma n ca c dc u kc n () va(). 1a scxa y du ng47kho ngganconMp1saochono cu ngtho ama n()va (). 1ruo chc t takc m traca c dc u kc n sau da y:(a)u(Mp) Kp1;(b)u(Mp) Kp2 = 0.Casux Mp. \i Mp Kp nc n up(x) = 0, hay up1(u(x)) = 0,nghia la u(x) Kp1. 1adachu ng mnh (a).a y go , gasuy u(Mp)Kp2. Kh do , ta coy = u(x), x Mpva up2(y)=0, kc othcoup1(x)=0, nghiala x Kp1. \a y,x Kp1 Mp= 0, suyrax = 0, kc othcoy=u(x) = 0. 1adachu ng mnh(b). Kc t ho p (a)va(b)ta coKp2u(Mp) Kp1.Iodo to nta Gp1la pha nbu cu aKp2 u(Mp)trongKp1,ngh|ia laKp1 = Kp2u(Mp) Gp1.a y gonc uda tMp1 = u(Mp) Gp1thi tatha yMp1tho a ca c dc u kc n () va().Bode11.7. \a /M1, M2, . . . , Mo/a /rco} 8cde 11.6, /a ccV= M1M2M.Chu ng minh. 1acoV= K = K1 M48= K2 M1M= . . . ............................= M1 +M2. . . M.1c pthco, tasc xa ydu ngmo tcoso cu aV ba ngca chxa ydu ngco socho mo kho ng gan con Mp, sau dosa p xc p la ca c vc c to co sota pha n hoa ch notha nh ca c co soxyclc cu a ca c kho ng gan con cu aV . 1rongcoso cuo cu ngna ymatra nbc udc nfsc co da ng kho duo ngchc oma trc nduo ngchc ochnhla ca ckho Jordan. \i mo kho Jordanchichu adu ngmo tvc ctorc ngnc nso ca ckho Jordandu ng ba ng sochc u cu a kho ng ganconrc ngE().Bode11.8. 4 o/ ca a mc / ca saca a Mp /amc / /cdc c /a p /ae o //o//rco}Mp1.Chu ng minh. Casu(v1, . . . , vr) lamo t co socu a Mp va1, r Ksao cho1u(v1) + . . . +ru(vr) = 0.Kh do , u(1v1+. . .+rvr) = 0, hay 1v1+. . .+rvr Keru =K1 Kp1. 1u dosuy ra1v1 +. . . +rvr Mp Kp1 = 0, kc othco1 = . . . = r = 0.a y go tasc la nluo txa ydu ng coso choMpba t da utu p =.1rongM(ma taco nky hc ula G), la ymo tcoso ba tky na odo .1c p thco, ba ng qu na p lu ta scxa y du ng co sotrongMp1 nhu sau:1aco MP1= u(Mp) Gp1. La y a nhquaucu acoso da xa ydu ngtrongMpho pvo mo tcoso cu aGp1tasc co mo tcoso cu aMp1. a ngca chna ytaxa ydu ngduo cca ccosoB, B1. . . , B1tuongu ng cu aM, M1. . . , M1:49B = (v1, . . . , vn). .G,B1 = (u(v1), . . . , u(vn), 1, . . . , n1. .G1),B2 = (u2(v1), . . . , u2(vn), u(1), . . . , u(n1), z1, . . . , zn2. .G1),B1 = (u1(v1), . . . , u1(vn), . . . , x1, . . . , xn1. .G1).\o k 1, va0 ,= x Gk, da tIk(x) := x, u(x), . . . , uk1(x)).Bode11.9. k 1, /a cc//dimIk(x) = k./// Ik(x) /a//c o} }/ao cco ba / b/e o dc / ca /u.//// u[Ik(x)/a/ca o /avc//c.Chu ng minh. () Casu 0, 1, . . . , k1 Ksaocho0x +1u(x) +. . . + k1uk1(x) = 0.1a cdo nguk1lc nha vc , nha nduo c0uk1(x)=0. Iox ,Kk1nc nuk1(x) ,= 0, kc othco0 = 0. Iodo1u(x) +. . . +k1uk1(x) = 0.b01a cdo nguk2lc nha vc , nha nduo c1=0. Cu tc ptu cquatrinhnhuva y, cuo cu ngtanha nduo c0=1 =. . . =k1= 0.\a ydimIk(x) = 0.() Hc n nhc n.() Iod}nh nghia toa ntu xyclc.0uaca chxa y du ng ca ckho ngganconIk(x) tanha ntha yra ngVlato ngtru c tc p cu a ca c kho ng ganconI(v1), . . . , I(vn), I1(1), . . . , I1(n1), . . . , I1(x1), . . . , I1(xn1).Nhuva ymatra nbc udc ntoa ntu f trongcoso cu aV nha nduo cba ngca ch ghc pca c coso xyclc cu aca ckho ngganconIk(x)sc coda ng kho duo ng chc omamo kho la mo t kho Jordan. 1adachu ng mnh Pha n 1) cu a I}nh ly11.J. Pha n 2) cu a d}nh lyna y duocchu ng mnhnhu sau:1ruo chc tpha ntchV tha nhto ngtru ctc pcu aca ckho ngganconda ctrung, saudo a pdu ng Pha n1)do vo mo kho ngganconda c trung. Nhu va y d}nh lydaduo c chu ng mnh hoa ntoa n.V du12. Cho matra nA =1 0 0 00 1 0 01 2 3 12 4 4 1.1im da ng chnh ta c Jordan cu a A. Xcm A nhu ma tra n bc u dc ntoa ntu tuyc ntnhf EndR(R4). Ha ychi ro coso trongdo fcob1da ng chnh ta c Jordandanc u.Cia i. PA() = ( 1)4. IoA I4 ,= 0va(A I4)2= 0nc nmA() = ( 1)2. 1atnhduo cdimE(1) = 3. Iodo da ngchnhta c Jordancu aA laA

=1 1 0 00 1 0 00 0 1 00 0 0 1.Co (u1, u2, u3, u4) la coso trongdo f co da ngchnhta cno trc n. 1atha yca cvc ctou1, u3va u4ca npha cho nsaochochu ngta o tha nh co socu a kho ng gan con rc ng E(1). Hon nu a u1 ca n pha duo c cho n sao cho ta cothctim duo c u2 tho a f(u1) = u1+u2. a ngca ch tnh toa ntru c tc p tacothcla y ca c vc c tosau:u1 = (0, 0, 1, 2), u2 = (1, 0, 0, 0), u3 = (0, 1, 0, 2), u4 = (1, 0, 0, 1).V du13. A =1 0 0 20 2 0 00 0 2 30 0 0 2.1aco PA() =( 1)( 2)3, dimE(2) = 2, suyraN(2)coha kho Jordan. IodimN(2) = 3 nc nN(2) como t kho Jordanca p1va mo tkho Jordanca p 2. 1u do suyradathu cto tc ucu aAlab2mA() = ( 2)2. Ia ng chnh ta c Jordancu aA laA

=1 0 0 00 2 0 00 0 2 10 0 0 2.BaitapBa i 1. ChoA, B, C, D Mn(R), vo CD = DC. Chu ng mnh ra ngdet_A BC D_= det(AD BC).Ba i2. ChoA, B, C, D Mn(R)la ca cmatra ntho aCD=DC.Chu ngmnhra ngmatra nX=_A BC D_kha ngh}chkh va chikhY= AD BCkhangh}ch.Ba i 3. Kyhc u Kn[t] lakho ng gan vc c to go m ca c da thu c cu aK[t]coba c n. Cho toa ntutuyc ntnhf: R2[t] R2[t],duo c xa c d}nh nhusau:f(Q) = (2t + 1)Q (t2 1)Q

, Q R2[t].bJHa y tnhfn(a0 + a1t +a2t2):Ba i 4. Cho Vlakho ng gan vc c to thu c go m ta t caca c ma tra n thu cca p 2 covc t ba ng 0.(a) 1im mo t cosovasochc u cu aV .(b) ChoB=_1 02 3_va f : VV duo cd}nhnghiabo f(X) =XB BX, X V . Chu ngmnhra ngfla mo ttoa ntutuyc n tnh trong kho ng ganVvatnhfn(A), vo A =_a bc a_.Ba i 5. CasuIbonac xa y du ng da y socu a minh vo F0 = 1, F1 = 3va Fk+2=Fk+1 + Fk, k 0. Ha ytnhca cso Ibonac mo vachu ng mnhra ng ti so Fk+1/Fkcu ng da n to ti lcva ng.Ba i 6. Ha y timdc u kc ndo vo ca c so thu ca, b, c saochomatra nsau da y chc oho a duo c:A =1 a b0 2 c0 0 2.Ba i7. Cho Rla truo ngso thu cva f: R3 R3la mo t toa ntutuyc ntnh trongkho ng gan vc c to R3duo c xa c d}nh bo co ng thu cf(x1, x2, x3) = (x1x2 +x3, 2x1 + 3x2, 2x1 +x2 + 2x3),do vo mo pha n tu(x1, x2, x3) R3.b4(a) Chu ng mnh ra ng toa n tu fchc o ho a duo c trc n R vatim mo tcoso cu a R3saochomatra nbc udc ntoa ntu ftrongcoso do lamo t ma tra n chc o.(b) \o mo songuyc nn 2,chu ng mnh ra ng to nta mo t toa ntu g : R3R3saochogn= f.Ba i8. Io vo mo matra nduo da yha yduavc da ngtamga cvachi roma tra n khangh}chPla m tamga c ho a no :(a)A =3 1 12 0 11 1 2; (b)A =3 2 21 0 11 1 0.Ba i 9. Ca hc phuongbtrinhv pha ndXdt= AX,vo Ala matra ntronga 8.Ba i 10. 1imda thu c to tc ucu a ca c matra n tronga 8.Ba i11. Ca su toa ntu tuyc ntnhf: R3 R3co matra nbc udc n trong cosochnh ta c laA =1 0 01 2 11 0 1.Ha ytimdathu cto tc ucu afvapha ntch R3tha nhto ngtru ctc p cu a ca c kho ng gan conda c trung.Ba i 12. 1imda ng chnh ta c Jordancu a ca c ma tra nbb(a)A =1 1 0 00 1 0 01 2 3 12 5 4 1; (b)1 3 0 32 6 0 130 3 1 31 4 0 8;(c) A =3 1 1 79 3 7 10 0 4 80 0 2 4; (d) A =1 1 1 . . . 10 1 1 . . . 10 0 1 . . . 1. . . . . . . . . . . . . . .0 0 0 . . . 1.bbChuong IIKH0NC CIAN FUCIID1. Tchvohuong|nh ngha 1.1. Cho Vlakho ng gan vc c to trc n truo ngK. A nh xa : VV Kduo c go lamo t da o} sco} /ae o //o/ trc n Vnc u x, y, z V, , K, taco()(x +y, z) = (x, z) + (y, z)()(x, y + z) = (x, y) +(x, z).|nh ngha 1.2. ChoVlakho nggan vc c totrc ntruo ngsothu c R.Ia ngsongtuyc ntnh , ): VVRduo cgo la mo t //c/cc/aa o} trongVnc u no la :b7() Io xu ng,nghiala x, y) = y, x), x, y R;() xa cd}nhduong, nghialax, x) 0, x V vax, x)=0 x = 0.Casu V la mo tkho ngganvc ctothu cvo tchvo huo ng. Nc uV hu uha nchc uthi tano V la //c o}}/ao/ac//d; nc uV vo ha nchc u thi tano Vla//c o} }/ao H//ber/.V du14. Chokho ngganvc ctoV= Rn,vo x = (x1, . . . , xn)vay = (y1, . . . , yn), d}nh nghiax, y) := x1y1 +. . . + xnyn.Khdo V la kho ngganLucld. 1chvo huo ngvu a d}nhnghiaduo cgo la//c/ cc/aa o} c//o/ /a c trong Rn.V du15. \o x = (x1, x2, x3), y = (y1, y2, y3) R3d}nh nghiax, y) := x1y1 + 2x2y2 + 3x3y3 + x1y2 + x2y1.Ia yla mo ta nhxa songtuyc ntnhdo xu ng. Honnu a, nhu ngtnh to anduo da y cho tha y noco nlaa nh xaxa c d}nh duong.x, x) = x21 + 2x22 + 3x23 + 2x1x2= x21 + 2x1x2 + x22 +x22 + 3x23= (x1 + x2)2+x22 + 3x23 0.1u do suyra x, x) = 0 x1 + x2=x2 =x3 = 0 x1=x2 = x3 = 0.V du16. Xc t kho ng gan vc c to M2(R) go m ca c ma tra n vuo ng ca p2 trc ntruo ng sothu c R. A nh xa A, B) := Tr(A

B) lamo t tch vohuo ng trongM2(R).b8V du17. P, Q R[x], d}nh nghiaP, Q) :=_10P(x)Q(x)dx.Hc n nhc nda y lamo t da ng song tuyc ntnh do xu ng trc n R[x].1asc chu ngmnhno xa cd}nhduong. 1ha t va y, tacoP, P) =_10P(x)2dx 0. Casu P, P) = 0. \iP(x) lamo t ha m lc n tu c vaP(x)2 0nc ntu dc ukc n _10P(x)2dx = 0suyraP(x)[[0,1] = 0.Iodathu cP(x)chi co thc co mo t so hu uha nnghc mnc ntu dosuy raP(x) 0.V du18. ChoWlamo t kho ng ganconcu a kho ng ganvc c toV .CasutrongVcotchvohuo ng , )V . x, y W, d}nh nghiax, y)W:= x, y)V.Ic tha y da y lamo t tch vohuo ng trongW.2. Da ngru t gon Causs1rongtc tna ytakha osa tva ndc la mthc na odc bc tduo cmo tda ng song tuyc n tnhdo xu ng trc nkho ng gan vc c toVlaxa c d}nhduonghaykho ng. Ic una yco nghiala la mthc na odc nha nbc tb9mo ta nhxa songtuyc ntnhdo xu ngco pha la mo ttchvo huo nghay kho ng.ChoV la kho ngganvc ctonchc utrc nKvo (e1, . . . , en)lamo tcoso cu aV . Ca su la mo tda ngsongtuyc ntnhtrc nV vax = ni=1xiei, y = ni=1yieila ca cvc ctoba tky cu aV duo cbc udc n tha nh to ho p tuyc ntnh cu a ca c vc c toco so . kh dota co(x, y) = _n

i=1xiei,n

i=1yiei_=n

i,j=1xiyj(ei, ej).Nc u da taij = (ei, ej) thi ta co(x, y) =n

i,j=1aijxiyj.\a y, nc u lamo t da ng song tuyc n tnhtrc nVthi trong mo t coso na odo cu aV bc uthu ccu aduo cvc tduo da ngto ngcu aca cdon thu c trong doxi vayjdc u xua t hc n duo da ng lu y thu a ba c nha t.Ia y lada u hc u dcchu ng ta dcda ng nha n da ng ca c da ng song tuyc ntnh. \ du(x, y) = x1y1 +2x2y2x3y +x1y3lada ng song tuyc ntnh trc nkho ng gan R3,trong kh(x, y) = 4x1y2+ . . . kho ng thclada ng song tuyc ntnh duo c.\c ckc mtraxcmmo t da ngsongtuyc ntnhco do xu nghaykho nglamo tvc c dc da ng. 1ha t va y, nc utrongbc uthu ccu ada ngsongtuyc ntnhtado va tro cu axva ychonhauma kho ngthaydo thino lada ng songtuyc ntnhdo xu ng. 1rongtruo ngho pnguo cla kho ng do xu ng.b0a ygo choK=Rva ta ca nkc mtra xcmmo t da ngsongtuyc ntnhdo xu ngtrc nkho ngganvc ctothu cco xa cd}nhduonghaykho ng. Iuo da yla mo tphuongpha p thuo cvc Causschophc pchu ngtala mduo cdc udo . Luuy la tada du ngphuongpha pna ytrongkh trinh ba y \ du2.Casu (x, x) duo c vc t duo da ng da thu c(x, x) = a11x21 + . . . +annx2n +. . . + aijxixj + . . .nghiala (x, x) duo cvc t duo da ngmo tdathu cthua nnha tba c 2thcoca ccbc nx1, . . . , xn. 1ago ca cdonthu cchu ax2ila ca cdao//a c cac o} vaca c don thu cchu a xixjlaca c dao //a c c/ao/a /.Ne nh de2.1. Ne a /amc / //c/ cc/aa o} /rco}V/// /a / caca c /esc aiica a ca c dao //a c cac o} de a /aca c scdaao}.Chu ng minh. 1ha t va y, nc uto nta mo ti 1, n sao choaii 0 thiaii = (ei, ei) 0, suy ra kho ng xa c d}nh duong.1hco mc nh dcvu a chu ng mnh, nc u trong bc u thu c cu a (x, x)kho ngco du ndonthu cvuo ngthi kho ngxa cd}nhduong. 1uynhc n, nc ula da ngsongtuyc ntnhdo xu ngsaochotrongbc uthu ccu a(x, y) co chu adu ndonthu cvuo ngthitachuaducosodc kc t lua nco pha la mo t tchvo huo nghaykho ng. Caussdadc xua t mo tphuongpha p chophc pchu ng taco thc d dc nmo t kc tlua n nhu va y. 1hu c cha t cu a phuong pha p Causs ladua bc u thu c cu avc da ngto ngcu anhu ngdonthu cvuo ngba ngca cphc pbc ndo tuyc ntnh. Ic dc hc utasc la mdc una ytho ngquamo tvdu cuthc . 1ruo ngho pto ngqua tcu ngduo cla mhoa ntoa ntuongtu nhuva y.V du19. 1rong R3xc t da ng songtuyc n tnhdo xu ng(x, y) = x1y1 + 2x2y2 + 5x3y3 +x1y22x2y3 +x2y1 2x3y2.b11a co(x, x) = x21 + 2x22 + 5x23 + 2x1x24x2x3.1uy(x, x)chu adu 3so ha ngvuo ng, nhungnhutrc nda no ,tachuathc kc tlua nco xa cd}nhduonghaykho ng. Phuongpha pCauss tra quaca c buo csau:1. Xc tmo t donthu cvuo ng, x21cha ngha n. Sa pxc pta t ca ca cdon thu ccochu ax1la vo nhau:(x, x) = (x21 + 2x1x2). .b/e a //a c c/a ax1+2x22 + 5x234x2x3.2. \c t bc uthu c cochu ax1tha nh mo t binh phuong:(x, x) = (x1 +x2)2. .b/e a //a c c/a ax1x2..b/e a //a c //e m ca c+2x22+5x234x2x3.J. \c t tha nhto ngcu amo t binhphuongva mo t bc uthu ckho ng chu ax1:(x, x) = (x1 +x2)2+ x22 + 5x23 4x2x3. .b/e a //a c //c o} c/a ax1.4. La p la quatrinh nhutrc ndo vo bc u thu c kho ngchu a x1:(x, x) = (x1 + x2)2+ x22 4x2x3. .b/e a //a c c/a a x2+5x3.b2= (x1+x2)2+ (x2 2x3)2. .b/e a //a c c/a a x24x3. .b/e a//a c //e m ca c+5x3.1c ptu cla mnhuva y chodc nkhna ota t ca ca c donthu cnha nduo c dc u vuo ng; kh doquatrinhsckc t thu c:(x, x) = (x1 +x2)2+ (x2 2x3)2+x23.c u thu c cuo cu ng scduo c go lada o} ra / }c o Caass cu a (x, x).Luu yra ng da ng ru t go n Causs kho ng duy nha t vi thay vi ba t da utu x1tacothcba t da u ba ng ba t kymo t bc nna o kha c.a y gota cothcdcda ng tralo ca u ho lc u coxa c d}nh duonghay kho ng. 1ha t va y, ta co(x, x) = 0 (x1 + x2)2= (x22x3)2= x23 = 0x1 + x2 = 0;x22x3 = 0;x3 = 0.x1 = x2 = x3 = 0.\a y xa c d}nh duong, do donolamo t tch vohuong trong R3.V du20. 1rong R3xc t da ng songtuyc n tnhdo xu ng(x, y) = x1y1 + 3x2y2 + 5x3y3 + 2x1y24x1y3 + 6x2y3.1a co(x, x) = x21 + 3x22 + 5x23 + 2x1x24x1x3 + 6x2x3= (x21 + 2x1x24x1x3) + 3x22 + 5x23 + 6x2x3bJ= [x1 + (x2 2x3)]2 (x22x3)2+ 3x22 + 5x23 + 6x2x3= (x1 + x2 2x3)2+ 2x22 +x23 + 10x2x3= (x1 + x2 2x3)2+ (x3 + 5x2)2 23x22.Ca hcphuong trinh tuyc ntnh_x1 +x2 2x3 = 0;5x2 +x3 =23x2.Chox2= 1, nha nduo cx1= 9 + 223, x3= 5 +23. \a y,vo x = (9 + 223, 1, 5 +23) ,= 0,taco (x, x) = 0. Iodo kho ng xa c d}nh tch vohuo ng trong R3.1rong\ du 10, da ngru tgo nCaussgo m3donthu cvuo ngvo da u co ng (ta no mo t dao //a c cac o} ca / da a cc o} kh hcsocu a nola so duong). 1rong\ du 11, so ca cdonthu cvuo ngcu ngba ngsochc ucu akho ng ganvc cto, nhungkho ngpha ta tcachu ngdc uladonthu cvuo ngvo da uco ng. 1rongtruo ngho pda utatha yda ngsongtuyc ntnhxa cd}nhduong, co ntrongtruo ngho pthu ha thikho ng. 1o ngqua t,ta cothc chu ng mnhd}nh lyduo da y.|nh ly2.2. C/c V/a//c o} }/ao ce c /a //a c scc//e a n ca /amc /da o}sco}/ae o//o/dc / va o}/re oV . K// dc , va cd/o/daao}oe acac// oe ada o} ra / }c o Caassca a cc ndao //a c cac o}ca /da a cc o}.Chu ng minh. Casuda ng ru t go n Causs cu a con don thu c vuo ngvo da u co ng. 1udc u kc n(x, x) = 0 suy ra suy ra x1, . . . , xn tho amo thc go mnphuongtrinhtuyc ntnhthua nnha tco matra nhc cso o da ngba cthang(dc una ytha yro quaca cphc pbc ndo ma tathu chc nba ngphuongpha pCauss). No thc phuongtrinhnhuva ychi conghc m ta mthuo ng,suy rax = 0.b4Nguo c la , gasuda ng ru t go nCauss cot honn don thu c vuo nghoa c cot nha t mo t donthu c vuo ng vo da u tru . a ng ca ch da nh sothutula nc u ca nthc t,ta cothc gathc t(x, x) = a11(x)2+ a22(x)2+ . . . +ann(x)2,trongdo ila ca cbc uthu ctuyc ntnh, a1> 0va a2 0. Xc thcphuong trinh tuyc ntnh_|a2|a12(x);3(x) = 0;. . . . . . . . .n(x) = 0.Ia y lahcphuong trinh tuyc n tnh thua n nha t cosoohuong trinht honso a nnc nco nghc mkho ngta mthuo ng. Io vo nghc mkho ng ta m thuo ngx ,= 0 nhuva y, ta co (x, x) = 0, do do kho ngxa c d}nh duong.3. Chuan cua vec to|nh ngha 3.1. Cho Vlakho ng gan vc c to thu c vo tch vohuo ng, ). \o mo vc c tox Vda t[[x[[ :=_x, x),vago nolac/aa o cu ax.bbBode3.2. /Ba t da ng thu c Cauchy-Schwarz)\a / mc / x, y V/a ccx, y)2 [[x[[2.[[y[[2.Haooa a, da a =va ra/// ca c///// xca yp/a //ac c/ae o//o/.Chu ngminh. Nc u [[x[[ = [[y[[ = 0thi x =y = 0va ba tda ng thu chc n nhc n duo c tho ama n.Casu [[y[[ ,= 0 va R lamo t sothu c ba t ky . 1a co[[x + y[[2 0=[[x[[2+[[y[[2+ 2x, y) 0=2.[[y[[2+ 2x, y) +[[x[[2 0.\ctra cu a ba t da ng thu c sau cu ng lamo t tam thu c ba c ha thco. Ic tamthu cna yluo nnha nga tr}kho nga mdo vo mo Rthi dc u kc n ca nvadulabc t so

0, nghialax, y)2[[x[2[[y[[2 0hayx, y)2 [[x[2[[y[[.a ygo , ga su da u =xa yra, nghialax, y)2= [[x[2[[y[[2. Khdotamthu c ba chano trc nconghc mkc p,nghialato nta Rsao cho2.[[y[[2+ 2x, y) + [[x[[2hay [[x + y[[2= 0. 1u dosuyrax + y = 0 hayx va yla ca c vc ctophuthuo ctuyc ntnh.Ne nh de3.3. 4 o/ va[[ [[ : V R+bbva c d/o/ba / [[x[[ =_x, x) //c a ma o ca c //o/ c/a / saada .// [[x[[ = [[.[[x[[, x V, R./// [[x[[ = 0 x = 0.//// [[x + y[[ [[x[[ + [[y[[, x, y V /ba /da o}//a c /am }/a c.Hao oa a, da a = va ra /// cac// /// /c o /a / 0 sacc/c y = x/ca cx = y.Chu ng minh. 1aco[[x + y[[2= [[x[[2+ [[y[[2+ 2x, y) [[x[[2+ [[y[[2+ 2[x, y)[ (ba t da ng thu c C-S)[[x[[2+ [[y[[2+ 2[[x[[.[[y[[ = ([[x[[ +[[y[[)2.Suy ra [[x +y[[ [[x[[ +[[y[[.Nc uy = x, vo 0 thi ta co[[x + y[[ = [[x +x[[ = [[(1 + x)x[[ = (1 + )[[x[[= [[x[[ +.[[x[[ = [[x[[ +[[x[[ = [[x[[ + [[y[[.Nguo c la , gasu[[x +y[[ = [[x[[ +[[y[[.Kh do[[x + y[[2= [[x[[2+ [[y[[2+ 2x, y)= [[x[[2+[[y[[2+ 2[[x[[.[[y[[.1u do suyra x, y)= [[x[[.[[y[[, kc othco x, y)2= [[x[[2[[y[[2.1hco odcJ.2, x va y phuthuo c tuyc n tnh. Casu , cha ng ha n x ,= 0vay = x. Kh dotuba t da ng thu c C-S ta co n co [x, y)[ = [[x[[.[[y[[,suy ra x, y) = [x, y)[. 1hayy = x va o da ng thu c cuo cu ng,nha nduo c.[[x[[ = [[.[[x[[. 1udosuy ra 0.b7|nh ngha 3.4. Casux vaylaha vc c to kha c kho ng cu aa V . A pdu ng ba t da ng thu c C-S,ta co[x, y)[[[x[[.[[y[[ 1.1udosuy ra to n ta duy nha t mo t go c [0, ] sao chocos =x, y)[[x[[.[[y[[ 1.1a go la}c c (kho ng d}nh huo ng)gu aca c vc c tox va y.Cuo cu ng, dc kc t thu ctc t na y,luuyra ng tchvo huo ng co thcduo c bc u dc n quachua n bo co ng thu cduo da y:x, y) =12([[x + y[[2[[x[[2[[y[[2).4. Sutruc giao|nh ngha 4.1. ChoVla mo t kho ngganvc c tovo tchvohuo ng, ).(a)1ano ca cvc ctox, y V /ra c }/acvo nhauva vc tx y,nc u x, y) = 0.b8(b)Nc u A Vlamo t ta p conkha c cu aVthi tada tA := x V [x, a) = 0, a A.Khdo Ala mo tkho ngganconcu aV vatago Ala //c o}}/ao cco ca aV/ra c }/ac vo A.Icda ng nha ntha y 0 = Vva V = 0.a ygo ga su V la kho ngganvc ctotrc ntruo ngKva Vlakho ng gan do nga u cu a no . Nc u Wlakho ng gan con cu a Vthi da tW0:= f V[f(v) = 0, v W.Ictha yW0lakho ng gan con cu a Vvata go nola//o/ /c a /acu aW. Hc n nhc n,nc u v1, . . . , vp lamo t cosocu aWthiW0= f V[f(v1) = . . . = f(vp) = 0.Ne nh de4.2. Ne a V/a//c o} }/ao ce c /a /a a /a o c//e a /re o KcaW/a//c o} }/ao cco ca a V///dimV= dimW +dimW0.Chu ngminh. Ca su dimV =nvav1, . . . , vpla mo t coso cu aW. otu c thc m ca c vc c to cu a Vva o ta p ho p no trc n dcnha n duo cmo t co socu aV :B = v1, . . . , vp, vp+1, . . . , vn.Co B = 1, . . . , p, p+1, . . . , n la coso do nga ucu a B. 1ascchu ng mnh p+1, . . . , n lacosocu aW0.b9k p + 1, n ta co k(v1) = . . . k(vp) = 0, suy rak W0. Iop+1, . . . , nlaca cvc c todo c la p tuyc ntnhnc ntachi ca nchu ngmnh chu ng snh raW0ladu . \a y, xc t f W0va x V . 1acox = x1v1 + . . . +xpvp +xp+1vp+1 + . . . +xnvn.Kh do f(x) = xp+1f(vp+1)+. . .+xnf(vn). Ia t k = f(vk), k p + 1, n, tacof(x) = p+1p+1(x) + . . . +nn(x).1udosuy raf = p+11 + . . . +nn.1ro la vo kho ngganLucldnchc uV . Nhutrc nda nha nxc t,V V . Iuo da y tasc xa y du ng mo t da ng ca u tu nhc n gu aVvaV.Ne nh de4.3. C/cV /a //c o}}/ao/ac//dca / //c/cc /aa o} , ).4 o/ va : V Vy (y),/rco} dc(y) : V Rx x, y)/amc / da o} ca a }/a a VcaV. Hao oa a, oe a W/amc / //c o} }/aocco ca a V///(W) = W0.Chu ngminh. Ic da ngkc mtrala mo t a nhxa tuyc ntnh. Iodim(V )=dim(V)nc ndc chu ngmnhla da ngca utachi ca n70chu ng mnh ladon ca u ladu . \a y, gasuy Vsao cho(y) = 0.Ic una yco nghialax, y) = 0, x V . No rc ng, la yx = ytacoy, y) = 0, kc o thcoy = 0. \a y ladon ca u, kc o thco lada ng ca u.1c p thcotaco1(W0) = y V [(y) W0= y V [(y)(x) = 0, x W= y V [ x, y) = 0, x W = W.Iolada ng ca u nc ntudosuy ra(W) = W0.Hequa4.4. Ne a W/a//c o} }/ao cco ca a //c o} }/ao /ac//dV///dim(W = dim(V ) dim(W).Ne nh de4.5. Ne aW/a //c o}}/aoccoca a//c o}}/ao/ac//dV/////V= W W./// W := (W) = W.Chu ng minh. () duo c suy ra ngaytuHc qua4.4.() x W, y Wtacox, y) = 0, suyrax W. \a yW W. A p du ng Hc qua4.4, tacodim(W) = dim(V ) dim(W)= dim(V ) (dim(V ) dim)(W) = dim(W).1udosuy radim(W) = dim(W).5. Co sotruc giaovaco so truc chuan71|nh ngha 5.1. Cho V la kho ngganLucldnchc uvaB=(e1, . . . , en) lamo t cosocu aV .() 1a no Blaca sa/ra c }/ac nc uei, ej) = 0, i ,= j.() 1a no Blaca sa/ra c c/aa o nc uei, ej) = ij,trongdo ijlakyhc u Kroncckcr.Hc n nhc n nc u (e1, . . . , en) laco sotru c gao thi (e1||e1||, . . . ,en||en||)laco sotru c chua n.|nh ly5.2. Trco} mc / //c o} }/ao /ac//d ba / //ac o /c o /a / ca c casa/ra c c/aa o.Chu ng minh. Ionha n xc tpha trc nnc ntachi ca nchu ng mnh suto n ta co sotru c gao ladu . Ic u na y scduo c chu ng mnh ba ng quna p thcon. Nc un = 1 thi kho ngcodc u gidcchu ng mnh. Casudc u kha ngd}nh la du ng chonhu ngkho ngganso chc ubc thuan.Xc t mo tvc cto 0 ,=v V va da tW= v). Khdo V= v) Wva dim(W) = n 1. 1hco gathc t qu na p trongWta tim duo c coso tru cgao, cha ngha n (u1, . . . , un1). Ia tun=v, hc nnhc ntacomo t co sotru c gaocu aVla(u1, . . . , un1, un).CasuB = (e1, . . . , en)la coso tru cchua ncu aV . \o mo ca p72vc c tox =

ni=1xieiva y =

ni=1yieicu aVta cox, y) = n

i=1xiei,n

i=1yiei) =n

i,j=1xiyjei, ej) =n

i=1xiyi.1udosuy ra hcquasau da y cu a I}nhlyb.2.Hequa5.3. 8a o} ca c/ c/c o ca sa/ra c c/aa o B = (e1, . . . , en) /rco}//c o} }/ao /ac//dV/a ccp/e p da o} ca a saa da }/a a Vca//c o}}/ao /ac//d Rnca / //c/ cc/aa o} c//o/ /a c.B : V Rnx =

ni=1xiei (x1, . . . , xn).Quatrnh tru c giaoho a Cram-Schmidt0uaHc qua b.Jtatha yra ngco thc do ngnha tmo tkho ngganLucldnchc uV vo kho nggan Rncu ngtchvo huo ngchnhta c.1uynhc nkh do ca npha xa ydu ngduo ctrongV mo t coso tru cchua n. Iuo da ytasc mo ta mo t thua t toa nchophc pnha nduo cmo t coso tru cgaotu mo t coso ba t ky cu aV (nhuda no phatrc n,tumo t co sotru c gaota dcda ng nha n duo c co sotru c chua n).No tthua ttoa nnhuva ythuo ngduo cgo la aa /r/o//ra c}/ac/c aCram-Sc/m/d/.|nh ly5.4. C/c (v1, . . . , vp) /amc / /cca c ce c /a dc c /a p /ae o //o/ca a //c o} }/ao /ac//d VcaW= v1, . . . , vp) /a//c o} }/ao cco ca aVs/o/ ba / ca c ce c /a oc / /re o. K// dc , /aca c ce c /a v1, . . . , vp /a cc//eva da o}mc / ca sa/ra c c/aa o c/cW.7JNc / r/e o}, /a mc / casa ba / / ca aV /acc //e va da o}daa cmc / ca sa/ra c c/aa o ca a V .Chu ngminh. Nhuda nha nxc to trc n, tachica nxa ydu ngmo tcosotru c gao (u1, . . . , up) choWladu .Ia tu1 := v1u2 := v2 + u1, vo R sao chou2 u1.\o dc u kc nna y taco0 = u2, u1) = v2 +u1, u1) = v2, u1) + u1, u1).Iou1 ,= 0 nc ntudosuy ra = v2, u1)[[u1[[2.1c p thco,timu3duo da ngu3 = v3 +u1 +u2,vo , R sao chou3 u1va u3 u2.1im nhu sau:0 = u3, u1) = v3 + u1 + u2, u1)= v3, u1) +[[u1[[2(do u2, u1) = 0).1u do suy ra =v3,u1

||u1||2 . Hoa ntoa ntuong tu , nha nduo c = v3,u2

||u2||2.Casu da timduo cca cvc ctotru cgaou1, . . . , up1. 1asc tim74vc c toupduo da ng sauup = vp + 1u1 +. . . +p1up1.1u dc ukc nup =uitatimduo ci = vp,ui

||ui||2 . Nhuva y tadaxa y du ngduo cmo tho ca cvc ctotru cgao (u 1, . . . , up). a y gotachi ca n chu ng mnhu1, . . . , up) = v1, . . . , vp).1acou1) = v1). Ca su 1 0nc nsuy ra1 = 1.() 1ruo ng ho p na y duo c chu ng mnh hoa n toa ntuongtu .Bode9.4. Ne adctA=1/// dimE(1)=1. Ne adctA= 1///dimE(1) = 1.Chu ngminh. 1achixc t truo ngho pdctA = 1. 1ruo ngho p dctA =1duo cchu ngmnhho antoa ntuongtu . \a y, ga su dctA=1.Kh do , thcoo dc 9.J, =1la tr} rc ngdonhoa cbo 3. Nc udimE(1) = 3 thiE(1) = R3, kc o thcoA = I3, tra vo gathc t. CasudimE(1) = 2. Kh do = 1 latr} rc ng bo 3. Co (v1, v2) lamo tco socu aE(1) va0 ,= E(1). 1acof(), vi) = f(), f(vi)) = , vi) = 0, i = 1, 2.Iodo f() E(1), nghiala f() ). \a y, to nta Rsaochof() =, nghiala la tr} rc ngthu ccu af. Iodo , thcoo dc 9.J, = 1va f() =, nghiala E(1), suyra = 0ladc u ma u thua n. \a y, dimE(1) = 1.Bode9.5. Ne adctA=1//aao}a o}, dctA= 1 /// ma / p/a o} = E(1) //aao} a o}, = E(1) ba / b/e o dc / ca / fca/a o c/ef/e o/amc / p/e p aa.Chu ng minh. CasudctA = 1, E(1) = ) va = E(1). x taco x, ) = 0, suy ra f(x), f()) = f(x), ) = 0. Iodo f(x) ,nghiala ba t bc ndo vo f. Nc u dctA = 1thi chu ng mnh hoa n87toa ntuongtu . a y go da tf= f[laha nchc cu aflc nma t pha ng. Hc nnhc nflatoa ntu tru cgaotrongkho nggan. 1achu ngmnhdctf= 1. Co (v1, v2)la coso cu a. Kh doB = (v1, v2, )laco socu a R3va[f]B =a b 0c d 00 0 ,trongdo_a bc d_ = [f](v1,v2).Nc u dctA = 1 thi E(1), kc othco = 1. Nc u dctA = 1 thi E(1), kc othco = 1. IododctA = . Na t kha c,dctA =a b 0c d 00 0 =a bc d =a bc ddctA.IododctA = 1 hayflaphc p quay.Ic nda ytatha yo dc da duo cchu ngmnhxong. I}nhly 9.1nhodocu ng daduo c chu ng mnh hoa n toa n.1rongca c chu ngmnhvu a trinhba y ta nha ntha yra ng, nc udctA = 1thi fla mo t phc pquayquanhtru cE(1). \i vc tcu amatra nkho ngthaydo kh tathaydo coso nc ngo ccu aphc pquay88duo c xa c d}nh tu co ng thu cTrA = 2 cos + 1.Nc u dctA = 1 thi pha n tch matra nA nhusau:A =cos sin 0sin cos 00 0 1=1 0 00 1 00 0 1cos sin 0sin cos 00 0 1.1atha yflaho pno cu a mo t phc pquayquanhtru cE(1) mo tgo cvamo t phc pdo xu ngtru cgaoquama t pha ng =E(1).Co c quayduo c xa c d}nh tuco ng thu cTrA = 2 cos 1.Nhu ng dc u vu a no otrc n cothcto ng kc t tha nh mo t d}nh lyduo da y:|nh ly9.6. C/cA O(3, R), A,= I3. K// dc .//Ne adctA = 1/// A/a ma/ra ob/e ad/e o/rco}casa c//o//a c ca a R3ca amc / p/e paaaao//ra cE(1) mc /}c cma }/a/r/ ca a ocdaa c va c d/o//acc o} //a cTrA = 2 cos + 1./// Ne a dctA = 1 /// A /ama /ra o b/e a d/e o /rco} ca sac//o//a c ca a R3ca a /a p oc /mc / p/e paaaao//ra cE(1) mc /}c ccamc / p/e p dc / va o} /ra c }/ac aa ma / p/a o} = E(1). C/a/r/ ca a }c cdaa c va c d/o/ /acc o} //a cTrA = 2 cos 1.89No rc ng, nc utrongI}nhly9.b, TrA = 1thi = 0. Khdo Abc u dc n mo t phc p do xu ng tru c gao qua ma t pha ng E(1). No tphc p do xu ng tru c gaonhu va y duo c go lamo t p/e p p/a o va .10. Cheohoa toan tutulien hop trongkhong gian Fuclid1a kc t thu cchuong na y ba ng mo t u ng du ng quantro ng cu a tchvohuo ng. Cuthcta scchu ng mnh ra ng mo ma tra n do xu ng thu cdc u chc o ho a duo c trc n R.|nh ngha 10.1. 1ano toa ntu f trongkho ngganLucldV lamo t /ca o /adc / va o} hay /ca o /a/a//e o /a p nc uf(x), y) = x, f(y)), x, y V.Ic tha yynghiacu akha nc mtoa ntu do xu ng, taha y xc tmo tco sotru c chua n (e1, . . . , en) cu a V . CasuA lama tra n cu a ftrongcoso no trc n. Ic ukc nnc utrongI}nhnghia10.1duo cvc tduo da ng ma tra nnhu sau:(AX)TY= XTAY, X, Y Mn1(R)hayXTATY= XTAY, X, Y Mn1(R).Ic u na y da n dc nAT= A hayA lama tra ndo xu ng.\a y flatoa n tudo xu ng kh vach kh ma tra n bc u dc n ftrongmo t co so tru c chua n (dodotrongmo co so tru c chua n) lamatra ndo xu ng.90|nh ly10.2. C/cf /a /ca o/a dc / va o}/rco}//c o}}/ao/ac//d.K// dc/a cco/a o} d/e asaa da .// Hc / /r/ r/e o} ca a fde a/asc//a c./// fc/e c /c a daa c.//// Ca c //c o} }/ao cco r/e o} ca a fdc / mc / /ra c }/ac ca / o/aa.Chu ng minh. () Co A lama tra n bc udc n toa ntu ftrong mo t cosotru cchua n. Khdo ,thconha n xc tpha trc nthi Alamatra ndo xu ng thu c. Xc t da thu c da c trungPA(t) vago lamo t nghc mba tkytrong C cu a no . Kh dohcphuong trinh tuyc ntnh thua nnha t(AIn)X = 0 (1)conghc mkho ngta mthuo ngX Mn1(C). Casux1x2...xn. Ia tX =x1x2...xn,trong do xilasophu c lc n ho p cu axi. Kh doXTX =n

i=1xixi =n

i=1[xi[2> 0.1a coAX = X =AX = X =AX = X =AX = X. (2)91\iAT= A nc nta co(AX)TX = XTATX = XT(AX). (3)Kc t ho p (1),(2) va(J), nha nduo c(X)TX = XT(X).1udosuy ra(XTX) = (XTX).Nhungnhutada tha yXTXla mo tso thu cduongnc ntu da ngthu c cuo cu ng suy ra = ,nghia la R.() 1a scchu ng mnh dc u kha ng d}nh ba ng qu na p thco sochc uncu aV ra ng to nta trongV mo t coso go mtoa nca cvc c torc ng.Nc u n = 1 thi kho ng cogi dcchu ng mnh. \a y, gasu n > 1 vadc ukha ng d}nh du ng do vo nhu ng kho nggancoso chc u ba ngn 1.Xc t mo t tr} rc ng cu a fvax lamo t vc c to rc ng u ng vo tr} rc ng .Ia tH := x). Kh do dim(H) = n 1. 1ruo chc t tachu ng mnhHlakho ng ganconba t bc n do vo f. 1ha t va y, y H, tacox, f(y)) = f(x), y) = x, y) = x, y) = 0,nghiala f(y) H. Iofla kho ngganconba tbc ndo vo fnc nf:= f[Hlatoa n tutuyc n tnh trong kho ng gan LucldH. Hon nu a,hc nnhc nfcu ngla toa ntu do xu ng. \a y, thcoga thc tqu na p,toa n tufchc o hoaduo c. Suy ra to n ta mo t co so B cu a Hgo m toa nca c vc c to rc ng cu af(cu ng lacu af). Kh do x Blamo t co socu aVgo m toa nca c vc c torc ng cu af. \a yfchc o ho a duo c.()Ca su ,=la ha tr} rc ngkha cnhaucu af, xla vc ctorc ng u ng vo , y lavc c torc ng u ng vo . Kh dof(x), y) = x, f(y))92=x, y) = x, y)=x, y) = x, y)=x, y) = 0.Cuo cu ngluuy ra ngmo t matra nphu cdo xu ngkho ngnha tthc t chc o ho a duo c trc n R hoa c tha m ch trc n C. Sau da y lamo t vdumnhho a.V du24. A =_0 _,vo , C. 1a coPA() = [AI2[ = = 2 2.Nc u = 2+42= 0 (ladc u cothcxa y ra do vo ca c sophu cva ) thi PA()co mo t nghc mkc p=2. Kh do PA()=_ 2_2. \a yAchc oho aduo ckh va chi kh mA()= 2.Nhungkhdo mA(A) =A 2I2= 0,suyraA =2I2la dc uma uthua n. \a yA kho ng chc oho a duo c.BaitapBa i 1. \o gatr}na ocu a Rca c da ng songtuyc ntnhduo da ytrc n R lamo t tch vohuo ng:9J(a)f(x, y) = x1y1 + 10x2y2 + 6x1y2 +x3y3x2y3x3y2.(b)f(x, y) = 2x1y1 +7x1y2 +7x2y1 +8x2y23x3y3 +x2y3 +x3y2.Ba i 2. ChoSlamo t ta p ho p ba t ky . 1a no a nh xad : S S R+:= x R[x 0lamo t //ca o} ca c/ /rco}Snc u no tho a ma n ca c tnhcha t:(a)d(x, y) = 0 x = y;(b)d(x, y) = d(y, x);(c)d(x, z) d(x, y) + d(y, z).1) Chu ng mnh ra ng, nc u Vlamo t kho ng gan Lucld thi a nh xad : VV R+xa c d}nh bo d(u, v) = [[u v[[ lamo t khoa ngca ch trongV . 1ago nola//ca o} ca c/ //e o /e / ca / c/aa o.2) Cuthc ho a khoa ng ca ch na y do vo kho ng ganLucld R3vo tch vohuo ng chnh ta c.Ba i 3. ChoWla kho ngganconcu akho ngganLucldV . Ia tpW: V Wlaphc p chc u tru c gaotu Vxuo ngW.(a)Chu ngmnhra ng, nc uv V tho ama nw =pW(v) ,= 0thiwlamo t trongnhu ng vc c to cu aWta o vo vmo t go cnhonha t.(b) \o v V , //c ao}ca c/tu vdc nWduo cd}nhnghianhusau:d(v, W) = infwWd(v, w).Chu ng mnh ra ngd(v, W) = [[v pW(v)[[.94Ba i4. ChoV la kho ngganLucldva pWla phc pchc utru cgaotu Vlc n kho ng gan conW. Chu ng mnh ra ng, nc u (u1, . . . , um) lamo t co sotru c chua ncu a WthipW(v) = v, u1)u1 +. . . +v, um)um.Ba i 5. \o n 0, xc t tch pha n suy ro ngIn =12_xnex22dx.(a) Chu ng mnh ra ng tch pha n na y luo n ho tuva I2k+1 = 0, k 0.(b)Chu ng mnh co ng thu c truy ho In = (n 1)In2, n 2.A p du ng dctnhI2k.(c) I}nhnghia a nh xa, ) : R[x] R[x] Rnhu sau:P, Q R[x], P, Q) =12_ex22P(x)Q(x)dx.Chu ng mnh ra ng a nh xano trc n lamo t tch vohuo ng.9b(d) Xc t kho ng gan con R2[x] cu a R[x]. Ha y tnh khoa ng ca ch tux3dc n R2[x].Ba i 6. Cho lada ng song tuyc n tnh trc n R4mabc u thu c trong cosochnh ta c duo c vc t nhusau:(x, y) = x1y1 +2x2y2 +4x3y3 +18x4y4 +x1y3 +x3y1 +2x2y4+2x4y2 + 6x3y4 + 6x4y3.(a) Chu ng mnhra ng lamo t tch vohuo ng trong R4.(b)\c t ma tra nbc u dc n trongco sochnh ta c.(c) ChoWla kho ngganconcu a R4xa cd}nhbo hc phuongtrinh tuyc ntnh thua nnha t_x1 x2 + x3x4 = 0;x22x4 = 0.Ha y tim mo t co socu aW.Ba i 7. 1rong kho ng gan Lucldc vo tch trong tho ng thuo ng cho ca cvc c to u1 = (2, 1, 2, 4), u2 = (2, 1, 1, 6), u3 = (2, 3, 4, 8).Co W= u1, u2, u3)la kho ngganconcu a R4snhrabo ca cvc ctou1, u2, u3va Wlakho ng gan concu a R4tru c gaovo W.(a) 1im mo t cosocho mo kho ng ganconWva W.(b) Cho u = (5, 5, 3, 1) R4. 1im hinh chc u tru c gao prW(u)cu au xuo ngWvakhoa ng ca chd(u, W) tu u dc nW.Ba i 8. ChoA O(n, R). Chu ng mnhra ng, nc udetA = 1 thimo pha n tu aijcu aA dc uba ng pha n buda socu a no .Ba i 9. Cho flamo t phc p bc n do tru c gao trong kho ng gan Lucld9bV .(a) Chu ng mnhra ngKer(f IdV ) = Im(f IdV ).(b)Chu ng mnh ra ng,nc u (f IdV )2= 0 thi f = IdV .Ba i 10. Xa ydu ngmo tcoso tru cchua ncu a R3tu ca cvc ctorc ngcu atoa ntu f: R3 R3co matra nbc udc ntrongcoso chnhta c laA =5 1 21 5 22 2 2.Ba i 11. 1oa ntu f: R3R3comatra n trongco sochnh ta c laA =131 2 22 1 22 2 1.Ha ychu ngmnhra ngf la toa ntu tru cgaotrongkho ngganLucld R3vo tch vohuo ng chnh ta c.97Ch mucchua n, bbco sotru cchua n, 72co sotru cgao,72co soxyclc, 4bda ng chnh ta c Jordan, 4Jda ng ru t go nCauss,bJda ng song tuyc ntnh,b7da y Ibonac,20kho Jordan,44kho ng ganconrc ng, 11kho ng gancontru c gao,b9kho ng ganconda c trung, JJkho ng ganLucld, b8kho ng ganHlbcrt, b8lnh ho a tu ,b9ma tra ntam ga c duo , 2Jma tra ntam ga c trc n, 2Jma tra ntru c gao,82nho m ca c phc p quay, 82nho m tru cgao,82nho m tru cgao da c bc t,82phc p pha n xa , 90qua trinhtru cgaoho aCram-Schmdt,7Jtoa ntutru cgao,8Jtoa ntutru cgao,80toa ntuxyclc, 4btoa ntudo xu ng,90tru c gao, b8ti lc va ng, 22tch vohuo ng,b7tch vohuo ng chnh ta c,b8I}nhlyHamlton- Callcy, 29da thu c to tc u, Jbda thu c trc t tc u,28, 2998