ak phty thuec /Xiai 1=1 I j=1 <=> i=1 ( k .i=1 j=1 pi =o =0 <=> EauXi =0 j=1 (2)
vdi j = 1 do he pi leriic lap tuyen tinh.
Vi he phuong trinh (2) la he phuong trinh thuan nhat, có s$ an nhigu hon sg phuong trinh, nen no co ve s6 nghiem, nhu vay
ton tai nghigm (x„..., x k) khac khong. Tn d6 suy ra he la,, ...,a k phu thuOc tuydn tinh. Dieu nay trai vai gia thi6t. Do do k < /.
Vi 4 2.4. Xet hai khong gian con E I va E2 cua khong giar vac to E. Gia e>i E/E 2 IA khong gian thacing va h: E, —> E/E 2 li han chd caa anh xa chiou chInh tde troll E,.
1) Tim didu kien can va chi dd a) h 1a toan anh
b) h la don anh
2) Chung tO khong gian (E, + E,)/E 2 (fang eau v6i kholu gian E i /E i nE2 .
L of Rich:
1) a) X& E —> E/E2 la anh xa chidu chinh cac, h = h toan anh a ME,/ =11(E).
E, + Kern = E + Ker n = E E, + E2 = E (vi kerb = E2)
b) Ta en Kerh = E, r1 Kern = E, n E2
Nhtt vey h don anh a Kerh = 0 a E, E, = {O} 2) GiA. se F = E, + E2;
Xet F F/E 2 va k = 1-1/E, •
Do phAn 1) k la toan anh va kerk = E l C Kern = E l 11 E2.
Vi du 2.5. GM. sit V la khong gian vec td tren truing K. Ty Bong can p: V -+ V ducic goi la met phep chien ngu p 2 = p.
1) Chung to rang ngu p, q la hai phep chigu, thi p + q la phep chigu khi va chi khi pq + qp = 0.
2) Chiang t6 rang p.q IA phep chi gu khi va chi khi [p,qI = qp - pq la anh xa tuyern tinh chuye'n Imq van Kerp.
3) Vol p,q Fa hai phep chigu sao cho p+q la phep chigu, hay cluing to Im (p+q) = Imp + Imq va Ker(p+q) = Kerp n Kerq.
Lei gied: 1) p+q la phep chigu ra (p+q) 2 = p+q <=> p 2 + p.q + q.p + = p+q <=> p.q + q.p = 0 2) p.q la phep chigu <=> (p.q) 2 = p.q.p.q = p.q = p 2 q 2 Er> p (qp — pq) . q q (pq — pq) (Imq) = 0 .(=> (qp — pq) (Imq) c Kerp.
3) Vi mai phep chigu p co rang p = trace p,
va vet cim tong hai anh xa tuygn tinh bang tong cac vet cim no, cho nen ngu p + q la Agri chigu thi:
trace (p + q) = trace p + trace q. va rang (p + q) = rang p + rang q.
Tit do dim Im (p + q) = dim Imp + dim Imq, suy ra Im (p + q) = Imp ED Imq.
D6 °hang minh Ker (p+q) = Kerp n Kerq to nhan thay kerp nKerq c Ker(p+q).
Nguuc lai vdi x e Ker (p + q) thi p(x) + q(x) = 0. Do Im (p.+ q) = Imp $ Imq nen MI p(x) + q(x) = 0 suy ra p(x) = q(x) = 0, hay x E Kerp n Kerq.
Vi du as. Gis sit E va Fla hai khong gian vec td tren trulang K; Horn (E, F) IA tap cac anh xa tuyeal tinh tit F Mn ]F; Ft la khong gian vec to con cna F.
a) Chiang to rang £ = e Horn (E, F) / Imf c F 1) la khong gian con cria Hom (E, F).
b) Gia. s5 F= K la mot phAn tich cua F thanh t"o"ng tryc tip
i=1
ciaa khong gian F. Vbi mdi F, xet 2, = if E Horn (E, F) I Imf c Fd.
Chung to rang Horn (E, F) = 2.
WL gia'i:
a) Do g ding kin vai phep town tong anh xa va nhan anh xA vdi mat vet hriong thuac K, nen hieln nhien la khdng gian vec to con crIm Horn (E, F).
b) Vdi h E Horn (E, F), vbi m6i x e E, to phan tich h(x) = y theo eec thinh ph'An y = h(x) = E yi , yi e F1
id
Va vdi moi x e E, thi h(x) = Eh; (x) i=1 Do 05 h= , e
Gia six Eh ; = 0 nghia la vial moi x e E,
( ) (x) = h i (x) = 0, h,(x) e F, 1=1 1=1
to do suy ra 11,(x) = 0 voi moi i vi F = e K. I=1 Nhu \ray Horn (E, F) = e g.
=1
Vi du a 7. GO Q la Huang s6 h> u ti va WI la mot tkp hap khong rkng. Ki hiku E la tap cac anh xa tif c32 vao Q. E la Q - khong gian vac to viii hai phep than: f, g e E, a e cc9, k E Q the
(f + g) (a) = f(a) + g (a) K. f) (a) = k. f (a). Gia sif V la khong gian vac to con ciaa E.
1) Chang to rang: nku f e V va co n dim x„ x„E Ca sao cho
ingui=j . . —
k(x,) = = ej=1,n
Ongui#j
Goi W la khong gian con cea V sinh bed f„), khi do m81 g E V 11611 viet dude met each duy nhat dudi clang: g = h, +112, a do h, e W, h 2 E V va. h = 2(x)= 0 yea mai i = 1,n.
2) Chung CO rang hai tinh chat eau day la Wong clueing: a) dim V n
b) Ten tai n ham g„ thuec V va n diem x„ x„ ciaa W( sao cho g, (x,) = 8,; vet moi i, j = I,n .
Lai gidi:
1) Xet to hop tuyen tfnh EXif, = 0, c Q. 1-1
Khi do vdi moi x E cam, LW; (x) = 0;
Chon x= x„ thi (xi) = 7.j = 0 ( = 1,2, n) suy ra he {f„ , f„) doe lap tuyen tinh.
Val g e V, ham h, e W cAn tim phai thaa man g(x) = (h, + h 2) (xi) = h r (x,) vdi moi
Do If„ f„) la co sa cem W nen h, :11 1 (x, = E gfrA . lei 1=1
Dat ha =g - h, eV
2) Theo phan 1) neu b) clime th6a man, thi he (g„ g„) doe lap tuyen tinh, do \ay dim V 2 n; nglila la menh de b) a) thing.
Ta chung minh a) suy ra b) bang phydng phap quy nap theo n.
Vdi n = 1; n6u dim V 2 1, tan tai fi x 0 va vi vAy có x, e c14
de fi (x,) = A x 0. Chon g, = thi (x,) = 1. Gia sii menh de dung vdi moi k = 1,2, n. Ta chiing minh nd dung vdi n 1.
Theo gia thi6t quy nap dim V n + 1 dim V ?. n nen tan tai n ham g, E V va n diem e A de gi (x,) = 8,, (i, j = 1,2, ..., n).
Goi W la khong gian con sinh bid {g„ gn}, to cd. dim W< dim V.
Theo cau 1), vdi f e V \ W, to cd f = h, + h 2, vi h, e W nen h2 0; vi th6 c6 x,,, d'e' h1 0; (x,„., x x, vdi i = 1, 2, ..., n).
h 2
hi -
) t bin+, vdi MOt i = 1,2, ..., n+1. Bay gia = g - =1, ..., n),
thi vdi j = 1, n to co gi (xj )= (x J )= s i; va gi (xn+1)
=g. (x„.1) - x; = 0 n6u 2, 1 = gi (x„,)_
Nhu Nty co n4-1 ham {g-i } th6a man digu kien bai town.
Vi du 2.8. Gia sit 98 la ho dem doe cac anh xa tuy6n tinh tit R" (16n R"I; vdi mai a e R" xet 0(a) = {f(a) / f e :43}. GM sit g la Anh xa tuyein tinh to den âR" sao cho g(a) E :0304 vdi moi a e K. Chung minh rang g c PC.
Lo gidi:
Ccieh 1. \TM m6i x e R, xet vec to dx (1, x, x2 , H9 eac vee to d x c6 tinh chat:
a) vdi x„ doi mat phan biet, thi ta xi dPe ld tuy6n tinh, vi Binh
# 0 (Dinh thite Vandermonde)
b) V6i m6i d x e R', tan tai f E sao cho g(Ci;( ) = f( ). Ta phai chiing minh g c 93, nghia la c6 f e d f g@t„ i ) tren bb la co so cua R".
Gia sii node Lai, vdi m6i f E c6 khong qui (n - 1) sd the phan biet xi cl6 g(d xi ) = * xi ).
Do hop &dm Mtge eac tap bop hfiu han phan to 1a mat to hap kh8ng qua deM dude, nen sod cle vee to d x e ma g(d x
e g3(a x ) le killing qua d6m dude. Digo nay trai vdi gia thie Nhu vay phai co co sa Vt xi ,...,a x ) oda an vti 06 f E tgi d = f(di xi ) vOi moi i= 1,n; nhu vay g=f e
(rich 2. (Mtn}, cho doe der hied mot sa sri kien cilia khong gnin metric).
Theo gia thiet vdi moi a e der f e de" f(a) = g(a). nhu vay — g) rad= 0 => a e Ker (rig).
NInt vay U (Ker (f — g) f E gq} = R.
Hia sir g vSy vdi mai f E thi f — g x O. do vay dim Key (1— g) n — i. Vi vay Ker (1 - g) lie met kliong gian con thing, khbng dau tra mat cent R". Dieu nay gay nen man thuan do RS la khong gian metric day vdi metric dieing Hwang vii dinh 19 Haire aid rang: met khong gian metric day kheng the bang hop dem dude mita nheing rap hop khong dim Hu mat.
III. du 2.9. Cho hai so nguyen dining r, n ma r < n: (it c Mat (Th R). rg." = ran k rang cl = r m& (CP =
Hay chdng minh vet ()f = a,, + a i„ + + a„,, = r. (Vet can ma Man red thudng duo( ki Heti bdi trace cel).
Xet f e End (IR") co ma trhn trong cd ed tly &nen e = le,, e„). i(e,)= apie
Theo gia Hired to cd f2 = f.
Bat Y = - f(a) I a e khi de Y c Kerf, nen Yea a e thi x = a - 1(a) e Keil, do vay a = f(a) + x e IMF+ Karl
Ta co Imf n Kerf = {0), that Gay, gia sit p e Imf Kerf, thi co a e R" d f(a) = p va do 13 E Kerf nen f(0)= 0 suy ra:
f(p) = 12(a) = f(a) = p = 0, vay Imf n Kerf = {0). Nhu vay R" Imf O Kern dim Imf = r va f I Imf = id, Chon ed sa s Rua R" sac) cho {E,, £,.} c Imf, {E„,„ E„) e Kerf, thi trace f= trace al= r.
Cho Nhaat lai rang n6u f la anh toa tuyeat tinh cie'n
Jrco ma Ran oat = (ad trong cd sa e = (e,, nao do ciaa Tthi so' a„ +,...,+ a„„ khong phu thuoc vita vice chop cd so oda °L va clack goi la vat ciaa anh xa tuyan tinh f va chiac kf hieu la trace f.
Vi du 2d0. Gia sa °W la hai khbng gian vac td treat twang K, f c HomK CP; °Tf).
Xet anh xa f: Kerr Gil"
[a]-a f [al = f (a),
Hay chttng to f la dun eau tuyen tinh va Imf = Imf , ta do f la dang cau to /Kerf len Imf.
ianh xa f lh sac dinh, khong phu thuR vao dai dien. That way, vdi [al = thi a - a' e Kerf, W do f(a) = f(a") hay II-al = f
thd thay f la Maya tinh va Imf =Imf.Ta chung to f la ddn cau. That Ray RR [a] # [a'] thi a - a Kerf f(a - = f(a) - f(a) # 0 do do f(a) # f(a) suy ra f [a] # f Nhu way f don cau va do do f la clang cau tit cliKerf len Imf.
Chu" y: neat so chieu Irau hen, thi tfx vi du Hen to dim tillierf = dim Imf say ra dim "P= dim Kerf + dim Imf.
du 2.11. Giei he phudng trtnh +2x 2 +3x 3 +4x 4 =30 -x i +2x., - 3x 3 + 4x 4 -10 x, - X3 +x 4 =3 x i + +x3 + X 4 =10 1 2 3 4 -1 2 -3 4 D = 0 1 -1 1 1 1 1 1 Day 1a he phudng Huh Cramer.
gthi: = -4 80 2 3 4 1 30 3 4 10 2 -3 4 -1 10 -3 4 1), = D,= = -8 3 1 -1 1 0 3 -1 1 10 1 1 I 1 10 1 1 1 2 30 4 1 2 3 30 - -1 2 10 4 -1 2 -3 10 D a = -12 D'= = -16 0 1 3 1 0 1 -1 3 1 1 10 1 1 1 1 10 NMI vey , = 1, x 2 = 2, x 3 = 3, x 4 = 4.
Vi dii 2.12. Gihi va bran luhn theo thaw s6
A.X1 x, + x• 1 Xx, + x. 2,
+ 4x 3
I) =
a) Veil X s 1 vic a x -2. Day la he Cramer.
it +1 1
va ta ce, - , x -
X +2 2, + 2 +2
b) vdi A = 1, to c6 he Wring during vdi:
x i + x2 + x,, =1 hay x 3 = I - x, -x., do x i , x, lay trtyl, c) Vol X = -2. He co dang: -2x 1 + x., + x. = 1 x i 2x, + x 3 g - 2 x i + x, - 2x 3 1
GUng vg still ve curt ba planing trinh Ln c6 0 = 3, nhn vhy he
Vi du 2.13. Dung phitong phap khn. hay giai he phuang trinh: x t + 3x, + x 3 + 2x 1 )(5 = 2 3x 1 + 10x 9 + 5x 3 + 7)( 4 + 5)( 5 = 6 2x 1 + 8x, + 6x 3 + 8x 1 + 10x 5 = 6 (I) 2x 1 + 9x„ + Sx 3 + 8)41 + 10)( 5 = 2 2x 1 + 8x, + 6x 3 + 9x + 12x 5 = 1 Lm giai:
Nhan hai ye aim phtiong trinh (tau vdi Inning so/ thich hop, I.()) acing vac) eac phtaing trinh khae, ta clia; he pinning trinh ()rang throng vdi he (I):
x + :3x 2 + x,, + 2x + X5 - 2 X9 + 2X3 x 4 + 2x5 = 0
2x, + 4x 3 + 4x 4 + 8x 5 = -2 (II) 3x, + 6x3 + 4)( 1 + 8x 5 = -1
2x, + 4x 3 + 5x 1 + 10x 5 = -3
Nhan pIntong trinh thit hai cna he (II) vdi gag s6 thich hop roi tong viva du; phuong trinh kink cita he, ta &toe he Wong dyeing - x 1 + 3x 2 + x 3 + 2x 1 + x 5 = 2 (I) x, + 2x 1 + x, + 2x 5 = 0 (2) 2x 4 + 4x 5 =-2 (3) (Iii) x 1 + 2x5 = -1 (4) ( 3x 1 + 6x 4 -3 (5)
Car phtfting trinh thit (3), (4). (5) trong he (III) la tudng during. Vi 04 he (III) tudng during vol. he
Giai x i + 3x 2 y 2 + he (IV), to (little xn = —1 — X2 = 1 — x i = 1 + 2x 3 2x 5 2x. 5x i + + + 2x 4 x - I x 4 3x 5 + 2x 5 = + 2x 3 = a do x4. x 2 0 -1 Inv 31 (IV)
Vi dv 2.14. Cho hai ma lien A, B thuOr Mat (n, K), A = (ad, B = 09.
Khi do A + B =(a i +1)0 (bloc goi la tang hai ma tren A va B. Chang minh rang:
1) I rang A -rang B l < rang (A + B) a rang A + rang B. 2) rang A + rang B -n a rang (A B) a min (rang A, rang B) 3) Nan A' = E, tin rang (E + A) + rang (E - A) = n. (3 do E la ma tran don vi cap n).
1) Gia stir f, g la hai ph&n tit cilia End (â "), co ma 'Iran A, B
Wong fing trong m(llt cc; se( s = (s 1) di( cho. Khi do f + g c6 ma tren A + B. VI Im(f + g) c Imf Img, nen:
Tit de suy ra:
rang(A + B) < rang A + rangB. Mat khdc:
rangA = rang(A+B-B) < rang(A+B) + rang(-B) suy ra: rangA < rang(A + B) + rangB
Tit do:
ningA - rangB < rang (A +
ta: rangB - rangA < rang(A + B). VI yay rangA - rangB 15 rang (A + B)
2) Ta co f: K" , g: K" la hai anh xa tuy6n tinh, Im(f o g) = Im(f img) c Imf nen rang(AB) < rangA.
Mat khan: rang(AB) = dim(lm fog) < dimlm f = rangB. Do yay rang(A o B) < min(rangA, rangB).
Bay gio ta churig minh
rangA rangB - n < rang(AB). Ta co dim lle = n = dimIm(f o g) + dimKer(f o g). Mat khde 26t anh xa Kor(f g)/Kerg Kerf (1 Img
a do tI/4xl = g(2), yin X E Ker(f 0 g). De they (To la (Tang au tuyeal tinh. Vi
dimKer(f o g) = dimKerg + dim(Kerf fl Img). Tit do dim Im (f 0 g) = n - dim Ker(f o g)
Nhu vay:
rang(AB) = rangB - dim(Kerf (I Iing) je rangB - dimKerf rang(AB) -e rangB+rangA- n.