DANG TOAN PHUONG - KHONG GIAN VEC TO OCLIT VA KHONG GIAN VEC TO UNITA
D. NCTONO DAN HOC DAP SO
9 32
3.1. a) H = (3, + x2 + 2302 - [ 1- 3-x2 + 73 1 x 3 ) -(a33
y i = +x 2 +2x 3 fat 2 =- 2 + I 3
Y 3 3x3
Taco H= 2 3 b) H -37F2 - 3 1
c) H = _2 2 32 Y3
3.2.
a) Khong co 1. nao th6a man b) Kh6ng nao th6a man c) A> 13.
3.3.
a) Q-602 -i612)2 + 1 673)2 +---+ n+1 67 :32 6 2n That \ray, dat y, = x, + —(x2 + 1
2 x„), thi
-3-x,x ) ;2 2
Dat y2 = x2 + 1 (x3 + + xn), tu do
f
= +71 1374) + -
6 Ex; +—x i x ; 3 <1
;>3 4 .i<j Gia si c1641 blidc thtt p:
ExVE 2 Ex i x j j ,
i=p+1 P ± 2 i<j
Q = 671) +
adO(p.4i<j)
p+1(_ , )2 + p+2
VP! 2p + 2 2p
Da, t = x,,, 1
, + (xl,„2 + + x„) ta dude tong thiic tren, p + 2
dng voi p+ I_ Do do sau (hsj) buoc ta co ket qua. Ta ce the' viet gon quy tacd6i bie'n nhu sau:
1 -- 1 1 1
2 2 2
0 1 1 1
3 3
(Yu, 0 0 0 b) Dua ve a) nhu sau
Q= x i x 2 +x l Ex; +x2 Ex i +Exix i ; 3<i<j i=3 i=3 i<j
= (x, + x s + + xs) (xs + x, + + xs) -
1 t Q =-1
+ x s + 2Z x,1
4 >3 4
Dang (n-2) bin x3, ..., xi, E .1
3.5+=j
dttqc xet trong phein a). Nhu vay ta co
Q=(Y1)2 -6,7)2 L 1(23 (k 2k -1 —2) (Y1J2 1 s
V = —kx, + x 2 ),- Lx,
2 =3
YI
y2
/ x1
9
X9
Y2=2 1 + x9) 1
Y3 = x3 + --kx 1 + 1 ' 2 Y4 = X4 + —21
/ +.
x i
x n
y fi =
3.5 Gr (a l , an ) = det (<a1, ak>); 1 < j, k 5 n
a.j =Ea-e• a k = Ea ik ei . Do ha {ea trite chuan nen
,=1 1=1
< a ja k >=Ea,a ;,, ; Dat A= (ao).
Ta có ma train (<ci, ak>) = At A.
Tit do Gr (a„ an) = (det A)2 =
3.6. Gia s8 doan [a, hi chita tat ca cac = 1, n). Xet
e [a, hi la khong gian cac ham lien tuc (co the trii mat s6 frau han diem) tren loan [a, N. WI* [a, 131 la kheing gian vec to vo han chiau Den truang sathac K.
Goi bla,b1= C a,b de "—" la quart he Wong throng f g caf= g tren [a, hi tril mat s6 hvu han x E [a, b].
NMI T,i e C [(Lb], (tat = cf(t)g(t)dt la tich vo httdng tren C la,b1 (khong phu thuec vao dai dien f, g x;
{.1n6uteA;
la ham dac trdng caa tile la X i(t) =
Oneitto6„
Thl > va det A = Gr (R I %^) >_ o.
3.7. Thong to bat 3.6.
3.10. Vi ma tran A d6i xfing, xac dinh &mug nen ton tai ma trail true giao C C-1 . A . C co clang cheo:
'a do x,, ... la citc gia
X2
0 tri rieng caa ma tran A.
1 xt 1
12 0
k„
0
= A-1 C.
Ta có: 131 =
NMI vay A-' có cac gia tr rieng —,..., —1 1 ki
n
Va trace A = Ek ; trace A-' E -
i=1
n 1 2
Ta có (tr A) (tr 21/2- ') = E-
LI Xi)
183
3.11. I-ID. Ta thky rang: rigu I la nghiem dac trong cda ma tran A thi - X (,) la nghiem dac trong ciia A - X0I. (xem giai bei 2.45, each 2). Nhu vay A Ia nghiem dac tning cim A, X e [a, b], thi k - ?to IA nghiem dac cna A - - X0 e [a - k, b - X].
Neu moi nghiem dac trong cda A deli thuec [a, b] va I, < a thi do a - 10 > 0 nen moi nghiem dac trong caa A - X0I, deo duong, vay A - Xid xic dinh duong vdi X < a (xem vi du 3.7). N6u b <10
thi b - < 0, nhu vay moi nghiem (lac thing cna A - 7.01 dgu am. Do vay A - XeI xac dinh am. Phan ngvdc Iai chting minh Wong tv.
3.12. a) Vol B, = A -1 11dUI, La c6 B I U, = AU, -1,15 1 .U1 .U1
= X 1U, - X,U 1 • IlUd1 2 = X(1 -111-11 1112)Up Do do, n6u 7L1 = 0 thi
= 0 vdi mm vec to rieng U 1 ; ngu X1 # 0 thi B I U, = 0 khi va chi khi U, la vec to rieng co chua'n dclit bang 1.
b) Gi sii X la vec to true giao vdi U, trong (vdi tich ye Inking chinh tic). Khi do .X = 0 va B,X = AX - X1U1 .1.1 X = AX.
Nhu vay, n6u U2 la tree to Hong don vi Gila A trip giao vdi U 1 , thi
U2 cling la vac to rieng cim B 1 . Xet clang eau 112 = A - a,U 1 . Ui - 12U2 .M2 , ta co KerB2 chtla khong gian vec to sinh bai U2};
va thu hep cita B2 tren khong gian con true giao vdi Vect(IJ I , U2)
trong vdi thu hep cila A tren do (coi ma trait A la ma trail cua to clang eau trong khong gian vec to dclit R5. Sau n bade ta nhan dude B,1=A-EX .111 , c6 tinh chat Ker(Ba) Vect(IJI,
i=1
Nhu vay B„ = 0 hay A = E)< lU i .U1 . i=1
U0)= an.
3.13. Gia sit co xx 0, yx 0 ma f(x, y) = 0 va f(y, y) >0 nhung (x, x) 0. Ta thky x, y dec lap tuy6n tinh va vat z thuec khong
;tan vec to sinh bat x, y thi fez, z) = f(ax I by, ax + by) = a2 f(x, x) + f(y, y) + 2ab f(x,y) ?0, (").
Theo gia thiet n6u f(x, a) =0 thi x cling phitong n6u f(y,a) = 0 hi y cane phtiong voi a. Nhung x, y khong tang phuong, nen (x, a) va f(y, khong deng that bang khong (**). Do do co hai
6 thvc k, I cK sao cho k' +1' > 0 va k f(x, + I f(y,a) = 0.
TU do f(kx + ly, = 0. Theo gia thiet kx + ly ding phucing 'di a, trudng hop f(kx + ly, kx + ly) < 0 Coat do nhan 'cot (*). Do to a= 'K ix + ltyx 0. Theo gia thiet f(a, a) = 0 k, 2f(x, x)-11,1 f(y, y) = O.
)o f(y, y) > 0, f(x, x) '2. 0 nen to co 1, = 0 Ira f(x, x) = 0. Nhtt
t = k, x va ter do [(a, y) = 0, f(a, x) = O. Mau thuan 'got (*").
3.14. Xet A e M:11. (n, c Mat (n, C). Vi A phan di51. ximg . len ma Iran iA Hemnit - Min vt ao). Ta co det (A - kin) = 0 <=>
let (iA - ixIn) = 0. Nhung mot nghiem da' c trung caa ma tran iecmit den thuc. Tif do suy ra mkti agh*n dac trung cila ma trail A
a thuAn ao hac bang khong. Gia this cat nghiem khac khong
as da thud dee trung PA la: ja i , ..., tat , -iak, (cti E K, ai x 0).
Chi do PA N=
PA W ) + a ?)
{0 vain >2k Do do det A = PA(0) = k 2
II ot i nefun=2k t
jo '
Do do det A a 0. TU day, de thAy detA = 0 ngu n le. Di&
nay cling có the suy ra ngay bang cluing mirth trkic tigp.
3.15. Bo dg: Cho V la khong gian vec to tren truang C, U li anh xa nem tuygn tinh: V —> V, nghia la u (ax + py) = u (x)+ 5 u (y) vol mgi x, y e V, a, p E C. Khi do u2 IA huh 3u.
tuygn tinh. Gia sit X la mgt gia tri rieng thkic, am cua u 2 , khi do X la nghiem bOi than cim da thfic dac trung Put .
Chung mink be, dg: De thy u 2 e End (V). Gin). six 11 mg gia tri rieng time < 0 cim u 2 va a x 0 la vec td rieng cua u2 vdi u2 (a) = Aa. Khi do u(a) va a la dec lap tuygn tinh trong V That vay n'elk u(a) = -4 a, 4 e C thi u 2(a) = u(4a) = 4 u(a) = 141 2
a=Xa.
-
Do do X = 141 2 a. 0, trai vdi X< O.
G9i W la khong gian vec td con hai chigu cna V, sinh bai a u (a). De thAy moi vec td cim W den la vec to rieng
vol gia tri rieng X va u (W) c W. Dal V 1 = V/W, xet anh xa can sinh.
u:V c —>
[x] —> u,[x] = [u(x)]
Ta c6 u, IA anh xa nUa tuygn tinh, tit do u 12 la anh xa tuygr tinh va u1. 2[x] = [u2(x)]. VI vay, ki higu PIO va Pu, 2 la cac dz thtc dac trung cliatt 2 c End (V) va u1 2 c End (V1) ttiOng ling thi Pu2(t) = - A.) 2 Pu 1 2(t). Niu y lai IA nghigm cf.a da thug 4(
thing Pil l', lap lai qua trinh tren to có (t - )0 2 la ink Gila Pu l a Qua trinh nay huu han vi dim V = n huu hart. Nhti vay
s6 mu
na. (t - A) trong phan b.& Put la s6 than. B6 de' dude cluing ainh.
Chung mink bai Man:
Xet u: C° —> C11, u (x) = Al; u la Anh xa n&a tuyen tinh;
t2(x) = u(u00) = u (Al) = K Ax voi moi x e Ta co Pu2 (t) =
let (A. A — tIn). Viii moi t e R, ta en det (A.A -t In) = let (A A - t In) = det k -t In) (xem bat 2.51). Nhu \Tay Pu 2 (t)
a da dine vol he so' thve.
Gin) Pu2(t) =(%14)a' • 0.2 - 0'2 - oak x dO sn ,sk muyen during, ..., Irk e R;
Q e R Q khong co nghiem thne
Vi Q khong có nghiem thile nen deg Q = n -(s, + s 2 + ...+ sk) Alan. He se cao nhat cua Q(t) la (-1)"" -- sk), nghia la bang 1.
Do do Q(t) > 0 vdi moi t e R, to do Q(-1) > 0.
Bay gid ta xet the nghiem (i = 1, 2, ..., k). N6u CO < 0, thl boi s, ena nghiem ?9 chart, khi do (Xi + nsi z 0. N6u 7u, z 0, thi re rang (X,+1) 31 >0.Nhv vay Pu2(-1) a 0, nghia la det (A A+ In) a a
Chu $: Deu bang co th6 x637 ra, chamg han xet A =
3.17. Vol f e End (U). Neu f la t0 Tang caM tor lien h0p thi vdi moi x e U; ta co:
< f(x), x >=<x,f(x)>=<x,f*(x)>=<x,f(x)>
Nhu vay <x, f(x)> thole hay <f(x), x> thitc \TM moi x e U.
Node lai n6u moi x e U, to có <f(x), x> thoc, to cheini minh f to lien hOp. Than Lich f thanh tting can hai to deing cM to lien hop:
f = ft + i . ft,, khi do
<f(x), x> = <f,(x), x> + i <ft,(x), x>.
Nhung ft, 1, la nhang to thing chin to lien hop nen <Ii(x), xi va <f,(x), x> thoc. Vi vay <fa (x), x> = Im <f(x), x>. N'eal <1(x), xi thitc thi <fa(x), x> = 0 yen moi x, ta do If„ = 0 va f, = 0. Do vay
= f1 nghia la f la to lien hqp.
3.18. Gia sa y la gia tri rieng c>ia f e End (U), ado f la tit long din to lien hop caa khong gian vec in Unita U. Khi do co
e U, z # 0 d' f(z) = Az.
D4t x
z khi do f(x) = ax, x = 1 va <f(x), x> = <Xx,
=GI4=X.
3.19. a) va b) hir3n nhi'en
c) 1(x) = ek > e k = E< x, f(e k )>e k
k=1 k=1
n
= Ex], < ek > ek = Xk Pk (X)
k=1 k=1
\ray f = Ek kPk .
k=1
d) Theo con trot) to co f = Xi, Pk, do vgy f 2 = 14 pk va
k=1 t=1
Is = En Ask Pk k=1
Nhu vgy Q(f)= iOi k )Pk
k=1
3.20. Gia sit f e End (C") la titan tit tit lien hdp trong kh8ng
;Mit vec to Unita C" vdi tfch vo hthing Hecmit tti nhien, f c6 ma rdn A trong c6 so chinh tag cria C. Gia sit X„ la cac gid ri Hong (thtic) cua f, tfic 2., la nghiem mad da thitc dgc trung det A - = P(X). Goi (e t , co so trite chugn gam nhiing vec 0 Hong ring voi cac gid tri rieng f(g) = ).1 e1 (i = 1, n).
Hat Pk: C" C", Pk(x) = <x, e k > ek. Theo bat 3.19. Ta co
P(f) = I P(21k)• Pk • Nhung P(1.k) =0 171c nen P(f)= 0, to do P(A) = 0.
k=1
3.22. Platong trinh flag trung ciia ma trdn c1 c6 dang:
1 x
2 2
.F3 1
2 2
Cac toa do cud vec to ring tim dupe tit he:
= 0 c5i — 1 = 0 <=> = -±1
o -1
(2 - X)X ' 2x 2 =0 1
2 1 2
1761 k = 1 to co x1 = f x 2 , churl vec to rieng
2 1 .2
e E2
=
\TM X = -1 to co x 2 =-Z xl , chon vec to rieng (
X2= 2 e 2
1/3 2 N6u lKy cd sd true chan
,,75 ix, =
2 2 2 2
—e 1 + -1e ; x 2 = le l -‘1 e 2 thi trong cd so do, r tran cim f co dqng eh&