f e Endx(V) ma da that ddc trung 94(X) cd dang:
i(X) = (>1 (X
, -X)`2 ... -X) sk
(cac &Ai mat phAn biet, i = 1, 2, ..., k), khi do V la King true tiEp tha cac khong gian riAng suy rang V = x G3.? X2 e Xk
va trong V co mot co sa (ei) = 1, n; (n = dimV) dA trong co ad
do ma trdn din f tao bdi the khung Jordan
nam doe (Wang cheo chfnh so khung jordan cAp s vai phan eh& X, bang:
rang(f - - 2rang(f - kildv)s + rang(f -
Ma tram clang tren cua f xac dinh duy nhAt sai khae each sap xap cac khung Jordan. Ma tran do dude goi la ma trail dang chuan tac Jordan cila tp (tang eau f.
B- VI DV
Vi d4 2.1: Xet R", R!" la the khOng gian vec td tren R. Cho x„ x 2 , , x„, la m vec td thuee R". ]E la met khong gian vac td con eila ChUng minh rang tap IF the vec td ena K" dang
Zt i x i (t 1 , , e E Fa khong gian vec to con ena K', dimF = dimE - dim(E nN), trong de N la khbng gian cac nghiem tha phtlong trinh:
E ti ., =0 (a do t i , tn, la cac An). i=1
Lai gidi:
Xet anh xa tuyan tinh u c Hom(R m, R") a do u(t,, t„,) =
EtiXj . Khi do F = u(E), vi vay F la khong gian vec to con cna R'. dimF = dimE - dim(Keru'), u' = uI E.
Kern' = (KerU) fl E = E (1 N.
Vi du 2.2. Cho x i , , xi, la nhUng vec to khac khong trong khong gian tuy6n tinh V. Gia sit c6 Oleg bi6n d6i f e End V sao cho f(x 1) = x 1 , f(xk) = xk + vdi moi k = 2, ... , n.
Chung L6 rang he {x 1, , xj la doe lap tuy6n tinh.
Lb gidi:
Ta chiing minh quy nap theo n.
Via n = 1, thi {x 1} dee lap tuy6n tinh do gia thiat x 1 # 0. Gia sit menh de dung vdi moi he fx„ xki; k < n - 1. Ta chimg minh menh d6 dung vdi k = n.
Xet t6 hqp tuyen tinh
ECiX i = 0 . (1)
i=1
Khi do:
) = CIX ] Zcif(xi) = c,x, + Ec i (x i +xi_1 ) =
i=1 i=2 1=2
= ZeiXi . (2)
i=1 i=2 n-1
Tit (1) va (2) to suy ra: Ec i x,„ = = 0. (adsbygoogle = window.adsbygoogle || []).push({});
i=2 i=1
Do gin thik quy nap hee, fx,, , doe lap tuyen tinh nen
C2 =e2 = = cn = 0.
Tit do suy ra c,x, = 0 c i = = = c„ = 0 Nhu vSy he {x1, , xj doe lap tuy6n tinh.
Vi du 2.3. Trong khong gian vec to V cho he ak} cac vec to doe lap tuygn tinh ma m8i vec to dm no la t6 hop tuyen tinh cua cac vec to cim he 113„ , Hay ehung minh k < 1.
Ta eó the' gia thigt ••• • la doe lap tuy6n tinh, neu khong to se lay he con doe lap tuyen tinh t6i dal eim {p,, , pi}
g6m m vec to va se elnYng minh k s m < 1.
Theo gia thigt a„ , a k bieu thi tuyern tinh qua D i , p, nen ten tai cac vS hudng a.„ e K sao cho:
a ; ; 0 =1 k) (1)
Ta gia sit k > 1. Ta se cluing minh tuygn tinh. Xet t6 hop euyen tinh