(b)设a∈V∩V,则因a∈V1,持Aa=0,由a∈V2,持Aa=a.于是a=0,即vnv2=0. 因此Kn=V1⊕V 11.设K"=V⊕V,其中V1,V为K的两个非平凡的子空间 证明:斐定存在唯斐的幂等矩阵(即42=A的矩阵A∈Mn(K),使 V1={X∈Kn|AX=0},v={X∈K"|AX=X} 察明:取Ⅵ的斐个基a1,……,ar以及v的斐个基ar+1,…,an.则a1,…,an是K"的基定义 Kn上的线性变换为 a2,r+1≤i≤ 把线性变换在K的自然基下的矩阵记为A.由的定义可得2=,相应地持A2=A 对任意的X∈K,持X=∑a2a则 AX=9 因此 1X=04∑aa1=0a=0+1≤i≤n台X∈v AX=X台∑aa1=∑aa1a=0M1≤i≤r台X∈V 所以A是满足条件的幂等矩阵 再证唯斐性:如果B∈Mn(K),使得 BX=XwX∈V, 则因Kn=V⊕V,可得 BX vX∈Kn 所以A-B=0,从而A=B 12.设A∈Mn(K),E为n阶单位方阵.波 V1={X∈K"|(A-E)X=0},v={X∈K"|(4+E)X=0} 证明:Kn=Ⅵ⊕V2←→A2 察明:(→)K=V1V→n=dim+dimV→n=(n-rank(A-E)+(n-rank(A+ E)→n=rak(A-E)+rank(A+E)=A2=E(习题48.12) (←)对任意的a∈Kn 因为 -E)|(A+Ea=(A2-E) 所以(4+Ea∈Ⅵ.又因 所以-(4-E∈V 因此Kn=V1+V
(b) α ∈ V1 ∩ V2, J! α ∈ V1, G Aα = 0, N α ∈ V2, G Aα = α. < α = 0, V1 ∩ V2 = 0. !O Kn = V1 ⊕ V2. 11. Kn = V1 ⊕ V2, < V1, V2 " Kn 7fn xpq. ST: H 1k,HRV]^ ( A2 = A ]^) A ∈ Mn(K), ' V1 = {X ∈ Kn | AX = 0}, V2 = {X ∈ Kn | AX = X}. : z V1 Hfz α1, · · · , αr $h V2 Hfz αr+1, · · · , αn. J α1, · · · , αn Kn z. M Kn yt&=J A ": A(αi) = ( 0, 1 6 i 6 r αi , r + 1 6 i 6 n Nt&=J A k Kn gz ]^ " A. N A M>P A 2 = A, e,mG A2 = A. X ∈ Kn, G X = Pn i=1 aiαi . J AX = A ÃXn i=1 aiαi ! = Xn i=1 aiA(αi) = Xn i=r+1 aiαi . !O AX = 0 ⇐⇒ Xn i=r+1 aiαi = 0 ⇐⇒ ai = 0∀r + 1 6 i 6 n ⇐⇒ X ∈ V1, AX = X ⇐⇒ Xn i=r+1 aiαi = Xn i=1 aiαi ⇐⇒ ai = 0∀1 6 i 6 r ⇐⇒ X ∈ V2. #$ A -.12RV]^. S,H&: B ∈ Mn(K), 'P BX = 0 ∀X ∈ V1, BX = X ∀X ∈ V2, J! Kn = V1 ⊕ V2, >P (A − B)X = 0, ∀X ∈ Kn . #$ A − B = 0, C% A = B. 12. A ∈ Mn(K), E " n y/@^. I V1 = {X ∈ Kn | (A − E)X = 0}, V2 = {X ∈ Kn | (A + E)X = 0}. ST: Kn = V1 ⊕ V2 ⇐⇒ A2 = E. : (⇒) Kn = V1 ⊕ V2 =⇒ n = dim V1 + dim V2 =⇒ n = (n − rank(A − E)) + (n − rank(A + E)) =⇒ n = rank(A − E) + rank(A + E) =⇒ A2 = E (`a 4–8.12). (⇐) α ∈ Kn, α = 1 2 (A + E)α − 1 2 (A − E)α. !" (A − E) · 1 2 (A + E)α ¸ = 1 2 (A 2 − E)α = 0, #$ 1 2 (A + E)α ∈ V1. Q! (A + E) · − 1 2 (A − E)α ¸ = − 1 2 (A 2 − E)α = 0, #$ − 1 2 (A − E)α ∈ V2. !O Kn = V1 + V2. · 6 ·
当a∈Ⅵ∩V2时又有 + e)o 因此V∩V=0.从而Kn=V⊕V2 习题: 1.在线性空间R2中对任意两个向量a=(a1,a2),B=(b1,b2),定义 (a,B)=5a1b+2a1b2+2a2b1+a2b 验证在此定义下R2构成一个欧几里得空间 2.在线性空间Mn(R)中定义 f(A, B)=Tr(A B)VA, BE Mn(R) 试问:f是否Mn(R)的一个内积? 解:是设A=(a1),B=(by),则 (a)f(4,B)=Tr(4B)=∑∑akbk=∑∑bkak=f(B,A) =1i=1 k=1 (b)f(A+B, C)=Tr((A+B)C)=Tr(AC+BC)=Tr(AC)+Tr(BTC)=f(A, C)+f(B, C) (c)f(k A, B)=Tr((k A)B)= Tr(kA B)=k Tr(A B)=kf(A, B) (d)f(A,A)=Tr(4A)=∑∑a2≥0,且 f(A,A)=0←→aki=0.k,i=1,……,n←>A=0. 所以f是Mn(R)的一个内积 3.设 0 0 规定 (X,Y)= X AY VX,Y∈Rn (1)证明:Rn关于此定义构成一个欧几里得空间 (2)求向量:1=(1,0.,…,0),E2=(0,1,0,…,0),…,En=(0.,0,…,0,1)的度量矩阵 (3)具体写出这个空间的柯西布涅柯夫斯基不等式 解:(1)略 (2)度量矩阵为A (3)设a=(a1,…,an),B=(b1,…,bn),则 kabul≤ k=1 4.设C是一个n阶实可逆矩阵在R中,对任意两个列向量x,Y,规定 (X,Y=XTCTCY 证明:R关于此定义构成 里得空间
b α ∈ V1 ∩ V2 RQG α = 1 2 (A + E)α − 1 2 (A − E)α = 0 + 0 = 0, !O V1 ∩ V2 = 0. C% Kn = V1 ⊕ V2. 5–3 1. kt&pq R 2 , 7f α = (a1, a2), β = (b1, b2), M (α, β) = 5a1b1 + 2a1b2 + 2a2b1 + a2b2. }SkO M R 2 u*HfS' Ppq. : i. 2. kt&pq Mn(R) , M f(A, B) = Tr(A TB) ∀A, B ∈ Mn(R). : f ) Mn(R) Hf{? : . A = (aij ), B = (bij ), J (a) f(A, B) = Tr(ATB) = Pn k=1 Pn i=1 akibki = Pn k=1 Pn i=1 bkiaki = f(B, A). (b) f(A+B, C) = Tr((A+B) TC) = Tr(ATC +BTC) = Tr(ATC)+Tr(BTC) = f(A, C)+f(B, C). (c) f(kA, B) = Tr((kA) TB) = Tr(kATB) = k Tr(ATB) = kf(A, B). (d) f(A, A) = Tr(ATA) = Pn k=1 Pn i=1 a 2 ki > 0, ? f(A, A) = 0 ⇐⇒ aki = 0, k, i = 1, · · · , n ⇐⇒ A = 0. #$ f Mn(R) Hf{. 3. A = 1 0 · · · 0 0 2 · · · 0 . . . . . . . . . . . . 0 0 · · · n . T (X, Y ) = XTAY ∀X, Y ∈ R n . (1) ST: R n *<O Mu*HfS' Ppq; (2) s ε1 = (1, 0, · · · , 0), ε2 = (0, 1, 0, · · · , 0), · · · , εn = (0, 0, · · · , 0, 1) w ]^, (3) ~%wfpqUV–WXUYdzUV). : (1) i. (2) w ]^" A. (3) α = (a1, · · · , an), β = (b1, · · · , bn), J ¯ ¯ ¯ ¯ ¯ Xn k=1 kakbk ¯ ¯ ¯ ¯ ¯ 6 vuutXn k=1 ka2 k × vuutXn k=1 kb2 k . 4. C Hf n y2> ]^. k R n , 7f X, Y , T (X, Y ) = XTC TCY ST: R n *<O Mu*HfS' Ppq. · 7 ·
证明:略. 5.在标准欧几里得空间内计算给下向量价内积,并求它的之间价夹角 (3)a=(3,-1,1,-1),B=( (4)a=(-1,1,-1,2,1),B=(3,1,-1,0,1) #f:(1)(a, B)=8,o, B)=arc cos (2)(a,B)=2,(a,B)=arco0 (3)(a,3)=-12,(a,B)= (4)(a,)=0.(a,)= 6.设x2+y2+2=1,(x,y,z∈R),试求 价最小值 解魔式=-3+1-x+1-y+1-2x,而魔标西布涅标夫斯基不等式 1-x2)2+(√1-y2)2+(√1-y v-2-ⅵ1-)(-z2)=3 明 2≥3, 所以 ≥-3+ 又当x=y=8=当时上式取等号因魔式价最小值为是 7.设a,b,c,x,,z∈R,列a2+b2+c2=25,x2+y2+2=36,am+by+cz=30.求旦+b+c价 值 解:魔标西-布涅标夫斯基不等式 30=ar +by +ca=(a b c) ≤√a2+b2+cm2+y2+2=30. 因等号成立时,(a,b,c)与(x,y,z)成相例设(a,b,c)=t(x,3,2),又入得 解而t 从而a+b+c
: i. 5. kUS' Ppq{xg {, Ws89q5: (1) α = (1, 1, 1, 1), β = (−1, 2, 4, 3); (2) α = ³ 1 2 , −1, 1 3 , 1 6 ´ , β = (3, −1, 2, 2); (3) α = (3, −1, 1, −1), β = (−2, 2, −2, 2); (4) α = (−1, 1, −1, 2, 1), β = (3, 1, −1, 0, 1). : (1) (α, β) = 8, hα, βi = arc cos 2 √ 30 15 . (2) (α, β) = 7 2 , hα, βi = arc cos 7 10 . (3) (α, β) = −12, hα, βi = 5π 6 . (4) (α, β) = 0, hα, βi = π 2 . 6. x 2 + y 2 + z 2 = 1, (x, y, z ∈ R), s x 2 1 − x 2 + y 2 1 − y 2 + z 2 1 − z 2 \. : K) = −3 + 1 1 − x 2 + 1 1 − y 2 + 1 1 − z 2 . %NUV–WXUYdzUV), vuut µ 1 √ 1 − x 2 ¶2 + Ã 1 p 1 − y 2 !2 + µ 1 √ 1 − z 2 ¶2 · q ( p 1 − x 2) 2 + (p 1 − y 2) 2 + (p 1 − y 2) 2 > Ã 1 √ 1 − x 2 , 1 p 1 − y 2 , 1 √ 1 − z 2 ! √ 1 − x 2) p 1 − y 2 p 1 − y 2 = 3, s 1 1 − x 2 + 1 1 − y 2 + 1 1 − z 2 · √ 2 > 3, #$ 1 1 − x 2 + 1 1 − y 2 + 1 1 − z 2 > 9 2 . x 2 1 − x 2 + y 2 1 − y 2 + z 2 1 − z 2 > −3 + 9 2 = 3 2 . Qb x = y = z = √ 3 3 Ry)zVY. !K)\" 3 2 . 7. a, b, c, x, y, z ∈ R, a 2 + b 2 + c 2 = 25, x 2 + y 2 + z 2 = 36, ax + by + cz = 30. s a + b + c x + y + z . : NUV–WXUYdzUV), 30 = ax + by + cz = (a b c) x y z 6 p a 2 + b 2 + c 2 · p x 2 + y 2 + z 2 = 30. !VY*+R, (a, b, c) B (x, y, z) *ej. (a, b, c) = t(x, y, z), QRP 30 = t(x 2 + y 2 + z 2 ) = 36t, -% t = 5 6 . C% a + b + c x + y + z = 5 6 . · 8 ·