B ci∈ 0→B∈W 明W 8.设Ⅵ1与V分别是齐一线性必程组 的满空间 证明 V1⊕V 交明:(a)对任问的a=(a1,……,an)∈K",令 1 B 则∈V,T∈v,且a=B+1.所以K"=+V (b)如果a=(a1,……,an)∈v∩V2,则 az=0, 满得a1= an=0,明a=0.所以V∩V=0 综上可得K"=V⊕V 9.设W1={A∈Mn(K)|41=4},W2={A∈Mn(K)|A=-4} 证明:Mn(K)=W1W2 交明:(a)对任问的n上矩阵A∈Mn(K),有 A=5(4+4)+(4-AT) 而2(4+AT)∈W1, 0(A-41)∈W2,所以Mn(K)=W1+W2 (b)设A∈W1∩W2,则 A=A=4 把2A=0可得A=0.所以W1∩W2=0. 最题得何Mn(K)=W1W2 10.设A∈Mn(K)且A2=A,令 V1={X∈Kn|AX=0},V2 LAX=X. 交明:(a)设a∈K",则a=(a-Aa)+Aa.而 A(a-Aa)=Aa-A2a=Aa-Aa=0,所以 a,所以Aa∈V 从而Kn=Ⅵ1+V2.J! β = Xn i=1 aiαi ∈ Vj ⇐⇒ aj = 0, β = Xn i=1 aiαi ∈ \n j=r+1 Vj ⇐⇒ ar+1 = · · · = an = 0 ⇐⇒ β ∈ W.  W = Tn j=r+1 Vj . 8.  V1 B V2 Ht&@AB x1 + x2 + · · · + xn = 0 B x1 = x2 = · · · = xn -pq. ST: Kn = V1 ⊕ V2. : (a) ￾
 α = (a1, · · · , an) ∈ Kn, I β = Ã a1 − 1 n Xn i=1 ai , a2 − 1 n Xn i=1 ai , · · · , an − 1 n Xn i=1 ai ! , γ = Ã 1 n Xn i=1 ai , 1 n Xn i=1 ai , · · · , 1 n Xn i=1 ai ! , J β ∈ V1, γ ∈ V2, ? α = β + γ. #$ Kn = V1 + V2. (b)  α = (a1, · · · , an) ∈ V1 ∩ V2, J Xn i=1 ai = 0, a1 = a2 = · · · = an -P a1 = a2 = · · · = an = 0,  α = 0. #$ V1 ∩ V2 = 0. Qy>P Kn = V1 ⊕ V2. 9.  W1 = {A ∈ Mn(K) | AT = A}, W2 = {A ∈ Mn(K) | AT = −A}. ST: Mn(K) = W1 ⊕ W2. : (a) ￾
 n y]^ A ∈ Mn(K), G A = 1 2 (A + A T) + 1 2 (A − A T), % 1 2 (A + AT) ∈ W1, 1 2 (A − AT) ∈ W2, #$ Mn(K) = W1 + W2. (b)  A ∈ W1 ∩ W2, J −A = A T = A, N 2A = 0 >P A = 0. #$ W1 ∩ W2 = 0. P Mn(K) = W1 ⊕ W2. 10.  A ∈ Mn(K) ? A2 = A, I V1 = {X ∈ Kn | AX = 0}, V2 = {X ∈ Kn | AX = X}. ST: Kn = V1 ⊕ V2. : (a)  α ∈ Kn, J α = (α − Aα) + Aα. % A(α − Aα) = Aα − A 2α = Aα − Aα = 0, #$ α − Aα ∈ V1, A(Aα) = A 2α = Aα, #$ Aα ∈ V2, C% Kn = V1 + V2. · 5 ·