7.解取R2的基:e1,e2,其中e1= Tk(e1)=e1+0e2 0 T(e2)=ken+e2=ene」1 因此,变换Tk在基e1,e2下对应的矩阵为 1 k 01 8.解:因x1,x2,x3线性独立,x1=Ba(i=1,2,3),及y=Bb(i=1,2,3,因 a(xi)=yi =0(Bai)=Bo(B)g ai=Bb (a)所以,σ在基B下的矩阵{(B)k满足下列关系 a (B)I bi bo b 求得: o(B)B= bi b2 b3 a a2 as 112|201 2-116 1-12520 (b)o(B)=o(BM)=o(B)M a(B)=Bo(B)]B=BM(o(B)Ig, a(B)M= Bo(B)IBM 因此M(B)B={a(BgM得: 14344 lo(B)g=M- [a(B)l7. 解: 取 R 2 的基: e1, e2, 其中 e1 = " 1 0 # , e2 = " 0 1 # , 则 τk (e1) = e1 + 0e2 = h e1 e2 i " 1 0 # τk (e2) = ke1 + e2 = h e1 e2 i " k 1 # 因此, 变换 τk 在基 e1, e2 下对应的矩阵为: " 1 k 0 1 # . 8. 解: 因 x1, x2, x3 线性独立, xi = Bai(i = 1, 2, 3), 及 yi = Bbi(i = 1, 2, 3), 因 σ (xi) = yi ⇒ σ (Bai) = B [σ (B)]B ai = Bbi (a) 所以, σ 在基 B 下的矩阵 [σ (B)]B 满足下列关系: [σ (B)]B h a1 a2 a3 i = h b1 b2 b3 i 求得: [σ (B)]B = h b1 b2 b3 i h a1 a2 a3 i−1 =    1 1 2 1 1 1 1 −1 2       2 0 1 3 1 0 5 2 0    −1 =    2 −11 6 1 −7 4 2 −1 0    (b) σ (B 0 ) = σ (BM) = σ (B)M σ (B 0 ) = B 0 [σ (B 0 )]B0 = BM[σ (B 0 )]B0 = σ (B)M = B [σ (B)]B M 因此 M[σ (B 0 )]B0 = [σ (B)]B M 得: σ ￾ B 0
B0 = M−1 [σ (B)]B M =    14 3 44 −3 −5 −2 −4 0 −14   