8.在标准欧几里得空间R3中求基a1=(1,0,1),a2=(1,1,0),a3=(0,1,1)的度量矩阵 解:A 9.在4维欧几里得空间V中,设基a1=(1,1,-1,-1),a2=(1,1,1,0),a3=(-1,1,1,1),a4= (1,0,0,-1)的度量矩阵为 210 A 012 0012 (1)求基E1=(1,0,0,0),E2=(0,1,0,0),3=(0,0,1,0),E4=(0,0.0,1)的度量矩阵; (2)求向量A1=(1,-1,1,-1),A2=(0,1,1,0)的内积 (3)求一单位向量与a1,a2,a3正交 解:(1)(a1,a2,a3,a4)=(∈1,E2,E3,E4)B, 11-11 B l11 0 12002 2-2112 因此基ε1,ε2,E3,E4得度量矩阵应为 G=BTAB (3)设a=∑x0;与a1,a2,a3正交,则 从而 0. 解得(x1,x2,x3,x4)=(1,-2,3,-4),所以 a=a1-202+3a3-4a4 (a,2)=(a.a)=20.因此所求的单位向量为±当(-1.0.3 10.设a1,a2,…,αm是欧几里得空间V的m个向量,称矩阵 (a1,a1)(a1,a2) (a1, am) (a2,a1)(a2,a2) (a2, am) 为向量组a1,a2,…,am的格拉姆(Gram)矩阵 证明:a1,a2,……,am线性无关当且仅当|G(a1,a2,…,am川≠0.8. kUS'Ppq R 3 , sz α1 = (1, 0, 1), α2 = (1, 1, 0), α3 = (0, 1, 1) w ]^. : A =   2 1 1 1 2 1 1 1 2  . 9. k 4 FS'Ppq V , z α1 = (1, 1, −1, −1), α2 = (1, 1, 1, 0), α3 = (−1, 1, 1, 1), α4 = (1, 0, 0, −1) w ]^" A =   2 1 0 0 1 2 1 0 0 1 2 1 0 0 1 2   (1) sz ε1 = (1, 0, 0, 0), ε2 = (0, 1, 0, 0), ε3 = (0, 0, 1, 0), ε4 = (0, 0, 0, 1) w ]^; (2) s β1 = (1, −1, 1, −1), β2 = (0, 1, 1, 0) {; (3) sH/ B α1, α2, α3 r. : (1) (α1, α2, α3, α4) = (ε1, ε2, ε3, ε4)B, B =   1 1 −1 1 1 1 1 0 −1 1 1 0 −1 0 1 −1   , B−1 = 1 2   0 1 −1 0 2 0 0 2 −2 1 1 −2 −2 0 2 −4   . !Oz ε1, ε2, ε3, ε4 Pw ]^," G = B −TAB−1 =   6 − 1 2 − 9 2 9 − 1 2 1 1 2 −1 − 9 2 1 2 4 −7 9 −1 −7 14   . (2) (β1, β2) = 6. (3)  α = P 4 i=1 xiαi B α1, α2, α3 r, J Ã αj , X 4 i=1 xiαi ! = 0, j = 1, 2, 3. C%   2 1 0 0 1 2 1 0 0 1 2 1     x1 x2 x3 x4   = 0. -P (x1, x2, x3, x4) = (1, −2, 3, −4), #$ α = α1 − 2α2 + 3α3 − 4α4 = (−8, 2, 0, 6). (α, α) = (α4, α) = 20. !O#s/ " ± √ 5 5 (−4, 1, 0, 3). 10.  α1, α2, · · · , αm S'Ppq V  m f , u]^ G(α1, α2, · · · , αm) =   (α1, α1) (α1, α2) · · · (α1, αm) (α2, α1) (α2, α2) · · · (α2, αm) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (αm, α1) (αm, α2) · · · (αm, αm)   " B α1, α2, · · · , αm Z`[ (Gram) ]^. ST: α1, α2, · · · , αm t&,*b?cb |G(α1, α2, · · · , αm)| 6= 0. · 9 ·