8.在标准欧几里得空间R3中，求基1=(1,0,1)，a2=(1,1,0),g=(0,1,1)的度量矩阵 /211Y 解：A=121 112 9.在4维欧几里得空间V中，设基a1-(1.1-1,-1),a2-(1,1,1,0),Q3-(-1,1,1,1),a4- (1,0,0,-1)的度量矩阵为 /2100 、 0121 0012 (1)求基1=(1,0,0,0)，2=(0,1,0,0)，e3=(0,0,1,0),e4=(0,0,0,1)的度量矩阵 (②)求向量3=(1，-1,1，-1)，3=(0,1,1,0)的内积 (3)求一单位向量与a1,a2,g正交 解()(1,2，a3a4)=(1,e2,3,e4B 01-10 B= -1110 -101-1/ 因此基61,2,3,4得度量矩阵应为 6--号9 G=B-TAB (2)(,32)=6. (③设a=4a与a4,a2,as正交，则 (店-a j-1,2,3. 从而 =0. 解得(e1,z2,x3,x4)-(1.-2,3,-4),所以 a=a1-2a2+3ag-4a4=(-8,2,0,6 (a,a)=(a4,a)=20.因此所求的单位向量为(-4,1,0,3) 10.设1,2，…，am是欧几里得空间V的m个向量，称矩阵 G(a1,a2,…,am）= (02,1) (a2,a2) (a2,am (am:a1)(am:a2)··(am+am） 为向量组a1,a2,…,am的格拉姆(Gram)矩阵 证明：a1,a2,…,m线性无关当且仅当1G(a1,a2,…,am≠0 98. 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 ·