解:(1)代入验证即可.证略 (2)我们有 010 (mh1n2n3n)=(1E2E3E4) 而f在基1,E2,E3,E4下的度量矩阵为 因此f在基m,m,T3,7下的度量矩阵为 2010 0210 1210334-1 11-21 11-21 (3)取a=(1,1,1,1),显然有f(a,a)=0. 10.设V=K4,a=(x1,x2,x3,x4),B=(y1,y2,y3,y4) f(a, B)=3T192-5T291+r3y (1)求∫在基 mh1=(2,1,-1,1),m=(1,2,1,-1) 7=(-1,1,2,1),n4=(1,-1,1,2) 下的度量矩阵 (2)另取V的基E1,E2,E3,E4 (1,E2,E3,E4)=(m,m2,m3,n)T, 其中 111 l11 1-1-1 11 求∫在E1,E21E3,∈4下的度量矩阵 解:(1)把∫在自然基下的度量矩阵记为B,把由自然基到基m1,m2,73,n4的过渡矩阵记为A,则 0300 5000 121 1-121 00-40 于是f在基m,m,T,7下的度量矩阵为 2 C=A BA✼ : (1) ⑩❶❷✦❸✐. ✦❹. (2) ❺❻✵ (η1 η2 η3 η4) = (ε1 ε2 ε3 ε4)   2 0 1 0 1 2 1 0 −1 1 −2 1 1 0 1 2   ● f ❏✺ ε1, ε2, ε3, ε4 ❳✢❦❧♠♥❍   1 1 −1 −1   , ❞❡ f ❏✺ η1, η2, η3, η4 ❳✢❦❧♠♥❍   2 1 −1 1 0 2 1 0 1 1 −2 1 0 0 1 2     1 1 −1 −1     2 0 1 0 1 2 1 0 −1 1 −2 1 1 0 1 2   =   3 3 0 −1 3 3 4 −1 0 4 −3 0 −1 −1 0 −5   . (3) ① α = (1, 1, 1, 1), ✬✭✵ f(α, α) = 0. 10. ✍ V = K4 , α = (x1, x2, x3, x4), β = (y1, y2, y3, y4), f(α, β) = 3x1y2 − 5x2y1 + x3y4 − 4x4y3. (1) ✻ f ❏✺ η1 = (2, 1, −1, 1), η2 = (1, 2, 1, −1), η3 = (−1, 1, 2, 1), η4 = (1, −1, 1, 2) ❳✢❦❧♠♥; (2) ❼① V ✢✺ ε1, ε2, ε3, ε4: (ε1, ε2, ε3, ε4) = (η1, η2, η3, η4)T, ✿❀ T =   1 1 1 1 1 1 −1 −1 1 −1 1 −1 1 −1 −1 1   , ✻ f ❏ ε1, ε2, ε3, ε4 ❳✢❦❧♠♥. ✼ : (1) ❽ f ❏ ❾✭✺❳✢❦❧♠♥❿❍ B, ❽ ❆ ❾✭✺✮✺ η1, η2, η3, η4 ✢➀➁♠♥❿❍ A, ✾ B =   0 3 0 0 −5 0 0 0 0 0 0 1 0 0 −4 0   , A =   2 1 −1 1 1 2 1 −1 −1 −1 2 1 1 −1 1 2   , ❝ ✎ f ❏✺ η1, η2, η3, η4 ❳✢❦❧♠♥❍ C = A TBA =   −1 4 2 −17 −20 −1 22 −7 −7 −17 −4 −2 22 2 −17 −4   . · 4 ·