七、计算题(16分求二维射影变换m=-1+2+x3的不变元素 解题给射影变换矩阵的特征方程为 1-入1 12-入 入 即 (1-)(1+A)(2-入)=0 于是特征根为A1=1,A2=-1,A3=2 题给射影变换的不变点方程组为 r1+(2-x2+x3=0 将A1=1代入(I),得 2x3=0 解之,得不变点坐标为(3,2,1) 将A2=-1代 2x1+x2-2x3=0 r1+3x2+x3=0 解之,得不变点坐标为(1,0,1) 将A3=2代入(I),得 0 解之,得不变点坐标为(1,3,1). 题给射影变换的不变直线方程组为 1-入 a1+(2-)u2+3=0 +u2-(1+)u3=06 ❃✿❄❅❆ (16 ❇). ❈❉❊❋●❍■    ρx0 1 = x1 + x2 − 2x3 ρx0 2 = −x1 + 2x2 + x3 ρx0 3 = x2 − x3 ❏❑❍▲▼✧ ★ ◆❖P◗✣ ❘❙❚☞ ❯❱❲❳☛         1 − λ 1 −2 −1 2 − λ 1 0 1 −1 − λ         = 0. ❨ (1 − λ)(1 + λ)(2 − λ) = 0. ☎✑❯❱❩☛ λ1 = 1, λ2 = −1, λ3 = 2. ◆❖P◗✣ ❘ ☞✢✣￾❲❳❬☛    (1 − λ)x1 + x2 − 2x3 = 0 −x1 + (2 − λ)x2 + x3 = 0 x2 − (1 + λ)x3 = 0 (I) ❭ λ1 = 1 ❪❫ (I), ❴    x2 − 2x3 = 0 −x1 + x2 + x3 = 0 x2 − 2x3 = 0 ❵❛ü ❴✢✣￾❜❝☛ (3, 2, 1). ❭ λ2 = −1 ❪❫ (I), ❴    2x1 + x2 − 2x3 = 0 −x1 + 3x2 + x3 = 0 x2 = 0 ❵❛ü ❴✢✣￾❜❝☛ (1, 0, 1). ❭ λ3 = 2 ❪❫ (I), ❴    −x1 + x2 − 2x3 = 0 −x1 + x3 = 0 x2 − 3x3 = 0 ❵❛ü ❴✢✣￾❜❝☛ (1, 3, 1). ◆❖P◗✣ ❘ ☞✢✣✱✓❲❳❬☛    (1 − λ)u1 − u2 = 0 u1 + (2 − λ)u2 + u3 = 0 −2u1 + u2 − (1 + λ)u3 = 0 (I 0 )