对于6≤ n、 ν=0。。。,bn公司-6c,bn-6c+1ν-向量((0)3ν,1,-2,1,(0)n-6.-6ν, 1, -2,1,(0)3ν)T∈ U是λ向量的正交替代。θ向量((0)bn)-3c,1,-2,1,(0)亿-3c)T∈ U表示奇数3≤ n=2l+1,l=1。与η向量正交。示例:n n- 10亿-2c bnc bn-6cηλνUT2 1 0不适用不适用0不适用不适用η:(1,-1) 3 2 0不适用不适用0不适用不适用η:(1,0,-1)θ : (1, -2,1)4 3 1 1不适用0不适用η:(1,0,0,-1)1 η : (0, 1, -1, 0)0 λ : (1, -1.-1,1)5 4 1 1不适用0不适用η:(1,0,0,0,-1)1 η : (0, 1, 0, -1, 0)0 λ : (1, -1, 0, -1, 1)θ : (0, 1, -2, 1, 0)6 5 2 1 0 0 η : (1, 0, 0, 0, 0, -1)1 η : (0, 1, 0, 0, -1, 0)2 η : (0, 0, 1, -1, 0, 0)0 λ : (1, -1, 0, 0, -1, 1)0 ν : (1, -2, 1, 1, -2, 1)7 6 2 1 0 0 η : (1, 0, 0, 0, 0, 0, -1)1 η : (0, 1, 0, 0, 0, -1, 0)2 η : (0, 0, 1, 0, -1, 0, 0)0 λ : (1, -1, 0, 0, 0, -1, 1)0 ν : (1, -2, 1, 0, 1, -2, 1)θ : (0, 0, 1, -2, 1, 0, 0)8 7 3 2 0 0 η : (1, 0, 0, 0, 0, 0, 0, -1)1 η : (0, 1, 0, 0, 0, 0, -1, 0)2 η : (0, 0, 1, 0, 0, -1, 0, 0)3 η : (0, 0, 0, 1, -1, 0, 0, 0)0 λ : (1, -1, 0, 0, 0, 0, -1, 1)1 λ : (0, 0, 1, -1.-1, 1, 0, 0)0 ν : (1, -2, 1, 0, 0, 1, -2, 1){η} ⊥ {λ}, {η} ⊥ {ν}, {θ} ⊥ {η}, {θ} ⊥ {λ} ,但{λ}6⊥ {ν}, {θ} 6⊥ {ν}.引理5.1从上面限制了U的秩。对于n=2、3、4,则为n- 1 = 1, 2, 3. 证明是U:{η:(1,-1) T},{η:(1,0,-1) T,θ:(1,-2,1)T},{η:(1,0,0,-1) T,η:(0,1,-1,0)T,λ:(1,-1.-1,1)T}。对于n=6,秩为n- 1 = 5: {η : (1, 0, 0, 0, 0, -1) T,η:(0,1,0,0,-1,0)T,η:(0,0,1,-1,0,0)T,(1,0,-1.-1,0,1)T,(1,-2, 1, 1, -2,1)T}。后两种策略不是η、λ、ν、θ-策略。对于n=8,秩为n- 1 = 7:{η : (1, 0, 0, 0, 0, 0, 0, -1) T,η:(0,1,0,0,0,0,-1,0)T,η:(0,0,1,0,0,-1,0,0)T,η:(0,0,0,1,-1,0,0,0)T,(0,1,-1, 0, 0, -1,1,0)T,(1,-1.-1, 1, 1, -1.-1,1)T,(1,0,0,-1.-1,0,0,1)T}。