4 A 95
96 4 1 n {1, 2,,n} n n σ ( ) 1 2 n σ(1) σ(2) σ(n) σ σ 2 1 n 1 2 {1, 2,,n} n n! n S n σ, τ S n {1, 2,,n} τ σ {1, 2,,n} n τ σ σ, τ τσ σ n σ 1 n σ 1 ( σ σ ) 1 σ = σσ 1 = ι 1 2 n ι 1 2 n 4.1. 4 σ = ( ) 1 2 3 4, τ = 2 4 1 3 ( 1 2 3 ) 4 4 1 3 2 τσ σ 1 σ 1 2 τ 2 1 τσ 1 τσ τσ = ( 1 2 3 ) 4 1 2 4 3 4.1 σ σ 1 2 σ 1 2 1 2, 3, 4 σ 1 = ( 1 2 3 ) 4 3 1 4 2
A 97 σ τ σ σ 1 τσ 4.1: σ S n i, j 1 i<j n γ σ (i, j) 0 σ(i) <σ(j) γ σ (i, j) = 1 σ(i) >σ(j) Γ σ = 1 i<j n γ σ (i, j) ( Γ ) σ σ = 1 2 3 4 1 4 4 3 1 2 P 1, P 2, P 3, P 4 4 Q 1, Q 2, Q 3, Q 4 σ σ 1 4 P 1 Q 4 P 1 Q 4 σ 2 3 P 2 Q 3 i P i Q σ(i) 3 1 4.2 5 Γ σ σ Γ σ
98 4 P 1 P 2 P 3 P 4 Q 1 Q 2 Q 3 Q 4 ( ) 4.2: Γ σ 1 2 3 4 σ = 5 4 3 1 2 Γ σ =5 σ S n sgn σ =( 1) Γσ sgn σ σ (sign) Γ σ sgn σ ( =1 ) Γ σ sgn σ = 1 1 2 3 4 σ = 4.2 Γ 4 3 1 2 σ =5 sgn σ = 1 4.2. σ, τ S n sgn(τσ)=sgn τ sgn σ (4.1) 4.3. 2N p N 2 1 q q p 2 r p p = q +2r p q
A 99 p, q 4.3 p r q 4.3: 4.2 γ σ (i, j) σ S n i >j γ σ (i, j) =γ σ (j, i) i, j σ(i), σ(j) γ σ (i, j) γ σ (i, j) = 0 (i j)(σ(i) σ(j)) > 0 1 (i j)(σ(i) σ(j)) < 0 i j σ, τ S n i, j {1, 2,,n} i j. γ σ (i, j) =0 γ τ (σ(i),σ(j)) =0 = γ τσ (i, j) =0 γ σ (i, j) =0 γ τ (σ(i),σ(j)) =1 = γ τσ (i, j) =1 (4.2) γ σ (i, j) =1 γ τ (σ(i),σ(j)) =0 = γ τσ (i, j) =1 γ σ (i, j) =1 γ τ (σ(i),σ(j)) =1 = γ τσ (i, j) =0 1 γ σ (i, j) = γ τ (σ(i),σ(j)) = 0 (i j)(σ(i) σ(j)) (σ(i) σ(j))((τσ)(i) (τσ)(j)) (i j)(σ(i) σ(j)) 2 ((τσ)(i) (τσ)(j)) (i j)((τσ)(i) (τσ)(j)) > 0 γ τσ (i, j) =0 1 i<j n i, j (i, j) 4.3 N = n(n 1)/2 (i, j) i, j 1 i<j n γ σ (i, j) =1 1 γ σ (i, j) =0 1
100 4 (i, j) γ τ (σ(i),σ(j)) =1 1 γ τ (σ(i),σ(j)) = 0 1 (i, j) 2 (4.2) 1 γ τσ (i, j) =Γ τσ 1 i<j n γ σ (i, j)+ γ τ (σ(i),σ(j)) 1 i<j n = 1 i<j n γ σ (i, j)+ 1 i<j n 1 i<j n γ τ (i, j) =Γ σ +Γ τ 4.3 Γ τσ Γ σ +Γ τ sgn(τσ)=( 1) Γτσ =( 1) Γσ+Γτ =( 1) Γσ ( 1) Γτ = sgn σ sgn τ, (4.1) 4.2 4.4. σ σ 1 sgn(σ 1 )=sgn σ 4.2 σ 1 σ = ι ι 1 σ 4.2 σ sgn σ 4.2 σ n σ p, q 1 p<q n p, q 1 p<q n σ(q) i = p σ(i) = σ(p) i = q i
A 101 σ p, q q = p +1 σ 1 σ τ 1,τ 2,,τ 2k+1 σ = τ 1 τ 2 τ 2k+1 4.2 sgn σ = sgn τ 1 sgn τ 2 sgn τ 2k+1 =( 1) 2k+1 = 1 1 4.4 4.4: 1 σ σ σ = τ 1 τ k τ 1,,τ k 4.2 sgn σ = sgn τ 1 sgn τ k =( 1) k 4.5. σ sgn σ =1 sgn σ = 1 4.6. 3 3 4.5 ( ) 1 2 3 1 2 3 ( ) ( ) 1 2 3 1 2 3, 1 3 2 3 1 3 1 2 1 4.5
102 4 ( ) ( ) 1 2 3 1 2 3 1 2 3 1 3 2 ( ) ( ) 1 2 3 1 2 3 2 1 3 2 3 1 ( ) ( ) 1 2 3 1 2 3 3 2 1 3 1 2 4.5: 3 B n A =(a ij ) det A = σ S n sgn σ a 1σ(1) a 2σ(2) a nσ(n) (4.3) det A A n n ( ) ( ) 1 2 1 2 n =2 2 1 2 2 1 1, 1 ( ) a b det = ad bc (4.4) c d n =3 4.7. 3 a b c p q r (4.5) x y z
B 103 4.6 det A = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 a 13 a 22 a 31 a 12 a 21 a 33 a 11 a 23 a 32, a b c det p q r = aqz + brx + cpy cqx bpz ary (4.6) x y z (4.6) (4.5) 4.6 aqz, brx, cpy 3 (4.6) (4.6) cqx, bpz, ary 4.6 (4.6) 3 4 4 a b c a b c a b c p q r p q r p q r x y z x y z x y z 4.6: 3 A =(a ij ) n
104 4 i <j i, j a ij =0 A i>j i, j a ij =0 A λ 1 λ 2 0....... λ n λ 1. λ.. 2.. 0.. λ n (4.7) (4.8) 4.8. (4.7) (4.8) A =(a ij ) (4.7) det A det A = σ S n sgn σ a 1σ(1) a 2σ(2) a nσ(n) (4.9) A 1 A a 12 = a 13 = = a 1n =0 (4.9) a 1σ(1) a 2σ(2) a nσ(n) σ(1) =1 det A (4.9) σ(1) =1 σ det A = σ Sn sgn σ a 11 a 2σ(2) a nσ(n) (4.10) σ(1)=1 A 2 a 2j j =1, 2 (4.10) σ(2) = 1 2 σ(1) = 1 σ(2) 1 σ(2) = 2 σ
B 105 det A = σ Sn σ(1)=1,σ(2)=2 sgn σ a 11 a 22 a 3σ(3) a nσ(n) (4.9) a 1σ(1) a 2σ(2) a nσ(n) σ(1) =1,σ(2) =2,,σ(n) =n σ det A = a 11 a 22 a nn = λ 1 λ 2 λ n A (4.8) A 1 n det A = a 11 a 22 a nn = λ 1 λ 2 λ n 4.8 4.9. A t A det t A = det A (4.11) (4.11) det A = sgn σ a σ(1)1 a σ(2)2 a σ(n)n (4.12) σ S n 4.9. σ S n a σ(1)1 a σ(2)2 a σ(n)n = a 1σ 1 (1)a 2σ 1 (2) a nσ 1 (n) 4.4 sgn(σ 1 )=sgn σ sgn σ a σ(1)1 a σ(2)2 a σ(n)n = sgn(σ 1 ) a 1σ 1 (1)a 2σ 1 (2) a nσ 1 (n) σ S n σ S n
106 4 = sgn τ a 1τ(1) a 2τ(2) a nτ(n) τ S n = det A 2 τ = σ 1 σ S n τ = σ 1 S n S n (4.12) 4.9 4.10. A n i j A i j B I A i κ B II B III A i j κ A A A i j i i κ i i + j κ B I, B II, B III. B I, B II, B III A A A i j i i κ i i + j κ B I, B II, B III. (1) det B I = det B I = det A, (2) det B II = det B II = κ det A, (3) det B III = det B III = det A.
B 107 4.11. 1 2 2 2 A = 2 1 2 2 2 2 1 2 2 2 2 1 4.10 1 2 2 2 det 2 1 2 2 2 2 1 2 2 2 2 1 1 = det 2 3 2 2 2 2 3 2 2 2 2 3 1 = det 2 3 2 2 2 2 3 2 2 2 2 3 1 = det 2 1 2 2 2 1 3 2 2 2 3 1 = det 2 1 2 1 5 4 2 2 3 1 = det 2 1 2 1 1 4 2 1 3 1 = det 2 1 2 1 1 2 1 7 { 2 2 1 2 3 3 1 2 4 4 1 2 { 2 2 ( 1) 3 3 ( 1) 4 4 ( 1) 2 2 3 { 3 3 2 2 4 4 2 2 3 3 4 4 4 3 4 4.8 7 det A = 7 det A = 7 4.10. (1) B I A t A i j 4.9 det B I
108 4 = det A A det B II = det A det B I = det A ( ) 1 i j n B I (p, q) b pq α 1 j i n σ S n b pσ(p) = a p (ασ)(p) p =1,,n 4.2 sgn(ασ) = sgn σ det B I = σ S n sgn σ b 1σ(1) b nσ(n) = σ S n sgn(ασ) a 1(ασ)(1) a n(ασ)(n) = τ S n sgn τ a 1τ(1) a nτ(n) = det A 2 σ S n τ = ασ S n S n (2) (3) (1) det B III = det A det B III = det A 4.9 1 i 1 i n c 1,, c i,, c n c i n det(c 1 c i + c i c n )=det(c 1 c i c n )+det(c 1 c i c n ) (4.13) (c 1 c i c n ) c 1,, c i,, c n (4.13) n C i n c i C C i c i (4.13) c q p c pq c i p c pi (4.12) det(c 1 c i + c i c n )= σ S n sgn σ c σ(1)1 (c σ(i)i + c σ(i)i ) c σ(n)n = σ S n sgn σ c σ(1)1 c σ(i)i c σ(n)n + σ S n sgn σ c σ(1)1 c σ(i)i c σ(n)n = det(c 1 c i c n )+det(c 1 c i c n ).
B 109 (4.13) det B III = det A A q q = 1,,n a q A = (a 1 a i a j a n ), B III =(a 1 a i + κa j a j a n ) (4.13) (2) det B III = det(a 1 a i + κa j a j a n ) = det(a 1 a i a j a n )+κ det(a 1 a j a j a n ) 1 det A 2 C =(a 1 a j a j a n ) i j a j i j C C C = C. (1) det C = det C. det C = det C det C =0 det B III = det A 4.10 4.12. A det A 0 4.12 4.12. A n B 4.10 det A 0 det B 0 (4.14)
110 4 B B λ 1 λ 2 0 B =....... λ n 4.8 det B = λ 1 λ 2 λ n B i λ i =0 p>i λ p =0 det B = λ 1 λ 2 λ n =0 λ n =0 λ n 0 = det B 0 (4.15) ( ) rank A = n = λ n 0 (4.16) det B = λ 1 λ 2 λ n 0 ( ) rank A = n det B 0 = rank A = n. (4.17) (4.15) (4.17) det B 0 rank A = n (4.14) det A 0 rank A = n 3.14 rank A = n A 4.12 4.13. A, B n A det(ba) =det B det A det A 1 = 1 det A
B 111 a 1,, a n n n (a 1 a n ) det(a 1 a n ) D(a 1,, a n ) a 1,, a n R n 4.10 (1) (2) (4.13) D D(a 1,, a j,, a i,, a n ) = D(a 1,, a i,, a j,, a n ) (i j), (4.18) D(a 1,,κa i,, a n )=κ D(a 1,, a i,, a n ), (4.19) D(a 1,, a i + a i,, a n ) = D(a 1,, a i,, a n )+D(a 1,, a i,, a n ). (4.20) (4.18) D(a 1,, a i,, a j,, a n ) a i a j i j (4.19) (4.20) a 1,, a i,, a n a i D(a 1,, a i,, a n ) a i R n 4.13 D(a 1,, a n )=det(a 1 a n ) 3 (4.18) (4.20) 4.14. D n n a 1,, a n R n D(a 1,, a n ) D (4.18) (4.20) D(a 1,, a n )=D(e 1,, e n ) det(a 1 a n ) (4.21) a 1,, a n R n {e 1,, e n } R n (4.21) (4.18) (4.20) D(a 1,, a n ) det(a 1 a n ) D D(e 1,, e n ) (4.21) 4.14. a 1,, a n R n p, q =1,,n a q p a qp a q = n p=1 a qpe p (4.19) (4.20) D(a 1,, a n )=D( n p 1 =1 a 1p 1 e p1, n p 2 =1 a 2p 2 e p2,, n p n=1 a np n e pn )
112 4 n = a 1p1 a 2p2 a npn D(e p1, e p2,, e pn ) p 1,p 2,,p n=1 (4.18) e p1, e p2,, e pn D(e p1, e p2,, e pn ) 4.10 (3) 2 p 1,p 2,,p n n a 1p1 a 2p2 a npn D(e p1, e p2,, e pn ) p 1,p 2,,p n=1 = a 1σ(1) a 2σ(2) a nσ(n) D(e σ(1), e σ(2),, e σ(n) ). σ S n e σ(1), e σ(2),, e σ(n) e 1, e 2,, e n (4.18) 4.5 D(e σ(1), e σ(2),, e σ(n) )=sgn σ D(e 1, e 2,, e n ) (4.21) 4.13. A =(a 1 a n ) BA =(Ba 1 Ba n ) det(ba) = det(ba 1 Ba n ) D(a 1,, a n ) D(a 1,, a n )=det(ba 1 Ba n ) D (4.18) (4.20) 4.14 det(ba) =D(a 1,, a n )=D(e 1,, e n ) det(a 1 a n ) det(a 1 a n )=det A D D(e 1,, e n )=det(be 1 Be n )=det B 4.13 C A n i j (n 1) Â ij A (i, j)- i, j =1,,n A (i, j)- ij A = ij =( 1) i+j det Âij 1 2 3 4 12 13 14 5 11 16 15 6 10 9 8 7
C 113 Â 32 = 1 3 4 12 14 5 10 8 7, 32 =( 1) 3+2 det 1 3 4 12 14 5 10 8 7 = 220, 4.15. A n i, j =1,,n det A = a i1 i1 + a i2 i2 + + a in in, (4.22) det A = a 1j 1j + a 2j 2j + + a nj nj (4.23) (4.22) (4.23) A i j. 4.15. 4.9 (4.23) (4.22) 4.10 (4.22) i =1 (4.22) i =1 j =1,,n T j = {σ S n : σ(1) = j} S n = T 1 T 2 T n T j T k = φ (j k) n S n T 1,T 2,,T n det A = sgn σ a 1σ(1) a 2σ(2) a nσ(n) σ S n = sgn σ a 11 a 2σ(2) a nσ(n) + sgn σ a 12 a 2σ(2) a nσ(n) σ T 1 σ T 2 + + sgn σ a 1n a 2σ(2) a nσ(n) σ T n
114 4 1j = σ T j sgn σ a 2σ(2) a nσ(n) (j =1,,n) (4.24) j =1,,n A (1,j) Â 1j (p, q) b pq 1j 1j =( 1) 1+j τ S n 1 sgn τ b 1τ(1) b n 1 τ(n 1) Â ij a p+1 q q<j b pq = a p+1 q+1 q j τ S n 1 ˇτ S n ˇτ(1) = j τ(i 1) τ(i 1) <j ˇτ(i) = τ(i 1)+1 τ(i 1) j (i =2,,n) 1j =( 1) 1+j τ S n 1 sgn τ a 2ˇτ(2) a nˇτ(n) (4.25) ˇτ sgn ˇτ τ sgn τ τ S n 1 τ S n n (i =1) τ(i) = τ(i 1) (2 i n) ( ) 1 j 1 j j +1 n 1 n α = S 1 j 1 j +1 j +2 n j n ˇτ = α τ sgn τ =( 1) n 1 sgn τ sgn α =( 1) n j 4.2 sgn ˇτ =( 1) 1+j sgn τ (4.25) 1j = sgn ˇτ a 2ˇτ(2) a nˇτ(n) τ S n 1 τ S n 1 ˇτ S n S n 1 T j (4.24) (4.24)
C 115 4.11 4.11 det = det 1 2 2 2 2 1 2 2 2 2 1 2 2 2 2 1 = 1 det 1 2 3 2 2 2 2 3 2 2 2 2 3 3 2 2 2 3 2 2 2 3 = 3 3 3 2 2 2 2 2 2 1 +2 3 2+2 2 3+3 2 2 = 7 4.16. n A = 0 1 1 0 1 0 1......... 0 1 0 1 0 n 2 n n A n det A n = det 1 1 0 0 1 1......... 0 1 1 0 1
116 4 = det 1 0 0 0 1 1......... 0 1 1 0 2 2 1 = det A n 2 1 ( 0 1 det A 2 = det 1 0 ) = 1, det A 3 = det 0 1 0 1 0 1 0 1 0 =0 det A = 1 (n =4m) 1 (n =4m +2) 0 (n =4m +1, 4m +3)