linearal1.dvi
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- けいざぶろう あさま
- 5 years ago
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5 1 11 m, n a ij (1 i m, 1 j n) a 11 a 12 a 13 a 1n a 21 a 22 a 23 a 2n A = a 31 a 32 a m1 a m2 a mn m n (m, n) a 11,a 12,a 13,,a 1n a 11 a 21 a 31 a ij A a 11 a 12 a ij a m1 (1, 1) (1, 2) (i, j) A =(a ij ) i=1,,m; j=1,,n A =(a ij ) 11 (1) (4, 2) (2) ( A = ) 1
6 (2, 3) (3) a ij = 2j i + j A =(a ij ) i=1,2,3; j=1, A, B A =(a ij ) i=1,,m; j=1,,n, B =(b ij ) i=1,,m; j=1,,n A + B A + B := (a ij + b ij ) i=1,,m; j=1,,n = c ca := (ca ij ) 12 ( ) 11 A B ( ) A B = A +( B) 1 m, n 2
7 11 ( ) A, B, C (m, n) c, d (1) A + B = B + A (2) (A + B)+C = A +(B + C), c(da) =(cd)a (3) c(a + B) =ca + cb, (c + d)a = ca + da ( ) 0 O 0( ) (2, 2) ( )( ) ( ) a b α β aα + bγ aβ + bδ := c d γ δ cα + dγ cβ + dδ (m, n) A (n, l) B AB A B 13 ( )
8 13 (1) (m, n) A (n, l) B AB (m, l) (2) AB BA 14 AB BA AB, BA = n = j n=1 j= = (2m 1) = m=1 5 (2l +1) l= = (4n +2) n=0 n n a 1 + a a n = a k = a j k=1 k=1 k=1 j=1 n n c( a k )= ca k m m m a k + b j = (a k + b k ) k=1 j=1 k=1 m a 11 + a a 1m = a 1j j=1 4
9 a 11 + a a 1n +a 21 + a a 2n + = = m ( i=1 +a m1 + a m2 + + a mn n a ij ) (i j ) j=1 n m ( a ij ) j=1 i=1 (j i ) 133 A =(a ij ) (m, n) B =(b ij ) (n, l) a 11 a 12 a 1n b 11 b 12 b 1l a 21 a 22 a 2n b 21 b 22 b 2l AB = a m1 a m2 a mn b n1 b n2 b nl AB (1, 1) c 11 c 11 = a 11 b 11 + a 12 b a 1n b n1 = n a 1k b k1 k=1 AB (2, 1) c 21 c 21 = a 21 b 11 + a 22 b a 2n b n1 = n a 2k b k1 k=1 AB (1, 2) c 12 c 12 = a 11 b 12 + a 12 b a 1n b n2 = n a 1k b k2 k=1 AB (i, j) c ij c ij = n a ik b kj k=1 5
10 2 ( n ) AB = a ik b kj k=1 i=1,,m; j=1,,l A, B (m, n) C (n, l) D (r, m) c R (1) (A + B)C = AC + BC ( ) (2) D(A + B) =DA + DB ( ) (3) c (AB) = (ca)b = A (cb) ( ) (4) (AB)C = A(BC) ( ) (5) AB BA (1) A =(a ij ), B =(b ij ), C =(c ij ) A + B =(a ij + b ij ) ( n (A + B)C = = = k=1 (a ik + b ik )c kj ) ( n ) (a ik c kj + b ik c kj ) k=1 ( n a ik c kj + k=1 = AC + BC ) n b ik c kj ( ) k=1 (2) (1) (3) c R n c a ik b kj = k=1 n (ca ik )b kj = k=1 n a ik (cb kj ) k=1 2 AB (m, l) 6
11 (4) A =(a ij ) (m, n) B =(b ij ) (n, l) C =(c ij ) (l, p) AB (i, j) (AB) ij (AB) ij = n a ik b kj k=1 (AB)C (i, j) ((AB)C) ij = = = = = l (AB) ir c rj r=1 l n ( a ik b kr )c rj r=1 l k=1 r=1 k=1 n k=1 r=1 n a ik b kr c rj l a ik b kr c rj n a ik ( k=1 r=1 ( ) ( ) l b kr c rj ) ( ) l r=1 b krc rj BC (k, j) A(BC) 14 (n, n) n A, B n AB, BA n AB BA 17 2 AB = BA AB BA n 1 0 E n E n 1( ) 11 n B E n B = BE n = B 7
12 E =(e ij ) e ij = δ ij (E n B) ij = n e ik b kj = e ii b ij = b ij k=1 BE n 1 1 A n AX = XA = E n n X A X A A 1 18 X 19 (A 1 ) 1 = A 13 A, B n AB (AB) 1 = B 1 A 1 X = B 1 A 1 (AB)X =(AB)(B 1 A 1 )=A(BB 1 )A 1 = E n X(AB) =E n 11 A j 110 (A 1 A 2 A k ) 1 = A 1 k A 1 k 1 A 1 2 A 1 1 8
13 (1) 0 c eg ( ) ( ) (2) c eg ( ) ( ( 2) ) (3) eg ( ) ( ) (1)-(3) 21 n E n 21 E n A E n 9
14 ( 3) ( 2) (1, 1) / and ( 3) ( 1) ( 2) and ( 2) E 3 22 E
15 22 ( )( ) ( ) = = ( )( ) ( ) = = ( 0 1 ) ( 3 0 ) 1 0, 0 1 E n E 3 (1) 1 c (c 0) c =: M(1; c) = (2) c c =: A(2, 1; c) = (3) =: P (1, 2) 11
16 = 22 M(i; c) E n M(1; 1/c)M(1; c)e n = M(1; c)m(1; 1/c)E n = E n c 0 A(i, j; c) A(2, 1; c) A(2, 1; c)e n = A(2, 1; c) A(2, 1; c)e n = E n P (i, j) P (2, 1)P (1, 2)E n = E n A E n 23 n A A E n B k (k =1, 2,m) B m B m 1 B 2 B 1 A = E n B k P := B m B 1 PA = E n AP = P 1 P (AP )=P 1 (PA)P = E n A
17 ( ) n A 1 (1) E n = 1 (2) ( Er O O ), E r r 21 (1) E r r A rank ( ) r = rank A (2) E n rank A = n (3) rank (4) E n A E n (4) E n A A 1 E n (2) A 3 13
18 (2) ( Er ) A = O O AX = E n n A ( Er ) A O O 24 n P j,q j ( Er ) P k P 1 AQ 1 Q l = O O ) ) ( )( = ( B := P k P 1 AQ 1 Q l A A E n ( ) PA = E n (P ) P E n E n X PE n = P X = P 14
19 26 n A (n, 2n) (A E n ) A E n (A E n ) (E n P ) P A 1 A E n A (1) (2) ( ) (m, n) A 1 (1) 1 (2) ( Er O O ) (3) ( E m ) 15
20 (4) ( En O ) x +2y +3z =1 4x +5y +6z = 1 7x +8y +9z = 3 (21) a 11 x a 1n x n = b 1 a m1 x a mn x n = b m n (22) (21) A = A =(Ab) = (22) A =
21 24 x + y =1 2x 3y =1 2x +2y =2 5y =1 y =1/5 x =4/5 28 ( ) A B B A 211 (21) 22 (A b) A = (A b ) A (A b ) ( b) x +2z =2 4x y +5z =1 2x + y +3z =
22 (A b) x +3z =1 y + z =1 0=3 29 ranka = rank(a b) b = 0 0 (i) (1) (A b) x =1 y = 1 z =3 (ii) (2) (A b) z = t x = t +4 y = 2t +1 18
23 x y z = t +4 2t +1 t = t t (iii) (2) (A b) z = t, w = s x = t +3s +1 y =4t + s 1 x y z w = t +3s +1 4t + s 1 t s = t s (1) (2) ( ) (3) (4)
24 214 x + y z = 2 (1) 2x +3y 4z = 9 x +2y 3z = 7 x +2y +3z =1 (2) 2x 2y +3z = 1 x 2y +5z =2 3x 3y z =1 (3) 2x + y + bz = 1 2x +5y 3z = a (a, b) 224 (A b) Ax = b b = 0 b 0 ranka = rank(a 0) x = 0 0 x (A b) x x = x 0 + x 1, x
25 3 31 n n =2 ( ) a b A = c d A = ad bc A 0 A ( ) A 1 = 1 d b A c a n 3 n n =2 n n (1, 2,,n) n (1, 2, 3) (1, 3, 2) (2, 3, 1) (1, 2, 3) (1, 2, 3) (2, 3, 1) (1, 3, 2) (1, 2, 3) n σ (1, 2,,n) sign σ =1 21
26 sign σ = 1 k (1, 2,,n) sign σ =( 1) k 31 sign (2, 3, 1) = (1) (2, 1) (2) (3, 2, 1) (3) (2, 4, 1, 3) (4) (5, 1, 2, 3, 4) n S n 32 S 3 = {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)} 32 S 2,S 4 S n n! (n ) 33 n A =(a ij ) i,j=1,,n A ( det A) A = sign (p 1,,p n )a p1 1a p2 2 a pnn (p 1,,p n) S n 22
27 33 n =2 A = (p 1,p 2 ) S 2 sign (p 1,p 2 )a p1 1a p2 2 = sign (1, 2)a 11 a 22 + sign (2, 1)a 21 a 12 = a 11 a 22 a 21 a n =3 4 A n! n! n B O a nn = a nn B n =3 a 11 a 12 0 a 21 a 22 0 a 31 a 32 a 33 = a 33 a 11 a 12 a 21 a 22 (n =3) A = sign (p 1,p 2,p 3 )a p1 1a p2 2a p3 3 (p 1,p 2,p 3 ) S 3 a 13 = a 23 =0 p 3 p 3 =3 a
28 a 11 a 12 0 a 21 a 22 0 = sign (p 1,p 2, 3)a p1 1a p2 2a 33 a 31 a 32 a 33 (p 1,p 2,3) S 3 = a 33 sign (p 1,p 2, 3)a p1 1a p2 2 (p 1,p 2,3) S 3 sign (p 1,p 2, 3) = sign (p 1,p 2 ) (p 1,p 2, 3) (p 1,p 2 ) a 11 a 12 0 a 21 a 22 0 = a 33 sign (p 1,p 2 )a p1 1a p2 2 a 31 a 32 a 33 (p 1,p 2 ) S = = 9(5 8) = (m, n) A =(a ij ) B =(b ij ) A B (n, m) b ij = a ji t A 35 t = ( ) 1 4 t =
29 32 (1) t ( t A)=A (2) t (A + B) = t A + t B (3) t (ca) =c t A, c R (4) t (AB) = t B t A (5) A t A ( t A) 1 = t (A 1 ) (1) (3) (4) A =(a ij ) (m, n) B =(b ij ) (n, l) AB (i, j) (AB) ij n (AB) ij = a ik b kj k=1 ( t (AB)) ij = n a jk b ki k=1 ( t A) ij = a ji, ( t B) ij = b ji ( t B t A) ij = n ( t B) ik ( t A) kj = n b ki a jk k=1 k=1 34 (5) 33 ( ) A =(a ij ) t A = A 25
30 t A (i, j) ( t A) ij ( t A) ij = a ji t A = sign(p 1,p 2,p 3 )( t A) p1 1( t A) p2 2( t A) p3 3 (p 1,p 2,p 3 ) S 3 = sign(p 1,p 2,p 3 )a 1p1 a 2p2 a 3p3 (p 1,p 2,p 3 ) S 3 a 1p1 a 2p2 a 3p3 a q1 1a q2 2a q3 3 a 1p1 a 2p2 a 3p3 = a q1 1a q2 2a q3 3 (p 1,p 2,p 3 ) (1, 2, 3) k (q 1,q 2,q 3 ) k (1, 2, 3) sign(p 1,p 2,p 3 ) = sign(q 1,q 2,q 3 ) (p 1,p 2,p 3 ) (q 1,q 2,q 3 ) (p 1,p 2,p 3 ) S 3 (q 1,q 2,q 3 ) S 3 sign(p 1,p 2,p 3 )a 1p1 a 2p2 a 3p3 = sign(q 1,q 2,q 3 )a q1 1a q2 2a q3 3 = A (p 1,p 2,p 3 ) S 3 (q 1,q 2,q 3 ) S 3 31 B O a nn = a nn B =( 3) = ( ) 33 26
31 343 n A =(a ij ) A =(a 1, a 2,,a n ) 1 a 1 = A = , a 2 = 2 5 8, a 3 = a 1, a 2,, a n = sign (p 1,,p n )a p1 1a p2 2 a pnn (p 1,,p n) S n 31 (1) a 1,,λa j + µb,,a n = λ a 1,,a j,,a n + µ a 1,,b,,a n (2) a 1,,a j,,a k,,a n = a 1,,a k,,a j,,a n (1) n =3 a 1, a 2 + b, a 3 = sign (p 1,p 2,p 3 ) a p1 1(a p2 2 + b p2 )a p3 3 (p 1,p 2,p 3 ) S 3 + α R a 1,αa 2, a 3 (2) a 1, a 3, a 2 = (p 1,p 2,p 3 ) S 3 sign (p 1,p 2,p 3 ) a p1 1a p2 3a p3 2 27
32 a p1 1a p2 3a p3 2 a p1 1a p2 3a p3 2 = a p1 1a p3 2a p2 3 (p 1,p 2,p 3 ) (p 1,p 3,p 2 ) sign (p 1,p 2,p 3 )= sign (p 1,p 3,p 2 ) (p 1,p 2,p 3 ) S 3 (p 1,p 3,p 2 ) S 3 34 (1) 31 (2) 0 a 1,,a j,,a j,,a n =0 (2) 31 (1) a 1,,a j + λa k,,a n = a 1,,a j,,a n (3) = (1, 4) = =
33 = (3, 2) = = ( ) = A =(a 1, a 2, a 3 ) A =3 (1) a 3, 5a 2, a 1 (2) a 2, a 1 + a 3, a 2 a 1 (3) 2a 2 + a 3, 3a 3 + a 1, 4a 1 + a 2 (1), (2) (3) index choice (2, 3, 1) (3, 1, 2) n A, B AB = A B 35 (1) (2) A A 0 A AX = E n ( ) X AX = E n =1 AX = A X A 0 29
34 n =3 AB = a 11 a 12 a 13 a 21 a 22 a 23 b 11 b 12 b 13 b 21 b 22 b 23 a 31 a 32 a 33 b 31 b 32 b 33 = A(b 1, b 2, b 3 ) = (Ab 1,Ab 2,Ab 3 ) AB b Ab 1 = A 0 + A b 21 A b = b 11 A 0 + b 21A 1 + b 31A = b 11 a 1 + b 21 a 2 + b 31 a 3, A =(a 1, a 2, a 3 ) Ab j = b 1j a 1 + b 2j a 2 + b 3j a 3 AB = b 11 a 1 + b 21 a 2 + b 31 a 3,b 12 a 1 + b 22 a 2 + b 32 a 3,b 13 a 1 + b 23 a 2 + b 33 a 3 AB = b p1 1a p1,b p2 2a p2,b p3 3a p3 (p 1,p 2,p 3 ) S 3 AB = b p1 1b p2 2b p3 3 a p1, a p2, a p3 (p 1,p 2,p 3 ) S 3 a p1, a p2, a p3 a 1, a 2, a 3 k a p1, a p2, a p3 =( 1) k a 1, a 2, a 3 (p 1,p 2,p 3 ) (p 1,p 2,p 3 ) (1, 2, 3) k 30
35 sgn(p 1,p 2,p 3 )=( 1) k AB = b p1 1b p2 2b p3 3 sgn(p 1,p 2,p 3 ) a 1, a 2, a 3 (p 1,p 2,p 3 ) S 3 = a 1, a 2, a 3 sgn(p 1,p 2,p 3 )b p1 1b p2 2b p3 3 (p 1,p 2,p 3 ) S 3 36 A =5, B =3 (1) A 2 (2) A 1 (3) (AB) 1 (4) B 1 AB n =3 n 3 A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 3 A (1, 1) a 11 ã 11 ã 11 =( 1) 1+1 a 22 a 23 a 32 a 33 A (1, 1) a 11 31
36 a 23 ã 23 5 ã 23 =( 1) 2+3 a 11 a 12 a 31 a A 36 n 38 (1) (2) ( ) 3 A =(a ij ) A = a 11 ã 11 + a 21 ã 21 + a 31 ã 31 A = a 11 ã 11 + a 12 ã 12 + a 13 ã 13 ( ) ( ) 38 ( ) = 2( 1) = 2( 6 20) = 52 ( ) 5 a 23 32
37 ( ) A n A = a 11 1 a 12 a 1n 0 0 a n2 a nn + a 21 0 a 12 a 1n 1 0 a n2 a nn + + a n1 0 a 12 a 1n 0 1 a n2 a nn ã 11,,ã n1 1 (n, n) ã ij 0 B 11 B B 21 B 22 0 =( 1) n j ( 1) n i B 11 B 12 B 21 B 22 =( 1) 2n (i+j) B 11 B 12 B 21 B 22 =( 1) i+j B 11 B 12 B 21 B 22 =ã ij resp resp 0=a 12 ã 11 + a 22 ã 21 + a 32 ã 31 ( ) 33
38 0=a 11 ã 31 + a 12 ã 32 + a 13 ã 33 ( ) a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 = a 11 ã 11 + a 21 ã 21 + a 31 ã 31 a 11,a 21,a 31 a 12 a 12 a 13 a 22 a 22 a 23 = a 12 ã 11 + a 22 ã 21 + a 32 ã 31 a 32 a 32 a A 3 A =(a ij ) ã 11 ã 12 ã 13 t ã 21 ã 22 ã 23 ã 31 ã 32 ã 33 = ã 11 ã 21 ã 31 ã 12 ã 22 ã 32 ã 13 ã 23 ã 33 = Ã A Ã A 32 ã 11 ã 21 ã 31 a 11 a 12 a 13 A 0 0 ÃA = ã 12 ã 22 ã 32 a 21 a 22 a 23 = 0 A 0 ã 13 ã 23 ã 33 a 31 a 32 a A AÃ ÃA = AÃ = A E 3 A 0 1/ A ( 1 1 = A( A Ã)A A Ã) =E 3 34
39 36 ( ) n A A A 0 A 0 A A 1 A 1 = 1 A Ã, Ã A 38 n =2 A =(a ij ) ( A 1 1 a22 a 12 = a 11 a 22 a 12 a 21 a 21 a 11 ) a 11 x 1 + a 12 x 2 + a 13 x 3 = b 1 a 11 a 12 a 13 A = a 21 a 22 a 23 a 31 a 32 a 33 a 21 x 1 + a 22 x 2 + a 23 x 3 = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 = b 3, x = x 1 x 2 x 3, b = b 1 b 2 b 3 Ax = b 37 A n A 0 A 35
40 Ax = b n =3 b 1 a 12 a 13 b 2 a 22 a 23 b 3 a 32 a 33 x 1 = b A a 11 b 1 a 13 a 21 b 2 a 23 a 31 b 3 a 33 x 2 = b A a 11 a 12 b 1 a 21 a 22 b 2 a 31 a 32 b 3 x 3 = b A A (x =)A 1 Ax = A 1 b Ãb A 1 = 1 Ã A A Ã, Ãb = x = 1 A Ãb ã 11 ã 21 ã 31 ã 12 ã 22 ã 32 ã 13 ã 23 ã 33 b 1 ã 11 + b 2 ã 21 + b 3 ã 31 b A b 1 a 12 a 13 b 1 ã 11 + b 2 ã 21 + b 3 ã 31 = b 2 a 22 a 23 b 3 a 32 a 33 x 2,x 3 36 b 1 b 2 b 3
41 39 4x +8y = 13 3x +11y = x = y =
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ad bc A A A = ad bc ( d ) b c a n A n A n A A det A A ( ) a b A = c d det A = ad bc σ {,,,, n} {,,, } {,,, } {,,, } ( ) σ = σ() = σ() = n sign σ sign(
I n n A AX = I, YA = I () n XY A () X = IX = (YA)X = Y(AX) = YI = Y X Y () XY A A AB AB BA (AB)(B A ) = A(BB )A = AA = I (BA)(A B ) = B(AA )B = BB = I (AB) = B A (BA) = A B A B A = B = 5 5 A B AB BA A
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LINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University
LINEAR ALGEBRA I Hiroshi SUZUKI Department of Mathematics International Christian University 2002 2 2 2 2 22 2 3 3 3 3 3 4 4 5 5 6 6 7 7 8 8 9 Cramer 9 0 0 E-mail:hsuzuki@icuacjp 0 3x + y + 2z 4 x + y
II 2 3.,, A(B + C) = AB + AC, (A + B)C = AC + BC. 4. m m A, m m B,, m m B, AB = BA, A,, I. 5. m m A, m n B, AB = B, A I E, 4 4 I, J, K
II. () 7 F 7 = { 0,, 2, 3, 4, 5, 6 }., F 7 a, b F 7, a b, F 7,. (a) a, b,,. (b) 7., 4 5 = 20 = 2 7 + 6, 4 5 = 6 F 7., F 7,., 0 a F 7, ab = F 7 b F 7. (2) 7, 6 F 6 = { 0,, 2, 3, 4, 5 },,., F 6., 0 0 a F
行列代数2010A
a ij i j 1) i +j i, j) ij ij 1 j a i1 a ij a i a 1 a j a ij 1) i +j 1,j 1,j +1 a i1,1 a i1,j 1 a i1,j +1 a i1, a i +1,1 a i +1.j 1 a i +1,j +1 a i +1, a 1 a,j 1 a,j +1 a, ij i j 1,j 1,j +1 ij 1) i +j a
x V x x V x, x V x = x + = x +(x+x )=(x +x)+x = +x = x x = x x = x =x =(+)x =x +x = x +x x = x ( )x = x =x =(+( ))x =x +( )x = x +( )x ( )x = x x x R
V (I) () (4) (II) () (4) V K vector space V vector K scalor K C K R (I) x, y V x + y V () (x + y)+z = x +(y + z) (2) x + y = y + x (3) V x V x + = x (4) x V x + x = x V x x (II) x V, α K αx V () (α + β)x
A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6
1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67
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