Ch4 Determiats 行列式 (The determiat) 是個 special scalar-valued fuctio 定義在 M ( F). 雖然其重要性因 umerical liear algebra 的成長, 而減低. 但行列式仍然在線性書中不可或缺. 這本書行列式主要用途在計算一個矩陣的 eigevalues 和建立這些 eigevalues 的一些性質. 4.1 介紹 矩陣之行列式. 並解釋其幾何特性. 4., 4. & 4.3 定義 矩陣行列式和其性質. 4.5 引進行列式另一等價定義, 也即行列式可定義為 從 M ( F) F 的函數 並滿足三個性質.(i) -liear (ii) alteratig (iii) ( I) 1. ( 這個觀點可以幫助我們了解 det A 的幾何性質 本章許多主要定理的證明常使用以下兩技巧 : (i) Mathematical Iductio( 數學歸納法 ) o the size of the matrix. 這是因為行 列式的定義利用降階的遞迴式來表達. 和 (ii) 先證明結論對 elemetary matrices 是適用的再推廣到一般矩陣. 一般可逆 矩陣可表成 elemetary matrices 的連乘積. 1
4.1 Determiats of Order H.W. 1, 11 a b Def: Let A. The determiat of A det( A) A ad bc. c d uvwf F Thm4.1,,, u v u v i) = + (1) w w w w w w ii) = + () u v u v (The determiat is a liear fuctio of each row whe the other rows are fixed Thm4. A 0 A: ivertible. Moreover, if A is ivertible, the A 1 1 d b =. A c a AA A A I 1 1 ( ). A b b b A e.r.o.'s 11 1 ( ) 0 det( ) 0. 0 b pf: ( ) 直接驗證. A A 已知 11 1 Let A ra( A) A11, A1不能同為 0. A1 A
A11 A1 e.r.o. 設 A11 0 A1 A1 ra ( e.r.o. ra preservig 0 A 的也是因為是 A 11 A A - A A 0 det( A) 0. 若 A 0, 同理我們也可得 det( A) 0 11 1 1 1 Def: Let uv, is a ordered basis for R, we defie the orietatio of to be the real umber u det u v 0 ( 分母不為 0 v u det v u u Note that is a full ta matrix. Hece, det 0. v v Clearly u 0 1. v u 0 1 uv, forms a right-haded coordiate system. (Why?) v ( u ca be rotated i a couterclocwise directio through a agle 10< < to coicide with v u 0 1 uv, forms a left-haded coordiate system. (Why?) v u For coveiece, we also defie 0 1. if uv, is liearly depedet. v 3
Facts: (Ex11) (I): Suppose : M F satisfies. (a) (1),() i the Theorem 4.1. u (b) Let A. The ( A) 0. u (c) ( I ) 1. The ( A) det( A) for all AM ( F (This result is geeralized i sectio 4.5, which ca be see as a equivalet defiitio of det( A) Pf: Let u ( a, b) ae be, ad v ( c, d) ce de. Usig (a), we have that 1 1 u ae1 be e1 e a b. v v v v e1 e a b. ce de ce de 1 1 e1 e1 e e ac d bc d. e e e e 1 1 e1 e e ad bc ad bc. e e e 1 1 To complete the proof, we eed to show that e -1. e1 4
To this ed, we have 0 1 0 1 1 0 1 1 +0= +. 1 0 1 0 1 0 1 0 ad 1 1 1 1 1 1 1 1 1 1 0. 1 0 0 1 1 1 1 0 0 1 However, 1 0 1 0 0 1 1 1 1 0. 0 1 0 1 0 1 0 1 e 0 1 e 1 1 0 1. u (II) The area of parallelogram determied by u ad v A v u u u A 0 det. v v v Pf: It is equivalet to showig that u u u det 0 A. v v v 5
u u u We oly eed to show that 0 A = satisfies the assumptios i Facts (I v v v u Pf: 第二和第三性質 ca be easily checed. To see is 1-liear, we first show that v u u c. Now, cv v u u u c u u u (i) For c 0, 0 A 0 c A c. cv cv cv c v v v u u c 0, =0. 0v v u u (ii) b. au bw w u u u u A =A. w u+ w w u w 若 a 0, 則 (ii) 對. 若 a 0, 則 u u u u b a b a b b. au ( w) u w w w a a a u u u (iii) ( b1 b) v1 v ( a1 a) u( b1 b) w w. 6
choose w so that u, w is a basis for R. v aubw. 1 1 1 v a ub w. u u u u u u = b + b = + = +. bw bw aubw aubw 1 w w 1 1 1 u u = +. v v 1 Remar: 以上的作法雖比直接作看似麻煩, 但可作為 維的平行 邊體體積的推廣方法. a1 Let AM( F) ad let A, ai為列向量, 則 det( A) 可代表由 a1, a,... a所製造的 a 平行 邊體體積. 7
4. Determiats of Order w H.W. 1-5, 3, 8, 9, 30. Def: (i) Let AM ( F If 1, so tha t A( A ), we defie det( A) A. For, 11 11 we defie det( A) recursively as i j det( A) ( 1) A det( A ) j1 ij ij = A. Here A deotes the ( -1) ( -1) matrix obtaied from A by deletig row i ij ad colum j. i j (ii) (-1) det( A ) the cofactor of the etry of A i row i,colum j C. ij ij The det( A) A C A C... A C. 11 11 1 1 1 1 = cofactor expasio alog the first row of A. Thm4.3 Let a uv. 1 r. r a1 a1 a1 a r1 a r1 a r1 det uv det u det v. a a a r1 r1 r1 a a a [det( A) det( b) det( C)] pf: Iductio o. 8
Pf: If 1, the result holds. Assume for, the result holds true for -1. (case i) r 1. Let u ( u, u,..., u ), v (( v, v,..., v 1 1 u1 v1 u v u v a1 a a A a 1 a a A B C 1j 1j 1j for 1 j. 1 j det( A) ( 1) ( u v )det( A j1 j j 1 j 1 j 1 j = ( 1) u det( B ) ( 1) v det( C j 1j j 1j j1 j1 det( B) det( C (case ii) r 1. Let u ( u, u,..., u ), v (( v, v,..., v 1 1 a11 a1 a1 ar 1,1 ar 1, a r1, A u1 v1 u v u v. ar 1,1 ar 1, a r1, a 1 a a 9
1 j det( A) a ( 1) det( A j1 1j 1j iductio 1 j = a ( 1) (d et( B ) det( C ) j1 1j 1j 1j det( B) det( C Cor. If AM has a row cosistig etirely of zeros, the det( A) 0. Pf: Let u - v ad 1 o Theorem4.3. Lemma. BM ( F), 0. if row i of B equals e for some, 1, the i det( B) ( 1) det( B i pf: Iductio o. pf: Iductio o I. The lemma is easily proved for. Assume for 3, the lemma is true for -1. (i) If i 1, the, clearly, the lemma holds. (ii) Let i 1. b1 b1 b1 bi 1,1 b i1, B 0 0 1 0 0 e. bi 1,1 bi 1, b i1, b 1 b b 1 j det B b ( 1) det( B (*) 1j 1j j1 10
註 : e 1 j B1 j的第 i-1 列 = 0 j e j, 1 where e 1,0, e F. i-1-1 ( 1) det( Cij ) j iductio det( B1 j ) = 0 j i-1 ( 1) det( Cij ) j, Cij =the matrix obtaied from B by deletig rows 1, i ad colums j ad. 將註代入 (*) i i j 1 ( j1) det B ( 1) ( 1) b1j det( Cij) ( 1) b1j det( Cij). j j B 1 i i =( 1) det( B i ( 也即上面的大中括號是將 B的第 i列, 第 行殺掉後得 B, 然後將 B 沿著第一列展開的 cofactor.) i i Thm4.4 i j det( A) ( 1) A det( A j1 ij ij ca be obtaied by cofactor expasio alog ay row. 11
Pf: This is a direct cosequece of Theorem4.3 ad Lemma. Cor. If AM ( F) has two idetical rows, the det( A) 0. Pf: Iductio o. For, it is trivial. For 3, if we expad A alog a row that is ot from these two idetical rows, the the iductio would carry the proof. Thm4.5. A M ( F) ad B is obtaied from A by iterchagig ay two rows of A, the det( B) -det( A Proof: Let a1 a1 a r a s A ad B. a s a r a a Let a1 ar a s det Da ( r as, as ar as a r a Thm4.3 = Da ( a) Da ( a) Da ( a) Da ( a r s r r s s s r 0 Da ( a) Da ( a r s s r Da ( a )= D( a a r s s r 1
Thm4.6. Let B be obtaied by a e.r.o. of type 3 o A. The det( B) det( A Thm4.3 Proof: Da (, a a) = Da (, a) Da (, a) Da (, a r r 3 r r r 3 r 3 Cor. If A M ( F) is ot full ra, the det( A) 0. Proof: Sice A is ot full ra, a ca The rth row of the matrix obtaied e.r.o.'s r i i i1 type3 ir Thm4.6 from A by suitable e.r.o.'s of type 3 cosists etirely of zeros det( A) 0. Facts: (i) A B det( B) det( A e.r.o. type1 (ii) A B det( B) det( A e.r.o. type (multiplyig a row of A by a ozero scalar (iii) A B det( B) det( A e.r.o. type3 (iv) *-4.3通常行列式的展開是利用 e.r.o.'s 將 A變成上三角形 3 矩陣, 這樣的計算量只需要 O ( ), 但利用定義需要! 的計算量. 13
Ex: 1 3 1 3 1 3 A 1 3 ( ) 0 5 1 0 5 1. 3 1 ( 3) 0 7 5 18 0 0 5 18 det( A) 1 ( 5) ( ) 18. 5 Ex. Let 1 1 A 1 C C 0 0 1 1 1 C C C C C C a Vadermode matrix. Copute det( A Sol: Let 1 C0 C0 C 0 F( x; C0, C1,..., C 1) det. 1 C 1 C 1 C 1 1 x x x The F( x; C, C,..., C ) F( C ; C, C,..., C )( xc )( xc..( xc 0 1 1 1 0 1 0 1 0 1 Here FC ( 1; C0, C1,..., C) det 1 C C C 0 0 0 C C C 1 1 1. det( A) F( C ; C, C,..., C ) F( C ; C, C,..., C )( C C ) ( C C 0 1 1 1 0 1 0 0 Iductively,we have that det( A) ( C C a i j j i 14
4.3 Properties of Determiats H.W. 1,, 10, 11, 1, 13, 15, 19, 1,. 因為任何一個可逆矩陣可表成 a product of elemetary matrices. 因此要研究一個矩陣行列式 的性質可先從 elemetary matrices 出發. 由上節末可知 (a) I E det( E ) 1. e.r.o. type1 1 1 (b) I E det( E ). e.r.o. type (c) I E e.r.o. type3 det( E ) 1. 3 3 Thm4.7 For ay AB, M ( F), det( AB) det( A) det( B Pf: (step1) Let A be a elemetary matrix. (i) I Adet( A) 1. Sice e.r.o. type1 B AB det(ab) -det(b)=det(a) det(b e.r.o. type1 (ii) type & type3 ca be similarly obtaied (step Let A be ay matrix with ra less tha. The ra( AB) ra( A). Hece det( AB) 0 det( A Ad so, det( AB) det( A)det( B (step3): Let A be ivertible. The A E... E, where E, 1 i p, are p 1 i elemetary matrice. 15
Hece ( step1) det( AB) ( E... E B) det( E )det( E... E B p 1 p p1 1 ( step1) det( E... E )det( B) det( A)det( B p 1 Cor. A matrix AM ( F) is ivertible det( A) 0. 1 1 Moreover, if A is ivertible, the det( A ). det( A) 1 1 Pf: ( ) AA I det( A)det( A ) 1. 1 det( A) 1 det( A) 0. Moreover, det( A ). ( ) If A is ot ivertible, the ra( A). Ad so det( A) 0. t Thm4.8 For ay AM ( F), det( A) det( A t Pf: (case i): A is ot ivertibe. The ra( A ) ra( A). Hece t t A is ot ivertible. det( A) 0 det( A (case ii): If A is ivertible, the A E... E, E, 1 i p, are elemetary p 1 i matrices. A E... E. t t t 1 p t Sice det( E ) det( E ),(Ex. 9 of 4. i i t det( A) det( A 16
Remars: 由定理 4.8可知 (i) det( A) ca be evaluated by the cofactor expassio alog ay colum. (ii) 由 e.r.o.'s 和 e.c.o.'s 運算所產生的行列式性質是相對應相同的. x1 b1 Thm4.9 (Cramer's Rule) AM( F), x, b. x b Ax b det( A) 0. The x det( M ), where det( A) M is the matrix obtaied by replacig colum of A by b. Pf: (i) Sice det(a) 0, Ax b has a uique solutio. Let such uique solutio be x. (ii) Let X ( e..., e, x, e,... e 1 1 1 A X ( Ae..., Ae, Ax, Ae,... Ae 1 1 1 =( a1..., a 1, b, a 1,... a) M. det( A) det( X ) det( M Clearly det( X ) X. det( M ) x =. det( A) 17
Fact: Let A=( a, a,... a ) M ( R), we may iterpret det( A) as the 1 -dimesioal volume of the parallelepiped havig the vectors a, a,... a 1 as adjacet sides. 如何將 4.1 中之 處理 det( A) 之幾何意義推廣. 其中兩件事需要處理. i ) 即 矩陣行列式的等價推廣, 這部分在 4.5 有處理. 也即滿足 (i) -liear, (ii) alteratig,(iii) ( I) 1 的 函數即為行列式. ii) ordeed base, orietatio. (, 3 給一個如何決定即左右座標系統以為例, R { e, e, e } right-haded coodiate system, let r R 3 1 3 為一個為的另一 ordered base, 則如何定義 r 的 orietatio 呢? 以下定義為 : r is right-haded coodiate system if ad oly if det( Q) 0, where Q [ I]. 則可證明若 a, a, a liearly idepedet i R 1 3 3 a1 a1 a1 det a sig( Q) V a, where V a a 3 a 3 a 3 is the volume of parallelepiped geerated by a, a, a. r a, a, a ad 1 3 1 3 1 if det Q 0. sig Q= -1 if det Q 0. 18
4.4 Summary-Importat Facts About Determiats H.W. 1, 6. 4.5 A Characterizatio of the Determiat. 我們在 4.1 提到一個 : M F. 若滿足 (I)-a,b,c 的性質, 則此函數 ( A) det( A) 這個結果在可推廣當成一個 矩陣行列式的等價性質. (a) 即是 is a -liear fuctio. (b) 即是 is alteratig. (c) det( I ) 1. 19