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1 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

2 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

3 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) 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

4 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

5 To this ed, we have = ad However, e 0 1 e 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

6 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

7 choose w so that u, w is a basis for R. v aubw v a ub w. u u u u u u = b + b = + = +. bw bw aubw aubw 1 w w 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

8 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, 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 = 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

9 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

10 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 e. bi 1,1 bi 1, b i1, b 1 b b 1 j det B b ( 1) det( B (*) 1j 1j j1 10

11 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

12 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

13 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

14 Ex: A 1 3 ( ) ( 3) det( A) 1 ( 5) ( ) Ex. Let 1 1 A 1 C C 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 Here FC ( 1; C0, C1,..., C) det 1 C C C C C C det( A) F( C ; C, C,..., C ) F( C ; C, C,..., C )( C C ) ( C C Iductively,we have that det( A) ( C C a i j j i 14

15 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 ) 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

16 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) 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

17 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 A X ( Ae..., Ae, Ax, Ae,... Ae =( a1..., a 1, b, a 1,... a) M. det( A) det( X ) det( M Clearly det( X ) X. det( M ) x =. det( A) 17

18 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 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 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 if det Q 0. sig Q= -1 if det Q 0. 18

19 4.4 Summary-Importat Facts About Determiats H.W. 1, 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

EC(2013-1 4)13 第 2 頁 (b) 把總目 100 在 2013-14 年度常額編制內所有非首長級職位按薪級中點估計的年薪總值上限提高 12,480,540 元, 即由 461,070,000 元增至 473,550

EC(2013-1 4)13 第 2 頁 (b) 把總目 100 在 2013-14 年度常額編制內所有非首長級職位按薪級中點估計的年薪總值上限提高 12,480,540 元, 即由 461,070,000 元增至 473,550 EC(2013-1 4)13 財務委員會人事編制小組委員會討論文件 2014 年 1 月 8 日總目 100- 海事處分目 000 運作開支總目 92- 律政司分目 000 運作開支總目 158- 政府總部 : 運輸及房屋局 ( 運輸科 ) 分目 000 運作開支請各委員向財務委員