A Simple Proof of Sylvester s (Determinants) Identity

Transcription

1 Appled Mathematcal Scences, Vol 2, 2008, no 32, A Smple Proof of Sylvester s (Determnants) Identty Abdelmalek Salem Department of Mathematcs and Informatques, Unversty Centre Chekh Larb Tebess of Tebessa Algera asalem@gawabcom Kouach Sad Unversty Centre of Khenchela Algera kouachsad@caramalcom Abstract In ths paper we gve a smple proof of Sylvester Identty that s based on ermnant propertes and that s obtaned nductvely Mathematcs Subject Class caton [2000]Prmary 05A,15A Keywords Dodgson s Algorthm, Matrx, Determnant, Sylvester s Identty 1 Introducton We can wrte the well-known algorthm of Dodgson, concernng a square matrx A (a ;j ) 1;jn as follows h(a ;j ) 1;jn (a ;j ) 6k;l j6k;l (a ;j ) 6l j6l (a ;j ) 6k j6l (a ;j ) 6l j6k (a ;j ) 6k j6k (1) for all k; l 1; ; n consderng k < l (see SKouach, SAbdelmalek and BReba [4]) ;

2 1572 S Abdelmalek and S Kouach Ths formula enables us condense the ermnant of n square matrx to the ermnant of 2 square matrx The elements of 2 square matrx are the ermnants of (n 1) square matrx In an other artcle (see SAbdelmalek and SKouach [1]), we condense the ermnant of n square matrx to the ermnant of (n 1) square matrx The elements of (n 1) square matrx are the ermnants of 2 square matrx as t s shown n the followng formula (a k;l ) n 2 h(a ;j ) 1;jn [ (A ;j)], 1 k; l n (2) 1;jn 1 when 8 >< A (;j) > a;j a k;j a;l a ;l a k;l a ;j+1 a k;l a k;j+1 ak;j a k;l a +1;j a +1;l ak;l a k;j+1 a +1;l a +1;j+1 f j < l; < k f j l; < k f j < l; k f j l; k We know that the generalzaton of formulas (1) and (2) lets us condense the ermnant of any n square matrx to the ermnant of m square matrx (m < n) The elements of m square matrx are the ermnants of (n (m 1)) square matrx And t s well-known by Sylvester s Identty n 1951 wth no proof see [5] p-193 After hm, others have gven proofs see [2] we gve an other clear proof We may clarfy t by the followng examples Example 1 For example n 7 and m A p

3 Proof of Sylvester s Identty p ; nally (89) p p Example 2 For example n 7 and m 3 0 B p 3 1 C C A p

4 1574 S Abdelmalek and S Kouach p p p p p p p p p p p p p p p p p p p p ; nally 2 Notatons For ths purpose we need some notatons Notaton 1 The (n k) (n l) matrx obtaned from A by removng the th 1 ; th 2 th k rows and the jth 1 ; j2 th ; jl th columns s denoted by (a ;j ) 61 ; 2 ; k j6j 1 ;j 2 ;j l Notaton 2 We denote by (a ;j ) 61 ; 2 ; k, k to the ;j j6j 1 ;j 2 ;j k ermnant of the ( k + 1) square matrx obtaned from A by removng the( 1) rst rows and columns, by removng the (n ) last rows and columns and by removng th 1 ; th 2 th k rows and the jth 1 ; j2 th ; jk th columns Notaton 3 We denote by S m (k; l) to the (n (m 1)) square matrx obtaned from A by removng the (m 1) last rows and columns, and then by replacng the last row wth the (n m + k) row and replacng the last column

5 Proof of Sylvester s Identty 1575 wth the (n m + l) column, e S m (k; l) n (m 1) z 0 } 1{ (a ;j ) 1;jn m a 1;n a 2;n a n m;n a n m+k;1 a n m+k;2 a n m+k;n m a n m+k;n also t can be wrtten as a block matrx when S m (k; l) D (k) n (m 1) 0z } 1{ n m a 1;n a 2;n a n m;n a n m+k;n z } { (a ;j ) D (k) 1;jn m n (m 1) a n m+k;1 a n m+k;2 a n m+k;n m C A ; (3) C A Notaton 4 We denote by H m (k; l) to the (n (m 1)) square matrx obtaned from A by deletng the (m 1) rows whch are ( 1 ; k 1 ; k+1 ; m ) and by deletng (m 1) columns whch are (j 1 ; j 2 ; l 1 ; l+1 ; j m ), e H m (k; l) 3 Man Results We need a lemma (a ;j ) 61 ; k 1 ; k+1 ; m j6j 1 ;j 2 ; l 1 ; l+1 ;j m Lemma 1 Let the n square matrx A (a ;j ) 1;jn 8m 2 N where 2 m < n, If h(a ;j ) 1;jn m 0, thus we get the followng formula [ [S m (k; l)]] 0 (4)

6 1576 S Abdelmalek and S Kouach Proof To prove formula (4), we let (a ;j ) 1;jn m 0 But (a ;j ) 1;jn m 0 t means the rows of the matrx (a ;j ) 1;jn m are dependent lnearly Consequently, 9 ( ) n m 1 ; 9r 2 f1; 2; ; n mg where [S m (k; l)] a 11 a 12 a 13 a 1k a 1n m a 1;n a 21 a 22 a 23 a 2k a 2n m a 2;n a 31 a 32 a 33 a 3k a 3n m a 3;n n Pm a ;n 1 a n 2;1 a n 2;2 a n 2;3 a n 2;k a n 2;n m a n m;n a n m+k;1 a n m+k;2 a n m+k;3 a n m+k;k a n m+k;;n m a n m+k;n And smply, we get n m [S m (k; l)] ( 1) n m r+1 P a ;n (D(k)) 6r, whch we can smplfy t as follows n [ [S m (k; l)]] ( 1) n m r+1 P m a ;n 1 ( 1) n m r+1 m n P m a ;n 1 1 n m (D(k)) 6r (D(k)) 6r ( 1) n m r+1 m mq P a ;n l mq (D(k)) 6r k But as t s known Thus formula (4) s realsed One of the man results of the paper s the followng

7 Proof of Sylvester s Identty 1577 Theorem 2 (Sylvester s Identty) Let the n square matrx A (a ;j ) 1;jn 8m 2 N where 2 m < n, thus the followng formula s realsed n h o m 1 h(a ;j ) 1;jn (a ;j ) 1;jn m [ [S m (k; l)]] (5) We notce that ths formula enables us condense the ermnant of n square matrx to the ermnant of m square matrx The elements of m square matrx are the ermnants of (n (m 1)) square matrx Proof To prove formula (5) there are two cases The rst case when h(a ;j ) 1;jn m 0, the proof of formula (5) s the same proof of lemma1 The second case when h(a ;j ) 1;jn m 6 0 we prove formula (5) nductvely For m 2, formula (5) s the same as formula (1) where k n 1; l n, and t has been proved (see SKouach, SAbdelmalek and BReba [4]) For m > 2 We suppose the formula (5) s correct for (m 1) n h o m 2 h(a ;j ) 1;jn (a ;j ) 1;jn (m 1) [ [S m 1 1 (k; l)]] (6) and we prove t for m We apply formula (1) where k n 1; l n on matrx S m 1 (k; l), we get [S m 2 h(a ;j ) 1;jn m S m 1 (k; l) 6n m+1 j6n m+2 1 (k; l)] h(a ;j ) 1;jn m S m 1 (k; l) 6n m+2 j6n m+1 S m 1 (k; l) 6n m+1 j6n m+1 By usng formula (7), formula (6) wll be as follows (7) n o m 2 n o m 1 h(a ;j ) 1;jn h(a ;j ) 1;jn (m 1) h(a ;j ) 2 2 1;jn m h(a ;j ) 1;jn m+1 S m 1 (k; l) 6n m+2 6 j6n m S m 1 (k; l) 6n m+1 S m 1 (k; l) 6n m+1 j6n m+2 j6n m+1 (8)

8 1578 S Abdelmalek and S Kouach We apply formula (2) on [ S m (k; l)] to get 1;jm h m 2 (a ;j ) 1;jn m+1 [ S m (k; l)] 1;jm h(a ;j ) 1;jn m+1 S m 1 (k; l) 6n m+2 6 j6n m+1 4 S m 1 (k; l) 6n m+1 S m 1 (k; l) 6n m+1 j6n m+2 j6n m+1 1 By usng formula (9), formula (8) wll be as follows (9) n h(a ;j ) 1;jn o m 2 n o m 1 h(a ;j ) 1;jn (m 1) h(a ;j ) 1;jn m h m 2 (a ;j ) 1;jn m+1 [ S m (k; l)] 1;jm Accordng to the prncple of recurrence h(a ;j ) 1;jn m+1 get formula (5) We can generalze theorem 2 by the followng theorem 6 0, we Theorem 3 Let the n square matrx A (a ;j ) 1;jn 8m 2 N where 2 m < n, we can generalze formula (5) as follows h m 1 (a ;j ) 1;jn (a ;j ) 61 ; 2 ; m [ [H m (k; l)]] (10) j6j 1 ;j 2 ;j m Proof We move row m to the poston of row n ; we replace row m and row ( m + 1) by each other Then, the new row ( m + 1) and row ( m + 2) by each other and so on tll row m n matrx A wll be the last row Thus, we have done (n m ) replacngs In the same way, to move row (m k) to the poston of row (n k), we need (n k) (m k) replacngs where k 0; ; m 1 We move column j m to the poston of column n ; we replace column j m and column (j m + 1) by each other Then, the new column (j m + 1) and column (j m + 2) by each other and so on tll column j m n matrx A wll be the last column Thus, we have done (n j m ) replacngs In the same way, to move column j (m l) to the poston of column (n l), we need (n l) j (m l) replacngs where l 0; ; m 1We get a new matrx B that realses A ( 1) ( 1) mp 1 ((n k) m k )+ m P 1 k0 mp (n m+k P k )+ m k1 k0 ((n l) j m l ) B l1(n j l ) B (11)

9 Proof of Sylvester s Identty 1579 We apply formula (5) on matrx B, we get when m 1 [B] (a ;j ) 61 ; 2 ; m [ [S m (k; l)]] ; (12) j6j 1 ;j 2 ;j m S m (k; l) n (m 1) z 0 } 1{ a (1;jl ) a (2;jl ) C (a ;j ) 61 ; 2 ; m j6j 1 ;j 2 ;j m a (n m;jl ) a (k ;1) a (k ;2) a (k ;n m) a (k ;j l ) By usng the ermnant propertes, we get C A S m (k; l) ( 1) ((n m+k) k)+((n ) j l ) H m (k; l) (13) By usng formula (13), formula (12) wll be as follows m 1 B (a ;j ) 61 ; 2 ; m j6j 1 ;j 2 ;j m m (k; l)]] h ( 1) ((n m+k) k)+((n ) j l ) [H m (k; l)] mp ( 1) P (n m+k k )+ m (n j l ) k1 l1 Ths equalty can be wrtten as follows " ( 1) mp (n m+k P k )+ m k1 l1(n j l ) B # [ [H m (k; l)]] [ [H m (k; l)]] m 1 (a ;j ) 61 ; 2 ; m j6j 1 ;j 2 ;j m By usng formula (11), we get formula (10) Ths nshes the proof Remark 1 By usng last theorem, we can wrte Dodgson s Algorthm n amore general way than formula (1) It wll be as follows h(a ;j ) 1;jn (a ;j ) 61 ; 2 j6l 1 ;l 2 (a ;j ) 61 j6j 1 (a ;j ) 62 j6j 2 (a ;j ) 61 j6j 2 (a ;j ) 62 j6j 1 ; for all 1 1 < 2 ; j 1 < j 2 n

10 1580 S Abdelmalek and S Kouach References [1] S Abdelmalek, and S Kouach, Condensaton of ermnants, arxv [2] A G Akrtas, E K Akrtas, and G I Malaschonok, Varous proofs of Sylvester s (ermnant) dentty, Math and Computers n Smulaton 42 (1996) [3] CL Dodgson, Condensaton of Determnants, Proceedngs of the Royal Socety of London, 15 (1866), [4] S Kouach, S Abdelmalek, and B Reba, A Mathematcal Proof of Dodgson s Algorthm, arxv [5] T Mur, The Theory of ermnants n The Hstorcal Order of Development, vol II London (1911) Receved December 14, 2007