A NOTE ON PASCAL S MATRIX. Gi-Sang Cheon, Jin-Soo Kim and Haeng-Won Yoon

Transcription

1 J Korea Soc Math Educ Ser B: Pure Appl Math 6(1999), o A NOTE ON PASCAL S MATRIX Gi-Sag Cheo, Ji-Soo Kim ad Haeg-Wo Yoo Abstract We ca get the Pascal s matrix of order by takig the first rows of Pascal s triagle ad fillig i with 0 s o the right I this paper we obtai some well kow combiatorial idetities ad a factorizatio of the Stirlig matrix from the Pascal s matrix 1 Itroductio The umbers ( k) are the so-called biomial coefficiets which cout the umber of k-combiatios of a set of elemets They have may fasciatig properties ad satisfy a umber of iterestig idetities Moreover, the biomial coefficiets are ope displayed i a array kow as Pascal s triagle Each etry i the triagle, other tha those equal to 1 occurrig o the left side ad hypoteuse, is obtaied by addig together two etries i the row above: the oe directly above ad the oe immediately to the left We defie the Pascal s matrix P [p ij ] of order by takig the first rows of Pascal s triagle ad fillig i with 0 s o the right (cf Call ad Vellema [5]) That is, p ij { ( i 1 ) j 1 if i j Received by the editors Jue 8, Mathematics Subject Classificatio Primary 05A19, Secodary 05A10 Key words ad phrases Biomial coefficiets, Pascal s matrix Stirlig umber, Stirlig matrix 121

2 122 GI-SANG CHEON, JIN-SOO KIM AND HAENG-WON YOON Thus we have O P I [3], it is show that the Pascal s matrix P ca be factorized by followig: P G G 1 G 1 where for each k 1,,, [ ] I k O G k T O T k where T k [t ij ] is the lower triagular matrix of order k defied by { 1 if i j t ij Cleary the determiat of P is 1, ad the iverse of P is obtaied (cf Brawer ad Pirovio [3]) I fact, P 1 [p ij ] where { ( ) ( 1) i j i 1 p j 1 if i j ij Thus P 1 is the same as P except that the mius sigs appear at (i, j)-positios with i j 1 (mod 2) I this paper, we obtai some well kow combiatorial idetities ad a factorizatio of the Stirlig matrix from the Pascal s matrix 2 Results We cosider a ordiary chessboard which is divided ito ( 1) 2 squares i rows ad colums Let c ij be the umber of paths with the legth i + j 2 from (1, 1)-positio to (i, j)-positio The it is easy to show that ( ) (i + j 2)! i + j 2 c ij (i 1)!(j 1)! (21) j 1

3 A NOTE ON PASCAL S MATRIX 123 For each c ij (1 i, j ) i (21), defie C [c ij ] to be the matrix of order For example, C , C I the followig theorem, by a combiatorial argumet we show that C ca be expressed by the Pascal s matrix Theorem 21 Let C [c ij ] be the matrix of order with etries i (21) The C P P T Proof We may assume that i j The we ca write each path from (1, 1)-positio to (i, j)-positio o the chessboard as a sequece of the form x 1 R, y 1 D, x 2 R, y 2 D,, x k R, y k D where R deotes right, D deotes dow ad, for some 1 k j, x 1 + x x k j 1 (x 1 0; x 2,, x k > 0), (22) y 1 + y y k i 1 (y 1,, y k 1 > 0; y k 0) (23) For a fixed k with 1 k j, clearly the umber of such sequeces is 1 2 where 1 ad 2 are the umber of solutios to (22) ad (23) respectively We claim that 1 ( ) j 1 k 1 ad 2 ( ) i 1 k 1 To show 1 ( j 1 k 1), we choose k 1 elemets of the umbers 1, 2,, j 1 The there is a 1-1 correspodece betwee solutios to (22) ad (k 1)-subsets of {1, 2,, j 1} Namely, if {x 1, x 2,, x k } is a solutio to (22), the {x 1, x 1 + x 2,, x 1 + x x k 1 } is a (k 1)-subset of {1, 2,, j 1} Also, if {z 1, z 2,, z k 1 } is a (k 1)-subset of {1, 2,, j 1} with 0 z 1 < z 2 < < z k 1, the x 1 z 1, x 2 z 2 z 1,, x k 1 z k 1 z k 2, x k j 1 z k 1 solves (22) also Thus 1 ( j 1 k 1)

4 124 GI-SANG CHEON, JIN-SOO KIM AND HAENG-WON YOON Similarly, we ca show that 2 ( i 1 k 1) Hece, if we ote that ( r) 0 for r >, the the umber c ij of paths from (11)-positio to (i, j)-positio is c ij j ( )( ) i 1 j 1 k 1 k 1 k1 ( )( ) i 1 j 1 k 1 k 1 k1 p ik p jk (P P T ) ij (24) where (P P T ) ij is the (i, j) etry of the matrix P P T Therefore C P P T, which completes the proof Note that it is kow that P P T is the Cholesky factorizatio of C [3] I particular, if i j i (24) the we ca establish the followig idetity from (21): t0 ( ) 2 t k1 ( ) 2 (25) More geerally, the followig Vadermode Covolutio ca be derived from Theorem 21 Corollary 22 t0 ( )( ) m t t ( m + ) Proof From (21) ad (24), we get j ( )( ) ( ) i 1 j 1 i + j 2 k 1 k 1 j 1 k1 Thus if we take i 1 m, j 1 ad k 1 t the Vadermode covolutio follows immediately from ( ) ( t ) ( t ad +1) 0 Vadermode covolutio ca be exteded as followig: ( t 1, 2,, t 1 ) t k0 ( t k )( ) ( ) t k 1, 2,, t (26) where t ad ( ) 1, 2,, t! 1! 2! t! Note that det C 1 ad A 1 (P 1 ) T P 1 Next, we cosider a famous coutig problem which is called Stirlig umber Let S(, k) deote the Stirlig umber for itegers ad k with 1 k The

5 A NOTE ON PASCAL S MATRIX 125 the umber S(, k) couts the umber of partitios of a set X of elemets ito k idistiguishable boxes i which o box is empty Example Let X {a, b, c, d} the we get the partitios for each k 1, 2, 3, 4: k 1 : X; k 2 : [{a}, {b, c, d}], [{b}, {a, c, d}], [{c}, {a, b, d}], [{d}, {a, b, c}], [{a, b}, {c, d}], [{a, c}, {b, d}], [{a, d}, {b, c}]; k 3 : [{a}, {b}, {c, d}], [{a}, {c}, {b, d}], [{a}, {d}, {b, c}], [{c}, {d}, {a, b}], [{b}, {d}, {a, c}], [{b}, {c}, {a, d}]; k 4 : [{a}, {b}, {c}, {d}] Thus we have 1 if k 1 7 if k 2 S(4, k) 6 if k 3 1 if k 4 It is well kow [1] that the Stirlig umbers S(, k) have a Pascal-like recurrece relatio as followig: 1 if k 1 S(, k) 1 if k S( 1, k 1) + ks( 1, k) if 2 k 1 As we did for the Pascal s triagle we ca obtai a Pascal-like matrix S of order for these Stirlig umbers S(, k) Defie S [s ij ] to be the matrix of order where { S(i, j) if i j s ij Thus for i ad j with i j, each etry s ij i the matrix S, other tha iitial values, is obtaied by multiplyig the etry i the row directly above it by j ad addig the result to the etry immediately to its left i the row directly above it We call S the Stirlig matrix of order For example, S 4, S

6 126 GI-SANG CHEON, JIN-SOO KIM AND HAENG-WON YOON By a simple computatio, we ca easily show the followig lemma Lemma 23 Let S be the Stirlig matrix of order ad let P be the Pascal s matrix of order The S P ([1] S 1 ) where deotes a direct sum Corollary 24 Let S(, k) be a Stirlig umber The 1 ( ) 1 S(, k) S(r, k 1), (k 1) (27) r r1 Proof Let S [s ij ] ad P [p ij ] The from Lemma 23, if k 1, we get 1 1 ( ) 1 S(, k) s k p r+1 s r k 1 S(r, k 1), r which completes the proof r1 r1 For the Pascal s matrix P k of order k, 1 k, defie [ ] I k O P k T O to be the matrix of order Thus P : P ad P 1 is the idetity matrix of order Corollary 25 Let S be the Stirlig matrix of order The S ca be factorized by the P k s: S P P 1 P2 P1 (28) Proof If we apply (27) recursively we obtai (28) Remark Note that S 1 P 1 1 P k P 1 2 P P Example S

7 A NOTE ON PASCAL S MATRIX 127 S Refereces 1 R A Brualdi, Itroductory Combiatorics, 2d Ed, North-Hollad Publishig Co, New York, 1992 MR 93g: L Comtet, Advaced Combiatorics, The art of fiite ad ifiite expasios Revised ad elarged editio, D Reidel Pub Co, Dordrecht, 1974 MR 57#124 3 R Brawer ad M Pirovio, The liear algebra of the Pascal matrix, Liear Algebra Appl 174 (1992), MR 93g: A T Bejami ad J J Qui, Paths to a multiomial iequality, Ars Combi 55 (2000), MR 2000m: G S Call ad D J Vellema, Pascal s matrices, Amer Math Mothly 100 (1993), MR 94a: Z Zhag, The liear algebra of the geeralized Pascal matrix, Liear Algebra Appl 250 (1997), MR 97g:15014 (Gi-Sag Cheo) Dept of Mathematics, Daeji Uiversity, Pocheo, Gyeoggi-do , Korea address: gscheo@roaddaejiackr (Ji-Soo Kim) Dept of Mathematics, Sugkyukwa Uiversity, Suwo, Gyeoggido , Korea (Haeg-Wo Yoo) Dept of Mathematics, Daeji Uiversity, Pocheo, Gyeoggido , Korea address: hwyoo@roaddaejiackr