EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 8, No. 1, 2015, 135-151 ISSN 1307-5543 – www.ejpam.com

Observations on Some Special Matrices and Polynomials Moawwad El-Mikkawy∗, Faiz Atlan Mathematics Department, Faculty of Science, Mansoura University, Mansoura 35516, Egypt

Abstract. In the current paper we focus on the study of three special matrices and two symmetric polynomials. As a consequence, a recurrence relation satisfied by the entries of the n × n inverse matrix, Qn of the n × n symmetric Pascal matrix, Pn is obtained. Moreover, a new proof for El-Mikkawy conjecture [14] is investigated. Finally, some identities are discovered. 2010 Mathematics Subject Classifications:15A23,15Bxx,11B73,05E05,68W30,65F05 Key Words and Phrases: Pascal matrix, Vandermonde matrix, Stirling matrix, Permutation matrix, Divided difference, Symmetric polynomials, MAPLE

1. Introduction and Basic Definitions

Matrices play an important role in all branches of science, engineering, social science and management. Matrices are used for data classification by which many problems are solved us- ing computers. There are many special types of matrices such as the Pascal [1], Vandermonde [10], Stirling [12], and others. These matrices are of specific importance in many scientific and engineering applications. For example the Pascal matrix, which has been known since 1303, but has been studied carefully only recently [1], appears in combinatorics, image pro- cessing, signal processing, numerical analysis, probability and surface reconstruction. Much researches has been devoted to deal with such matrices and relations between them (see for instance, [3–5, 7, 9, 19, 21–25, 29–37]). The main objectives of the current paper is to intro- duce a recurrence relation for the inverse of the symmetric Pascal matrix, Pn of order n, some new identities and to give another proof for El-Mikkawy conjecture [14]. The paper is orga- nized as follows: The main results are given in the next section. In Section 3, we present a new proof for El-Mikkawwy conjecture [14]. Moreover, new identities are obtained. Throughout this paper, δij is the Kronecker delta which is equal to 1 or 0 according as i = j or not. Also n denotes the binomial coefficient and X denotes the set {x , x ,...,x }. r 1 2 n ! "

∗Corresponding author.

Email addresses: m−[email protected], (M. El-Mikkawy), [email protected] (F. Atlan) http://www.ejpam.com 135$ c 2015 EJPAM All rights reserved. M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 136

Definition 1. [6]. For integer numbers n and k with n ≥ k ≥ 0, the Stirling numbers of the first kind, s(n, k) and of the second kind, S(n, k) are defined respectively by:

n k (x)n =[x − n + 1]n = s(n, k)x , (1) k=0 # and n n x = S(n, k)(x)k, (2) k=0 # where the falling factorial of x, (x)n and the rising factorial of x, [x]n are given respectively by:

1 if n = 0, (x) = (3) n x(x 1)(x 2) ...(x n + 1) if n 1, $ − − − ≥ and 1 if n = 0, [x] = . (4) n x(x + 1)(x + 2) ...(x + n 1) if n 1 $ − ≥ It is well known that for integers n, k ≥ 0, the s(n, k) and S(n, k) satisfy the following Pascal-type recurrence relations:

s(n, k)=s(n − 1, k − 1) − (n − 1)s(n − 1, k), (5) and S(n, k)=S(n − 1, k − 1)+kS(n − 1, k), (6) subject to: s(k, k)=S(k, k)=1, 1 ≤ k ≤ n (7) and s(n, k)=S(n, k)=δnk,ifk = 0orn = 0. (8) Replace x by −x in (1) gives n k [x]n = c(n, k)x , (9) k=0 # where c(n, k)=(−1)n−ks(n, k) is called the unsigned or signless Stirling number of the first kind. It can be shown that the rising factorial, [x]n and the falling factorial, (x)n are also related by: n

[x]n = H(n, k)(x)k, (10) k=0 # where n! n − 1 H(n, k)= . (11) k! k − 1 ! " M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 137

The coefficients, H(n, k) in (10) are called the Lah numbers. These numbers satisfy the fol- lowing Pascal-type recurrence relation:

H(n, k)=H(n − 1, k − 1)+(n + k − 1)H(n − 1, k). (12)

The Lah numbers also satisfy:

n H(i, j)= (−1)i+ks(i, k)S(k, j),1≤ i, j ≤ n. (13) k=1 # Definition 2 ([6]). The Stirling matrix of the first kind, sn and of the second kind, Sn are defined respectively by: s(i, j) for i ≥ j, s = , (14) n 0 otherwise $ and S(i, j) for i ≥ j, S = . (15) n 0 otherwise $ For example

10000 10 0 00 −11 0 00 11 0 00     s5 = 2 −31 00and S5 = 13 1 00 .  −6 11 −610  17 6 10      24 −50 35 −10 1   1 15 25 10 1          It is known that snSn = Snsn = In,whereIn is the n × n identity matrix. In other words, the i matrices, sn and Sn are inverse to each other. If fi(x)=x ,1≤ i ≤ n,then k ∆ fi (0) S(i, k)=fi[0, 1, 2, . . . , k]= k! ,wherefi[0, 1, 2, . . . , k] is the k-th divided difference of the i function fi(x)=x which lies on the top of each column and ∆ is the forward difference 5 operator. For example, for f5(x)=x , we have:

Table 1: Divided Differences and Stirling Numbers of the Second Kind

x f5(x) First DD Second DD Third DD Fourth DD Fifth DD 0 0 1=S(5,1) 15=S(5,2) 25=S(5,3) 10=S(5,4) 1=S(5,5) 1 1 31 90 65 15 2 32 211 285 125 3 243 781 660 4 1024 2101 5 3125

Thus, 5 f5(x)=x =(x)1 + 15(x)2 + 25(x)3 + 10(x)4 +(x)5 M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 138 and 2 3 4 ∆f5(0)=1, ∆ f5(0)=30, ∆ f5(0)=150, ∆ f5(0)=240, and 5 ∆ f5(0)=120. (n) Definition 3 ([16]). The elementary symmetric polynomial σr and the complete symmetric (n) polynomial τr in x1, x2,...,xn are defined respectively by:

0 if r > norn< 0 or r < 0, (n) 1 if r = 0, σr (x) :=  (16)  x x ...x if 1 ≤ r ≤ n.  i1 i2 ir 1≤i1 0), (n) 1 if r = 0, τr (X) :=  (17)  x x ...x if r ≥ 1.  i1 i2 ir 1≤i1≤i2≤...≤ir ≤n  /  n n + r − 1 It should be noticed that each σ(n) has terms and each τ(n) has terms. r r r r Moreover these polynomials can be expressed! as:" ! "

(n) k1 k2 kn σr (x) := x1 x2 ...xn ,0≤ r ≤ n. (18) k1+k2+...+kn=r k1,k2,...,#kn∈{0,1} and (n) d1 d2 dn τr (X) := x1 x2 ...xn , r ≥ 0. (19) d1+d2+...+dn=r d1,d2,...,#dn∈{0,1,...,r}

The falling factorial (x)n in (3) can be written in the form: n ( ) = ( )n−k (n) ( ) k x n −1 σn−k 0, 1, . . . , n − 1 x . (20) k=0 # Comparing the coefficients of x k in (1) and (20), yields

( )= (n) ( )= (n−1)( ) c n, k σn−k 0, 1, . . . , n − 1 σn−k 1, 2, . . . , n − 1 . (21) It can also be shown that ( )= (k) ( ) S n, k τn−k 1, 2, . . . , k . (22) Also, k 1 k S(n, k)= (−1)k−j jn. (23) k! j j=0 # ! " M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 139

For any x j ∈ X, we have

(n) (n−1) σi (x1, x2,...,xn)=σi (x1, x2,...,x j−1, x j+1,...,xn) (n−1) + x jσi−1 (x1, x2,...,x j−1, x j+1,...,xn). (24)

Partial differentiation for both sides of (24) with respect to x j, j ∈{1, 2, . . . , n} gives

∂ (n) (n) (n−1) σi = σi,j = σi−1 (x1, x2,...,x j−1, x j+1,...,xn). (25) ∂ x j

Therefore by using (16), we obtain

0ifi > n or i < 0orn < 0, (n) 1ifi = 1, σij (X)= (26)  x x ...x if 2 ≤ i ≤ n.  r1 r2 ri−1  1≤r1

n V = x i−1 . (27) j i,j=1 0 1 The Vandermonde determinant formula is well known, see for instance [26], in many text books and articles. It is given by:

n i−1 det(V )= x j = (xi − x j). (28) i,j=1 1 j

(n) σ (x1, x2,...,xn) n−j n−j+1,i αij =(−1) (29) Fi (n−1) σn−j (x1, x2,...,xi−1, xi+1,...,xn) =(−1)n−j ,1≤ i, j ≤ n, (30) Fi where n

Fi = (xi − xr ), i = 1, 2, . . . , n. (31) r=1 3r(=i Consequently, the cost of the solution of the Vandermonde linear system

T T V [u1u2 ...un] =[d1d2 ...dn] (32) M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 140

2 is O(n ). Setting xk = k,1≤ k ≤ n, in (30), yields:

i+j (−1) n − 1 ( ) α = σ n−1 (1, 2, . . . , i − 1, i + 1, . . . , n),1≤ i, j ≤ n. (33) ij (n − 1)! i − 1 n−j ! " The Vandermonde matrix, V in (27) satisfies:

V = LU, (34)

n where L =(Lij)i,j=1 is an n × n lower triangular matrix given by:

(j) τi−j(x1, x2,...,x j) for i ≥ j Lij = ,1≤ i, j ≤ n (35) 40fori < j

n and U =(Uij)i,j=1 is an n × n upper triangular matrix given by:

0fori > j, i−1 Uij = ,1≤ i, j ≤ n (36)  (x j − xr ) for i ≤ j r=1 5 The inverse matrices L−1 andU−1 of the matrices L and U in (35) and (36) are given respec- tively by: n −1 i−j (i−1) L = (−1) σi−j (x1, x2,...,xi−1) (37) i,j=1 6 7 and 1 n U−1 = . (38) j i,j=1 6 (xi − xr )7 r=1 r(=i 5 In particular, if xk = k,1≤ k ≤ n, then we have [6]:

T V = Sn˜L , ˜ ˜ n where Sn is the Stirling matrix of the second kind and L =(Lij)i,j=1 is an n×n lower triangular matrix given by:

[i − 1]j−1 for i ≥ j ˜Lij = ,1≤ i, j ≤ n (39) 0fori < j $ The inverse of the matrix ˜L is given by:

1 i − 1 n ˜L−1 = (−1)i−j . (i − 1)! j − 1 i,j=1 6 ! " 7 Definition 5 ([18]). An n × n matrix A is called totally positive if all its minors of all sizes are positive. M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 141

n Definition 6 ([8]). The symmetric matrix A =(aij)i,j=1 is called positive definite if and only if

xT Ax > 0, for all x ∈ !n, x (= 0.

Definition 7. The n × n permutation matrix, J is defined by:

J = en, en−1,...,e1 , (40) T 8 9 where ei = δi1, δi2,...,δin , and δij is the kronecker delta. The permutation8 matrix9J of order n enjoys the following properties:

• J = J T = J −1. I k even • J k = n ,whereI is the identity matrix of order n. Jkodd n $

n(n−1) 1ifn ≡ 0or1mod(4) • det(J)=(−1) 2 = . 1ifn 2or3mod(4). $− ≡ Lemma 1. For nonnegative integer numbers n and k, we have

n r n + 1 = − δ . (41) k k + 1 k0 r=1 # ! " ! " Proof. By using the following Pascal’s rule:

r r + 1 r = − , k k + 1 k + 1 ! " ! " ! " we get:

n n r r + 1 r = − k k + 1 k + 1 r=1 r=1 # ! " # 6 ! " ! " 7 n + 1 1 = − k + 1 k + 1 ! " ! " n + 1 = − δ , k + 1 k0 ! " having used the telescoping sum.

n Definition 8 ([20]). Arealn× n matrix A =(aij)i,j=1 is called row stochastic matrix if

(i) aij ≥ 0,for1 ≤ i, j ≤ n n (ii) j=1 aij = 1,for1 ≤ i ≤ n / M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 142

Note that (ii) is equivalent to AE = E,whereE =[1, 1, . . . , 1]T . n Definition 9 ([15]). The symmetric Pascal matrix Pn =(Pij)i,j=1 of order n is a matrix of integers defined by: i + j − 2 P = ,1≤ i, j ≤ n. (42) ij i − 1 ! " The Pascal matrix Pn enjoys the following properties:

(i) Pn is totally positive;

(ii) Pn is positive definite;

(iii) The eigenvalues of the matrix Pn are real and positive; 1 (iv) If λ (= 0 is an eigenvalue of Pn,then λ is also an eigenvalue of Pn;

(v) det(Pn)=1;

(vi) The Cholesky’s factorization [2] of the matrix, Pn is given by: T Pn = AA , where A is the Pascal matrix defined by: i − 1 n A = ; (43) j − 1 i,j=1 6 ! " 7 −1 n (vii) The explicit form of the inverse matrix, Pn = Qn =(βij)i,j=1 is given by [15]: n k − 1 k − 1 β =(−1)i+j ; (44) ij i − 1 j − 1 k=max(i,j) # ! "! " (viii) The matrix, Pn satisfies: −1 Pn = AB snV = TV, (45) n where T =(Tij)i,j=1 is an n × n lower triangular stochastic matrix given by: 1 T = c(i − 1, j − 1),1≤ i, j ≤ n, i ≥ j, (46) ij (i − 1)! n = i−1 = ( ( ) ) V j i,j=1, A is given in (43) and B diag 0!, 1!, 2!, . . . , n − 1 ! (for more details, see [13, 14, 27, 28]); : ; (ix) The entries of the matrix, Pn satisfy the recurrence relation:

Pi1 = P1j = 1, (47) and

Pij = Pi,j−1 + Pi−1,j. (48)

(x) Let Rn be the matrix obtained from the Pascal matrix Pn by subtracting one from the element in position (n, n) of Pn,thendet(Rn)=0. M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 143

2. Main Results

This section is mainly devoted to study the n × n inverse matrix Qn, in (44), of the n × n symmetric Pascal matrix Pn in (42). The main object is to find a recurrence relation satisfied by the entries of Qn. Setting j = 1 in (44), we obtain:

n β =(−1)i+1 = β ,1≤ i ≤ n, (49) i1 i 1i ! " having used Lemma 1. Putting j = n in (44), yields:

n − 1 β =(−1)i+n = β ,1≤ i ≤ n. (50) in i − 1 ni ! " At this point we may formulate the following result.

−1 n Theorem 1. The entries of the inverse matrix Qn = Pn =(βij)i,j=1, in (44), satisfy:

n n β = β + β +(−1)i+j ,1≤ i, j ≤ n. (51) i,j i,j+1 i+1,j i j ! "! " Proof. It is well-known that

m m − 1 m − 1 = + . (52) n n − 1 n ! " ! " ! " From (44), we have n k − 1 k − 1 β =(−1)i+j . (53) i,j i − 1 j − 1 k=1 # ! "! " By using the identity (52), we get

n k k − 1 k k − 1 β =(−1)i+j − − i,j i i j j 6 k=1

n n n n k − 1 k − 1 k k − 1 =(−1)i+j + − i j i j i j 6 ! "! " k=1 ! "! " k=1 ! "! " n #n # k k − 1 k − 1 k − 1 − + j i i j k=1 ! "! " k=1 ! "! " 7 # #n n n k − 1 k k − 1 =(−1)i+j − − i j j i i 6 ! "! " k=1 ! "

Using (49), (50) and (55), we see that in order to obtain Qn, we only need to compute 1 the 2 (n − 1)(n − 2) elements β22, β32,...,βn−1,n−1 and taking into account the fact that Qn is symmetric. Also, the solution of the linear system of the Pascal type

T T Pn[u1u2 ...un] =[f1 f2 ...fn] (56) may be obtained in O(n2) operations by using (49), (50) and (55). The following is a MAPLE code to compute the inverse matrix of the Pascal matrix, Pn for n = 6, as an example. MAPLE Code for Inverting Pascal Matrix. > restart : > n:=6: Q:= array (1..n,1..n,symmetric): > for i to n do Q[ i,n]:=(−1)^(n+i)∗ binomial(n−1,i −1): Q[ i,1]:=(−1)^( i +1)∗ binomial(n, i ): od : > for i from n−1by−1to2do for j from i by −1to2do Q[ i,j]:=Q[ i,j+1]+Q[ i +1,j]+(−1)^( i+j)∗ binomial(n, i )∗ binomial(n, j ): od : od : > Q:=evalm(Q): >#Checkusingthelinalgpackage. > P:= linalg[ inverse](Q): M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 145

The results are printed as shwon.

6 −15 20 −15 6 −1 −15 55 −85 69 −29 5   20 −85 146 −127 56 −10 Q = −15 69 −127 117 −54 10     6 −29 56 −54 26 −5     −15−10 10 −51   11 1 1 1 1  12 3 4 5 6   1 3 6 10 15 21 P = 1 4 10 20 35 56    1 5 15 35 70 126   1 6 21 56 126 252     3. Applications

In this section we are going to give a new proof for El-Mikkawy conjecture [14].Moreover, some new identities will be obtained.

3.1. A New Proof for El-Mikkawy Conjecture [14]

(n) Theorem 2. Let G be the n × n matrix whose (i, j) entry is given by σi,j (x1, x2,...,xn) and let V be the n × n Vandermonde matrix defined by (27).Thenwehave

n(n−1) det(V ) if n ≡ 0 or 1 mod(4), det(G)=(−1) 2 det(V )= (57) det(V ) if n 2 or 3 mod(4). $− ≡

Proof. For i = 1, 2, . . . , n,letFi as given in (31), D and D˜ are n × n diagonal matrices given by: D = diag( 1 , 1 ,..., 1 ) and D˜ = diag((−1)n−1, (−1)n−2,...,(−1)0). F1 F2 Fn It can be shown that for the matrices D, J and D˜, we have 1 1 ( )= = det D n 2 , (58) n(n−1) Fi (−1) 2 det(V ) ) i=1 6 7 having used (28) and (31), 5 :

n(n−1) n(n−1) det(J)=(−1) 2 and det(D˜)=(−1) 2 . (59)

By noticing that the matrix V −1 in (30) can be written in the form:

V −1 = DGJD˜, (60) the result follows. M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 146

3.2. New Identities

i−1 n In [13] it is shown that the Pascal matrix, Pn and the Vandermonde matrix, V =(j )i,j=1 are related by:

Pn = TV. (61) From (61), we get: n −1 −1 −1 Qn =(βij)i,j=1 = Pn = V T , (62) where −1 n T =(γij)i,j=1, (63) with i+j γij =(−1) (j − 1)!S(i − 1, j − 1),1≤ i, j ≤ n, i ≥ j. (64) Thus

n n i+k (−1) n − 1 ( ) β = α γ = σ n (−1)j+k(j − 1)!S(k − 1, j − 1) ij ik kj (n − 1)! i − 1 n−k+1,i k=1 k=1 6 ! " 76 7 # # n (65) (j − 1)! ( ) =(−1)i+j σ n S(k − 1, j − 1). (i − 1)!(n − i)! n−k+1,i k=1 # Rewriting (44) in the form:

n k − 1 k − 1 β =(−1)i+j , (66) ij i − 1 j − 1 k=1 # ! "! " then using (65) and (66), yields:

n n k − 1 k − 1 (j − 1)! ( ) = σ n S(k − 1, j − 1). (67) i − 1 j − 1 (i − 1)!(n − i)! n−k+1,i k=max (i,j) k=j # ! "! " # Setting i = n on both sides of (67) gives

n n − 1 (j − 1)! ( ) = σ n S(k − 1, j − 1) j − 1 (n − 1)!0! n−k+1,n ! " k=j n# (j − 1)! ( ) = σ n−1 (1, 2, . . . , n − 1)S(k − 1, j − 1) (n − 1)! n−k k=j #n (j − 1)! (68) = c(n, k)S(k − 1, j − 1) (n − 1)! k=j # n 1 = c(n, k)S(k − 1, j − 1). n − 1 (n − j)! k=j j − 1 # ! " M. El-Mikkawy, F. Atlan / Eur. J. Pure Appl. Math, 8 (2015), 135-151 147

Hence

n 2 n − 1 c(n, k)S(k − 1, j − 1)=(n − j)! j − 1 k=j # ! " (69) n − 1 =(n − 1) . " n−j n − j ! " From (61), we see that

n i + j − 2 1 = c(i − 1, k − 1) jk−1 . i − 1 (i − 1)! k=1 ! " # > ?> ? Then n i + j − 2 jk−1 = c(i − 1, k − 1). (70) i − 1 (i − 1)! k=1 ! " # Setting j = 2 in (70), we obtain

n i 2k−1c(i − 1, k − 1)=(i − 1)! = i!. (71) i − 1 k=1 # ! " Therefore, for any positive integers n and 1 ≤ i ≤ n, we have

n 2k−1c(i − 1, k − 1)=i!. " (72) k=1 # We list the following additional identities, without proof, for the sake of space require- ments. In all cases n is a positive integer and i, j are integers such that 1 ≤ i, j ≤ n.

n i + r − 2 n • (−1)r+1 = δ . i − 1 r i1 r=1 ! "! " / n n • (−1)k+1ki−1 = δ . k i1 k=1 ! " /n ( )k+1 (n−1)( )=( ) • −1 σn−k 2, 3, . . . , n n − 1 !. k=1 /n k−1 • s(n, k)j =(n − 1)!δnj. k=1 / n n − 1 • (n − 1) s(r − 1, j − 1)=c(n, j). n − r n−r r=j ! " / n n n − 1 ( ) • (−1)j+k σ n S(r − 1, j − 1)=0. k − 1 n−r+1,k r=j k=1 ! " / / REFERENCES 148

−(n − 1)!ifi = 1 n n n − 1 i + k − 2 ( 1)k σ(n) = (n 1)!ifi = 2 . • − n−r+1,k  − r=1 k=1 k − 1 k − 1 ! "! " 0otherwise / /  n n n (n)  • (−1)kk2 σ = n!. k n−r+1,k r=1 k=1 ! " / / k n n  i   i  ( )= (n−1)( + ) •  k!  c k, j  n!  σn−j 1, 2, . . . , i − 1, i 1, . . . , n ; k=max(i,j)

/n (n−1) • (n − i)n−kc(k, j)=σn−j (1, 2, . . . , i − 1, i + 1, . . . , n); k=max(i,j) / n − i n   n n − k (n i) • = − n−k = n ;   (n−1)n−k i k=i n − 1 k=i /  n − k  /   k n n  i  n  i  = (k−1)i−1 = •  k  i!  i  ; k=i k=i /n / (k)i (n)i • k = i . k=i /n • ks(n + 1, k + 1)=(−1)n−1(n − 1)! = s(n,1). k=1 / ACKNOWLEDGEMENTS The authors are extremely grateful to the editorial team of EJPAM and anonymous referees for their useful comments and suggestions that have helped in im- proving the readability of this paper.

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