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Matrix factorization of the n × n shift Bell matrix
Linear Algebra and its Applications 577 (2019) 214–239
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Linear Algebra and its Applications www.elsevier.com/locate/laa
Matrix factorization of the n × n shift Bell matrix Liming Zhang, Zhihua Sun, Xiaoying Qu, Xiqiang Zhao ∗ School of Mathematical Science, Ocean University of China, Qingdao 266100, PR China
a r t i c l e
i n f o
Article history: Received 4 October 2018 Accepted 25 April 2019 Available online 3 May 2019 Submitted by R. Brualdi MSC: 05A15 05A05 15B36 15A06 05A19
a b s t r a c t Let Bn = [Bn,k ]n,k0 be the Bell matrix. Define the n × n shift Bell matrix Pn,k by (Pn,k )i,j = Bk+i−1,k+j−1 for i, j = 1, 2, · · · n and k = 0, 1, 2 · · · . In this paper, matrix factorizations of the n × n shift Bell matrix and the n × n generalized Riordan matrix are studied. As a result, many lower triangular matrices related to Bell polynomials can be factorized by the corresponding matrices and some identities are derived from the matrix representations. In addition, some harmonic number identities are obtained from the Riordan array method. © 2019 Elsevier Inc. All rights reserved.
Keywords: Bell polynomials n × n shift Bell matrix n × n shift iteration matrix Generalized Riordan array
1. Introduction In 1934, Bell [3] extensively introduced the important combinatorial tools-Bell polynomials. Some useful combinatorial sequences can be obtained from the Bell polynomials by choosing the variables x1 , x2 · · · . For example, all the Stirling numbers of both kinds, the * Corresponding author. E-mail address: [email protected] (X. Zhao). https://doi.org/10.1016/j.laa.2019.04.037 0024-3795/© 2019 Elsevier Inc. All rights reserved.
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Lah numbers and the idempotent numbers are the special cases of the Bell polynomials (see [6, p. 135, Theorem B]): Bn.k (1, 1, 1, · · · ) = S(n, k), (Stirling number of the second kind) n − 1 n! Bn.k (1!, 2!, 3!, · · · ) = , (Lah number) k − 1 k! Bn.k (0!, 1!, 2!, · · · ) = s(n, k), (unsigned Stirling number of the first kind) n Bn.k (1, 2, 3, · · · ) = k n−k . (idempotent number) k Recently, the authors [1,6,7] studied some identities for Bell polynomials, which have found many applications in combinatorics. Some relations between the Bell matrix and the Fibonacci matrix are given by W. Wang and T. Wang [17]. The factorization of the Bell matrix and iteration matrix are also demonstrated by them and some combinatorial identities are obtained from the matrix representations (see [18]). In 1991, the Riordan array is introduced by Shapiro et al. [10,11,14], which is a generalization of the renewal array [12]. Riordan array is a useful tool to study combinatorial sum and identities. Many applications of the Riordan array have been studied in [2,19,24–26] and some characterization of Riordan array can be found in [9,10,14]. Gi-Sang Cheon et al. [4] studied the relations between the Stirling numbers of both kinds and other generalized harmonic numbers. They also obtained some generalized harmonic number identities by using the approach of the Riordan arrays. Moreover, the r t 1 Riordan array ( −ln(1−t) (1−t) , 1−t ) and ( (1−t)r+1 , −ln(1 − t)) are generated by the number P (r, n, k) (see [19]). It can be seen that the Riordan array is very important for studying combinatorial identities. The exponential partial Bell polynomials [6, p. 133] and the Riordan array are defined as follows. Definition 1.1. The exponential partial Bell polynomials are the polynomials Bn,k = Bn,k (x1 , x2 , · · · , xn−k+1 ) , in an infinite number of variables x1 , x2 , · · · , defined by the formal double series expansion ⎛ Φ = Φ(t, u) = exp ⎝u tn =1+ n! n1
m1
n
k=1
⎞ tm ⎠ tn = xm Bn,k uk m! n! n,k0
u Bn,k (x1 , x2 , · · · ) , k
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or by the series expansion ⎛ Φk (t) :=
1 ⎝ k!
⎞k
t ⎠ tn = Bn,k , m! n! m
xm
m1
k = 0, 1, 2, · · · .
(1.1)
nk
The exponential complete Bell polynomials Yn = Yn (x1 , x2 , · · · , xn ) are defined by
Yn =
n
Bn,k ,
Y0 := 1.
(1.2)
k=1
Definition 1.2. The generalized Riordan array with respected to the sequence cn is a pair (g(t), f (t)), where g(t) = k0 gk tk /ck and f (t) = k1 fk tk /ck with f1 = 0. It defines an infinite lower triangular array (dn,k )n,k∈N according to the rule dn,k =
f (t)k tn g(t) , cn ck
where g(t)f (t)k /ck is the generating function of the k column, and dn,k is the general term of the generalized Riordan array. By the definition, the classical Riordan arrays are corresponded to the case of cn = 1, and the exponential Riordan arrays are corresponded to the case of cn = n!. The rest of this paper is structured as follows. Section 2, we study the matrix factorization of the n ×n shift Bell matrix. For instance, some lower triangular matrices can be factorized by the corresponding matrices and some identities are gotten from the matrix representations. Section 3, we give the factorization of the n × n shift iteration matrix and the n × n shift generalized Riordan matrix, and some combinatorial identities are obtained from the Riordan array method. 2. Matrix factorization of the n × n shift Bell matrix Lemma 2.1. The partial Bell polynomials satisfy the following relations (see [7]) k1 !k2 ! = (k1 + k2 )! n
Bn,k1 +k2
l=0
Bn,k
n−k 1 = x1 (n − k) l=1
n Bl,k1 Bn−l,k2 , l
(n k1 + k2 , k1 , k2 0)
n+1 n xl+1 Bn−l,k . (k + 1) − l+1 l
(2.1)
(2.2)
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We now define the n × n shift Bell matrix Pn,k and the n × n matrices Qn,k , Mn,k by (Pn,k )i,j = Bk+i−1,k+j−1 , (Qn,k )i,j = Mn,k = diag
k+i−1 j−1
1!k! (n − 1)!k! 0!k! , ,··· , (0 + k)! (1 + k)! (n − 1 + k)!
Bk+i−j,k , ,
for i, j = 1, 2, · · · n and k = 0, 1, 2, · · · . If k = 1, the matrix Pn,1 = Bn is the n × n Bell matrix (see [17]). From identity (2.1), we can obtain the following theorem. Theorem 2.2. The matrix Pn,k can be factorized as Pn,k = Qn,k Pn,0 Mn,k .
(2.3)
Proof. It follows immediately from identity (2.1). 2 For example, if n = 4, k = 4, we have ⎡
P4,4
⎤ x41 0 0 0 ⎢ x51 0 0 ⎥ 10x31 x2 ⎢ ⎥ =⎢ ⎥ 3 2 2 4 6 ⎣ 15x1 x2 x1 0 ⎦ 20x1 x3 + 45x1 x2 35x31 x4 + 210x21 x2 x3 + 105x1 x32 35x41 x3 + 105x31 x22 21x51 x2 x71 ⎡ ⎤ x41 0 0 0 ⎢ 5x41 0 0 ⎥ 10x31 x2 ⎢ ⎥ =⎢ ⎥ 3 2 2 3 4 ⎣ 60x1 x2 15x1 0 ⎦ 20x1 x3 + 45x1 x2 35x31 x4 + 210x21 x2 x3 + 105x1 x32 140x31 x3 + 315x21 x22 210x31 x2 35x41 ⎤ ⎡ ⎤⎡ 1 0 0 0 1 0 0 0 ⎢ ⎢0 x 0 0 ⎥ 0 0 ⎥ ⎥ ⎢ ⎥ ⎢ 0 1/5 1 ×⎢ ⎥. ⎥⎢ 2 ⎣ 0 x2 0 ⎦ x1 0 ⎦ ⎣ 0 0 1/15 0 0 0 1/35 0 x3 3x1 x2 x31
Particularly, if k = 1, we have Pn,1 = Qn,1 ([1] ⊕ Pn−1,1 )Mn,1 ,
(2.4)
which is equivalent to Bn = Qn,1 ([1] ⊕ Bn−1 )Mn,1 . The notation ⊕ denotes the direct sum of two matrices. For instance, if n = 4 and k = 1, we can factorize 4 × 4 Bell matrix as
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⎡
x1 ⎢x ⎢ 2 B4 = ⎢ ⎣ x3 x4 ⎡ 1 ⎢0 ⎢ ×⎢ ⎣0 0
⎤ ⎡ x1 0 0 0 ⎢ ⎥ 0 0 ⎥ ⎢ x2 2x1 ⎥=⎢ x31 0 ⎦ ⎣ x3 3x2 6x21 x2 x41 x4 4x3 ⎤⎡ ⎤ 0 1 0 0 0 ⎢ 0 ⎥ 0 ⎥ ⎥ ⎢ 0 1/2 0 ⎥ ⎥⎢ ⎥. 0 1/3 0 ⎦ 0 ⎦⎣0 0 0 0 1/4 x31
0 x21 3x1 x2 4x1 x3 + 3x22 0 x1 x2 x3
0 0 x21 3x1 x2
0 0 3x1 6x2
⎤ 0 0 ⎥ ⎥ ⎥ 0 ⎦ 4x1
For any l × l matrices Xl,1 , we define the n × n matrices X l,1 by X l,1 =
In−l o
o Xl,1
,
where In−l is the (n − l) × (n − l) unit matrix. Then we can further factorize the Bell matrix. Theorem 2.3. The n × n Bell matrix Bn can be factorized as Bn = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 .
(2.5)
For example, if n = 3 and k = 1, we can factorize B3 as ⎤ ⎡ ⎤⎡ ⎤ x1 1 0 0 x1 0 0 0 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ B3 = ⎣ x2 x21 0 ⎦ = ⎣ x2 2x1 0 0 ⎦ ⎣ 0 x1 ⎦ x3 3x1 x2 x31 x3 3x2 3x1 0 x2 2x1 ⎤ ⎤⎡ ⎤⎡ ⎡ ⎤⎡ 1 0 0 1 0 0 1 0 0 1 0 0 ⎥ ⎥⎢ ⎥⎢ ⎢ ⎥⎢ ×⎣0 1 0 ⎦⎣0 1 0 ⎦⎣0 1 0 ⎦ ⎣ 0 1/2 0 ⎦. 0 0 1/3 0 0 1/2 0 0 x1 0 0 1 ⎡
Based on Theorem 2.2 and Theorem 2.3, the matrix factorization of some lower triangular matrices related to Bell polynomials can be obtained directly. Example 2.1. The unsigned Stirling numbers of the first kind s(n, k) and the Stirling numbers of the second kind S(n, k) satisfy the following equations [6, p. 135, Eqs. (3g) and (3i)] Bn,k (0!, 1!, 2!, · · · ) = s(n, k),
Bn,k (1, 1, 1, · · · ) = S(n, k).
We can obtain (2.6), (2.7) from the identity (2.1)
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n s(l, k1 )s(n − l, k2 ), l l=0 n k1 !k2 ! n S(n, k1 + k2 ) = S(l, k1 )S(n − l, k2 ), (k1 + k2 )! l
k1 !k2 ! s(n, k1 + k2 ) = (k1 + k2 )!
219
n
(2.6)
(2.7)
l=0
for n k1 + k2 , k1 , k2 0. Setting k1 = k − 1, k2 = 1 in (2.6) and (2.7) gives n s(l, k − 1)(n − l − 1)!, l l=0 n−1 1 n S(l, k − 1). S(n, k) = k l
n−1 1 s(n, k) = k
l=0
We define the n × n matrices sn,k and Sn,k by (sn,k )i,j = s(k + i − 1, k + j − 1), (Sn,k )i,j = S(k + i − 1, k + j − 1), for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . From Theorem 2.2, we have sn,k = qn,k sn,0 Mn,k
(2.8)
Sn,k = Qn,k Sn,0 Mn,k
(2.9)
where qn,k , Qn,k and Mn,k are defined by
k+i−1 k+i−1 (qn,k )i,j = s(k + i − j, k), (Qn,k )i,j = S(k + i − j, k), j−1 j−1 1!k! (n − 1)!k! 0!k! , ,··· , . Mn,k = diag (0 + k)! (1 + k)! (n − 1 + k)! For instance, if n = 4 and k = 4, the matrix s4,4 can be factorized as ⎡
s4,4
1 0 0 ⎢ 10 1 0 ⎢ =⎢ ⎣ 85 15 1 735 175 21 ⎡ 1 0 0 ⎢ 10 5 0 ⎢ =⎢ ⎣ 85 60 15 735 595 210
⎤ 0 0⎥ ⎥ ⎥ 0⎦ 1
⎤⎡ 0 1 ⎢ 0 ⎥ ⎥⎢0 ⎥⎢ 0 ⎦⎣0 35 0
0 1 1 2
0 0 1 3
⎤⎡ ⎤ 0 1 0 0 0 ⎢ 0⎥ 0 0 ⎥ ⎥ ⎢ 0 1/5 ⎥ ⎥⎢ ⎥. 0 ⎦ ⎣ 0 0 1/15 0 ⎦ 1 0 0 0 1/35
If k = 1, the matrix sn = sn,1 is the Stirling matrix of the first kind (unsigned) and Sn = Sn,1 is the Stirling matrix of the second kind (see [5]). By Theorem 2.3, we have
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sn = qn,1 ([1] ⊕ sn−1 )Mn,1 = q n,1 q n−1,1 · · · q 1,1 M 1,1 · · · M n−1,1 M n,1 , Sn = Qn,1 ([1] ⊕ Sn−1 )Mn,1 = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 . For example, if n = 4, we can factorize S4 as ⎡
1 ⎢1 ⎢ S4 = ⎢ ⎣1 1
0 1 3 7
0 0 1 6
⎤ ⎡ 0 1 ⎢1 0⎥ ⎥ ⎢ ⎥=⎢ 0⎦ ⎣1 1 1
0 2 3 4
0 0 3 6
⎤⎡ 0 1 ⎢0 0⎥ ⎥⎢ ⎥⎢ 0⎦⎣0 4 0
0 1 1 1
0 0 1 3
⎤⎡ ⎤ 0 1 0 0 0 ⎢ 0⎥ 0 ⎥ ⎥ ⎢ 0 1/2 0 ⎥ ⎥⎢ ⎥. 0 ⎦ ⎣ 0 0 1/3 0 ⎦ 1 0 0 0 1/4
Example 2.2. In [20], the following equation was obtained
I
(a)
n (n, k) := Bn,k (1, 2a, 3a , · · · ) = (ka)n−k . k 2
From (2.1) and (2.3), we get k1 !k2 ! (n, k1 + k2 ) = (k1 + k2 )! n
I
(a)
l=0
(a)
n I (a) (l, k1 )I (a) (n − l, k2 ), l
(a)
IPn,k = Qn,k IPn,0 Mn,k ,
(2.10) (2.11)
(a)
where IPn,k , Qn,k and Mn,k are defined by (a) (IPn,k )i,j
=I
(a)
(k + i − 1, k + j − 1), (Qn,k )i,j =
Mn,k = diag
k+i−1 j−1
1!k! (n − 1)!k! 0!k! , ,··· , (0 + k)! (1 + k)! (n − 1 + k)!
I (a) (k + i − j, k), ,
for i, j = 1, 2, · · · n and k = 0, 1, 2, · · · . (a)
For example, if n = 4 and k = 4, the matrix IP4,4 can be factorized as ⎡
(a)
IP4,4
1 ⎢ 20a ⎢ =⎢ ⎣ 240a2 2240a3 ⎡ 1 0 ⎢0 1 ⎢ ×⎢ ⎣ 0 2a 0 3a2
0 1 30a 525a2
0 0 1 42a ⎤⎡ 0 0 1 ⎢ 0 0⎥ ⎥⎢0 ⎥⎢ 1 0⎦⎣0 0 6a 1
⎤ ⎡ 1 0 ⎢ 0⎥ ⎥ ⎢ 20a ⎥=⎢ 0 ⎦ ⎣ 240a2 1 2240a3
0 5 120a 1680a2 ⎤
0 0 0 1/5 0 0 ⎥ ⎥ ⎥. 0 1/15 0 ⎦ 0 0 1/35
⎤ 0 0 0 0 ⎥ ⎥ ⎥ 15 0 ⎦ 420a 35
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If k = 1, we have (a)
(a)
IPn,1 = Qn,1 ([1] ⊕ IPn−1,1 )Mn,1 = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 . (a)
For instance, if n = 4 and k = 1, we can factorize IP4,1 as ⎡
1 0 ⎢ 2a 1 ⎢ (a) IP4,1 = ⎢ 2 ⎣ 3a 6a 4a3 24a2 ⎡ 1 0 0 ⎢0 1 0 ⎢ ×⎢ ⎣ 0 2a 1 0 3a2 6a
⎤ ⎡ 0 0 1 0 0 ⎢ 0 0⎥ 2 0 ⎥ ⎢ 2a ⎥=⎢ 2 1 0 ⎦ ⎣ 3a 6a 3 4a3 12a2 12a 12a 1 ⎤ ⎤⎡ 1 0 0 0 0 ⎢ 0 ⎥ 0⎥ ⎥ ⎥ ⎢ 0 1/2 0 ⎥. ⎥⎢ 0 ⎦ ⎣ 0 0 1/3 0 ⎦ 0 0 0 1/4 1
⎤ 0 0⎥ ⎥ ⎥ 0⎦ 4
Example 2.3. Using the equation [1, remark 5] (a)
L
(n, k) := Bn,k (1!a , 2!a , 3!a , · · · ) = a 0
1
2
n−k
n−1 k−1
n! , k!
we find that k1 !k2 ! L(a) (n, k1 + k2 ) = (k1 + k2 )! n
l=0
(a)
n L(a) (l, k1 )L(a) (n − l, k2 ), l
(a)
Ln,k = Qn,k Ln,0 Mn,k ,
(2.12) (2.13)
(a) (Ln,k )i,j
k+i−1 j−1
= L(a) (k + i − 1, k + j − 1), (Qn,k )i,j = L(a) (k + i − j, k), (n−1)!k! 0!k! 1!k! Mn,k = diag (0+k)! , (1+k)! , · · · , (n−1+k)! , for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . For example, if n = 4 and k = 4, we have where
⎡
(a)
L4,4
1 ⎢ 20a ⎢ =⎢ ⎣ 300a2 4200a3 ⎡ 1 0 ⎢0 1 ⎢ ×⎢ ⎣ 0 2a 0 6a2 (a)
0 1 30a 630a2
0 0 1 42a ⎤⎡ 0 0 1 ⎢ ⎥ 0 0⎥⎢0 ⎥⎢ 1 0⎦⎣0 0 6a 1
⎤ ⎡ 1 0 ⎢ ⎥ 0 ⎥ ⎢ 20a ⎥=⎢ 0 ⎦ ⎣ 300a2 1 4200a3
0 5 120a 2100a2 ⎤
0 0 0 1/5 0 0 ⎥ ⎥ ⎥. 0 1/15 0 ⎦ 0 0 1/35
If k = 1, the matrix Ln,1 can be factorized as
⎤ 0 0 0 0 ⎥ ⎥ ⎥ 15 0 ⎦ 420a 35
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(a)
(a)
Ln,1 = Qn,1 ([1] ⊕ Ln−1,1 )Mn,1 = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 . Let L(i, j) be the Lah numbers, and if a = 1, we have (see [6, p. 156] and [15]) Ln,1 = sn Sn = Qn,1 ([1] ⊕ Ln−1,1 )Mn,1 , where Ln,1 is the Lah matrix with (Ln,1 )i,j = L(1) (i, j) = L(i, j), for i, j = 1, 2, · · · , n. From the above expression, we obtain i
s(i, l)S(l, j) =
l=j
i 1 l=j
j
=
i j−1
i (1 + i − l)!L(l − 1, j − 1) l−1 i 1 i − j + 1 (i − l + 1)!(l − 2)! l=j
j
(j − 2)!
l−j
.
Example 2.4. Let us study the following equation [20, Eq. (17)] n Bn,k (B0 (x), 2B1 (x), 3B2 (x), · · · ) = Bn−k (kx), k n where Bn (x) := i=0 S(n, j)xj are the single variable Bell polynomials. It can be found that Bn (1) are the Bell numbers (see [9, p. 210]). From (2.1) and (2.3), we get Bn−k1 −k2 ((k1 + k2 )x) =
n
l=0
n − k1 − k2 l − k1
Bl−k1 (k1 x)Bn−l−k2 (k2 x),
n,k = Qn,k B n,0 Mn,k , B
(2.14) (2.15)
where
k+i−1 Bi−j ((k + j − 1)x), k+j−1 k+i−1 k+i−j (Qn,k )i,j = Bi−j (kx), j−1 k 1!k! (n − 1)!k! 0!k! Mn,k = diag , ,··· , , (0 + k)! (1 + k)! (n − 1 + k)! n,k )i,j = (B
for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . Setting k1 = k − 1, k2 = 1 in (2.14) and letting m = n − k, we obtain Bm (kx) =
m l=0
m l
Bm−l (x)Bl ((k − 1)x).
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From the definition of Bn (x), we get m
n
S(m, n)(kx) =
n=0
m m−l l
l=0 i=0 j=0
m l
(k − 1)j S(m − l, i)S(l, j)xi+j .
By equating the coefficients of xn , we obtain S(m, n)k n =
m n
m l
l=0 j=0
(k − 1)j S(m − l, n − j)S(l, j).
n,1 as This identity can be found in [18]. If k = 1 in (2.15), we can factorize B n−1,1 )Mn,1 = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 . n,1 = Qn,1 ([1] ⊕ B B For instance, if n = 4, k = 1, we have ⎡
⎤ 1 0 0 0 ⎢ 2x 1 0 0⎥ ⎥ 4,1 = ⎢ B ⎢ ⎥ 2 ⎣ 3(x + x ) 6x 1 0⎦ 4(x + 3x2 + x3 ) 12(x + 2x2 ) 12x 1 ⎡ ⎤ 1 0 0 0 ⎢ 2x 2 0 0⎥ ⎢ ⎥ =⎢ ⎥ 2 ⎣ 3(x + x ) 6x 3 0⎦ 4(x + 3x2 + x3 ) 12(x + x2 ) 12x 4 ⎤ ⎡ ⎤⎡ 1 0 0 0 1 0 0 0 ⎢ ⎢0 0 ⎥ 1 0 0⎥ ⎥ ⎢ ⎥ ⎢ 0 1/2 0 ×⎢ ⎥. ⎥⎢ ⎣0 2x 1 0 ⎦ ⎣ 0 0 1/3 0 ⎦ 0 0 0 1/4 0 3(x + x2 ) 6x 1 We consider some generalizations of the n × n shift Bell matrix. We define the n × n matrices Pn,k [y], P n,k [y] and Pn,k [y, z] by (Pn,k [y])i,j = y i−j Bk+i−1,k+j−1 , (P n,k [y])n,k = y i+j−2 Bk+i−1,k+j−1 , (Pn,k [y, z])i,j = y i−j z i+j−2 Bk+i−1,k+j−1 , for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · where y, z are nonzero real number. The n × n matrices Gn,k [y], Qn,k [y], Rn,k [y, z] and Mn,k are given by (Gn,k [y])i,j = y i−j
i j−1
Bi−j+k,k ,
224
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i (Qn,k [y])i,j = y Bi−j+k,k , j−1 i i−j i+j−2 (Rn,k [y, z])i,j = y z Bi−j+k,k , j−1 1!k! (n − 1)!k! 0!k! , ,··· , . Mn,k = diag (0 + k)! (1 + k)! (n − 1 + k)! i+j−2
In similar fashion to [5,15,21–23], we can obtain the following theorem. Theorem 2.4. For any nonzero number y and z, we have Pn,k [y] = Gn,k [y]Pn,0 [y]Mn,k , 1 Mn,k , P n,k [y] = Qn,k [y]Pn,0 y y Pn,k [y, z] = Rn,k [y, z]Pn,0 Mn,k . z Proof. It can be proved by identity (2.1). 2 Particularly, if k = 1, we have Pn,1 [y] = Gn,1 [y]([1] ⊕ Pn−1,1 [y])Mn,1 = Gn,1 [y] · · · G1,1 [y]M 1,1 · · · M n,1 , 1 )Mn,1 P n,1 [y] = Qn,1 [y]([1] ⊕ Pn−1,1 y 1 1 · · · G1,1 = Qn,1 [y]Gn−1,1 M 1,1 · · · M n,1 , y y y Pn,1 [y, z] = Rn,1 [y, z]([1] ⊕ Pn−1,1 )Mn,1 z y y = Rn,1 [y, z]Gn−1,1 M 1,1 · · · M n,1 . · · · G1,1 z z At the end of this section, we will study another factorization of the Bell matrix. We define the n × n matrices Xn , Yn by ⎧ ⎪ ⎪ ⎨0, i j, (Xn )i,j= 1 i ⎪ ⎪ ⎩ x1 i − j xi−j+1 , i > j,
⎧ ⎪ ⎪ ⎨0, i j, (Yn )i,j= 1 i i+1 ⎪ ⎪ ⎩ x1 i − j i−j+1 xi−j+1 , i > j,
for i, j = 1, 2, · · · , n. We define a map Tn : Gn → Gn , where Gn denotes the set of all n × n lower triangular matrices. For any (An )i,j = aij ∈ Gn (i, j = 1, 2 · · · , n), let ai,j , i j, Tn (A) = Then we have the following theorem. 1 i > j. i−j ai,j ,
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Theorem 2.5. The matrix Bn can be factorized as
1 1 Xn (Bn−1 diag(2, 3, · · · , n) ⊕ [0])) − Yn (Bn−1 ⊕ [0]) x1 x1 ! + diag(x1 , x21 · · · , xn1 ) .
Bn = Tn
Proof. It follows immediately from identity (2.2). 2 For instance, if n = 4, we have ⎡
⎤ x1 0 0 0 ⎢x x21 0 0 ⎥ ⎢ 2 ⎥ B4 = ⎢ ⎥ 3 ⎣ x3 3x1 x2 x1 0 ⎦ x4 4x1 x3 + 3x22 6x21 x2 x41 ⎧ ⎡ ⎤ ⎤⎡ ⎪ 0 0 0 0 0 0 0 2x1 ⎪ ⎪ ⎢ ⎨ 1 ⎢ 2x 0 0 0⎥ 3x21 0 0⎥ ⎢ 2 ⎥ ⎥ ⎢ 2x2 = T4 ⎢ ⎥ ⎥⎢ 3 ⎪ x ⎣ 3x 0 0 9x x 4x 0 ⎦ 3x ⎦ ⎣ 2x 1 3 2 3 1 2 1 ⎪ ⎪ ⎩ 4x4 6x3 4x2 0 0 0 0 0 ⎡ ⎤⎡ ⎤ ⎡ 0 0 0 0 0 0 0 x1 x1 ⎢ ⎥ ⎢ ⎥ ⎢ 2 1 ⎢ 3x2 0 0 0 ⎥ ⎢ x2 x1 0 0⎥ ⎢ 0 − ⎢ ⎥⎢ ⎥+⎢ x1 ⎣ 4x3 6x2 0 0 ⎦ ⎣ x3 3x1 x2 x31 0 ⎦ ⎣ 0 5x4 10x3 10x2 0 0 0 0 0 0
0 x21 0 0
0 0 x31 0
⎤⎫ 0 ⎪ ⎪ ⎪ ⎬ 0 ⎥ ⎥ ⎥ . 0 ⎦⎪ ⎪ ⎪ x41 ⎭
Example 2.5. If xn = 1 (n 1) in the partial Bell polynomials, we can factorize Sn as Sn = Tn (Xn (Sn−1 diag(2, · · · , n) ⊕ [0])) − Yn (Sn−1 ⊕ [0]) + In ) . For example, if n = 4, we have ⎡
1 ⎢1 ⎢ S4 = ⎢ ⎣1 1 ⎡ 0 ⎢3 ⎢ −⎢ ⎣4 5
0 1 3 7 0 0 6 10
⎧⎡ ⎤ ⎪ 0 0 ⎪ ⎪ ⎨⎢ 2 0⎥ ⎥ ⎢ ⎥ = T4 ⎢ ⎪ ⎣3 0⎦ ⎪ ⎪ ⎩ 4 1 ⎤⎡ 0 0 1 0 0 ⎢1 1 0 0 0⎥ ⎥⎢ ⎥⎢ 0 0⎦⎣1 3 1 10 0 0 0 0
0 0 1 6
0 0 3 6
⎤⎡ 0 2 ⎢2 0⎥ ⎥⎢ ⎥⎢ 0⎦⎣2 0 0 ⎡ 0 1 0 ⎢0 1 0⎥ ⎥ ⎢ ⎥+⎢ 0⎦ ⎣0 0 0 0 0 0 0 0 4 ⎤
0 3 9 0 0 0 1 0
⎤ 0 0⎥ ⎥ ⎥ 0⎦ 0 ⎤⎫ 0 ⎪ ⎪ ⎪ ⎬ 0⎥ ⎥ ⎥ . 0 ⎦⎪ ⎪ ⎪ 1 ⎭
0 0 4 0
By Theorem 2.5, we can get the factorization of some famous lower triangular matrices by taking appropriate values to the sequence {xn }. Analogous to example 2.5, if xn = (n −1)!, we can obtain the factorization of the Stirling matrix of the first kind (unsigned). If xn = n!, we can get the factorization of the Lah matrix.
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3. Matrix factorization of the n × n shift iteration matrix and the n × n generalized Riordan matrix In this section, the formal series f is supposed to be of the form f=
Ωn fn tn ,
n≥1
where Ω1 , Ω2 , · · · is a reference sequence, Ω1 = 1 and Ωn = 0. With every formal series f , we associate the infinite lower iteration matrix with respect to Ω ⎡ ⎤ B1,1 0 0 ··· ⎢ ⎥ 0 ···⎥ ⎢ B2,1 B2,2 ⎢ ⎥ B = B(f ) := ⎢ B B3,2 B3,3 · · · ⎥ , ⎣ 3,1 ⎦ .. .. .. .. . . . . Ω where Bn,k = Bn,k (f1 , f2 , · · · ) is the Bell polynomial with respect to Ω [6, p. 137, 145], defined by
Ωk f k =
Bn,k Ωn tn .
(3.1)
n≥k
Now, analogous to Theorem 2.2, we study the factorization of the n × n shift iteration matrix. At first, we have the following result. Theorem 3.1. The Bell polynomials Bn,k with respect to Ω satisfy the following identity Bn,k1 +k2 =
n Ωk1 +k2 Ωl Ωn−l Bl,k1 Bn−l,k2 , for n k1 + k2 , k1 , k2 0. Ωk1 Ωk2 Ωn
(3.2)
l=0
Proof. Using equation (3.1), we have 1 n 1 Bn,k1 = [t ]f (t)k1 , Ωk1 Ωn 1 n 1 Bn,k2 = [t ]f (t)k2 . Ωk2 Ωn Thus 1 Ωk1 +k2
Bn,k1 +k2 = =
1 n 1 n [t ]f (t)k1 +k2 = [t ]f (t)k1 f (t)k2 Ωn Ωn n n 1 l 1 Ωl Bl,k1 Ωn−l Bn−l,k2 [t ]f (t)k1 [tn−l ]f (t)k2 = . Ωn Ωn Ωk1 Ωk2 l=0
We can obtain (3.2) from the above identity. 2
l=0
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If Ωn = 1/n! in (3.2) gives k1 !k2 ! = (k1 + k2 )! n
Bn,k1 +k2
l=0
n Bl,k1 Bn−l,k2 , l
which is just the identity (2.1). If Ωn = 1, we have Bn,k1 +k2 =
n
Bl,k1 Bn−l,k2 .
(3.3)
l=0
Now, we define the n × n shift iteration matrix Pn,k and the n × n matrices Qn,k , Mn,k by Ωj−1 Ωk+i−j Bk+i−j,k , Ωk+i−1 Ω(n−1+k) Ω0+k Ω1+k , = diag , ,··· , Ω0 Ωk Ω1 Ωk Ωn−1 Ωk
(Pn,k )i,j = Bk+i−1,k+j−1 , (Qn,k )i,j = Mn,k
for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · , where Ω0 = 1. Then we have the following theorem. Theorem 3.2. The n × n shift iteration matrix Pn,k can be factorized as Pn,k = Qn,k Pn,0 Mn,k .
(3.4)
Proof. From identity (3.2). 2 If k = 1, the matrix Pn,1 = Bn is the n × n iteration matrix, we can factorize Bn as Bn = Qn,1 ([1] ⊕ Bn−1 )Mn,1 = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 .
(3.5)
Corollary 3.3. If the reference sequence Ωn = 1 (n 1) for n × n iteration matrix Bn , we have Bn = Qn ([1] ⊕ Bn−1 ) = Qn Qn−1 · · · Q2 Q1 , where (Qn )i,j = B1+i−j,1 (i, j = 1, 2, · · · , n). √
Example 3.1. For the Bell matrix Bn with respect to f = 1− 21−4t = tC(t) = t n1 Cn tn and Ωn = 1 (n 1), where Cn is the Catalan number, we have nk
n
k
k
Bn,k t = t C(t) =
nk
k 2n − k
2n − k n
tn ,
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228
2n − k which implies that Bn,k = n matrix. From identity (3.3), we get
k 2n−k
k1 + k2 2n − k1 − k2
2n − k1 − k2 n
=
n l=0
and the iteration matrix B(f ) is the Catalan
k1 2l − k1
2l − k1 l
k2 2n − 2l − k2
2n − 2l − k2 n−l
,
(3.6) for n k1 + k2 and k1 , k2 0. Letting k1 = 1, k2 = k − 1 in (3.6) give identity (3.7). k 2n − k
2n − k n
=
n l=0
1 2l − 1
2l − 1 l
k−1 2n − 2l − k + 1
2n − 2l − k + 1 n−l
. (3.7)
We define the n × n shift Catalan matrix Cn,k and the n × n matrix Qn,k by
(Cn,0 )1,1
(Qn,k )i,j
k+j−1 = 1 and (Cn,k )i,j = k + 2i − j − 1 k k + 2i − 2j = , k + 2i − 2j k+i−j
k + 2i − j − 1 i−j
,
for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . From Theorem 3.2, we can factorize the n × n shift Catalan matrix as Cn,k = Qn,k Cn,0 . For example, if n = 4, k = 4, we obtain ⎡
C4,4
1 0 ⎢ 4 1 ⎢ =⎢ ⎣ 14 5 48 20
0 0 1 6
⎤ ⎡ 0 1 0 ⎢ 0⎥ ⎥ ⎢ 4 1 ⎥=⎢ 0 ⎦ ⎣ 14 4 1 48 14
0 0 1 4
⎤⎡ 0 1 ⎢ 0⎥ ⎥⎢0 ⎥⎢ 0⎦⎣0 1 0
0 1 1 2
0 0 1 2
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
If k = 1, the matrix Cn,1 is the n × n Catalan matrix, we can further factorize Cn,1 as Cn,1 = Qn,1 ([1] ⊕ Cn−1 ) = Qn,1 Qn−1,1 · · · Q1,1 , where (Qn,1 )i,j =
1 1+2i−2j
1 + 2i − 2j 1+i−j
. If n = 4, we have
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⎡
C4,1
1 ⎢1 ⎢ =⎢ ⎣2 5 ⎡ 1 ⎢1 ⎢ =⎢ ⎣2 5
0 1 2 5
0 0 1 3
0 1 1 2
0 0 1 1
⎤ ⎡ 0 1 ⎢ ⎥ 0⎥ ⎢1 ⎥=⎢ 0⎦ ⎣2 1 5 ⎤⎡ 0 1 0 ⎥ ⎢ 0⎥⎢0 1 ⎥⎢ 0⎦⎣0 1 1 0 2
0 1 1 2 0 0 1 1
⎤⎡ 0 1 ⎢ ⎥ 0⎥⎢0 ⎥⎢ 0⎦⎣0 1 0 ⎤⎡ 0 1 0 ⎥ ⎢ 0⎥⎢0 1 ⎥⎢ 0⎦⎣0 0 1 0 0
0 0 1 1
⎤ 0 0⎥ ⎥ ⎥ 0⎦ 1 ⎤⎡ 0 1 ⎥ ⎢ 0⎥⎢0 ⎥⎢ 0⎦⎣0 1 0
0 1 1 2
229
0 0 1 2
0 0 1 1
0 1 0 0
0 0 1 0
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
Let us introduce the following lemma [6]: Lemma 3.4. Let f , g, h be three sequences, then h = f ◦ g is equivalent to the matrix equality B(h) = B(g)B(f ). The matrix Pn,k (f ) denote the n × n shift iteration matrix associated with formal series f , we have Theorem 3.5. For three formal series f , g, h, h = f ◦g is equivalent to the matrix equality Pn,k (h) = Pn,k (g)Pn,k (f ).
(3.8)
Proof. The (n + k − 1) × (n + k − 1) iteration matrix Bn+k−1 (f ) can be written as Bn+k−1 (f ) =
O Bk−1 (f ) A Pn,k (f )
,
where A is n × (k − 1) matrix. From Lemma 3.4, we have
Bk−1 (f ) B
Bn+k−1 (g)Bn+k−1 (f ) =
O Pn,k (g)
=
O Bk−1 (f ) A Pn,k (f )
O Bk−1 (h) C Pn,k (h)
= Bn+k−1 (h).
From above expression, we have Pn,k (g)Pn,k (f ) = BO + Pn,k (g)Pn,k (f ) = Pn,k (h). By Theorem 3.2 and Theorem 3.5, we get the following result.
2
230
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Corollary 3.6. Let h0 , h1 , h2 be three formal series, with h0 = h2 ◦ h1 , and assume that the n × n shift iteration matrix associated with hi can be factorized as Pn,k (hi ) = (i) Qn,k Pn,0 (hi ) Mn,k , for i = 0, 1, 2. Then (1)
(2)
Pn,k (h0 ) = Qn,k Pn,0 (h1 )Mn,k Qn,k Pn,0 (h2 )Mn,k . Example 3.2. The n × n iteration matrix associated with the formal series f = t/(1 − t) and Ωn = 1 is the Pascal matrix. We define the n × n shift Pascal matrix Pn,k and the n × n matrix Qn,k by (Pn,k )i,j =
k+i−2 k+j−2
, (Qn,k )i,j =
k+i−j−1 k−1
,
for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . Using Theorem 3.2, we can factorize the n × n shift Pascal matrix as Pn,k = Qn,k Pn,0 . For example, if n = 4, k = 4, we have ⎡
P4,4
1 0 ⎢ 4 1 ⎢ =⎢ ⎣ 10 5 20 15
0 0 1 6
⎤ ⎡ 0 1 0 ⎢ 4 1 0⎥ ⎥ ⎢ ⎥=⎢ 0 ⎦ ⎣ 10 4 1 20 10
⎤⎡ 0 1 ⎢0 0⎥ ⎥⎢ ⎥⎢ 0⎦⎣0 1 0
0 0 1 4
0 1 1 1
0 0 1 2
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
Then f 2 is expressed as f 2 = f ◦ f =
t = 2n−1 tn , 1 − 2t n1
and nk
tk Bn,k tn = = (1 − 2t)k
nk
It can be found that
2 fn
= 2n−1 and Bn,k =
n−1 k−1
n−1 k−1
2n−k tn .
2n−k . By Corollary 3.6, we have
2 Pn,k (f 2 ) = Qn,k Pn,0 (f 2 ) = Pn,k = Qn,k Pn,0 Qn,k Pn,0 ,
k+i−2 k+i−j−1 i−j where (Pn,k (f ))i,j = 2 , (Qn,k )i,j = 2i−j for i, j = k+j−2 k−1 1, 2, · · · , n, and k = 0, 1, 2 · · · . For example, if n = 4, k = 4, we have 2
L. Zhang et al. / Linear Algebra and its Applications 577 (2019) 214–239
⎡
1 ⎢ 8 ⎢ Pn,k (f 2 ) = ⎢ ⎣ 40 160 ⎡ 1 ⎢ 4 ⎢ =⎢ ⎣ 10 20
0 0 1 0 10 1 60 12 0 1 4 10
0 0 1 4
⎤ ⎡ 0 1 ⎢ ⎥ 0⎥ ⎢ 8 ⎥=⎢ 0 ⎦ ⎣ 40 1 160 ⎤⎡ 0 1 0 0 ⎢ ⎥ 0⎥⎢0 1 0 ⎥⎢ 0⎦⎣0 1 1 1 0 1 2
⎤⎡ 0 0 0 1 ⎢ ⎥ 1 0 0⎥⎢0 ⎥⎢ 8 1 0⎦⎣0 40 8 1 0 ⎤⎡ 0 1 0 0 ⎢ ⎥ 0⎥⎢ 4 1 0 ⎥⎢ 0 ⎦ ⎣ 10 4 1 1 20 10 4
0 1 2 4
0 0 1 4 ⎤⎡
231
⎤ 0 0⎥ ⎥ ⎥ 0⎦ 1
0 1 ⎢ ⎥ 0⎥⎢0 ⎥⎢ 0⎦⎣0 1 0
0 1 1 1
0 0 1 2
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
Generalized Riordan arrays are generalization of the iteration matrices. We continue to study the matrix factorization of the n ×n shift generalized Riordan matrix. Analogous to Theorem 3.1, we have the following theorem. Theorem 3.7. Let D = (g(t), f (t)) = (dn,k )n,k0 and T = (1, f (t)) = (tn,k )n,k0 be two generalized Riordan arrays with respect to cn , where g(t) = n0 gn tn /cn and f (t) = n n1 fn t /cn . Then we have dn,k1 +k2 =
n ck1 ck2 cn dl,k1 tn−l,k2 , for n k1 + k2 , k1 , k2 0, ck1 +k2 cl cn−l
(3.9)
l=0
dn,k+1 =
n−1
1 ck+1
l 1 −1
···
l1 =k l2 =k−1
l k −1 lk+1 =0
cn fn−l1 fl1 −l2 · · · flk −lk+1 glk+1 cn−l1 cl1 −l2 · · · clk −lk+1 clk+1
(3.10)
(n k + 1, k = 0, 1, 2, · · · ) Proof. From the definition of the generalized Riordan array, we have ck dn,k = cn [tn ]g(t)f (t)k ,
ck tn,k = cn [tn ]f (t)k .
Thus ck1 +k2 dn,k1 +k2 = cn [tn ]g(t)f (t)k1 +k2 = cn
n
[tl ]g(t)f (t)k1 [tn−l ]f (t)k2
l=0
= cn
n ck1 dl,k1 ck2 tn−l,k2 . cl cn−l l=0
We can obtain (3.9) from the above expression. The identity (3.10) can be proved by setting k1 = k, k2 = 1 in (3.9). 2
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The n × n shift generalized Riordan matrix Rn,k and the n × n matrices Tn,k , Mn,k are defined by ck+i−1 tk+i−j,k , cj−1 ck+i−j c0 ck c1 ck cn−1 ck , = diag , ,··· , c0+k c1+k c(n−1+k)
(Rn,k )i,j = dk+i−1,k+j−1 , (Tn,k )i,j = Mn,k
for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . Then we have the following theorem. Theorem 3.8. The n × n shift generalized Riordan matrix can be factorized as Rn,k = Tn,k Rn,0 Mn,k .
(3.11)
Proof. From identity (3.9). 2 Example 3.3. The Riordan array related to the polynomials of the Jacobi case. From [16], it can be found that
where cn =
2t 1 , (1 − t)1+α+β (1 − t)2
22n 1+αn 1+α+β2n ,
=
cn k 2 ck
α+β+n+k n−k
, n,k0
the xn is defined by xn = x(x + 1)(x + 2) · · · (x + n − 1). And
2t 1, (1 − t)2
=
cn k 2 ck
n+k−1 n−k
. n,k0
Thus, the identity (3.9) gives
α + β + n + k1 + k2 n − k1 − k2
For the formal series g(t) =
=
n l=0
1 (1−t)1+α+β
α + β + l + k1 l − k1
and f (t) =
2t (1−t)2 ,
n − l + k2 − 1 n − l − k2
we have
1 tn α+β+n gn = = cn , cn (1 − t)1+α+β n n 2t t = 2cn n. fn = cn (1 − t)2
From Identity (3.10), we have
.
L. Zhang et al. / Linear Algebra and its Applications 577 (2019) 214–239
α+β+n+k+1 n−k−1
=
n−1
l 1 −1
l k −1
···
l1 =k l2 =k−1
233
(n − l1 ) · · · (lk − lk+1 )
lk+1 =0
α + β + lk+1 lk+1
By (3.11), the following matrix factorization holds ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
23 c4 c3 23 c5 c3 23 c6 c3
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ =⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡
3 2 α+β+7 1 α+β+8 2 α+β+9 3
23 6 23 c4 c3 1 7 23 c5 c3 2 8 23 c6 c3 3
1
0
0
24
24 c6 c4
α+β+9 1
0
23 c4 c1 c3
0
23 c6 c1 c3
6 1 7 2
α + β + 10 2
0
23 c5 c1 c3
23 c5 c2 c3 23 c6 c2 c3
⎥ ⎥ 0 ⎥ ⎥ ⎥ ⎥ ⎥ 0 ⎥ 25 ⎥ ⎥ ⎥ α + β + 11 ⎦ 26 1
6 1
25 c6 c5
0
⎤
⎥ ⎥ 0 ⎥ ⎥ ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥ ⎥ 23 c6 ⎦ c3 c3
0
⎤
0
24 c5 c4
0
⎢ α+β+1 ⎢ ⎢ c1 2 ⎢ 1 ⎢ ⎢ ×⎢ α+β+2 α + β + 3 2c2 ⎢ c2 c1 ⎢ 2 1 ⎢ ⎢ α+β+3 α+β+4 ⎣ 2c3 c3 c1 3 2 ⎡ ⎤ 1 0 0 0 ⎢ 0 c1 c3 0 0 ⎥ ⎢ ⎥ c4 ×⎢ ⎥. c2 c3 0 0 ⎣0 ⎦ c5 c3 c3 0 0 0 c6
0
0
0 4 4c3 c2
α+β+5 1
⎤
⎥ ⎥ 0⎥ ⎥ ⎥ ⎥ ⎥ 0⎥ ⎥ ⎥ ⎥ ⎦ 8
Example 3.4. The exponential Riordan array related to the Actuarial polynomials. From [16], the generic term of Riordan array R = (eβt , 1 − et ) is n (−1) β n−j S(j, k). j j=0 k
And
n
.
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234
%
1, 1 − e
t
!
tn (1 − et )k = (−1)k S(n, k). = n! k!
From identity (3.9), we have
k1 + k2 k1
l n n n l n−j β S(j, k1 + k2 ) = β l−j S(j, k1 )S(n − l, k2 ). j l j j=0 j=0
n
l=0
For the formal series g(t) = eβt and f (t) = 1 − et , we have
tn βt e = βn, gn = n! n t (1 − et ) = −1 (n = 1, 2, · · · ), and f0 = 0. fn = n!
Therefore, identity (3.11) gives n β n−j S(j, k + 1) j j=0
n
l n−1 1 −1 k −1 l 1 = ··· (k + 1)! l1 =k l2 =k−1
n l1
lk+1 =0
l1 l2
···
lk lk+1
β lk+1 .
Example 3.5. Some classical Riordan arrays related to the generalized harmonic number. Some harmonic number identities can be obtained from the Theorem 3.7. The harmonic numbers are defined by H0 = 0 and Hn =
n 1
l
l=1
, for n = 1, 2, · · · ,
and the generating function of the harmonic number is −ln(1−x) . The following general1−x ization of the harmonic number can be found in [4,8,13]. (k)
H0
= 0 and Hn(k) =
n 1 , for n, k 1, lk l=1
Hn0 = H(n, k) =
1 and Hnk = n
n
Hlk−1 , for n, k 1,
l=1
1≤n0 +n1 +···+nk ≤n
1 , for n 1, k 0. n0 n1 · · · nk
Letting D1 = ( −ln(1−t) t(1−t) , −ln(1 − t)) and T1 = (1, −ln(1 − t)), the general term of D1 k! is H(n + 1, k) (see [4]) and the general term of T1 is n! s(n, k). From (3.9), we have
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H(n + 1, k1 + k2 ) =
n
H(l + 1, k1 )
l=0
k2 ! s(n − l, k2 ). (n − l)!
235
(3.12)
Letting k1 = 0, k2 = k in (3.12), we get a following relation between H(n, k) and Hn
H(n + 1, k) =
n−k
Hl+1
l=0
k! s(n − l, k). (n − l)!
Setting k1 = k − 1 and k2 = 1 in (3.12) gives H(n + 1, k) =
n−1
H(l + 1, k − 1)
l=0
1 . n−l
From identity (3.10), we have
H(n + 1, k + 1) =
n−1
l 1 −1
lk−1 −1
···
l1 =k l2 =k−1
lk =0
Hlk +1 . (n − l1 )(l1 − l2 ) · · · (lk−1 − lk )
We define the n × n matrices Hn,k and Ln,k by (Hn,k )i,j = H(k + i, k + j − 1),
(Ln,k )i,j =
k! s(k + i − j, k), (k + i − j)!
for i, j = 1, 2, · · · , n. By (3.11), we can factorize Hn,k as Hn,k = Ln,k Hn,0 . For example, if n = 4, k = 3, we have ⎤ ⎡ 1 0 0 1 0 0 0 ⎢ ⎢ 3 1 0 1 0 0⎥ ⎢ ⎥ ⎢ 3/2 =⎢ ⎥=⎢ ⎣ 35/6 7/2 1 0 ⎦ ⎣ 7/4 3/2 1 45/24 7/4 3/2 28/3 23/3 4 1 ⎡
H4,3
⎤⎡ 1 0 0 0 ⎢ 1 0 0⎥ ⎥ ⎢ 3/2 ⎥⎢ 2 1 0 ⎦ ⎣ 11/6 25/12 35/12 5/2 1
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
k+1 t t Letting D2 = ( −ln(1−t) 2 = (1, 1−t ), the general term of D2 is Hn−k+1 t(1−t) , 1−t ) and T n−1 (see [4]) and the general term of T2 is . From (3.9), we obtain k−1
k1 +k2 +1 Hn−k = 1 −k2 +1
n l=0
k1 +1 Hl−k 1 +1
n−l−1 k2 − 1
Setting k1 = 0 and k2 = k in (3.13) gives the following identity
.
(3.13)
236
L. Zhang et al. / Linear Algebra and its Applications 577 (2019) 214–239
k+1 Hn−k+1
n−k
=
Hl+1
l=0
n−l−1 k−1
.
Letting k1 = 1 and k2 = k in (3.13) yields k+2 Hn−k =
n−k
Hl2
l=0
n−l−1 k−1
.
By identity (3.10), we have k+2 Hn−k =
n−1
l 1 −1
lk−1 −1
···
l1 =k l2 =k−1
Hlk +1 .
lk =0
n,k and Nn,k by We define the n × n matrices H n,k )i,j = H k+i+1 , (H i−j+1
(Nn,k )i,j =
k+i−j−1 k−1
,
n,k as for i, j = 0, 1, · · · , n − 1. Form (3.11), we can factorize H n,k = Nn,k H n,0 . H For instance, if n = 4, k = 1, we have ⎡
4,3 H
1 ⎢ 9/2 ⎢ =⎢ ⎣ 18 + 2/6 45 + 7/12 ⎡ 1 0 0 ⎢ 3 1 0 ⎢ =⎢ ⎣ 12 3 1 20 12 3
⎤ 0 0 0 1 0 0⎥ ⎥ ⎥ 11/2 1 0⎦ 23 + 5/6 13/2 1 ⎤⎡ 0 1 0 0 ⎥ ⎢ 0 ⎥ ⎢ 3/2 1 0 ⎥⎢ 0 ⎦ ⎣ 11/6 5/2 1 1 25/12 13/3 7/2
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
Example 3.6. The classical Riordan array related to the polynomials P (r, n, k). In [13], the polynomials Pr (x1 , x2 , · · · , xr ) and P (r, n, k) are defined by Pr (x1 , x2 , · · · , xr ) = (−1)r Yr (−0!x1 , −1!x2 , · · · , −(r − 1)!xr ), (1)
(r)
P (r, n, k) = Pr (Hn(1) − Hk , · · · , Hn(r) − Hk ), where Yr are the exponential complete Bell polynomials. If r = 1, 2, 3, it can be found that
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237
P1 (x1 ) = x1 , P2 (x1 , x2 ) = x21 − x2 , P3 (x1 , x2 , x3 ) = x31 − 3x1 x2 + 2x3 . t t Letting D3 = ( −ln(1−t) ), the general term of D3 is (1−t) , 1−t ) and T3 = (1, 1−t n n−1 P (r, n, k) (see [19]) and the general term of T3 is . From (3.9), we obtain k k−1 r
n k1 + k2
P (r, n, k1 + k2 ) =
n
l=0
l k1
P (r, l, k1 )
n−l−1 k2 − 1
.
(3.14)
Setting k1 = 0, k2 = k and k1 = k − 1, k2 = 1 in (3.14) yields n n n − l − 1 r! s(l + 1, r + 1), P (r, n, k) = l! k k−1 l=0 n−1 n l P (r, n, k) = P (r, l, k − 1). k k−1
(3.15)
(3.16)
l=0
The substitution k = 1 in (3.15) yields (3.17), setting r = 1 in (3.15) and (3.16) gives (3.18) and (3.19). n−1 1 r! s(l + 1, r + 1), n l! l=0 n n n−l−1 (Hn − Hk ) = Hl , k k−1 l=0 n−1 n l (Hn − Hk ) = (Hl − Hk−1 ). k k−1
P (r, n, 1) =
(3.17)
(3.18)
(3.19)
l=0
1 Letting D4 = ( (1−t) r+1 , −ln(1 − t)) and T4 = (1, −ln(1 − t)), the general term of D4 n+r k! is s(n, k). From (3.9), P (k, n + r, r) (see [19]) and the general term of T4 is n! r we obtain n k2 ! n+r l+r s(n − l, k2 ). P (k1 + k2 , n + r, r) = P (k1 , l + r, r) (n − l)! r r l=0
(3.20) Setting k1 = 0, k2 = k and k1 = k − 1, k2 = 1 in (3.20) yields (3.21) and (3.22).
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n+r r
P (k, n + r, r) =
l=0
n+r r
n
P (k, n + r, r) =
n−1
l+r r
l=0
k! s(n − l, k), (n − l)!
l+r r
P (k − 1, l + r, r)
1 . n−l
(3.21)
(3.22)
And setting r = 0 in (3.21) and (3.22) gives (3.23) and (3.24). The substitution k = 1 in (3.22) yields (3.25). k! k! s(n + 1, k + 1) = s(n − l, k), n! (n − l)! n
(3.23)
l=0
n−1 (k − 1)! k! s(n + 1, k + 1) = s(l + 1, k), n! l!(n − l) l=0 n−1 l+r 1 n+r . (Hn+r − Hr ) = n−l r r
(3.24)
(3.25)
l=0
The identities (3.23) and (3.25) can be found in [19]. Declaration of Competing Interest This article has no competing interest. Acknowledgements The work is supported by the National Natural Science Foundation of China (NO. 11271341). We would like to express our gratitude to the editor and the anonymous referee for their helpful comments and remarks. References [1] M. Abbas, S. Bouroubi, On new identities for Bell’s polynomials, Discrete Math. 293 (1–3) (2005) 5–10. [2] P. Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra Appl. 491 (2016) 343–385. [3] E.T. Bell, Exponential polynomials, Ann. Math. 35 (1934) 258–277. [4] G.-S. Cheon, M.E.A. El-Mikkawy, Generalized harmonic numbers with Riordan arrays, J. Number Theory 128 (2008) 413–425. [5] G.-S. Cheon, J.-S. Kim, Stirling matrix via Pascal matrix, Linear Algebra Appl. 329 (1–3) (2001) 49–59. [6] L. Comtet, Advanced Combinatorics, D. Reidel Publishing Co, Dordrecht, 1974. [7] D. Cvijovic, New identities for the partial Bell polynomials, Appl. Math. Lett. 24 (2011) 1544–1547. [8] A. Gertsch, Generalized harmonic numbers, in: Number Theory, in: C. R. Acad. Sci. Paris Ser. I, vol. 324, 1997, pp. 7–10. [9] T.X. He, R. Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math. 309 (2009) 3962–3974.
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[10] D. Merlini, D.G. Rogers, R. Sprugnoli, M.C. Verri, On some alternative characterizations of Riordan arrays, Canad. J. Math. 49 (2) (1997) 301–320. [11] J. Riordan, Combinatorial Identities, Wiley, New York, 1968. [12] D.G. Rogers, Pascal triangles, Catalan numbers and renewal arrays, Discrete Math. 22 (1978) 301–310. [13] J. Spiess, Some identities involving harmonic numbers, Math. Comp. 55 (192) (1990) 839–863. [14] L.W. Shapiro, S. Getu, W.J. Woan, L. Woodson, The Riordan group, Discrete Appl. Math. 34 (1991) 229–239. [15] M. Tan, T. Wang, Lah matrix and its algebraic properties, Ars Combin. 70 (2004) 97–108. [16] W. Wang, T. Wang, Generalized Riordan arrays, Discrete Math. 308 (2008) 6466–6500. [17] W. Wang, T. Wang, Identities via Bell matrix and Fibonacci matrix, Discrete Appl. Math. 156 (2008) 2793–2803. [18] W. Wang, T. Wang, Matrices related to the Bell polynomials, Linear Algebra Appl. 422 (2007) 139–154. [19] W. Wang, Riordan arrays and harmonic number identities, Comput. Math. Appl. 60 (2010) 1494–1509. [20] S. Yang, Some identities involving the binomial sequences, Discrete Math. 308 (2008) 51–58. [21] Z. Zhang, The linear algebra of the generalized Pascal matrix, Linear Algebra Appl. 250 (1997) 51–60. [22] Z. Zhang, M. Liu, An extension of the generalized Pascal matrix and its algebraic properties, Linear Algebra Appl. 271 (1998) 169–177. [23] Z. Zhang, T. Wang, Generalized Pascal matrix and recurrence sequences, Linear Algebra Appl. 283 (1–3) (1998) 289–299. [24] X. Zhao, Some identities related to reciprocal functions, Discrete Math. 265 (2003) 323–335. [25] X. Zhao, Some summation rules related to the Riordan arrays, Discrete Math. 281 (2004) 295–307. [26] X. Zhao, T. Wang, The algebraic properties of the generalized Pascal functional matrices associated with the exponential families, Linear Algebra Appl. 318 (1–3) (2000) 45–52.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Matrix factorization of the n × n shift Bell matrix Liming Zhang, Zhihua Sun, Xiaoying Qu, Xiqiang Zhao ∗ School of Mathematical Science, Ocean University of China, Qingdao 266100, PR China
a r t i c l e
i n f o
Article history: Received 4 October 2018 Accepted 25 April 2019 Available online 3 May 2019 Submitted by R. Brualdi MSC: 05A15 05A05 15B36 15A06 05A19
a b s t r a c t Let Bn = [Bn,k ]n,k0 be the Bell matrix. Define the n × n shift Bell matrix Pn,k by (Pn,k )i,j = Bk+i−1,k+j−1 for i, j = 1, 2, · · · n and k = 0, 1, 2 · · · . In this paper, matrix factorizations of the n × n shift Bell matrix and the n × n generalized Riordan matrix are studied. As a result, many lower triangular matrices related to Bell polynomials can be factorized by the corresponding matrices and some identities are derived from the matrix representations. In addition, some harmonic number identities are obtained from the Riordan array method. © 2019 Elsevier Inc. All rights reserved.
Keywords: Bell polynomials n × n shift Bell matrix n × n shift iteration matrix Generalized Riordan array
1. Introduction In 1934, Bell [3] extensively introduced the important combinatorial tools-Bell polynomials. Some useful combinatorial sequences can be obtained from the Bell polynomials by choosing the variables x1 , x2 · · · . For example, all the Stirling numbers of both kinds, the * Corresponding author. E-mail address: [email protected] (X. Zhao). https://doi.org/10.1016/j.laa.2019.04.037 0024-3795/© 2019 Elsevier Inc. All rights reserved.
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Lah numbers and the idempotent numbers are the special cases of the Bell polynomials (see [6, p. 135, Theorem B]): Bn.k (1, 1, 1, · · · ) = S(n, k), (Stirling number of the second kind) n − 1 n! Bn.k (1!, 2!, 3!, · · · ) = , (Lah number) k − 1 k! Bn.k (0!, 1!, 2!, · · · ) = s(n, k), (unsigned Stirling number of the first kind) n Bn.k (1, 2, 3, · · · ) = k n−k . (idempotent number) k Recently, the authors [1,6,7] studied some identities for Bell polynomials, which have found many applications in combinatorics. Some relations between the Bell matrix and the Fibonacci matrix are given by W. Wang and T. Wang [17]. The factorization of the Bell matrix and iteration matrix are also demonstrated by them and some combinatorial identities are obtained from the matrix representations (see [18]). In 1991, the Riordan array is introduced by Shapiro et al. [10,11,14], which is a generalization of the renewal array [12]. Riordan array is a useful tool to study combinatorial sum and identities. Many applications of the Riordan array have been studied in [2,19,24–26] and some characterization of Riordan array can be found in [9,10,14]. Gi-Sang Cheon et al. [4] studied the relations between the Stirling numbers of both kinds and other generalized harmonic numbers. They also obtained some generalized harmonic number identities by using the approach of the Riordan arrays. Moreover, the r t 1 Riordan array ( −ln(1−t) (1−t) , 1−t ) and ( (1−t)r+1 , −ln(1 − t)) are generated by the number P (r, n, k) (see [19]). It can be seen that the Riordan array is very important for studying combinatorial identities. The exponential partial Bell polynomials [6, p. 133] and the Riordan array are defined as follows. Definition 1.1. The exponential partial Bell polynomials are the polynomials Bn,k = Bn,k (x1 , x2 , · · · , xn−k+1 ) , in an infinite number of variables x1 , x2 , · · · , defined by the formal double series expansion ⎛ Φ = Φ(t, u) = exp ⎝u tn =1+ n! n1
m1
n
k=1
⎞ tm ⎠ tn = xm Bn,k uk m! n! n,k0
u Bn,k (x1 , x2 , · · · ) , k
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or by the series expansion ⎛ Φk (t) :=
1 ⎝ k!
⎞k
t ⎠ tn = Bn,k , m! n! m
xm
m1
k = 0, 1, 2, · · · .
(1.1)
nk
The exponential complete Bell polynomials Yn = Yn (x1 , x2 , · · · , xn ) are defined by
Yn =
n
Bn,k ,
Y0 := 1.
(1.2)
k=1
Definition 1.2. The generalized Riordan array with respected to the sequence cn is a pair (g(t), f (t)), where g(t) = k0 gk tk /ck and f (t) = k1 fk tk /ck with f1 = 0. It defines an infinite lower triangular array (dn,k )n,k∈N according to the rule dn,k =
f (t)k tn g(t) , cn ck
where g(t)f (t)k /ck is the generating function of the k column, and dn,k is the general term of the generalized Riordan array. By the definition, the classical Riordan arrays are corresponded to the case of cn = 1, and the exponential Riordan arrays are corresponded to the case of cn = n!. The rest of this paper is structured as follows. Section 2, we study the matrix factorization of the n ×n shift Bell matrix. For instance, some lower triangular matrices can be factorized by the corresponding matrices and some identities are gotten from the matrix representations. Section 3, we give the factorization of the n × n shift iteration matrix and the n × n shift generalized Riordan matrix, and some combinatorial identities are obtained from the Riordan array method. 2. Matrix factorization of the n × n shift Bell matrix Lemma 2.1. The partial Bell polynomials satisfy the following relations (see [7]) k1 !k2 ! = (k1 + k2 )! n
Bn,k1 +k2
l=0
Bn,k
n−k 1 = x1 (n − k) l=1
n Bl,k1 Bn−l,k2 , l
(n k1 + k2 , k1 , k2 0)
n+1 n xl+1 Bn−l,k . (k + 1) − l+1 l
(2.1)
(2.2)
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217
We now define the n × n shift Bell matrix Pn,k and the n × n matrices Qn,k , Mn,k by (Pn,k )i,j = Bk+i−1,k+j−1 , (Qn,k )i,j = Mn,k = diag
k+i−1 j−1
1!k! (n − 1)!k! 0!k! , ,··· , (0 + k)! (1 + k)! (n − 1 + k)!
Bk+i−j,k , ,
for i, j = 1, 2, · · · n and k = 0, 1, 2, · · · . If k = 1, the matrix Pn,1 = Bn is the n × n Bell matrix (see [17]). From identity (2.1), we can obtain the following theorem. Theorem 2.2. The matrix Pn,k can be factorized as Pn,k = Qn,k Pn,0 Mn,k .
(2.3)
Proof. It follows immediately from identity (2.1). 2 For example, if n = 4, k = 4, we have ⎡
P4,4
⎤ x41 0 0 0 ⎢ x51 0 0 ⎥ 10x31 x2 ⎢ ⎥ =⎢ ⎥ 3 2 2 4 6 ⎣ 15x1 x2 x1 0 ⎦ 20x1 x3 + 45x1 x2 35x31 x4 + 210x21 x2 x3 + 105x1 x32 35x41 x3 + 105x31 x22 21x51 x2 x71 ⎡ ⎤ x41 0 0 0 ⎢ 5x41 0 0 ⎥ 10x31 x2 ⎢ ⎥ =⎢ ⎥ 3 2 2 3 4 ⎣ 60x1 x2 15x1 0 ⎦ 20x1 x3 + 45x1 x2 35x31 x4 + 210x21 x2 x3 + 105x1 x32 140x31 x3 + 315x21 x22 210x31 x2 35x41 ⎤ ⎡ ⎤⎡ 1 0 0 0 1 0 0 0 ⎢ ⎢0 x 0 0 ⎥ 0 0 ⎥ ⎥ ⎢ ⎥ ⎢ 0 1/5 1 ×⎢ ⎥. ⎥⎢ 2 ⎣ 0 x2 0 ⎦ x1 0 ⎦ ⎣ 0 0 1/15 0 0 0 1/35 0 x3 3x1 x2 x31
Particularly, if k = 1, we have Pn,1 = Qn,1 ([1] ⊕ Pn−1,1 )Mn,1 ,
(2.4)
which is equivalent to Bn = Qn,1 ([1] ⊕ Bn−1 )Mn,1 . The notation ⊕ denotes the direct sum of two matrices. For instance, if n = 4 and k = 1, we can factorize 4 × 4 Bell matrix as
218
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⎡
x1 ⎢x ⎢ 2 B4 = ⎢ ⎣ x3 x4 ⎡ 1 ⎢0 ⎢ ×⎢ ⎣0 0
⎤ ⎡ x1 0 0 0 ⎢ ⎥ 0 0 ⎥ ⎢ x2 2x1 ⎥=⎢ x31 0 ⎦ ⎣ x3 3x2 6x21 x2 x41 x4 4x3 ⎤⎡ ⎤ 0 1 0 0 0 ⎢ 0 ⎥ 0 ⎥ ⎥ ⎢ 0 1/2 0 ⎥ ⎥⎢ ⎥. 0 1/3 0 ⎦ 0 ⎦⎣0 0 0 0 1/4 x31
0 x21 3x1 x2 4x1 x3 + 3x22 0 x1 x2 x3
0 0 x21 3x1 x2
0 0 3x1 6x2
⎤ 0 0 ⎥ ⎥ ⎥ 0 ⎦ 4x1
For any l × l matrices Xl,1 , we define the n × n matrices X l,1 by X l,1 =
In−l o
o Xl,1
,
where In−l is the (n − l) × (n − l) unit matrix. Then we can further factorize the Bell matrix. Theorem 2.3. The n × n Bell matrix Bn can be factorized as Bn = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 .
(2.5)
For example, if n = 3 and k = 1, we can factorize B3 as ⎤ ⎡ ⎤⎡ ⎤ x1 1 0 0 x1 0 0 0 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ B3 = ⎣ x2 x21 0 ⎦ = ⎣ x2 2x1 0 0 ⎦ ⎣ 0 x1 ⎦ x3 3x1 x2 x31 x3 3x2 3x1 0 x2 2x1 ⎤ ⎤⎡ ⎤⎡ ⎡ ⎤⎡ 1 0 0 1 0 0 1 0 0 1 0 0 ⎥ ⎥⎢ ⎥⎢ ⎢ ⎥⎢ ×⎣0 1 0 ⎦⎣0 1 0 ⎦⎣0 1 0 ⎦ ⎣ 0 1/2 0 ⎦. 0 0 1/3 0 0 1/2 0 0 x1 0 0 1 ⎡
Based on Theorem 2.2 and Theorem 2.3, the matrix factorization of some lower triangular matrices related to Bell polynomials can be obtained directly. Example 2.1. The unsigned Stirling numbers of the first kind s(n, k) and the Stirling numbers of the second kind S(n, k) satisfy the following equations [6, p. 135, Eqs. (3g) and (3i)] Bn,k (0!, 1!, 2!, · · · ) = s(n, k),
Bn,k (1, 1, 1, · · · ) = S(n, k).
We can obtain (2.6), (2.7) from the identity (2.1)
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n s(l, k1 )s(n − l, k2 ), l l=0 n k1 !k2 ! n S(n, k1 + k2 ) = S(l, k1 )S(n − l, k2 ), (k1 + k2 )! l
k1 !k2 ! s(n, k1 + k2 ) = (k1 + k2 )!
219
n
(2.6)
(2.7)
l=0
for n k1 + k2 , k1 , k2 0. Setting k1 = k − 1, k2 = 1 in (2.6) and (2.7) gives n s(l, k − 1)(n − l − 1)!, l l=0 n−1 1 n S(l, k − 1). S(n, k) = k l
n−1 1 s(n, k) = k
l=0
We define the n × n matrices sn,k and Sn,k by (sn,k )i,j = s(k + i − 1, k + j − 1), (Sn,k )i,j = S(k + i − 1, k + j − 1), for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . From Theorem 2.2, we have sn,k = qn,k sn,0 Mn,k
(2.8)
Sn,k = Qn,k Sn,0 Mn,k
(2.9)
where qn,k , Qn,k and Mn,k are defined by
k+i−1 k+i−1 (qn,k )i,j = s(k + i − j, k), (Qn,k )i,j = S(k + i − j, k), j−1 j−1 1!k! (n − 1)!k! 0!k! , ,··· , . Mn,k = diag (0 + k)! (1 + k)! (n − 1 + k)! For instance, if n = 4 and k = 4, the matrix s4,4 can be factorized as ⎡
s4,4
1 0 0 ⎢ 10 1 0 ⎢ =⎢ ⎣ 85 15 1 735 175 21 ⎡ 1 0 0 ⎢ 10 5 0 ⎢ =⎢ ⎣ 85 60 15 735 595 210
⎤ 0 0⎥ ⎥ ⎥ 0⎦ 1
⎤⎡ 0 1 ⎢ 0 ⎥ ⎥⎢0 ⎥⎢ 0 ⎦⎣0 35 0
0 1 1 2
0 0 1 3
⎤⎡ ⎤ 0 1 0 0 0 ⎢ 0⎥ 0 0 ⎥ ⎥ ⎢ 0 1/5 ⎥ ⎥⎢ ⎥. 0 ⎦ ⎣ 0 0 1/15 0 ⎦ 1 0 0 0 1/35
If k = 1, the matrix sn = sn,1 is the Stirling matrix of the first kind (unsigned) and Sn = Sn,1 is the Stirling matrix of the second kind (see [5]). By Theorem 2.3, we have
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sn = qn,1 ([1] ⊕ sn−1 )Mn,1 = q n,1 q n−1,1 · · · q 1,1 M 1,1 · · · M n−1,1 M n,1 , Sn = Qn,1 ([1] ⊕ Sn−1 )Mn,1 = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 . For example, if n = 4, we can factorize S4 as ⎡
1 ⎢1 ⎢ S4 = ⎢ ⎣1 1
0 1 3 7
0 0 1 6
⎤ ⎡ 0 1 ⎢1 0⎥ ⎥ ⎢ ⎥=⎢ 0⎦ ⎣1 1 1
0 2 3 4
0 0 3 6
⎤⎡ 0 1 ⎢0 0⎥ ⎥⎢ ⎥⎢ 0⎦⎣0 4 0
0 1 1 1
0 0 1 3
⎤⎡ ⎤ 0 1 0 0 0 ⎢ 0⎥ 0 ⎥ ⎥ ⎢ 0 1/2 0 ⎥ ⎥⎢ ⎥. 0 ⎦ ⎣ 0 0 1/3 0 ⎦ 1 0 0 0 1/4
Example 2.2. In [20], the following equation was obtained
I
(a)
n (n, k) := Bn,k (1, 2a, 3a , · · · ) = (ka)n−k . k 2
From (2.1) and (2.3), we get k1 !k2 ! (n, k1 + k2 ) = (k1 + k2 )! n
I
(a)
l=0
(a)
n I (a) (l, k1 )I (a) (n − l, k2 ), l
(a)
IPn,k = Qn,k IPn,0 Mn,k ,
(2.10) (2.11)
(a)
where IPn,k , Qn,k and Mn,k are defined by (a) (IPn,k )i,j
=I
(a)
(k + i − 1, k + j − 1), (Qn,k )i,j =
Mn,k = diag
k+i−1 j−1
1!k! (n − 1)!k! 0!k! , ,··· , (0 + k)! (1 + k)! (n − 1 + k)!
I (a) (k + i − j, k), ,
for i, j = 1, 2, · · · n and k = 0, 1, 2, · · · . (a)
For example, if n = 4 and k = 4, the matrix IP4,4 can be factorized as ⎡
(a)
IP4,4
1 ⎢ 20a ⎢ =⎢ ⎣ 240a2 2240a3 ⎡ 1 0 ⎢0 1 ⎢ ×⎢ ⎣ 0 2a 0 3a2
0 1 30a 525a2
0 0 1 42a ⎤⎡ 0 0 1 ⎢ 0 0⎥ ⎥⎢0 ⎥⎢ 1 0⎦⎣0 0 6a 1
⎤ ⎡ 1 0 ⎢ 0⎥ ⎥ ⎢ 20a ⎥=⎢ 0 ⎦ ⎣ 240a2 1 2240a3
0 5 120a 1680a2 ⎤
0 0 0 1/5 0 0 ⎥ ⎥ ⎥. 0 1/15 0 ⎦ 0 0 1/35
⎤ 0 0 0 0 ⎥ ⎥ ⎥ 15 0 ⎦ 420a 35
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If k = 1, we have (a)
(a)
IPn,1 = Qn,1 ([1] ⊕ IPn−1,1 )Mn,1 = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 . (a)
For instance, if n = 4 and k = 1, we can factorize IP4,1 as ⎡
1 0 ⎢ 2a 1 ⎢ (a) IP4,1 = ⎢ 2 ⎣ 3a 6a 4a3 24a2 ⎡ 1 0 0 ⎢0 1 0 ⎢ ×⎢ ⎣ 0 2a 1 0 3a2 6a
⎤ ⎡ 0 0 1 0 0 ⎢ 0 0⎥ 2 0 ⎥ ⎢ 2a ⎥=⎢ 2 1 0 ⎦ ⎣ 3a 6a 3 4a3 12a2 12a 12a 1 ⎤ ⎤⎡ 1 0 0 0 0 ⎢ 0 ⎥ 0⎥ ⎥ ⎥ ⎢ 0 1/2 0 ⎥. ⎥⎢ 0 ⎦ ⎣ 0 0 1/3 0 ⎦ 0 0 0 1/4 1
⎤ 0 0⎥ ⎥ ⎥ 0⎦ 4
Example 2.3. Using the equation [1, remark 5] (a)
L
(n, k) := Bn,k (1!a , 2!a , 3!a , · · · ) = a 0
1
2
n−k
n−1 k−1
n! , k!
we find that k1 !k2 ! L(a) (n, k1 + k2 ) = (k1 + k2 )! n
l=0
(a)
n L(a) (l, k1 )L(a) (n − l, k2 ), l
(a)
Ln,k = Qn,k Ln,0 Mn,k ,
(2.12) (2.13)
(a) (Ln,k )i,j
k+i−1 j−1
= L(a) (k + i − 1, k + j − 1), (Qn,k )i,j = L(a) (k + i − j, k), (n−1)!k! 0!k! 1!k! Mn,k = diag (0+k)! , (1+k)! , · · · , (n−1+k)! , for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . For example, if n = 4 and k = 4, we have where
⎡
(a)
L4,4
1 ⎢ 20a ⎢ =⎢ ⎣ 300a2 4200a3 ⎡ 1 0 ⎢0 1 ⎢ ×⎢ ⎣ 0 2a 0 6a2 (a)
0 1 30a 630a2
0 0 1 42a ⎤⎡ 0 0 1 ⎢ ⎥ 0 0⎥⎢0 ⎥⎢ 1 0⎦⎣0 0 6a 1
⎤ ⎡ 1 0 ⎢ ⎥ 0 ⎥ ⎢ 20a ⎥=⎢ 0 ⎦ ⎣ 300a2 1 4200a3
0 5 120a 2100a2 ⎤
0 0 0 1/5 0 0 ⎥ ⎥ ⎥. 0 1/15 0 ⎦ 0 0 1/35
If k = 1, the matrix Ln,1 can be factorized as
⎤ 0 0 0 0 ⎥ ⎥ ⎥ 15 0 ⎦ 420a 35
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(a)
(a)
Ln,1 = Qn,1 ([1] ⊕ Ln−1,1 )Mn,1 = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 . Let L(i, j) be the Lah numbers, and if a = 1, we have (see [6, p. 156] and [15]) Ln,1 = sn Sn = Qn,1 ([1] ⊕ Ln−1,1 )Mn,1 , where Ln,1 is the Lah matrix with (Ln,1 )i,j = L(1) (i, j) = L(i, j), for i, j = 1, 2, · · · , n. From the above expression, we obtain i
s(i, l)S(l, j) =
l=j
i 1 l=j
j
=
i j−1
i (1 + i − l)!L(l − 1, j − 1) l−1 i 1 i − j + 1 (i − l + 1)!(l − 2)! l=j
j
(j − 2)!
l−j
.
Example 2.4. Let us study the following equation [20, Eq. (17)] n Bn,k (B0 (x), 2B1 (x), 3B2 (x), · · · ) = Bn−k (kx), k n where Bn (x) := i=0 S(n, j)xj are the single variable Bell polynomials. It can be found that Bn (1) are the Bell numbers (see [9, p. 210]). From (2.1) and (2.3), we get Bn−k1 −k2 ((k1 + k2 )x) =
n
l=0
n − k1 − k2 l − k1
Bl−k1 (k1 x)Bn−l−k2 (k2 x),
n,k = Qn,k B n,0 Mn,k , B
(2.14) (2.15)
where
k+i−1 Bi−j ((k + j − 1)x), k+j−1 k+i−1 k+i−j (Qn,k )i,j = Bi−j (kx), j−1 k 1!k! (n − 1)!k! 0!k! Mn,k = diag , ,··· , , (0 + k)! (1 + k)! (n − 1 + k)! n,k )i,j = (B
for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . Setting k1 = k − 1, k2 = 1 in (2.14) and letting m = n − k, we obtain Bm (kx) =
m l=0
m l
Bm−l (x)Bl ((k − 1)x).
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From the definition of Bn (x), we get m
n
S(m, n)(kx) =
n=0
m m−l l
l=0 i=0 j=0
m l
(k − 1)j S(m − l, i)S(l, j)xi+j .
By equating the coefficients of xn , we obtain S(m, n)k n =
m n
m l
l=0 j=0
(k − 1)j S(m − l, n − j)S(l, j).
n,1 as This identity can be found in [18]. If k = 1 in (2.15), we can factorize B n−1,1 )Mn,1 = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 . n,1 = Qn,1 ([1] ⊕ B B For instance, if n = 4, k = 1, we have ⎡
⎤ 1 0 0 0 ⎢ 2x 1 0 0⎥ ⎥ 4,1 = ⎢ B ⎢ ⎥ 2 ⎣ 3(x + x ) 6x 1 0⎦ 4(x + 3x2 + x3 ) 12(x + 2x2 ) 12x 1 ⎡ ⎤ 1 0 0 0 ⎢ 2x 2 0 0⎥ ⎢ ⎥ =⎢ ⎥ 2 ⎣ 3(x + x ) 6x 3 0⎦ 4(x + 3x2 + x3 ) 12(x + x2 ) 12x 4 ⎤ ⎡ ⎤⎡ 1 0 0 0 1 0 0 0 ⎢ ⎢0 0 ⎥ 1 0 0⎥ ⎥ ⎢ ⎥ ⎢ 0 1/2 0 ×⎢ ⎥. ⎥⎢ ⎣0 2x 1 0 ⎦ ⎣ 0 0 1/3 0 ⎦ 0 0 0 1/4 0 3(x + x2 ) 6x 1 We consider some generalizations of the n × n shift Bell matrix. We define the n × n matrices Pn,k [y], P n,k [y] and Pn,k [y, z] by (Pn,k [y])i,j = y i−j Bk+i−1,k+j−1 , (P n,k [y])n,k = y i+j−2 Bk+i−1,k+j−1 , (Pn,k [y, z])i,j = y i−j z i+j−2 Bk+i−1,k+j−1 , for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · where y, z are nonzero real number. The n × n matrices Gn,k [y], Qn,k [y], Rn,k [y, z] and Mn,k are given by (Gn,k [y])i,j = y i−j
i j−1
Bi−j+k,k ,
224
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i (Qn,k [y])i,j = y Bi−j+k,k , j−1 i i−j i+j−2 (Rn,k [y, z])i,j = y z Bi−j+k,k , j−1 1!k! (n − 1)!k! 0!k! , ,··· , . Mn,k = diag (0 + k)! (1 + k)! (n − 1 + k)! i+j−2
In similar fashion to [5,15,21–23], we can obtain the following theorem. Theorem 2.4. For any nonzero number y and z, we have Pn,k [y] = Gn,k [y]Pn,0 [y]Mn,k , 1 Mn,k , P n,k [y] = Qn,k [y]Pn,0 y y Pn,k [y, z] = Rn,k [y, z]Pn,0 Mn,k . z Proof. It can be proved by identity (2.1). 2 Particularly, if k = 1, we have Pn,1 [y] = Gn,1 [y]([1] ⊕ Pn−1,1 [y])Mn,1 = Gn,1 [y] · · · G1,1 [y]M 1,1 · · · M n,1 , 1 )Mn,1 P n,1 [y] = Qn,1 [y]([1] ⊕ Pn−1,1 y 1 1 · · · G1,1 = Qn,1 [y]Gn−1,1 M 1,1 · · · M n,1 , y y y Pn,1 [y, z] = Rn,1 [y, z]([1] ⊕ Pn−1,1 )Mn,1 z y y = Rn,1 [y, z]Gn−1,1 M 1,1 · · · M n,1 . · · · G1,1 z z At the end of this section, we will study another factorization of the Bell matrix. We define the n × n matrices Xn , Yn by ⎧ ⎪ ⎪ ⎨0, i j, (Xn )i,j= 1 i ⎪ ⎪ ⎩ x1 i − j xi−j+1 , i > j,
⎧ ⎪ ⎪ ⎨0, i j, (Yn )i,j= 1 i i+1 ⎪ ⎪ ⎩ x1 i − j i−j+1 xi−j+1 , i > j,
for i, j = 1, 2, · · · , n. We define a map Tn : Gn → Gn , where Gn denotes the set of all n × n lower triangular matrices. For any (An )i,j = aij ∈ Gn (i, j = 1, 2 · · · , n), let ai,j , i j, Tn (A) = Then we have the following theorem. 1 i > j. i−j ai,j ,
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Theorem 2.5. The matrix Bn can be factorized as
1 1 Xn (Bn−1 diag(2, 3, · · · , n) ⊕ [0])) − Yn (Bn−1 ⊕ [0]) x1 x1 ! + diag(x1 , x21 · · · , xn1 ) .
Bn = Tn
Proof. It follows immediately from identity (2.2). 2 For instance, if n = 4, we have ⎡
⎤ x1 0 0 0 ⎢x x21 0 0 ⎥ ⎢ 2 ⎥ B4 = ⎢ ⎥ 3 ⎣ x3 3x1 x2 x1 0 ⎦ x4 4x1 x3 + 3x22 6x21 x2 x41 ⎧ ⎡ ⎤ ⎤⎡ ⎪ 0 0 0 0 0 0 0 2x1 ⎪ ⎪ ⎢ ⎨ 1 ⎢ 2x 0 0 0⎥ 3x21 0 0⎥ ⎢ 2 ⎥ ⎥ ⎢ 2x2 = T4 ⎢ ⎥ ⎥⎢ 3 ⎪ x ⎣ 3x 0 0 9x x 4x 0 ⎦ 3x ⎦ ⎣ 2x 1 3 2 3 1 2 1 ⎪ ⎪ ⎩ 4x4 6x3 4x2 0 0 0 0 0 ⎡ ⎤⎡ ⎤ ⎡ 0 0 0 0 0 0 0 x1 x1 ⎢ ⎥ ⎢ ⎥ ⎢ 2 1 ⎢ 3x2 0 0 0 ⎥ ⎢ x2 x1 0 0⎥ ⎢ 0 − ⎢ ⎥⎢ ⎥+⎢ x1 ⎣ 4x3 6x2 0 0 ⎦ ⎣ x3 3x1 x2 x31 0 ⎦ ⎣ 0 5x4 10x3 10x2 0 0 0 0 0 0
0 x21 0 0
0 0 x31 0
⎤⎫ 0 ⎪ ⎪ ⎪ ⎬ 0 ⎥ ⎥ ⎥ . 0 ⎦⎪ ⎪ ⎪ x41 ⎭
Example 2.5. If xn = 1 (n 1) in the partial Bell polynomials, we can factorize Sn as Sn = Tn (Xn (Sn−1 diag(2, · · · , n) ⊕ [0])) − Yn (Sn−1 ⊕ [0]) + In ) . For example, if n = 4, we have ⎡
1 ⎢1 ⎢ S4 = ⎢ ⎣1 1 ⎡ 0 ⎢3 ⎢ −⎢ ⎣4 5
0 1 3 7 0 0 6 10
⎧⎡ ⎤ ⎪ 0 0 ⎪ ⎪ ⎨⎢ 2 0⎥ ⎥ ⎢ ⎥ = T4 ⎢ ⎪ ⎣3 0⎦ ⎪ ⎪ ⎩ 4 1 ⎤⎡ 0 0 1 0 0 ⎢1 1 0 0 0⎥ ⎥⎢ ⎥⎢ 0 0⎦⎣1 3 1 10 0 0 0 0
0 0 1 6
0 0 3 6
⎤⎡ 0 2 ⎢2 0⎥ ⎥⎢ ⎥⎢ 0⎦⎣2 0 0 ⎡ 0 1 0 ⎢0 1 0⎥ ⎥ ⎢ ⎥+⎢ 0⎦ ⎣0 0 0 0 0 0 0 0 4 ⎤
0 3 9 0 0 0 1 0
⎤ 0 0⎥ ⎥ ⎥ 0⎦ 0 ⎤⎫ 0 ⎪ ⎪ ⎪ ⎬ 0⎥ ⎥ ⎥ . 0 ⎦⎪ ⎪ ⎪ 1 ⎭
0 0 4 0
By Theorem 2.5, we can get the factorization of some famous lower triangular matrices by taking appropriate values to the sequence {xn }. Analogous to example 2.5, if xn = (n −1)!, we can obtain the factorization of the Stirling matrix of the first kind (unsigned). If xn = n!, we can get the factorization of the Lah matrix.
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3. Matrix factorization of the n × n shift iteration matrix and the n × n generalized Riordan matrix In this section, the formal series f is supposed to be of the form f=
Ωn fn tn ,
n≥1
where Ω1 , Ω2 , · · · is a reference sequence, Ω1 = 1 and Ωn = 0. With every formal series f , we associate the infinite lower iteration matrix with respect to Ω ⎡ ⎤ B1,1 0 0 ··· ⎢ ⎥ 0 ···⎥ ⎢ B2,1 B2,2 ⎢ ⎥ B = B(f ) := ⎢ B B3,2 B3,3 · · · ⎥ , ⎣ 3,1 ⎦ .. .. .. .. . . . . Ω where Bn,k = Bn,k (f1 , f2 , · · · ) is the Bell polynomial with respect to Ω [6, p. 137, 145], defined by
Ωk f k =
Bn,k Ωn tn .
(3.1)
n≥k
Now, analogous to Theorem 2.2, we study the factorization of the n × n shift iteration matrix. At first, we have the following result. Theorem 3.1. The Bell polynomials Bn,k with respect to Ω satisfy the following identity Bn,k1 +k2 =
n Ωk1 +k2 Ωl Ωn−l Bl,k1 Bn−l,k2 , for n k1 + k2 , k1 , k2 0. Ωk1 Ωk2 Ωn
(3.2)
l=0
Proof. Using equation (3.1), we have 1 n 1 Bn,k1 = [t ]f (t)k1 , Ωk1 Ωn 1 n 1 Bn,k2 = [t ]f (t)k2 . Ωk2 Ωn Thus 1 Ωk1 +k2
Bn,k1 +k2 = =
1 n 1 n [t ]f (t)k1 +k2 = [t ]f (t)k1 f (t)k2 Ωn Ωn n n 1 l 1 Ωl Bl,k1 Ωn−l Bn−l,k2 [t ]f (t)k1 [tn−l ]f (t)k2 = . Ωn Ωn Ωk1 Ωk2 l=0
We can obtain (3.2) from the above identity. 2
l=0
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If Ωn = 1/n! in (3.2) gives k1 !k2 ! = (k1 + k2 )! n
Bn,k1 +k2
l=0
n Bl,k1 Bn−l,k2 , l
which is just the identity (2.1). If Ωn = 1, we have Bn,k1 +k2 =
n
Bl,k1 Bn−l,k2 .
(3.3)
l=0
Now, we define the n × n shift iteration matrix Pn,k and the n × n matrices Qn,k , Mn,k by Ωj−1 Ωk+i−j Bk+i−j,k , Ωk+i−1 Ω(n−1+k) Ω0+k Ω1+k , = diag , ,··· , Ω0 Ωk Ω1 Ωk Ωn−1 Ωk
(Pn,k )i,j = Bk+i−1,k+j−1 , (Qn,k )i,j = Mn,k
for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · , where Ω0 = 1. Then we have the following theorem. Theorem 3.2. The n × n shift iteration matrix Pn,k can be factorized as Pn,k = Qn,k Pn,0 Mn,k .
(3.4)
Proof. From identity (3.2). 2 If k = 1, the matrix Pn,1 = Bn is the n × n iteration matrix, we can factorize Bn as Bn = Qn,1 ([1] ⊕ Bn−1 )Mn,1 = Qn,1 Qn−1,1 · · · Q1,1 M 1,1 · · · M n−1,1 M n,1 .
(3.5)
Corollary 3.3. If the reference sequence Ωn = 1 (n 1) for n × n iteration matrix Bn , we have Bn = Qn ([1] ⊕ Bn−1 ) = Qn Qn−1 · · · Q2 Q1 , where (Qn )i,j = B1+i−j,1 (i, j = 1, 2, · · · , n). √
Example 3.1. For the Bell matrix Bn with respect to f = 1− 21−4t = tC(t) = t n1 Cn tn and Ωn = 1 (n 1), where Cn is the Catalan number, we have nk
n
k
k
Bn,k t = t C(t) =
nk
k 2n − k
2n − k n
tn ,
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228
2n − k which implies that Bn,k = n matrix. From identity (3.3), we get
k 2n−k
k1 + k2 2n − k1 − k2
2n − k1 − k2 n
=
n l=0
and the iteration matrix B(f ) is the Catalan
k1 2l − k1
2l − k1 l
k2 2n − 2l − k2
2n − 2l − k2 n−l
,
(3.6) for n k1 + k2 and k1 , k2 0. Letting k1 = 1, k2 = k − 1 in (3.6) give identity (3.7). k 2n − k
2n − k n
=
n l=0
1 2l − 1
2l − 1 l
k−1 2n − 2l − k + 1
2n − 2l − k + 1 n−l
. (3.7)
We define the n × n shift Catalan matrix Cn,k and the n × n matrix Qn,k by
(Cn,0 )1,1
(Qn,k )i,j
k+j−1 = 1 and (Cn,k )i,j = k + 2i − j − 1 k k + 2i − 2j = , k + 2i − 2j k+i−j
k + 2i − j − 1 i−j
,
for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . From Theorem 3.2, we can factorize the n × n shift Catalan matrix as Cn,k = Qn,k Cn,0 . For example, if n = 4, k = 4, we obtain ⎡
C4,4
1 0 ⎢ 4 1 ⎢ =⎢ ⎣ 14 5 48 20
0 0 1 6
⎤ ⎡ 0 1 0 ⎢ 0⎥ ⎥ ⎢ 4 1 ⎥=⎢ 0 ⎦ ⎣ 14 4 1 48 14
0 0 1 4
⎤⎡ 0 1 ⎢ 0⎥ ⎥⎢0 ⎥⎢ 0⎦⎣0 1 0
0 1 1 2
0 0 1 2
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
If k = 1, the matrix Cn,1 is the n × n Catalan matrix, we can further factorize Cn,1 as Cn,1 = Qn,1 ([1] ⊕ Cn−1 ) = Qn,1 Qn−1,1 · · · Q1,1 , where (Qn,1 )i,j =
1 1+2i−2j
1 + 2i − 2j 1+i−j
. If n = 4, we have
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⎡
C4,1
1 ⎢1 ⎢ =⎢ ⎣2 5 ⎡ 1 ⎢1 ⎢ =⎢ ⎣2 5
0 1 2 5
0 0 1 3
0 1 1 2
0 0 1 1
⎤ ⎡ 0 1 ⎢ ⎥ 0⎥ ⎢1 ⎥=⎢ 0⎦ ⎣2 1 5 ⎤⎡ 0 1 0 ⎥ ⎢ 0⎥⎢0 1 ⎥⎢ 0⎦⎣0 1 1 0 2
0 1 1 2 0 0 1 1
⎤⎡ 0 1 ⎢ ⎥ 0⎥⎢0 ⎥⎢ 0⎦⎣0 1 0 ⎤⎡ 0 1 0 ⎥ ⎢ 0⎥⎢0 1 ⎥⎢ 0⎦⎣0 0 1 0 0
0 0 1 1
⎤ 0 0⎥ ⎥ ⎥ 0⎦ 1 ⎤⎡ 0 1 ⎥ ⎢ 0⎥⎢0 ⎥⎢ 0⎦⎣0 1 0
0 1 1 2
229
0 0 1 2
0 0 1 1
0 1 0 0
0 0 1 0
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
Let us introduce the following lemma [6]: Lemma 3.4. Let f , g, h be three sequences, then h = f ◦ g is equivalent to the matrix equality B(h) = B(g)B(f ). The matrix Pn,k (f ) denote the n × n shift iteration matrix associated with formal series f , we have Theorem 3.5. For three formal series f , g, h, h = f ◦g is equivalent to the matrix equality Pn,k (h) = Pn,k (g)Pn,k (f ).
(3.8)
Proof. The (n + k − 1) × (n + k − 1) iteration matrix Bn+k−1 (f ) can be written as Bn+k−1 (f ) =
O Bk−1 (f ) A Pn,k (f )
,
where A is n × (k − 1) matrix. From Lemma 3.4, we have
Bk−1 (f ) B
Bn+k−1 (g)Bn+k−1 (f ) =
O Pn,k (g)
=
O Bk−1 (f ) A Pn,k (f )
O Bk−1 (h) C Pn,k (h)
= Bn+k−1 (h).
From above expression, we have Pn,k (g)Pn,k (f ) = BO + Pn,k (g)Pn,k (f ) = Pn,k (h). By Theorem 3.2 and Theorem 3.5, we get the following result.
2
230
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Corollary 3.6. Let h0 , h1 , h2 be three formal series, with h0 = h2 ◦ h1 , and assume that the n × n shift iteration matrix associated with hi can be factorized as Pn,k (hi ) = (i) Qn,k Pn,0 (hi ) Mn,k , for i = 0, 1, 2. Then (1)
(2)
Pn,k (h0 ) = Qn,k Pn,0 (h1 )Mn,k Qn,k Pn,0 (h2 )Mn,k . Example 3.2. The n × n iteration matrix associated with the formal series f = t/(1 − t) and Ωn = 1 is the Pascal matrix. We define the n × n shift Pascal matrix Pn,k and the n × n matrix Qn,k by (Pn,k )i,j =
k+i−2 k+j−2
, (Qn,k )i,j =
k+i−j−1 k−1
,
for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . Using Theorem 3.2, we can factorize the n × n shift Pascal matrix as Pn,k = Qn,k Pn,0 . For example, if n = 4, k = 4, we have ⎡
P4,4
1 0 ⎢ 4 1 ⎢ =⎢ ⎣ 10 5 20 15
0 0 1 6
⎤ ⎡ 0 1 0 ⎢ 4 1 0⎥ ⎥ ⎢ ⎥=⎢ 0 ⎦ ⎣ 10 4 1 20 10
⎤⎡ 0 1 ⎢0 0⎥ ⎥⎢ ⎥⎢ 0⎦⎣0 1 0
0 0 1 4
0 1 1 1
0 0 1 2
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
Then f 2 is expressed as f 2 = f ◦ f =
t = 2n−1 tn , 1 − 2t n1
and nk
tk Bn,k tn = = (1 − 2t)k
nk
It can be found that
2 fn
= 2n−1 and Bn,k =
n−1 k−1
n−1 k−1
2n−k tn .
2n−k . By Corollary 3.6, we have
2 Pn,k (f 2 ) = Qn,k Pn,0 (f 2 ) = Pn,k = Qn,k Pn,0 Qn,k Pn,0 ,
k+i−2 k+i−j−1 i−j where (Pn,k (f ))i,j = 2 , (Qn,k )i,j = 2i−j for i, j = k+j−2 k−1 1, 2, · · · , n, and k = 0, 1, 2 · · · . For example, if n = 4, k = 4, we have 2
L. Zhang et al. / Linear Algebra and its Applications 577 (2019) 214–239
⎡
1 ⎢ 8 ⎢ Pn,k (f 2 ) = ⎢ ⎣ 40 160 ⎡ 1 ⎢ 4 ⎢ =⎢ ⎣ 10 20
0 0 1 0 10 1 60 12 0 1 4 10
0 0 1 4
⎤ ⎡ 0 1 ⎢ ⎥ 0⎥ ⎢ 8 ⎥=⎢ 0 ⎦ ⎣ 40 1 160 ⎤⎡ 0 1 0 0 ⎢ ⎥ 0⎥⎢0 1 0 ⎥⎢ 0⎦⎣0 1 1 1 0 1 2
⎤⎡ 0 0 0 1 ⎢ ⎥ 1 0 0⎥⎢0 ⎥⎢ 8 1 0⎦⎣0 40 8 1 0 ⎤⎡ 0 1 0 0 ⎢ ⎥ 0⎥⎢ 4 1 0 ⎥⎢ 0 ⎦ ⎣ 10 4 1 1 20 10 4
0 1 2 4
0 0 1 4 ⎤⎡
231
⎤ 0 0⎥ ⎥ ⎥ 0⎦ 1
0 1 ⎢ ⎥ 0⎥⎢0 ⎥⎢ 0⎦⎣0 1 0
0 1 1 1
0 0 1 2
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
Generalized Riordan arrays are generalization of the iteration matrices. We continue to study the matrix factorization of the n ×n shift generalized Riordan matrix. Analogous to Theorem 3.1, we have the following theorem. Theorem 3.7. Let D = (g(t), f (t)) = (dn,k )n,k0 and T = (1, f (t)) = (tn,k )n,k0 be two generalized Riordan arrays with respect to cn , where g(t) = n0 gn tn /cn and f (t) = n n1 fn t /cn . Then we have dn,k1 +k2 =
n ck1 ck2 cn dl,k1 tn−l,k2 , for n k1 + k2 , k1 , k2 0, ck1 +k2 cl cn−l
(3.9)
l=0
dn,k+1 =
n−1
1 ck+1
l 1 −1
···
l1 =k l2 =k−1
l k −1 lk+1 =0
cn fn−l1 fl1 −l2 · · · flk −lk+1 glk+1 cn−l1 cl1 −l2 · · · clk −lk+1 clk+1
(3.10)
(n k + 1, k = 0, 1, 2, · · · ) Proof. From the definition of the generalized Riordan array, we have ck dn,k = cn [tn ]g(t)f (t)k ,
ck tn,k = cn [tn ]f (t)k .
Thus ck1 +k2 dn,k1 +k2 = cn [tn ]g(t)f (t)k1 +k2 = cn
n
[tl ]g(t)f (t)k1 [tn−l ]f (t)k2
l=0
= cn
n ck1 dl,k1 ck2 tn−l,k2 . cl cn−l l=0
We can obtain (3.9) from the above expression. The identity (3.10) can be proved by setting k1 = k, k2 = 1 in (3.9). 2
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232
The n × n shift generalized Riordan matrix Rn,k and the n × n matrices Tn,k , Mn,k are defined by ck+i−1 tk+i−j,k , cj−1 ck+i−j c0 ck c1 ck cn−1 ck , = diag , ,··· , c0+k c1+k c(n−1+k)
(Rn,k )i,j = dk+i−1,k+j−1 , (Tn,k )i,j = Mn,k
for i, j = 1, 2, · · · , n and k = 0, 1, 2 · · · . Then we have the following theorem. Theorem 3.8. The n × n shift generalized Riordan matrix can be factorized as Rn,k = Tn,k Rn,0 Mn,k .
(3.11)
Proof. From identity (3.9). 2 Example 3.3. The Riordan array related to the polynomials of the Jacobi case. From [16], it can be found that
where cn =
2t 1 , (1 − t)1+α+β (1 − t)2
22n 1+αn 1+α+β2n ,
=
cn k 2 ck
α+β+n+k n−k
, n,k0
the xn is defined by xn = x(x + 1)(x + 2) · · · (x + n − 1). And
2t 1, (1 − t)2
=
cn k 2 ck
n+k−1 n−k
. n,k0
Thus, the identity (3.9) gives
α + β + n + k1 + k2 n − k1 − k2
For the formal series g(t) =
=
n l=0
1 (1−t)1+α+β
α + β + l + k1 l − k1
and f (t) =
2t (1−t)2 ,
n − l + k2 − 1 n − l − k2
we have
1 tn α+β+n gn = = cn , cn (1 − t)1+α+β n n 2t t = 2cn n. fn = cn (1 − t)2
From Identity (3.10), we have
.
L. Zhang et al. / Linear Algebra and its Applications 577 (2019) 214–239
α+β+n+k+1 n−k−1
=
n−1
l 1 −1
l k −1
···
l1 =k l2 =k−1
233
(n − l1 ) · · · (lk − lk+1 )
lk+1 =0
α + β + lk+1 lk+1
By (3.11), the following matrix factorization holds ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
23 c4 c3 23 c5 c3 23 c6 c3
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ =⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡
3 2 α+β+7 1 α+β+8 2 α+β+9 3
23 6 23 c4 c3 1 7 23 c5 c3 2 8 23 c6 c3 3
1
0
0
24
24 c6 c4
α+β+9 1
0
23 c4 c1 c3
0
23 c6 c1 c3
6 1 7 2
α + β + 10 2
0
23 c5 c1 c3
23 c5 c2 c3 23 c6 c2 c3
⎥ ⎥ 0 ⎥ ⎥ ⎥ ⎥ ⎥ 0 ⎥ 25 ⎥ ⎥ ⎥ α + β + 11 ⎦ 26 1
6 1
25 c6 c5
0
⎤
⎥ ⎥ 0 ⎥ ⎥ ⎥ ⎥ ⎥ 0 ⎥ ⎥ ⎥ ⎥ 23 c6 ⎦ c3 c3
0
⎤
0
24 c5 c4
0
⎢ α+β+1 ⎢ ⎢ c1 2 ⎢ 1 ⎢ ⎢ ×⎢ α+β+2 α + β + 3 2c2 ⎢ c2 c1 ⎢ 2 1 ⎢ ⎢ α+β+3 α+β+4 ⎣ 2c3 c3 c1 3 2 ⎡ ⎤ 1 0 0 0 ⎢ 0 c1 c3 0 0 ⎥ ⎢ ⎥ c4 ×⎢ ⎥. c2 c3 0 0 ⎣0 ⎦ c5 c3 c3 0 0 0 c6
0
0
0 4 4c3 c2
α+β+5 1
⎤
⎥ ⎥ 0⎥ ⎥ ⎥ ⎥ ⎥ 0⎥ ⎥ ⎥ ⎥ ⎦ 8
Example 3.4. The exponential Riordan array related to the Actuarial polynomials. From [16], the generic term of Riordan array R = (eβt , 1 − et ) is n (−1) β n−j S(j, k). j j=0 k
And
n
.
L. Zhang et al. / Linear Algebra and its Applications 577 (2019) 214–239
234
%
1, 1 − e
t
!
tn (1 − et )k = (−1)k S(n, k). = n! k!
From identity (3.9), we have
k1 + k2 k1
l n n n l n−j β S(j, k1 + k2 ) = β l−j S(j, k1 )S(n − l, k2 ). j l j j=0 j=0
n
l=0
For the formal series g(t) = eβt and f (t) = 1 − et , we have
tn βt e = βn, gn = n! n t (1 − et ) = −1 (n = 1, 2, · · · ), and f0 = 0. fn = n!
Therefore, identity (3.11) gives n β n−j S(j, k + 1) j j=0
n
l n−1 1 −1 k −1 l 1 = ··· (k + 1)! l1 =k l2 =k−1
n l1
lk+1 =0
l1 l2
···
lk lk+1
β lk+1 .
Example 3.5. Some classical Riordan arrays related to the generalized harmonic number. Some harmonic number identities can be obtained from the Theorem 3.7. The harmonic numbers are defined by H0 = 0 and Hn =
n 1
l
l=1
, for n = 1, 2, · · · ,
and the generating function of the harmonic number is −ln(1−x) . The following general1−x ization of the harmonic number can be found in [4,8,13]. (k)
H0
= 0 and Hn(k) =
n 1 , for n, k 1, lk l=1
Hn0 = H(n, k) =
1 and Hnk = n
n
Hlk−1 , for n, k 1,
l=1
1≤n0 +n1 +···+nk ≤n
1 , for n 1, k 0. n0 n1 · · · nk
Letting D1 = ( −ln(1−t) t(1−t) , −ln(1 − t)) and T1 = (1, −ln(1 − t)), the general term of D1 k! is H(n + 1, k) (see [4]) and the general term of T1 is n! s(n, k). From (3.9), we have
L. Zhang et al. / Linear Algebra and its Applications 577 (2019) 214–239
H(n + 1, k1 + k2 ) =
n
H(l + 1, k1 )
l=0
k2 ! s(n − l, k2 ). (n − l)!
235
(3.12)
Letting k1 = 0, k2 = k in (3.12), we get a following relation between H(n, k) and Hn
H(n + 1, k) =
n−k
Hl+1
l=0
k! s(n − l, k). (n − l)!
Setting k1 = k − 1 and k2 = 1 in (3.12) gives H(n + 1, k) =
n−1
H(l + 1, k − 1)
l=0
1 . n−l
From identity (3.10), we have
H(n + 1, k + 1) =
n−1
l 1 −1
lk−1 −1
···
l1 =k l2 =k−1
lk =0
Hlk +1 . (n − l1 )(l1 − l2 ) · · · (lk−1 − lk )
We define the n × n matrices Hn,k and Ln,k by (Hn,k )i,j = H(k + i, k + j − 1),
(Ln,k )i,j =
k! s(k + i − j, k), (k + i − j)!
for i, j = 1, 2, · · · , n. By (3.11), we can factorize Hn,k as Hn,k = Ln,k Hn,0 . For example, if n = 4, k = 3, we have ⎤ ⎡ 1 0 0 1 0 0 0 ⎢ ⎢ 3 1 0 1 0 0⎥ ⎢ ⎥ ⎢ 3/2 =⎢ ⎥=⎢ ⎣ 35/6 7/2 1 0 ⎦ ⎣ 7/4 3/2 1 45/24 7/4 3/2 28/3 23/3 4 1 ⎡
H4,3
⎤⎡ 1 0 0 0 ⎢ 1 0 0⎥ ⎥ ⎢ 3/2 ⎥⎢ 2 1 0 ⎦ ⎣ 11/6 25/12 35/12 5/2 1
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
k+1 t t Letting D2 = ( −ln(1−t) 2 = (1, 1−t ), the general term of D2 is Hn−k+1 t(1−t) , 1−t ) and T n−1 (see [4]) and the general term of T2 is . From (3.9), we obtain k−1
k1 +k2 +1 Hn−k = 1 −k2 +1
n l=0
k1 +1 Hl−k 1 +1
n−l−1 k2 − 1
Setting k1 = 0 and k2 = k in (3.13) gives the following identity
.
(3.13)
236
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k+1 Hn−k+1
n−k
=
Hl+1
l=0
n−l−1 k−1
.
Letting k1 = 1 and k2 = k in (3.13) yields k+2 Hn−k =
n−k
Hl2
l=0
n−l−1 k−1
.
By identity (3.10), we have k+2 Hn−k =
n−1
l 1 −1
lk−1 −1
···
l1 =k l2 =k−1
Hlk +1 .
lk =0
n,k and Nn,k by We define the n × n matrices H n,k )i,j = H k+i+1 , (H i−j+1
(Nn,k )i,j =
k+i−j−1 k−1
,
n,k as for i, j = 0, 1, · · · , n − 1. Form (3.11), we can factorize H n,k = Nn,k H n,0 . H For instance, if n = 4, k = 1, we have ⎡
4,3 H
1 ⎢ 9/2 ⎢ =⎢ ⎣ 18 + 2/6 45 + 7/12 ⎡ 1 0 0 ⎢ 3 1 0 ⎢ =⎢ ⎣ 12 3 1 20 12 3
⎤ 0 0 0 1 0 0⎥ ⎥ ⎥ 11/2 1 0⎦ 23 + 5/6 13/2 1 ⎤⎡ 0 1 0 0 ⎥ ⎢ 0 ⎥ ⎢ 3/2 1 0 ⎥⎢ 0 ⎦ ⎣ 11/6 5/2 1 1 25/12 13/3 7/2
⎤ 0 0⎥ ⎥ ⎥. 0⎦ 1
Example 3.6. The classical Riordan array related to the polynomials P (r, n, k). In [13], the polynomials Pr (x1 , x2 , · · · , xr ) and P (r, n, k) are defined by Pr (x1 , x2 , · · · , xr ) = (−1)r Yr (−0!x1 , −1!x2 , · · · , −(r − 1)!xr ), (1)
(r)
P (r, n, k) = Pr (Hn(1) − Hk , · · · , Hn(r) − Hk ), where Yr are the exponential complete Bell polynomials. If r = 1, 2, 3, it can be found that
L. Zhang et al. / Linear Algebra and its Applications 577 (2019) 214–239
237
P1 (x1 ) = x1 , P2 (x1 , x2 ) = x21 − x2 , P3 (x1 , x2 , x3 ) = x31 − 3x1 x2 + 2x3 . t t Letting D3 = ( −ln(1−t) ), the general term of D3 is (1−t) , 1−t ) and T3 = (1, 1−t n n−1 P (r, n, k) (see [19]) and the general term of T3 is . From (3.9), we obtain k k−1 r
n k1 + k2
P (r, n, k1 + k2 ) =
n
l=0
l k1
P (r, l, k1 )
n−l−1 k2 − 1
.
(3.14)
Setting k1 = 0, k2 = k and k1 = k − 1, k2 = 1 in (3.14) yields n n n − l − 1 r! s(l + 1, r + 1), P (r, n, k) = l! k k−1 l=0 n−1 n l P (r, n, k) = P (r, l, k − 1). k k−1
(3.15)
(3.16)
l=0
The substitution k = 1 in (3.15) yields (3.17), setting r = 1 in (3.15) and (3.16) gives (3.18) and (3.19). n−1 1 r! s(l + 1, r + 1), n l! l=0 n n n−l−1 (Hn − Hk ) = Hl , k k−1 l=0 n−1 n l (Hn − Hk ) = (Hl − Hk−1 ). k k−1
P (r, n, 1) =
(3.17)
(3.18)
(3.19)
l=0
1 Letting D4 = ( (1−t) r+1 , −ln(1 − t)) and T4 = (1, −ln(1 − t)), the general term of D4 n+r k! is s(n, k). From (3.9), P (k, n + r, r) (see [19]) and the general term of T4 is n! r we obtain n k2 ! n+r l+r s(n − l, k2 ). P (k1 + k2 , n + r, r) = P (k1 , l + r, r) (n − l)! r r l=0
(3.20) Setting k1 = 0, k2 = k and k1 = k − 1, k2 = 1 in (3.20) yields (3.21) and (3.22).
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L. Zhang et al. / Linear Algebra and its Applications 577 (2019) 214–239
n+r r
P (k, n + r, r) =
l=0
n+r r
n
P (k, n + r, r) =
n−1
l+r r
l=0
k! s(n − l, k), (n − l)!
l+r r
P (k − 1, l + r, r)
1 . n−l
(3.21)
(3.22)
And setting r = 0 in (3.21) and (3.22) gives (3.23) and (3.24). The substitution k = 1 in (3.22) yields (3.25). k! k! s(n + 1, k + 1) = s(n − l, k), n! (n − l)! n
(3.23)
l=0
n−1 (k − 1)! k! s(n + 1, k + 1) = s(l + 1, k), n! l!(n − l) l=0 n−1 l+r 1 n+r . (Hn+r − Hr ) = n−l r r
(3.24)
(3.25)
l=0
The identities (3.23) and (3.25) can be found in [19]. Declaration of Competing Interest This article has no competing interest. Acknowledgements The work is supported by the National Natural Science Foundation of China (NO. 11271341). We would like to express our gratitude to the editor and the anonymous referee for their helpful comments and remarks. References [1] M. Abbas, S. Bouroubi, On new identities for Bell’s polynomials, Discrete Math. 293 (1–3) (2005) 5–10. [2] P. Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra Appl. 491 (2016) 343–385. [3] E.T. Bell, Exponential polynomials, Ann. Math. 35 (1934) 258–277. [4] G.-S. Cheon, M.E.A. El-Mikkawy, Generalized harmonic numbers with Riordan arrays, J. Number Theory 128 (2008) 413–425. [5] G.-S. Cheon, J.-S. Kim, Stirling matrix via Pascal matrix, Linear Algebra Appl. 329 (1–3) (2001) 49–59. [6] L. Comtet, Advanced Combinatorics, D. Reidel Publishing Co, Dordrecht, 1974. [7] D. Cvijovic, New identities for the partial Bell polynomials, Appl. Math. Lett. 24 (2011) 1544–1547. [8] A. Gertsch, Generalized harmonic numbers, in: Number Theory, in: C. R. Acad. Sci. Paris Ser. I, vol. 324, 1997, pp. 7–10. [9] T.X. He, R. Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math. 309 (2009) 3962–3974.
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