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Matrix methods for determinants of Pascal-like matrices
Accepted Manuscript Matrix Methods for determinants of Pascal-like matrices
De-Yin Zheng
PII: DOI: Reference:
S0024-3795(19)30181-8 https://doi.org/10.1016/j.laa.2019.04.023 LAA 14950
To appear in:
Linear Algebra and its Applications
Received date: Accepted date:
20 November 2016 22 April 2019
Please cite this article in press as: D.-Y. Zheng, Matrix Methods for determinants of Pascal-like matrices, Linear Algebra Appl. (2019), https://doi.org/10.1016/j.laa.2019.04.023
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MATRIX METHODS FOR DETERMINANTS OF PASCAL-LIKE MATRICES DE-YIN ZHENG Department of Mathematics, Hangzhou Normal University, Hangzhou 310018, P. R. China Email: [email protected]
Abstract. Using the LDU-like-factorization method, the evaluations of the determinants of some Pascal-like matrices are derived. The entries of these matrices satisfy the recurrence relation ai,j = αai−1,j +βai,j−1 +γai−1,j−1 , i, j ≥ 1, and the values of the first row a0,j and the column ai,0 have eight cases. Part of the results is due to C. Krattenthaler [Evaluations Of Some Determinants Of Matrices Related To The Pascal Triangle, S´eminaire Lotharingien de Combinatoire, 1 (2002): 1-19].
1. Introduction Pascal triangles or Pascal matrices are one of the most famous arithmetic triangles with many wonderful properties. However, the determinant of a Pascal-like matrix defined by a simple linear recursive relationship with constant coefficients has only been studied in detail recently, see [1–12]. Bacher [1] studies the determinant of the matrix whose entries satisfy the Pascal triangle recurrence relation ai,j = ai−1,j + ai,j−1 + αi βj or ai,j = ai−1,j + ai,j−1 + xai−1,j−1 with various choices for the first row a0,j and column ai,0 . He proved some results and proposed some conjectures about the determinant of such matrices. Some of these conjectures have been proved by Krattenthaler [3] through the “LUfactorization method”, which has been included in the review [2] on the “state of the art” of determinant calculations. Krattenthaler [3] proves and generalizes several evaluations of determinants conjectured in [1]. Specifically, evaluations are given in the cases when the entries of the matrix satisfy the recurrence relation ai,j = ai−1,j + ai,j−1 + xai−1,j−1 , i, j ≥ 1, and the initial values (i) ai,0 = ρi , a0,i = σ i , (ii) ak,k = 0, ai,0 = ρi−1 and a0,i = −ρi−1 for i ≥ 1, or (iii) ai,0 = i, a0,i = −i. Using the generating function method for a matrix product and the fact that the principal minors of every band-diagonal matrix satisfy a nontrivial recurrence relation, Zakraj˘sek and Petkov˘sek [12] studied the determinants of arrays with recurrence relation pi,j = λpi−1,j + μpi,j−1 + νpi−1,j−1 for i, j ≥ 1, and proved a generalization of Bacher’s conjecture [1, Conjecture 3.3] on principal minors of Pascal-like matrices. In 2008, Tan [10] gave the matrix product factorization of this Pascal-like matrix, and the value of the determinant of such a matrix whose initial values are two geometric sequences was gotten as a by-product. Furthermore, Date: April 25, 2019. 2000 Mathematics Subject Classification. Primary 05A19; Secondary 05A10; 05A15; 15A15; 11C20. Key words and phrases. determinant; Pascal matrix; LDU -factorization; tridiagonal matrix; bivariate recurrent sequence.
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
2
Neuwirth [9] and Moghaddamfar [5, 6] also considered the determinants of similar matrices whose entries satisfy linear recurrence relations with variable coefficients and generalized some of the results obtained. In addition, by finite differences, Wang [11] provides new proofs for two theorems due to Krattenthaler [3] and then proves a more general theorem that can be considered as a generalization of one determinant identity also due to Krattenthaler [3]. Here we study the generalized Pascal matrices (called Pascal-like matrices), whose entries are determined by the recursive relation ai,j = αai−1,j + βai,j−1 + γai−1,j−1 ,
i, j ≥ 1,
(1.1)
except for the first row and column. This recursive relation and initial values ai,0 and a0,j play an important role in constructing the matrix and calculating its determinant. Taking different values for the first row a0,j and column ai,0 will construct different matrices. In this paper, the determinants of these matrices will be calculated using the “LDU-like-factorization method”. The process is as follows: First, the Pascal-like matrix A is converted into a LDU-like matrix by generating function method, and then the matrix A is pre-multiplied by an appropriate lower triangular matrix, and is post-multiplied by an appropriate upper triangular matrix, so that A is converted into a diagonal matrix or a three-diagonal matrix. Further, the determinant of the matrix A can be calculated. The article is organized as follows: This section is preliminary. The “LDU-likefactorization” of the Pascal-like matrix is derived in Section 2. It will be used in Section 3 to calculate the determinants of Pascal-like matrices with eight kinds of initial values. They cover the part of the results of Krattenthaler [2] and Wang [11]. Finally, the paper is concluded in Section 4. To describe our results, we first give some basic notations and definitions, which will be required in the development of this paper. The Fibonacci sequence is a sequence Fn of natural numbers defined recursively: F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 , if n ≥ 2. Given a matrix A, we use the notation AT for the transpose of A. Conventionally, the zeroth power of the arbitrary matrix A is the identity matrix of the corresponding order. The (n + 1) × (n + 1) identity matrix is denoted by I. Write Dδ := diag(1, δ, δ 2 , . . . , δ n ) for a diagonal matrix whose diagonal entries starting in the upper left corner are 1, δ, δ 2 , . . . , δ n . Let F denote a square forward shift matrix of size n + 1, with ones only on the first sub-diagonal and zeros elsewhere, then B := F T is a backward shift matrix, i.e., ⎞ ⎞ ⎛ ⎛ 0 1 0 ⎟ ⎟ ⎜ ⎜ .. .. ⎟ ⎟ ⎜ ⎜1 . . . . . ⎟ ⎟ ⎜ , B = . F =⎜ ⎟ ⎟ ⎜ ⎜ . . . .. .. . . 1⎠ ⎠ ⎝ ⎝ 0 (n+1)×(n+1) 1 0 (n+1)×(n+1)
0
0
0
0
Obviously, F k and B k both are zero matrices, if k ≥ n + 1. Matrix A is premultiplied by F corresponds to all elements of A are shifted one position downward, and the elements in the first row are all set to zero, such as ⎞ ⎛ ⎞ ⎛ ⎞⎛ 0 0 0 0 0 0 0 0 a0,0 a0,1 a0,2 a0,3 ⎜1 0 0 0⎟ ⎜a1,0 a1,1 a1,2 a1,3 ⎟ ⎜a0,0 a0,1 a0,2 a0,3 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎝0 1 0 0⎠ ⎝a2,0 a2,1 a2,2 a2,3 ⎠ = ⎝a1,0 a1,1 a1,2 a1,3 ⎠ . 0 0 1 0 a3,0 a3,1 a3,2 a3,3 a2,0 a2,1 a2,2 a2,3 Matrix A is post-multiplied by B is equivalent to all elements of A is shifted one position towards the right, and the elements in the first column are all set to zero,
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
such as ⎛
a0,0 ⎜a1,0 ⎜ ⎝a2,0 a3,0
a0,1 a1,1 a2,1 a3,1
a0,2 a1,2 a2,2 a3,2
⎞⎛ 0 a0,3 ⎜ a1,3 ⎟ ⎟ ⎜0 a2,3 ⎠ ⎝0 0 a3,3
1 0 0 0
⎞ ⎛ 0 0 a0,0 ⎜ 0⎟ ⎟ = ⎜0 a1,0 1⎠ ⎝0 a2,0 0 0 a3,0
0 1 0 0
a0,1 a1,1 a2,1 a3,1
3
⎞ a0,2 a1,2 ⎟ ⎟. a2,2 ⎠ a3,2
For a non-zero unknown α, the (n + 1) × (n + 1) generalized Pascal matrix Pα is determined by its construction rule: Pα (i, j) := Pα (i−1, j −1) + αPα (i−1, j), Pα (i, 0) := αi ,
Pα (0, j) := 0,
Thus (see [13]) Pα (i, j) :=
i
j
αi−j
0
for i, j ≥ 1,
(1.2)
for i ≥ 0, j ≥ 1. if i ≥ j, if i < j.
It is easy to see that when α = 1, P1 = P is a Pascal matrix. Inverse of the −1 T generalized Pascal matrix is Pα−1 = P(−α) , and PαT = P(−α) . Lemma 1.1. For the sum of the powers of the shift matrices, we have (i)
(ii)
n i=0 n
ρi F i = (I − ρF)−1 ,
σ j B j = (I − σB)−1 ,
(1.3)
iF i = F(I − F)−2 ,
jB j = B(I − B)−2 ,
(1.4)
j=0 n
i=1
(iii)
n
j=1
n i−1
ρk F i = F(I − F)−1 (I − ρF)−1 ,
(1.5)
σ k B j = B(I − B)−1 (I − σB)−1 .
(1.6)
i=1 k=0 j−1 n j=1 k=0
Proof. (i) We only prove the first equation, because transposing the first equation in (1.3) and replacing ρ with σ is the second one. n
ρi F i · (I − ρF) =
i=0
n
(ρF)i − (ρF)i+1 = I.
i=0
Note that the conclusion used here is that the zeroth power and the (n + 1)-th power of the (n + 1) × (n + 1) forward shift matrix F are the identity matrix and the zero matrix, respectively. (ii) Because n
iF i · (I − F)2
=
i=1
n
iF i − (i + 1)F i+1 + iF i+2 − (i − 1)F i+1
i=1
=
F − (n + 1)F n+1 + nF n+2 = F,
the first equation in (1.4) holds, and the second is the transpose of the first. (iii) Since (1.6) is obtained by transposing (1.5) and replacing ρ with σ, thus we only prove (1.5). Similar to the proofs of (i) and (ii), we can easily verify the
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
4
following identities. i−1 n
ρk F i (I − ρF)(I − F) =
n−1
i=1 k=0
=
=
n−1
=
ρk F i (I − F − ρF + ρF 2 )
k=0 i=k+1
ρk
F i − F i+1 + ρ F i+2 − F i+1
n
k=0
i=k+1
n−1
k=0
n
ρk F k+1 − F n+1 + ρ F n+2 − F k+2
n−1
ρk F k+1 − ρk+1 F k+2 = F.
k=0
Therefore, (1.5) holds. This completes the proof of this lemma. Lemma 1.2. For any complex numbers α, β, ρ, σ, λ, τ , one has (a) (b) (c)
FPα = (I − αF)Pα F,
or
Pα F = FPα (I + αF),
(1.7)
PβT B = BPβT (I − βB),
or
BPβT = (I + βB)PβT B;
(1.8)
(I − (1 + α)F)Pα F = FPα (I − F),
(1.9)
BPβT (I − (1 + β)B) = (I − B)PβT B;
(1.10)
(I − (1 + α)F)Pα (I + αF) = Pα (I − F),
(1.11)
(I + (d) (e)
B)PβT ;
(1.12) 2
(I − (1 + α)F) Pα (I + αF)F = FPα (I − F) ,
(1.13)
B(I + βB)Pβ (I − (1 + β)B)2 = (I − B)2 PβT B;
(1.14)
Pα−1 (I
(1.15)
2
PβT (I (f )
− (1 + β)B) = (I −
βB)PβT (I
− ρF)(I − αF)
− βB)
−1
(I −
−1
Pα = I + (α − ρ)F,
−1 σB)PβT
= I + (β − σ)B;
(I + αF)Pα−1 (I + λF)Pα = I + (α + λ)F, −1 PβT (I + τ B)PβT (I + βB) = I + (β + τ )B.
(1.16) (1.17) (1.18)
Proof. We only prove (a) and (b) because the proofs of (c)-(f ) are similar. First prove the first formula of (1.7). It is clear that the (i, j)-entry of (I − αF)Pα equals 0 if i < j, and equals 1 if i = j, and it is easily verified that ((I − αF)Pα )(i, 0) = 0, for i ≥ 1. Now suppose i > j. By the definition of the matrix F and the recurrence relation (1.2) we get that for i > j ≥ 1 ((I − αF)Pα )(i, j) = Pα (i, j) − αPα (i − 1, j) = Pα (i − 1, j − 1). Further
((I − αF)Pα F)(i, j)
= =
0 Pα (i − 1, j)
FPα (i, j).
if i = 0, if i ≥ 1.
which leads to (I − αF)Pα F = FPα , and this equation can also be rewritten as Pα F = FPα (I + αF). This completes the proof of (1.7). Then (1.8) is obtained by transposing the matrix equations in (1.7) and replacing α with β. Next prove (b). By (1.7), we have (I − (1 + α)F)Pα F
= Pα F − (1 + α)FPα F = FPα (I + αF) − (1 + α)FPα F = FPα (I − F).
This is (1.9). Transpose this matrix equation and replace α with β, then (1.10) is obtained. This completes the proof of (a) and (b).
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
5
Lemma 1.3 (Determinant of the tridiagonal matrix). Let x z . .. y x , Jn := Jn (x, y, z) := . . . . . . z y x
0
0
(1.19)
n×n
then it satisfies the second-order linear recurrence relation Jn = xJn−1 − yzJn−2 ,
n≥2
(1.20)
and the initial values J0 = 1, J1 = x. Its explicit expressions are as follows. x n . (1.21) (i) If x2 − 4yz = 0, then Jn = (n + 1) 2 (ii) If x2 − 4yz = 0, then Jn = where λ1,2 =
1 2
x±
x2 − 4yz . In particular,
λn+1 − λn+1 1 2 λ1 − λ 2
(1.22)
1) when x = 1 and yz = −1, we have Jn = Fn+1 (Fibonacci numbers). 2) When x = 0, we get Jn = −yzJn−2 . So if n is an odd number, then Jn = 0. If n is an even number, then Jn = (−yz)n/2 .
2. LDU-like-factorization of Pascal-like Matrix Let A(u, v) = i,j≥0 ai,j ui v j be the bivariate generating function of the sequence {ai,j }i,j≥0 . If ai,j satisfy the recurrence relation (1.1), then we have ai,0 ui + a0,j v j + ai,j ui v j A(u, v) = −a0,0 + i≥0
j≥0
= −a0,0 + A(u, 0) + A(0, v) +
i,j≥1
(αai−1,j + βai,j−1 + γai−1,j−1 )ui v j
i,j≥1
= −a0,0 + A(u, 0) + A(0, v) + αu · A(u, v) − A(u, 0) + βv · A(u, v) − A(0, v) + γuv · A(u, v) where A(u, 0) = i≥0 ai,0 ui , A(0, v) = j≥0 a0,j v j . Thus (1 − αu − βv − γuv)A(u, v) = (1 − αu)A(u, 0) + (1 − βv)A(0, v) − a0,0 or A(u, v) =
(1 − αu)A(u, 0) + (1 − βv)A(0, v) − a0,0 . 1 − αu − βv − γuv
(2.1)
Next, expand 1/(1 − αu − βv − γuv) into a power series of u and v: 1 1 1 = · 1 − αu − βv − γuv (1 − αu)(1 − βv) 1 − (αβ+γ)uv (1−αu)(1−βv) (αβ + γ)k uk v k = , (1 − αu)k+1 (1 − βv)k+1 k≥0
where the last equality follows from the power series expansion Then use the binomial power series expansion i uk = αi−k ui , (1 − αu)k+1 k i≥k
1 1−x
=
i≥0
xi .
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
we get
6
i j 1 k = αi−k β j−k ui v j (αβ + γ) 1 − αu − βv − γuv k k k≥0
=
i
i,j≥k
i j i−k j−k α β (αβ + γ)k ui v j . k k
i,j≥0 k=0
For the sake of simplicity, we set Φ(i, j) :=
i i j i−k j−k α β (αβ + γ)k . k k
(2.2)
k=0
It is obviously that the LDU-factorization of the matrix Φ = (Φ(i, j))0≤i,j≤n is Φ = Pα Dδ PβT
(2.3)
where δ := αβ + γ. In addition, we also have
(1 − αu)A(u, 0) = (1 − αu) ai,0 − αai−1,0 ui , ai,0 ui = i≥0
(1 − βv)A(0, v) = (1 − βv)
i≥0
j
a0,j v =
j≥0
a0,j − αa0,j−1 v j .
j≥0
For convenience, we write a−1,0 = a0,−1 = 0. Therefore, (2.1) can be rewritten as A(u, v) =
i
ai−k,0 − αai−k−1,0 Φ(k, j)ui v j
i,j≥0 k=0
+
j
Φ(i, k) a0,j−k − βa0,j−k−1 ui v j − a0,0 Φ(i, j)ui v j .
i,j≥0 k=0
i,j≥0 i j
By extracting the coefficients of u v from the bivariate power series expansions, we obtain the following expression of ai,j with initial values ai,0 and a0,j : ai,j =
i
j
ai−k,0 − αai−k−1,0 Φ(k, j) + Φ(i, k) a0,j−k − βa0,j−k−1 − a0,0 Φ(i, j).
k=0
k=0
(2.4) This equation is expressed in the form of matrices, which is Theorem 2.1. The (n + 1) × (n + 1) matrix A = (ai,j )0≤i,j≤n , whose entries are given by the recurrence relation (1.1), can be decomposed into
A = I − αF C0 Φ + ΦR0 I − βB − a0,0 Φ
= I − αF C0 Pα Dδ PβT + Pα Dδ PβT R0 I − βB − a0,0 Pα Dδ PβT (2.5) n n i j where δ := αβ + γ, C0 := i=0 ai,0 F , R0 := j=0 a0,j B . This expression for A is called the LDU-like-factorization of A. 3. Determinants of Pascal-like matrices 3.1. Integer sequences as the initial values. Take ai,0 = i, a0,j = −j, i, j ≥ 0, then we have by (1.4) C0 =
n i=0
iF i = F(I − F)−2 ,
R0 = −
n
jB j = −B(I − B)−2 .
j=0
By (2.5), the matrix A can be rewritten as
A = I − αF F(I − F)−2 Pα Dδ PβT − Pα Dδ PβT B(I − B)−2 I − βB .
(3.1)
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
For (3.1) pre-multiply (I − (1 − α)F)2 Pα−1 and post-multiply PβT then there is
−1
7
(I − (1 − β)B)2 ,
−1
(I − (1 − α)F)2 Pα−1 · A · PβT (I − (1 − β)B)2
= (I − (1 − α)F)2 Pα−1 · I − αF F(I − F)−2 Pα Dδ (I − (1 − β)B)2
−1 − (I − (1 − α)F)2 Dδ PβT (I − B)−2 B I − βB · PβT (I − (1 − β)B)2 . By (1.13) and (1.14), this equation can be simplified as (I − (1 − α)F)2 Pα−1 · A · PβT
−1
(I − (1 − β)B)2
= FDδ (I − (1 − β)B)2 − (I − (1 − α)F)2 Dδ B = FDδ − Dδ B + 2(β − α)FDδ B + (1 − β)2 FDδ B 2 − (1 − α)2 F 2 Dδ B ⎞ ⎛ 0 −1 ⎟ ⎜1 2(β − α) (1 − β)2 − δ ⎟ ⎜ ⎟ ⎜ . .. ⎟ ⎜ δ − (1 − α)2 2(β − α)δ =⎜ ⎟. ⎟ ⎜ . . .. .. ⎜ 2 n−2 ⎟ ((1 − β) − δ)δ ⎠ ⎝ (δ − (1 − α)2 )δ n−2 2(β − α)δ n−1
0
0
Therefore,
2(β − α)δ ((1 − β)2 − δ)δ (δ − (1 − α)2 )δ 2(β − α)δ 2 det A = .. .
0
..
.
..
.
(δ−(1−α)2 )δ n−2
2 n−2 ((1 − β) − δ)δ 2(β − α)δ n−1
0
n = δ ( 2 ) Jn−1 2(β − α), 1 − (1 − α)2 /δ, (1 − β)2 − δ . Then, by Lemma 1.3, we obtain Theorem 3.1. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values ai,0 = i, a0,j = −j, i, j ≥ 0. Then
n (3.2) det (ai,j ) = δ ( 2 ) Jn−1 2(β − α), 1 − (1 − α)2 /δ, (1 − β)2 − δ . 0≤i,j≤n
In particular, when α = β, if n is an even number, then det0≤i,j≤n (ai,j ) = 0; if n is an odd number and n = 2k + 1, then det
0≤i,j≤2k+1
2
(ai,j ) = (2α + γ − 1)2k (α2 + γ)2k .
(3.3)
Furthermore, 1)
If
2)
If
3)
If
4)
If
5)
If
6)
If
5 (n2 ) 1 3 , γ = − , then det (ai,j ) = Fn , 0≤i,j≤n 2 4 4 5 (n2 ) 1 3 α = , β = 1, γ = , then det (ai,j ) = Fn , 0≤i,j≤n 2 4 4 n 3 det (ai,j ) = 5( 2 ) Fn , α = −2, β = − , γ = 2, then 0≤i,j≤n 2 n 7 det (ai,j ) = 5( 2 ) Fn , α = , β = 4, γ = −9, then 0≤i,j≤n 2 5 (n2 ) 1 3 1 det (ai,j ) = α = − , β = , γ = , then Fn , 0≤i,j≤n 4 4 8 16 5 (n2 ) 9 9 7 det (ai,j ) = α = , β = , γ = − , then Fn . 0≤i,j≤n 4 4 8 16
α = 1, β =
Note that when α = β = 1, (3.3) returns to Theorem 3 of Krattenthaler [3].
(3.4) (3.5) (3.6) (3.7) (3.8) (3.9)
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
8
3.2. The geometric sequences as initial values. Take ai,0 = bρi , a0,j = dσ j , i ≥ 0, j ≥ 1, then we have by (1.3) C0 = b
n
ρi F i = b(I − ρF)−1 ,
i=0
R0 = bI + d
n
σ j B j = (b − d)I + d(I − σB)−1 .
j=1
By (2.5), the matrix A can be rewritten as
A = b I − αF (I − ρF)−1 Pα Dδ PβT − dPα Dδ PβT
+ dPα Dδ PβT (I − σB)−1 I − βB − (b − d)βPα Dδ PβT B.
(3.10)
−1 −1 (I − σB)PβT Pre-multiply Pα−1 (I − ρF)(I − αF)−1 and post-multiply I − βB for (3.10), we have
−1 −1 (I − σB)PβT Pα−1 (I − ρF)(I − αF)−1 · A · I − βB
−1 −1 = bDδ PβT · I − βB (I − σB)PβT + dPα−1 (I − ρF)(I − αF)−1 · Pα Dδ
−1 −1 − dPα−1 (I − ρF)(I − αF)−1 · Pα Dδ PβT · I − βB (I − σB)PβT
−1 −1 − (b − d)βPα−1 (I − ρF)(I − αF)−1 · Pα Dδ PβT B · I − βB (I − σB)PβT . By (1.15) and (1.16), the equation can be simplified as
−1 −1 (I − σB)PβT Pα−1 (I − ρF)(I − αF)−1 · A · I − βB = bDδ (I + (β − σ)B) + d(I + (α − ρ)F)Dδ (I + (β − σ)B) − (b − d)β(I + (α − ρ)F)Dδ PβT BPβT
−1
(I + (β − σ)B)
− dβ(I + (α − ρ)F)Dδ (I + (β − σ)B) = bDδ − d(α − ρ)(β − σ)FDδ B − (b − d)β(α − ρ)FDδ B + · · ·
upper triangular matrices = bDδ − (α − ρ)(bβ − dσ)FDδ B + , with 0’s on their diagonals
(3.11)
where the second equality has used the following derivation: PβT BPβT
−1
by(1.8)
====== PβT · (I − βB)PβT
−1
B = B − βPβT B · PβT
−1
B.
(3.12)
The entries lying on the principal diagonal of the matrix in the right-hand side of the last equation in (3.11) are b, bδ − (α − ρ)(bβ − dσ), (bδ − (α − ρ)(bβ − dσ))δ, . . . , (bδ − (α − ρ)(bβ − dσ))δ n−1 . Thus
n n det A = b bδ − (α − ρ)(bβ − dσ) δ ( 2 ) . So we get Theorem 3.2 (Wang [11, Theorem 6]). Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values ai,0 = bρi , a0,j = dσ j , i ≥ 0, j ≥ 1. Then n
det (ai,j ) = b(bγ + bβρ + dασ − dρσ)n (αβ + γ)( 2 ) .
0≤i,j≤n
(3.13)
When α = β = 1 and b = d = 1, (3.13) returns to Theorem 1 of Krattenthaler [3] (see also Bacher [1, Theorem 1.5]).
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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3.3. Another geometric sequences as initial values. Take a0,0 = 0, ai,0 = bρi−1 , a0,j = dσ j−1 , i, j ≥ 1, then we have by (1.3) C0 = b
n
ρi−1 F i = bF(I − ρF)−1 ,
R0 = d
i=1
n
σ j−1 B j = dB(I − σB)−1 .
j=1
By (2.5), the matrix A can be rewritten as
A = b I − αF F(I − ρF)−1 Pα Dδ PβT + dPα Dδ PβT B(I − σB)−1 I − βB . (3.14) Using (1.15) and (1.16), for (3.14) pre-multiply Pα−1 (I − ρF)(I − αF)−1 and post
−1 −1 (I − σB)PβT , we have multiply I − βB
−1 −1 (I − σB)PβT Pα−1 (I − ρF)(I − αF)−1 · A · I − βB = bPα−1 FPα Dδ (I + (β − σ)B) + d(I + (α − ρ)F)Dδ PβT BPβT
−1
.
Again pre-multiply (I + αF) and post-multiply (I + βB) for the equation, by using (1.7) and (1.8), we get
−1 −1 (I − σB)PβT (I + βB) (I + αF)Pα−1 (I − ρF)(I − αF)−1 · A · I − βB = bFDδ (I + (β − σ)B)(I + βB) + d(I + αF)(I + (α − ρ)F)Dδ B = bFDδ + dDδ B + b(2β − σ) + d(2α − ρ) FDδ B + dα(α − ρ)F 2 Dδ B + bβ(β − σ)FDδ B 2 ⎛ 0 d ⎜b ε dδ + bβ(β − σ) ⎜ ⎜ .. ⎜ . εδ = ⎜ bδ + dα(α − ρ) ⎜ . . .. .. ⎜ ⎝
0
[bδ+dα(α−ρ)]δ n−2
⎞
0
⎟ ⎟ ⎟ ⎟ ⎟ ⎟ n−2 ⎟ [dδ+bβ(β−σ)]δ ⎠ εδ n−1
where ε := b(2β − σ) + d(2α − ρ). Then 0 d b ε dδ+bβ(β−σ) . . bδ+dα(α−ρ) . εδ det A = .. .. n−2 . . [dδ+bβ(β−σ)]δ [bδ+dα(α−ρ)]δ n−2 εδ n−1 [dδ+bβ(β−σ)]δ εδ .. 2 [bδ+dα(α−ρ)]δ . εδ = − bd . . .. .. n−2 [dδ+bβ(β−σ)]δ n−1 n−2 [bδ+dα(α−ρ)]δ εδ
0
0
0
0
n = − bdδ ( 2 ) Jn−1 ε, b + dα(α − ρ)/δ, dδ + bβ(β − σ) . Therefore, by Lemma 1.3, we obtain Theorem 3.3. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values a0,0 = 0, ai,0 = bρi−1 , a0,j = dσ j−1 , i, j ≥ 1. Then
n det (ai,j ) = −bdδ ( 2 ) Jn−1 ε, b + dα(α − ρ)/δ, dδ + bβ(β − σ) (3.15) 0≤i,j≤n
where ε = b(2β − σ) + d(2α − ρ). In particular, 1) when α = β, ρ = σ and d = −b, then ε = 0 and the elements on the principal diagonal are all zero. At this time, if n is an even number, then det0≤i,j≤n (ai,j ) = 0;
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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if n is an odd number and n = 2k + 1, then 2
det
0≤i,j≤2k+1
(ai,j ) = b2k+2 (αρ + γ)2k (α2 + γ)2k .
(3.16)
2) When β = −α, σ = −ρ and d = b, then ε = 0 and the elements on the principal diagonal are all zero. At this time, if n is an even number, then det0≤i,j≤n (ai,j ) = 0; if n is an odd number and n = 2k + 1, then 2
det
0≤i,j≤2k+1
(ai,j ) = (−1)k+1 b2k+2 (γ − αρ)2k (γ − α2 )2k .
(3.17)
Note that when α = 1 and b = 1, (3.16) becomes Theorem 2 of Krattenthaler [3]. 3.4. The difference between two geometric sequences as the initial values. Let ai,0 = b(αi − ρi ), a0,j = d(β j − σ j ), i, j ≥ 0, then we have by (1.3) C0 = b
n
(αi − ρi )F i = b(I − αF)−1 − b(I − ρF)−1 ,
i=0 n
R0 = d
(β j − σ j )B j = d(I − βB)−1 − d(I − σB)−1 .
j=0
By (2.5), the matrix A can be rewritten as
A = b I − αF (I − αF)−1 − (I − ρF)−1 Φ
+ dΦ (I − βB)−1 − (I − σB)−1 I − βB
= (b + d)Φ − b I − αF (I − ρF)−1 Φ − dΦ(I − σB)−1 I − βB . (3.18)
−1 −1 Pre-multiply Pα−1 (I − ρF)(I − αF)−1 and post-multiply I − βB (I − σB)PβT for (3.18), by using (1.15) and (1.16), we have
−1 −1 (I − σB)PβT Pα−1 (I − ρF)(I − αF)−1 · A · I − βB = (b + d)(I + (α − ρ)F)Dδ (I + (β − σ)B) − bDδ (I + (β − σ)B) − d(I + (α − ρ)F)Dδ = b(α − ρ)FDδ + d(β − σ)Dδ B + (b + d)(α − ρ)(β − σ)FDδ B ⎛ ⎞ 0 d(β − σ) ⎜b(α − ρ) ⎟ η d(β − σ)δ ⎜ ⎟ ⎜ ⎟ . . ⎜ ⎟ . b(α − ρ)δ ηδ =⎜ ⎟ ⎜ ⎟ .. .. ⎜ n−1 ⎟ . . d(β − σ)δ ⎝ ⎠ b(α − ρ)δ n−1 ηδ n−1
0
0
where η := (b + d)(α − ρ)(β − σ). Then 0 d(β − σ) b(α − ρ) η d(β − σ)δ . .. b(α − ρ)δ ηδ det A = .. .. n−1 . . d(β − σ)δ b(α − ρ)δ n−1 ηδ n−1 ηδ d(β − σ)δ 2 . . 2 b(α − ρ)δ 2 . ηδ = −bd(α − ρ)(β − σ) . . .. .. n−1 d(β − σ)δ n−1 n−1 b(α − ρ)δ ηδ
0
0
0
0
= − bd(α − ρ)(β − σ)δ ( ) Jn−1 η, b(α − ρ), d(β − σ)δ . n 2
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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Therefore, by Lemma 1.3, we obtain Theorem 3.4. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values ai,0 = b(αi − ρi ), a0,j = d(β j − σ j ), i, j ≥ 0. Then
n det (ai,j ) = −bd(α − ρ)(β − σ)δ ( 2 ) Jn−1 η, b(α − ρ), d(β − σ)δ (3.19) 0≤i,j≤n
where η = (b + d)(α − ρ)(β − σ). In particular, when d = −b, then η = 0 and the elements on the principal diagonal are all zero. At this time, if n is an even number, then det0≤i,j≤n (ai,j ) = 0; if n is an odd number and n = 2k + 1, then det
0≤i,j≤2k+1
(ai,j ) = b2(k+1) (α − ρ)k+1 (β − σ)k+1 (αβ + γ)2k(k+1) .
(3.20)
3.5. The arithmetic and geometric sequences as the initial values. Let ai,0 = bρi , a0,j = c + dj, i ≥ 0, j ≥ 1, then we have by (1.3) and (1.4) C0 = b
n
ρi F i = b(I − ρF)−1 ,
i=0
R0 = bI +
n (c + dj)B j = (b − c)I + c(I − B)−1 + dB(I − B)−2 . j=1
By (2.5), the matrix A can be rewritten as
A = b I − αF (I − ρF)−1 Φ − bΦ
+ Φ (b − c)I + c(I − B)−1 + dB(I − B)−2 I − βB
= b I − αF (I − ρF)−1 Φ + cΦ(I − B)−1 (I − βB) + dΦB(I − B)−2 (I − βB) − cΦ − (b − c)βΦB. Pα−1 (I
−1
Pre-multiply − ρF)(I − αF) and post-multiply (3.21), by using (1.15), (1.14) and (3.12), we have Pα−1 (I − ρF)(I − αF)−1 · A · PβT
−1
−1 PβT (I
(3.21) − (1 − β)B)2 for
(I − (1 − β)B)2
= bDδ (I − (1 − β)B)2 + c(I + (α − ρ)F)Dδ (I − (1 − β)B) + d(I + (α − ρ)F)Dδ B − c(I + (α − ρ)F)Dδ (I − (1 − β)B)2 −1
− (b − c)β(I + (α − ρ)F)Dδ · PβT BPβT · (I − (1 − β)B)2 = bDδ + d + c(1 − β) − (b − c)β (α − ρ)FDδ B + upper triangular matrices with 0’s on their principal diagonals. To the right of the last equal sign is an upper triangular matrix whose elements on the principal diagonal are b, bδ + (c + d − bβ)(α − ρ), bδ + (c + d − bβ)(α − ρ) δ, · · · , bδ + (c + d − bβ)(α − ρ) δ n−1 . Thus n n det A = b bδ + (c + d − bβ)(α − ρ) δ ( 2 ) . So we get Theorem 3.5. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values ai,0 = bρi , a0,j = c + dj, i ≥ 0, j ≥ 1. Then n n det (ai,j ) = b (c + d)(α − ρ) + b(βρ + γ) (αβ + γ)( 2 ) . (3.22) 0≤i,j≤n
3.6. Only two non-zero initial values. Take a0,0 = 0, a1,0 = b, a0,1 = d, ai,0 = a0,j = 0, i, j ≥ 2, then C0 =
n i=0
ai,0 F i = bF,
R0 =
n j=0
a0,j B j = dB.
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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By (2.5), the matrix A can be rewritten as
(3.23) A = b I − αF FPα Dδ PβT + dPα Dδ PβT B I − βB .
−1 T −1 Pre-multiply (I + αF)Pα−1 (I − αF)−1 and post-multiply I − βB Pβ (I + βB) for (3.23), using (1.7), (1.8), (1.15) in the case ρ = 0, and (1.16) in the case σ = 0, we have
−1 T −1 (I + αF)Pα−1 (I − αF)−1 · A · I − βB Pβ (I + βB) = bFDδ (I + βB)2 + d(I + αF)2 Dδ B = bFDδ + dDδ B + 2(bβ + dα)FDδ B + bβ 2 FDδ B 2 + dα2 F 2 Dδ B ⎞ ⎛ 0 d ⎟ ⎜ b 2(bβ + dα) dδ + bβ 2 ⎟ ⎜ ⎟ ⎜ .. 2 ⎟ ⎜ . bδ + dα 2(bβ + dα)δ =⎜ ⎟. ⎟ ⎜ . . .. .. ⎜ 2 n−2 ⎟ (dδ + bβ )δ ⎠ ⎝ 2 n−2 n−1 (bδ + dα )δ 2(bβ + dα)δ
0
0
Then
0 b det A =
d 2(bβ + dα)
0
dδ + bβ 2
bδ + dα2 2(bβ + dα)δ .. .
0
2 n−2 (dδ + bβ )δ 2(bβ + dα)δ n−1
..
.
..
.
(bδ + dα2 )δ n−2
2(bβ + dα)δ (dδ + bβ 2 )δ . .. (bδ + dα2 )δ 2(bβ + dα)δ 2 = −bd . . .. .. 2 n−2 (dδ + bβ )δ 2 n−2 n−1 (bδ + dα )δ 2(bβ + dα)δ
0
0
n = − bdδ ( 2 ) Jn−1 2(bβ + dα), b + dα2 /δ, dδ + bβ 2 . Therefore, by Lemma 1.3, we obtain Theorem 3.6. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values a0,0 = 0, a1,0 = b, a0,1 = d, ai,0 = a0,j = 0, i, j ≥ 2. Then
n (3.24) det (ai,j ) = −bdδ ( 2 ) Jn−1 2(bβ + dα), b + dα2 /δ, dδ + bβ 2 . 0≤i,j≤n
In particular, when α = β and d = −b, then 2(bβ + dα) = 0. At this time, if n is an even number, then det0≤i,j≤n (ai,j ) = 0; if n is an odd number and n = 2k + 1, then 2 det (ai,j ) = b2k+2 γ 2k (α2 + γ)2k . (3.25) 0≤i,j≤2k+1
3.7. Only three non-zero initial values. Let a0,0 = 1, a1,0 = α, a0,1 = −β, ai,0 = a0,j = 0, i, j ≥ 2, then C0 =
n i=0
ai,0 F i = I + αF,
R0 =
n
a0,j B j = I − βB.
j=0
By (2.5), the matrix A can be rewritten as
2 A = I − αF I + αF Pα Dδ PβT + Pα Dδ PβT I − βB − Pα Dδ PβT .
(3.26)
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
−1
13 −1
Pre-multiply (I + αF)Pα−1 (I − αF)−1 and post-multiply I − βB PβT (I + βB) for (3.26), using (1.15) in the case ρ = 0, (1.16) in the case σ = 0, (1.17) in the case λ = α, and (1.18) in the case τ = −β, we have
−1 T −1 Pβ (I + βB) (I + αF)Pα−1 (I − αF)−1 · A · I − βB
2
= I + 2αF Dδ (I + βB)2 + (I + αF)2 Dδ − I + αF Dδ (I + βB)2 = Dδ + 2αFDδ − α2 β 2 F 2 Dδ B 2 − 2α2 βF 2 Dδ B ⎛ 1 δ ⎜2α ⎜ ⎜ 2α(δ − αβ) δ 2 − α2 β 2 =⎜ .. .. ⎜ . . ⎝
0
2α(δ − αβ)δ n−2
0
⎞ ⎟ ⎟ ⎟ ⎟. ⎟ ⎠
(δ 2 − α2 β 2 )δ n−2
Then
n−1 2
det A = δ (
)+1 (δ 2 − α2 β 2 )n−1 .
Therefore, we obtain Theorem 3.7. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values a0,0 = 1, a1,0 = α, a0,1 = −β, ai,0 = a0,j = 0, i, j ≥ 2. Then
n−1
(n−1)+1 . αβ + γ 2 det (ai,j ) = γ n−1 2αβ + γ
0≤i,j≤n
(3.27)
3.8. First-order geometric sequences as initial values. Let a0,0 = 0, ai,0 = i−1 j−1 ρai−1,0 + b, a0,j = σa0,j−1 + d, i, j ≥ 1, then ai,0 = b k=0 ρk , a0,j = d k=0 σ k . By (1.5) and (1.6), we get
C0 = b
n i−1
ρk F i = bF(I − F)−1 (I − ρF)−1 ,
i=1 k=0
R0 = d
j−1 n
σ k B j = dB(I − B)−1 (I − σB)−1 .
j=1 k=0
By (2.5), the matrix A can be rewritten as
A = b I − αF F(I − F)−1 (I − ρF)−1 Φ + dΦB(I − B)−1 (I − σB)−1 I − βB . (3.28)
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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For (3.28) pre-multiply I − (1 − α)F Pα−1 (I − ρF)(I − αF)−1 and post-multiply
−1
−1 I − βB I − (1 − β)B , by using (1.9) and (1.10), we have (I − σB)PβT
I − (1 − α)F Pα−1 (I − ρF)(I − αF)−1 · A
−1
−1 · I − βB I − (1 − β)B (I − σB)PβT
= b I − (1 − α)F Pα−1 (I − F)−1 FPα Dδ (I + (β − σ)B) I − (1 − β)B
−1 + d I − (1 − α)F I + (α − ρ)F Dδ PβT B(I − B)−1 PβT I − (1 − β)B
= bFDδ I + (2β − σ − 1)B − (1 − β)(β − σ)B 2
+ d I + (2α − ρ − 1)F − (1 − α)(α − ρ)F 2 Dδ B = bFDδ + dDδ B − b(1 − β)(β − σ)FDδ B 2 − d(1 − α)(α − ρ)F 2 Dδ B
+ b(2β − σ − 1) + d(2α − ρ − 1) FDδ B ⎞ ⎛ 0 d ⎟ ⎜b ω dδ − bν ⎟ ⎜ ⎟ ⎜ .. ⎟ ⎜ bδ − dμ . ωδ =⎜ ⎟ ⎟ ⎜ . . .. .. ⎜ n−2 ⎟ (dδ − bν)δ ⎠ ⎝ (bδ − dμ)δ n−2 ωδ n−1
0
0
where ω := b(2β − σ − 1) + d(2α − ρ − 1), μ := (1 − α)(α − ρ), ν := (1 − β)(β − σ). Then 0 d b ω dδ − bν . . . bδ − dμ ωδ det A = . . .. .. n−2 (dδ − bν)δ n−2 n−1 (bδ − dμ)δ ωδ ωδ (dδ − bν)δ . . 2 (bδ − dμ)δ . ωδ = −bd .. .. n−2 . . (dδ − bν)δ (bδ − dμ)δ n−2 ωδ n−1
0
0
0
0
n = −bdδ ( 2 ) Jn−1 ω, b − dμ/δ, dδ − bν . Therefore, by Lemma 1.3, we obtain Theorem 3.8. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values a0,0 = 0, ai,0 = ρai−1,0 + b, a0,j = σa0,j−1 + d, i, j ≥ 1. Then
n (3.29) det (ai,j ) = −bdδ ( 2 ) Jn−1 ω, b − dμ/δ, dδ − bν 0≤i,j≤n
where ω = b(2β − σ − 1) + d(2α − ρ − 1), μ = (1 − α)(α − ρ), ν = (1 − β)(β − σ). In particular, when α = β, ρ = σ and d = −b, then ω = 0 and the elements on the principal diagonal are all zero. At this time, if n is an even number, then det0≤i,j≤n (ai,j ) = 0; if n is an odd number and n = 2k + 1, then det
0≤i,j≤2k+1
2
(ai,j ) = b2k+2 (α + γ − ρ + αρ)2k (α2 + γ)2k .
(3.30)
4. Conclusions We have already seen that the initial values play an important role in calculating the determinants of the Pascal-like matrices. Changing the initial values may affect the
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
15
process and results of the determinant calculation. Eight different cases of initial values are considered here, and the values of the determinants of their corresponding Pascal-like matrices have been derived. Interested readers can choose other initial values to calculate the determinant of such Pascal-like matrices. Acknowledgments: The authors thank the referee for the careful reading and useful suggestions that led to an improved version of the manuscript. References [1] R. Bacher, Determinants of matrices related to the pascal triangle, Journal de Th´eorie des Nombres de Bordeaux 14 (2002), no. 1, 19-41. [2] C. Krattenthaler, Advanced determinant calculus, S´ em. Lothar. Combin., 42 (1999), pp. Art. B42q, 67 pp. (electronic). [3] C. Krattenthaler, Evaluations Of Some Determinants Of Matrices Related To The Pascal Triangle, S´ eminaire Lotharingien de Combinatoire, 1 (2002), pp. 1-19. [4] C. Krattenthaler, Advanced determinant calculus: A complement, Linear Algebra and its Applications, 411 (2005), pp. 68-166. [5] A. R. Moghaddamfar, S. Navid Salehy and S. Nima Salehy, The determinants of matrices with recursive entries, Linear Algebra Appl., 428 (2008), no. 11-12, 2468-2481. [6] A. R. Moghaddamfar and S. M. H. Pooya, Generalized pascal triangles and toeplitz matrices, Electronic Journal of Linear Algebra 18 (2009), no. 1, 564-588. [7] A. R. Moghaddamfar, S. M. H. Pooya, S. Navid Salehy and S. Nima Salehy, More on evaluating determinants, Matematicki Vesnik 64 (2012), no. 3, 211-222. [8] A. R. Moghaddamfar, S. N. Salehy and S. N. Salehy, Determinant representations of sequences: A survey, Special Matrices 2 (2014), no. 1, 46-60. [9] E. Neuwirth, Treeway Galton arrays and generalized Pascal-like determinants, Technical Report, University ofVienna, (2004), pp. 1-16. [10] M. Tan, Matrices Associated to Biindexed Linear Recurrence Relations. Ars Combinatoria., 86(2008), 305-319. [11] X. Wang, Some determinants of Pascal-like matrices, Quaestiones Mathematicae (2012), pp. 37-41. [12] H. Zakraj˘sek and M. Petkov˘sek, Pascal-like determinants are recursive, Advances in Applied Mathematics, 33 (2004), pp. 431-450. [13] Z. Z. Zhang, The Linear Algebra of Generalized Pascal Matrix, Linear Algebra Appl. 250 (1997): 51-60.
De-Yin Zheng
PII: DOI: Reference:
S0024-3795(19)30181-8 https://doi.org/10.1016/j.laa.2019.04.023 LAA 14950
To appear in:
Linear Algebra and its Applications
Received date: Accepted date:
20 November 2016 22 April 2019
Please cite this article in press as: D.-Y. Zheng, Matrix Methods for determinants of Pascal-like matrices, Linear Algebra Appl. (2019), https://doi.org/10.1016/j.laa.2019.04.023
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MATRIX METHODS FOR DETERMINANTS OF PASCAL-LIKE MATRICES DE-YIN ZHENG Department of Mathematics, Hangzhou Normal University, Hangzhou 310018, P. R. China Email: [email protected]
Abstract. Using the LDU-like-factorization method, the evaluations of the determinants of some Pascal-like matrices are derived. The entries of these matrices satisfy the recurrence relation ai,j = αai−1,j +βai,j−1 +γai−1,j−1 , i, j ≥ 1, and the values of the first row a0,j and the column ai,0 have eight cases. Part of the results is due to C. Krattenthaler [Evaluations Of Some Determinants Of Matrices Related To The Pascal Triangle, S´eminaire Lotharingien de Combinatoire, 1 (2002): 1-19].
1. Introduction Pascal triangles or Pascal matrices are one of the most famous arithmetic triangles with many wonderful properties. However, the determinant of a Pascal-like matrix defined by a simple linear recursive relationship with constant coefficients has only been studied in detail recently, see [1–12]. Bacher [1] studies the determinant of the matrix whose entries satisfy the Pascal triangle recurrence relation ai,j = ai−1,j + ai,j−1 + αi βj or ai,j = ai−1,j + ai,j−1 + xai−1,j−1 with various choices for the first row a0,j and column ai,0 . He proved some results and proposed some conjectures about the determinant of such matrices. Some of these conjectures have been proved by Krattenthaler [3] through the “LUfactorization method”, which has been included in the review [2] on the “state of the art” of determinant calculations. Krattenthaler [3] proves and generalizes several evaluations of determinants conjectured in [1]. Specifically, evaluations are given in the cases when the entries of the matrix satisfy the recurrence relation ai,j = ai−1,j + ai,j−1 + xai−1,j−1 , i, j ≥ 1, and the initial values (i) ai,0 = ρi , a0,i = σ i , (ii) ak,k = 0, ai,0 = ρi−1 and a0,i = −ρi−1 for i ≥ 1, or (iii) ai,0 = i, a0,i = −i. Using the generating function method for a matrix product and the fact that the principal minors of every band-diagonal matrix satisfy a nontrivial recurrence relation, Zakraj˘sek and Petkov˘sek [12] studied the determinants of arrays with recurrence relation pi,j = λpi−1,j + μpi,j−1 + νpi−1,j−1 for i, j ≥ 1, and proved a generalization of Bacher’s conjecture [1, Conjecture 3.3] on principal minors of Pascal-like matrices. In 2008, Tan [10] gave the matrix product factorization of this Pascal-like matrix, and the value of the determinant of such a matrix whose initial values are two geometric sequences was gotten as a by-product. Furthermore, Date: April 25, 2019. 2000 Mathematics Subject Classification. Primary 05A19; Secondary 05A10; 05A15; 15A15; 11C20. Key words and phrases. determinant; Pascal matrix; LDU -factorization; tridiagonal matrix; bivariate recurrent sequence.
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
2
Neuwirth [9] and Moghaddamfar [5, 6] also considered the determinants of similar matrices whose entries satisfy linear recurrence relations with variable coefficients and generalized some of the results obtained. In addition, by finite differences, Wang [11] provides new proofs for two theorems due to Krattenthaler [3] and then proves a more general theorem that can be considered as a generalization of one determinant identity also due to Krattenthaler [3]. Here we study the generalized Pascal matrices (called Pascal-like matrices), whose entries are determined by the recursive relation ai,j = αai−1,j + βai,j−1 + γai−1,j−1 ,
i, j ≥ 1,
(1.1)
except for the first row and column. This recursive relation and initial values ai,0 and a0,j play an important role in constructing the matrix and calculating its determinant. Taking different values for the first row a0,j and column ai,0 will construct different matrices. In this paper, the determinants of these matrices will be calculated using the “LDU-like-factorization method”. The process is as follows: First, the Pascal-like matrix A is converted into a LDU-like matrix by generating function method, and then the matrix A is pre-multiplied by an appropriate lower triangular matrix, and is post-multiplied by an appropriate upper triangular matrix, so that A is converted into a diagonal matrix or a three-diagonal matrix. Further, the determinant of the matrix A can be calculated. The article is organized as follows: This section is preliminary. The “LDU-likefactorization” of the Pascal-like matrix is derived in Section 2. It will be used in Section 3 to calculate the determinants of Pascal-like matrices with eight kinds of initial values. They cover the part of the results of Krattenthaler [2] and Wang [11]. Finally, the paper is concluded in Section 4. To describe our results, we first give some basic notations and definitions, which will be required in the development of this paper. The Fibonacci sequence is a sequence Fn of natural numbers defined recursively: F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 , if n ≥ 2. Given a matrix A, we use the notation AT for the transpose of A. Conventionally, the zeroth power of the arbitrary matrix A is the identity matrix of the corresponding order. The (n + 1) × (n + 1) identity matrix is denoted by I. Write Dδ := diag(1, δ, δ 2 , . . . , δ n ) for a diagonal matrix whose diagonal entries starting in the upper left corner are 1, δ, δ 2 , . . . , δ n . Let F denote a square forward shift matrix of size n + 1, with ones only on the first sub-diagonal and zeros elsewhere, then B := F T is a backward shift matrix, i.e., ⎞ ⎞ ⎛ ⎛ 0 1 0 ⎟ ⎟ ⎜ ⎜ .. .. ⎟ ⎟ ⎜ ⎜1 . . . . . ⎟ ⎟ ⎜ , B = . F =⎜ ⎟ ⎟ ⎜ ⎜ . . . .. .. . . 1⎠ ⎠ ⎝ ⎝ 0 (n+1)×(n+1) 1 0 (n+1)×(n+1)
0
0
0
0
Obviously, F k and B k both are zero matrices, if k ≥ n + 1. Matrix A is premultiplied by F corresponds to all elements of A are shifted one position downward, and the elements in the first row are all set to zero, such as ⎞ ⎛ ⎞ ⎛ ⎞⎛ 0 0 0 0 0 0 0 0 a0,0 a0,1 a0,2 a0,3 ⎜1 0 0 0⎟ ⎜a1,0 a1,1 a1,2 a1,3 ⎟ ⎜a0,0 a0,1 a0,2 a0,3 ⎟ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎝0 1 0 0⎠ ⎝a2,0 a2,1 a2,2 a2,3 ⎠ = ⎝a1,0 a1,1 a1,2 a1,3 ⎠ . 0 0 1 0 a3,0 a3,1 a3,2 a3,3 a2,0 a2,1 a2,2 a2,3 Matrix A is post-multiplied by B is equivalent to all elements of A is shifted one position towards the right, and the elements in the first column are all set to zero,
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
such as ⎛
a0,0 ⎜a1,0 ⎜ ⎝a2,0 a3,0
a0,1 a1,1 a2,1 a3,1
a0,2 a1,2 a2,2 a3,2
⎞⎛ 0 a0,3 ⎜ a1,3 ⎟ ⎟ ⎜0 a2,3 ⎠ ⎝0 0 a3,3
1 0 0 0
⎞ ⎛ 0 0 a0,0 ⎜ 0⎟ ⎟ = ⎜0 a1,0 1⎠ ⎝0 a2,0 0 0 a3,0
0 1 0 0
a0,1 a1,1 a2,1 a3,1
3
⎞ a0,2 a1,2 ⎟ ⎟. a2,2 ⎠ a3,2
For a non-zero unknown α, the (n + 1) × (n + 1) generalized Pascal matrix Pα is determined by its construction rule: Pα (i, j) := Pα (i−1, j −1) + αPα (i−1, j), Pα (i, 0) := αi ,
Pα (0, j) := 0,
Thus (see [13]) Pα (i, j) :=
i
j
αi−j
0
for i, j ≥ 1,
(1.2)
for i ≥ 0, j ≥ 1. if i ≥ j, if i < j.
It is easy to see that when α = 1, P1 = P is a Pascal matrix. Inverse of the −1 T generalized Pascal matrix is Pα−1 = P(−α) , and PαT = P(−α) . Lemma 1.1. For the sum of the powers of the shift matrices, we have (i)
(ii)
n i=0 n
ρi F i = (I − ρF)−1 ,
σ j B j = (I − σB)−1 ,
(1.3)
iF i = F(I − F)−2 ,
jB j = B(I − B)−2 ,
(1.4)
j=0 n
i=1
(iii)
n
j=1
n i−1
ρk F i = F(I − F)−1 (I − ρF)−1 ,
(1.5)
σ k B j = B(I − B)−1 (I − σB)−1 .
(1.6)
i=1 k=0 j−1 n j=1 k=0
Proof. (i) We only prove the first equation, because transposing the first equation in (1.3) and replacing ρ with σ is the second one. n
ρi F i · (I − ρF) =
i=0
n
(ρF)i − (ρF)i+1 = I.
i=0
Note that the conclusion used here is that the zeroth power and the (n + 1)-th power of the (n + 1) × (n + 1) forward shift matrix F are the identity matrix and the zero matrix, respectively. (ii) Because n
iF i · (I − F)2
=
i=1
n
iF i − (i + 1)F i+1 + iF i+2 − (i − 1)F i+1
i=1
=
F − (n + 1)F n+1 + nF n+2 = F,
the first equation in (1.4) holds, and the second is the transpose of the first. (iii) Since (1.6) is obtained by transposing (1.5) and replacing ρ with σ, thus we only prove (1.5). Similar to the proofs of (i) and (ii), we can easily verify the
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
4
following identities. i−1 n
ρk F i (I − ρF)(I − F) =
n−1
i=1 k=0
=
=
n−1
=
ρk F i (I − F − ρF + ρF 2 )
k=0 i=k+1
ρk
F i − F i+1 + ρ F i+2 − F i+1
n
k=0
i=k+1
n−1
k=0
n
ρk F k+1 − F n+1 + ρ F n+2 − F k+2
n−1
ρk F k+1 − ρk+1 F k+2 = F.
k=0
Therefore, (1.5) holds. This completes the proof of this lemma. Lemma 1.2. For any complex numbers α, β, ρ, σ, λ, τ , one has (a) (b) (c)
FPα = (I − αF)Pα F,
or
Pα F = FPα (I + αF),
(1.7)
PβT B = BPβT (I − βB),
or
BPβT = (I + βB)PβT B;
(1.8)
(I − (1 + α)F)Pα F = FPα (I − F),
(1.9)
BPβT (I − (1 + β)B) = (I − B)PβT B;
(1.10)
(I − (1 + α)F)Pα (I + αF) = Pα (I − F),
(1.11)
(I + (d) (e)
B)PβT ;
(1.12) 2
(I − (1 + α)F) Pα (I + αF)F = FPα (I − F) ,
(1.13)
B(I + βB)Pβ (I − (1 + β)B)2 = (I − B)2 PβT B;
(1.14)
Pα−1 (I
(1.15)
2
PβT (I (f )
− (1 + β)B) = (I −
βB)PβT (I
− ρF)(I − αF)
− βB)
−1
(I −
−1
Pα = I + (α − ρ)F,
−1 σB)PβT
= I + (β − σ)B;
(I + αF)Pα−1 (I + λF)Pα = I + (α + λ)F, −1 PβT (I + τ B)PβT (I + βB) = I + (β + τ )B.
(1.16) (1.17) (1.18)
Proof. We only prove (a) and (b) because the proofs of (c)-(f ) are similar. First prove the first formula of (1.7). It is clear that the (i, j)-entry of (I − αF)Pα equals 0 if i < j, and equals 1 if i = j, and it is easily verified that ((I − αF)Pα )(i, 0) = 0, for i ≥ 1. Now suppose i > j. By the definition of the matrix F and the recurrence relation (1.2) we get that for i > j ≥ 1 ((I − αF)Pα )(i, j) = Pα (i, j) − αPα (i − 1, j) = Pα (i − 1, j − 1). Further
((I − αF)Pα F)(i, j)
= =
0 Pα (i − 1, j)
FPα (i, j).
if i = 0, if i ≥ 1.
which leads to (I − αF)Pα F = FPα , and this equation can also be rewritten as Pα F = FPα (I + αF). This completes the proof of (1.7). Then (1.8) is obtained by transposing the matrix equations in (1.7) and replacing α with β. Next prove (b). By (1.7), we have (I − (1 + α)F)Pα F
= Pα F − (1 + α)FPα F = FPα (I + αF) − (1 + α)FPα F = FPα (I − F).
This is (1.9). Transpose this matrix equation and replace α with β, then (1.10) is obtained. This completes the proof of (a) and (b).
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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Lemma 1.3 (Determinant of the tridiagonal matrix). Let x z . .. y x , Jn := Jn (x, y, z) := . . . . . . z y x
0
0
(1.19)
n×n
then it satisfies the second-order linear recurrence relation Jn = xJn−1 − yzJn−2 ,
n≥2
(1.20)
and the initial values J0 = 1, J1 = x. Its explicit expressions are as follows. x n . (1.21) (i) If x2 − 4yz = 0, then Jn = (n + 1) 2 (ii) If x2 − 4yz = 0, then Jn = where λ1,2 =
1 2
x±
x2 − 4yz . In particular,
λn+1 − λn+1 1 2 λ1 − λ 2
(1.22)
1) when x = 1 and yz = −1, we have Jn = Fn+1 (Fibonacci numbers). 2) When x = 0, we get Jn = −yzJn−2 . So if n is an odd number, then Jn = 0. If n is an even number, then Jn = (−yz)n/2 .
2. LDU-like-factorization of Pascal-like Matrix Let A(u, v) = i,j≥0 ai,j ui v j be the bivariate generating function of the sequence {ai,j }i,j≥0 . If ai,j satisfy the recurrence relation (1.1), then we have ai,0 ui + a0,j v j + ai,j ui v j A(u, v) = −a0,0 + i≥0
j≥0
= −a0,0 + A(u, 0) + A(0, v) +
i,j≥1
(αai−1,j + βai,j−1 + γai−1,j−1 )ui v j
i,j≥1
= −a0,0 + A(u, 0) + A(0, v) + αu · A(u, v) − A(u, 0) + βv · A(u, v) − A(0, v) + γuv · A(u, v) where A(u, 0) = i≥0 ai,0 ui , A(0, v) = j≥0 a0,j v j . Thus (1 − αu − βv − γuv)A(u, v) = (1 − αu)A(u, 0) + (1 − βv)A(0, v) − a0,0 or A(u, v) =
(1 − αu)A(u, 0) + (1 − βv)A(0, v) − a0,0 . 1 − αu − βv − γuv
(2.1)
Next, expand 1/(1 − αu − βv − γuv) into a power series of u and v: 1 1 1 = · 1 − αu − βv − γuv (1 − αu)(1 − βv) 1 − (αβ+γ)uv (1−αu)(1−βv) (αβ + γ)k uk v k = , (1 − αu)k+1 (1 − βv)k+1 k≥0
where the last equality follows from the power series expansion Then use the binomial power series expansion i uk = αi−k ui , (1 − αu)k+1 k i≥k
1 1−x
=
i≥0
xi .
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
we get
6
i j 1 k = αi−k β j−k ui v j (αβ + γ) 1 − αu − βv − γuv k k k≥0
=
i
i,j≥k
i j i−k j−k α β (αβ + γ)k ui v j . k k
i,j≥0 k=0
For the sake of simplicity, we set Φ(i, j) :=
i i j i−k j−k α β (αβ + γ)k . k k
(2.2)
k=0
It is obviously that the LDU-factorization of the matrix Φ = (Φ(i, j))0≤i,j≤n is Φ = Pα Dδ PβT
(2.3)
where δ := αβ + γ. In addition, we also have
(1 − αu)A(u, 0) = (1 − αu) ai,0 − αai−1,0 ui , ai,0 ui = i≥0
(1 − βv)A(0, v) = (1 − βv)
i≥0
j
a0,j v =
j≥0
a0,j − αa0,j−1 v j .
j≥0
For convenience, we write a−1,0 = a0,−1 = 0. Therefore, (2.1) can be rewritten as A(u, v) =
i
ai−k,0 − αai−k−1,0 Φ(k, j)ui v j
i,j≥0 k=0
+
j
Φ(i, k) a0,j−k − βa0,j−k−1 ui v j − a0,0 Φ(i, j)ui v j .
i,j≥0 k=0
i,j≥0 i j
By extracting the coefficients of u v from the bivariate power series expansions, we obtain the following expression of ai,j with initial values ai,0 and a0,j : ai,j =
i
j
ai−k,0 − αai−k−1,0 Φ(k, j) + Φ(i, k) a0,j−k − βa0,j−k−1 − a0,0 Φ(i, j).
k=0
k=0
(2.4) This equation is expressed in the form of matrices, which is Theorem 2.1. The (n + 1) × (n + 1) matrix A = (ai,j )0≤i,j≤n , whose entries are given by the recurrence relation (1.1), can be decomposed into
A = I − αF C0 Φ + ΦR0 I − βB − a0,0 Φ
= I − αF C0 Pα Dδ PβT + Pα Dδ PβT R0 I − βB − a0,0 Pα Dδ PβT (2.5) n n i j where δ := αβ + γ, C0 := i=0 ai,0 F , R0 := j=0 a0,j B . This expression for A is called the LDU-like-factorization of A. 3. Determinants of Pascal-like matrices 3.1. Integer sequences as the initial values. Take ai,0 = i, a0,j = −j, i, j ≥ 0, then we have by (1.4) C0 =
n i=0
iF i = F(I − F)−2 ,
R0 = −
n
jB j = −B(I − B)−2 .
j=0
By (2.5), the matrix A can be rewritten as
A = I − αF F(I − F)−2 Pα Dδ PβT − Pα Dδ PβT B(I − B)−2 I − βB .
(3.1)
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
For (3.1) pre-multiply (I − (1 − α)F)2 Pα−1 and post-multiply PβT then there is
−1
7
(I − (1 − β)B)2 ,
−1
(I − (1 − α)F)2 Pα−1 · A · PβT (I − (1 − β)B)2
= (I − (1 − α)F)2 Pα−1 · I − αF F(I − F)−2 Pα Dδ (I − (1 − β)B)2
−1 − (I − (1 − α)F)2 Dδ PβT (I − B)−2 B I − βB · PβT (I − (1 − β)B)2 . By (1.13) and (1.14), this equation can be simplified as (I − (1 − α)F)2 Pα−1 · A · PβT
−1
(I − (1 − β)B)2
= FDδ (I − (1 − β)B)2 − (I − (1 − α)F)2 Dδ B = FDδ − Dδ B + 2(β − α)FDδ B + (1 − β)2 FDδ B 2 − (1 − α)2 F 2 Dδ B ⎞ ⎛ 0 −1 ⎟ ⎜1 2(β − α) (1 − β)2 − δ ⎟ ⎜ ⎟ ⎜ . .. ⎟ ⎜ δ − (1 − α)2 2(β − α)δ =⎜ ⎟. ⎟ ⎜ . . .. .. ⎜ 2 n−2 ⎟ ((1 − β) − δ)δ ⎠ ⎝ (δ − (1 − α)2 )δ n−2 2(β − α)δ n−1
0
0
Therefore,
2(β − α)δ ((1 − β)2 − δ)δ (δ − (1 − α)2 )δ 2(β − α)δ 2 det A = .. .
0
..
.
..
.
(δ−(1−α)2 )δ n−2
2 n−2 ((1 − β) − δ)δ 2(β − α)δ n−1
0
n = δ ( 2 ) Jn−1 2(β − α), 1 − (1 − α)2 /δ, (1 − β)2 − δ . Then, by Lemma 1.3, we obtain Theorem 3.1. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values ai,0 = i, a0,j = −j, i, j ≥ 0. Then
n (3.2) det (ai,j ) = δ ( 2 ) Jn−1 2(β − α), 1 − (1 − α)2 /δ, (1 − β)2 − δ . 0≤i,j≤n
In particular, when α = β, if n is an even number, then det0≤i,j≤n (ai,j ) = 0; if n is an odd number and n = 2k + 1, then det
0≤i,j≤2k+1
2
(ai,j ) = (2α + γ − 1)2k (α2 + γ)2k .
(3.3)
Furthermore, 1)
If
2)
If
3)
If
4)
If
5)
If
6)
If
5 (n2 ) 1 3 , γ = − , then det (ai,j ) = Fn , 0≤i,j≤n 2 4 4 5 (n2 ) 1 3 α = , β = 1, γ = , then det (ai,j ) = Fn , 0≤i,j≤n 2 4 4 n 3 det (ai,j ) = 5( 2 ) Fn , α = −2, β = − , γ = 2, then 0≤i,j≤n 2 n 7 det (ai,j ) = 5( 2 ) Fn , α = , β = 4, γ = −9, then 0≤i,j≤n 2 5 (n2 ) 1 3 1 det (ai,j ) = α = − , β = , γ = , then Fn , 0≤i,j≤n 4 4 8 16 5 (n2 ) 9 9 7 det (ai,j ) = α = , β = , γ = − , then Fn . 0≤i,j≤n 4 4 8 16
α = 1, β =
Note that when α = β = 1, (3.3) returns to Theorem 3 of Krattenthaler [3].
(3.4) (3.5) (3.6) (3.7) (3.8) (3.9)
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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3.2. The geometric sequences as initial values. Take ai,0 = bρi , a0,j = dσ j , i ≥ 0, j ≥ 1, then we have by (1.3) C0 = b
n
ρi F i = b(I − ρF)−1 ,
i=0
R0 = bI + d
n
σ j B j = (b − d)I + d(I − σB)−1 .
j=1
By (2.5), the matrix A can be rewritten as
A = b I − αF (I − ρF)−1 Pα Dδ PβT − dPα Dδ PβT
+ dPα Dδ PβT (I − σB)−1 I − βB − (b − d)βPα Dδ PβT B.
(3.10)
−1 −1 (I − σB)PβT Pre-multiply Pα−1 (I − ρF)(I − αF)−1 and post-multiply I − βB for (3.10), we have
−1 −1 (I − σB)PβT Pα−1 (I − ρF)(I − αF)−1 · A · I − βB
−1 −1 = bDδ PβT · I − βB (I − σB)PβT + dPα−1 (I − ρF)(I − αF)−1 · Pα Dδ
−1 −1 − dPα−1 (I − ρF)(I − αF)−1 · Pα Dδ PβT · I − βB (I − σB)PβT
−1 −1 − (b − d)βPα−1 (I − ρF)(I − αF)−1 · Pα Dδ PβT B · I − βB (I − σB)PβT . By (1.15) and (1.16), the equation can be simplified as
−1 −1 (I − σB)PβT Pα−1 (I − ρF)(I − αF)−1 · A · I − βB = bDδ (I + (β − σ)B) + d(I + (α − ρ)F)Dδ (I + (β − σ)B) − (b − d)β(I + (α − ρ)F)Dδ PβT BPβT
−1
(I + (β − σ)B)
− dβ(I + (α − ρ)F)Dδ (I + (β − σ)B) = bDδ − d(α − ρ)(β − σ)FDδ B − (b − d)β(α − ρ)FDδ B + · · ·
upper triangular matrices = bDδ − (α − ρ)(bβ − dσ)FDδ B + , with 0’s on their diagonals
(3.11)
where the second equality has used the following derivation: PβT BPβT
−1
by(1.8)
====== PβT · (I − βB)PβT
−1
B = B − βPβT B · PβT
−1
B.
(3.12)
The entries lying on the principal diagonal of the matrix in the right-hand side of the last equation in (3.11) are b, bδ − (α − ρ)(bβ − dσ), (bδ − (α − ρ)(bβ − dσ))δ, . . . , (bδ − (α − ρ)(bβ − dσ))δ n−1 . Thus
n n det A = b bδ − (α − ρ)(bβ − dσ) δ ( 2 ) . So we get Theorem 3.2 (Wang [11, Theorem 6]). Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values ai,0 = bρi , a0,j = dσ j , i ≥ 0, j ≥ 1. Then n
det (ai,j ) = b(bγ + bβρ + dασ − dρσ)n (αβ + γ)( 2 ) .
0≤i,j≤n
(3.13)
When α = β = 1 and b = d = 1, (3.13) returns to Theorem 1 of Krattenthaler [3] (see also Bacher [1, Theorem 1.5]).
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
9
3.3. Another geometric sequences as initial values. Take a0,0 = 0, ai,0 = bρi−1 , a0,j = dσ j−1 , i, j ≥ 1, then we have by (1.3) C0 = b
n
ρi−1 F i = bF(I − ρF)−1 ,
R0 = d
i=1
n
σ j−1 B j = dB(I − σB)−1 .
j=1
By (2.5), the matrix A can be rewritten as
A = b I − αF F(I − ρF)−1 Pα Dδ PβT + dPα Dδ PβT B(I − σB)−1 I − βB . (3.14) Using (1.15) and (1.16), for (3.14) pre-multiply Pα−1 (I − ρF)(I − αF)−1 and post
−1 −1 (I − σB)PβT , we have multiply I − βB
−1 −1 (I − σB)PβT Pα−1 (I − ρF)(I − αF)−1 · A · I − βB = bPα−1 FPα Dδ (I + (β − σ)B) + d(I + (α − ρ)F)Dδ PβT BPβT
−1
.
Again pre-multiply (I + αF) and post-multiply (I + βB) for the equation, by using (1.7) and (1.8), we get
−1 −1 (I − σB)PβT (I + βB) (I + αF)Pα−1 (I − ρF)(I − αF)−1 · A · I − βB = bFDδ (I + (β − σ)B)(I + βB) + d(I + αF)(I + (α − ρ)F)Dδ B = bFDδ + dDδ B + b(2β − σ) + d(2α − ρ) FDδ B + dα(α − ρ)F 2 Dδ B + bβ(β − σ)FDδ B 2 ⎛ 0 d ⎜b ε dδ + bβ(β − σ) ⎜ ⎜ .. ⎜ . εδ = ⎜ bδ + dα(α − ρ) ⎜ . . .. .. ⎜ ⎝
0
[bδ+dα(α−ρ)]δ n−2
⎞
0
⎟ ⎟ ⎟ ⎟ ⎟ ⎟ n−2 ⎟ [dδ+bβ(β−σ)]δ ⎠ εδ n−1
where ε := b(2β − σ) + d(2α − ρ). Then 0 d b ε dδ+bβ(β−σ) . . bδ+dα(α−ρ) . εδ det A = .. .. n−2 . . [dδ+bβ(β−σ)]δ [bδ+dα(α−ρ)]δ n−2 εδ n−1 [dδ+bβ(β−σ)]δ εδ .. 2 [bδ+dα(α−ρ)]δ . εδ = − bd . . .. .. n−2 [dδ+bβ(β−σ)]δ n−1 n−2 [bδ+dα(α−ρ)]δ εδ
0
0
0
0
n = − bdδ ( 2 ) Jn−1 ε, b + dα(α − ρ)/δ, dδ + bβ(β − σ) . Therefore, by Lemma 1.3, we obtain Theorem 3.3. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values a0,0 = 0, ai,0 = bρi−1 , a0,j = dσ j−1 , i, j ≥ 1. Then
n det (ai,j ) = −bdδ ( 2 ) Jn−1 ε, b + dα(α − ρ)/δ, dδ + bβ(β − σ) (3.15) 0≤i,j≤n
where ε = b(2β − σ) + d(2α − ρ). In particular, 1) when α = β, ρ = σ and d = −b, then ε = 0 and the elements on the principal diagonal are all zero. At this time, if n is an even number, then det0≤i,j≤n (ai,j ) = 0;
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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if n is an odd number and n = 2k + 1, then 2
det
0≤i,j≤2k+1
(ai,j ) = b2k+2 (αρ + γ)2k (α2 + γ)2k .
(3.16)
2) When β = −α, σ = −ρ and d = b, then ε = 0 and the elements on the principal diagonal are all zero. At this time, if n is an even number, then det0≤i,j≤n (ai,j ) = 0; if n is an odd number and n = 2k + 1, then 2
det
0≤i,j≤2k+1
(ai,j ) = (−1)k+1 b2k+2 (γ − αρ)2k (γ − α2 )2k .
(3.17)
Note that when α = 1 and b = 1, (3.16) becomes Theorem 2 of Krattenthaler [3]. 3.4. The difference between two geometric sequences as the initial values. Let ai,0 = b(αi − ρi ), a0,j = d(β j − σ j ), i, j ≥ 0, then we have by (1.3) C0 = b
n
(αi − ρi )F i = b(I − αF)−1 − b(I − ρF)−1 ,
i=0 n
R0 = d
(β j − σ j )B j = d(I − βB)−1 − d(I − σB)−1 .
j=0
By (2.5), the matrix A can be rewritten as
A = b I − αF (I − αF)−1 − (I − ρF)−1 Φ
+ dΦ (I − βB)−1 − (I − σB)−1 I − βB
= (b + d)Φ − b I − αF (I − ρF)−1 Φ − dΦ(I − σB)−1 I − βB . (3.18)
−1 −1 Pre-multiply Pα−1 (I − ρF)(I − αF)−1 and post-multiply I − βB (I − σB)PβT for (3.18), by using (1.15) and (1.16), we have
−1 −1 (I − σB)PβT Pα−1 (I − ρF)(I − αF)−1 · A · I − βB = (b + d)(I + (α − ρ)F)Dδ (I + (β − σ)B) − bDδ (I + (β − σ)B) − d(I + (α − ρ)F)Dδ = b(α − ρ)FDδ + d(β − σ)Dδ B + (b + d)(α − ρ)(β − σ)FDδ B ⎛ ⎞ 0 d(β − σ) ⎜b(α − ρ) ⎟ η d(β − σ)δ ⎜ ⎟ ⎜ ⎟ . . ⎜ ⎟ . b(α − ρ)δ ηδ =⎜ ⎟ ⎜ ⎟ .. .. ⎜ n−1 ⎟ . . d(β − σ)δ ⎝ ⎠ b(α − ρ)δ n−1 ηδ n−1
0
0
where η := (b + d)(α − ρ)(β − σ). Then 0 d(β − σ) b(α − ρ) η d(β − σ)δ . .. b(α − ρ)δ ηδ det A = .. .. n−1 . . d(β − σ)δ b(α − ρ)δ n−1 ηδ n−1 ηδ d(β − σ)δ 2 . . 2 b(α − ρ)δ 2 . ηδ = −bd(α − ρ)(β − σ) . . .. .. n−1 d(β − σ)δ n−1 n−1 b(α − ρ)δ ηδ
0
0
0
0
= − bd(α − ρ)(β − σ)δ ( ) Jn−1 η, b(α − ρ), d(β − σ)δ . n 2
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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Therefore, by Lemma 1.3, we obtain Theorem 3.4. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values ai,0 = b(αi − ρi ), a0,j = d(β j − σ j ), i, j ≥ 0. Then
n det (ai,j ) = −bd(α − ρ)(β − σ)δ ( 2 ) Jn−1 η, b(α − ρ), d(β − σ)δ (3.19) 0≤i,j≤n
where η = (b + d)(α − ρ)(β − σ). In particular, when d = −b, then η = 0 and the elements on the principal diagonal are all zero. At this time, if n is an even number, then det0≤i,j≤n (ai,j ) = 0; if n is an odd number and n = 2k + 1, then det
0≤i,j≤2k+1
(ai,j ) = b2(k+1) (α − ρ)k+1 (β − σ)k+1 (αβ + γ)2k(k+1) .
(3.20)
3.5. The arithmetic and geometric sequences as the initial values. Let ai,0 = bρi , a0,j = c + dj, i ≥ 0, j ≥ 1, then we have by (1.3) and (1.4) C0 = b
n
ρi F i = b(I − ρF)−1 ,
i=0
R0 = bI +
n (c + dj)B j = (b − c)I + c(I − B)−1 + dB(I − B)−2 . j=1
By (2.5), the matrix A can be rewritten as
A = b I − αF (I − ρF)−1 Φ − bΦ
+ Φ (b − c)I + c(I − B)−1 + dB(I − B)−2 I − βB
= b I − αF (I − ρF)−1 Φ + cΦ(I − B)−1 (I − βB) + dΦB(I − B)−2 (I − βB) − cΦ − (b − c)βΦB. Pα−1 (I
−1
Pre-multiply − ρF)(I − αF) and post-multiply (3.21), by using (1.15), (1.14) and (3.12), we have Pα−1 (I − ρF)(I − αF)−1 · A · PβT
−1
−1 PβT (I
(3.21) − (1 − β)B)2 for
(I − (1 − β)B)2
= bDδ (I − (1 − β)B)2 + c(I + (α − ρ)F)Dδ (I − (1 − β)B) + d(I + (α − ρ)F)Dδ B − c(I + (α − ρ)F)Dδ (I − (1 − β)B)2 −1
− (b − c)β(I + (α − ρ)F)Dδ · PβT BPβT · (I − (1 − β)B)2 = bDδ + d + c(1 − β) − (b − c)β (α − ρ)FDδ B + upper triangular matrices with 0’s on their principal diagonals. To the right of the last equal sign is an upper triangular matrix whose elements on the principal diagonal are b, bδ + (c + d − bβ)(α − ρ), bδ + (c + d − bβ)(α − ρ) δ, · · · , bδ + (c + d − bβ)(α − ρ) δ n−1 . Thus n n det A = b bδ + (c + d − bβ)(α − ρ) δ ( 2 ) . So we get Theorem 3.5. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values ai,0 = bρi , a0,j = c + dj, i ≥ 0, j ≥ 1. Then n n det (ai,j ) = b (c + d)(α − ρ) + b(βρ + γ) (αβ + γ)( 2 ) . (3.22) 0≤i,j≤n
3.6. Only two non-zero initial values. Take a0,0 = 0, a1,0 = b, a0,1 = d, ai,0 = a0,j = 0, i, j ≥ 2, then C0 =
n i=0
ai,0 F i = bF,
R0 =
n j=0
a0,j B j = dB.
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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By (2.5), the matrix A can be rewritten as
(3.23) A = b I − αF FPα Dδ PβT + dPα Dδ PβT B I − βB .
−1 T −1 Pre-multiply (I + αF)Pα−1 (I − αF)−1 and post-multiply I − βB Pβ (I + βB) for (3.23), using (1.7), (1.8), (1.15) in the case ρ = 0, and (1.16) in the case σ = 0, we have
−1 T −1 (I + αF)Pα−1 (I − αF)−1 · A · I − βB Pβ (I + βB) = bFDδ (I + βB)2 + d(I + αF)2 Dδ B = bFDδ + dDδ B + 2(bβ + dα)FDδ B + bβ 2 FDδ B 2 + dα2 F 2 Dδ B ⎞ ⎛ 0 d ⎟ ⎜ b 2(bβ + dα) dδ + bβ 2 ⎟ ⎜ ⎟ ⎜ .. 2 ⎟ ⎜ . bδ + dα 2(bβ + dα)δ =⎜ ⎟. ⎟ ⎜ . . .. .. ⎜ 2 n−2 ⎟ (dδ + bβ )δ ⎠ ⎝ 2 n−2 n−1 (bδ + dα )δ 2(bβ + dα)δ
0
0
Then
0 b det A =
d 2(bβ + dα)
0
dδ + bβ 2
bδ + dα2 2(bβ + dα)δ .. .
0
2 n−2 (dδ + bβ )δ 2(bβ + dα)δ n−1
..
.
..
.
(bδ + dα2 )δ n−2
2(bβ + dα)δ (dδ + bβ 2 )δ . .. (bδ + dα2 )δ 2(bβ + dα)δ 2 = −bd . . .. .. 2 n−2 (dδ + bβ )δ 2 n−2 n−1 (bδ + dα )δ 2(bβ + dα)δ
0
0
n = − bdδ ( 2 ) Jn−1 2(bβ + dα), b + dα2 /δ, dδ + bβ 2 . Therefore, by Lemma 1.3, we obtain Theorem 3.6. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values a0,0 = 0, a1,0 = b, a0,1 = d, ai,0 = a0,j = 0, i, j ≥ 2. Then
n (3.24) det (ai,j ) = −bdδ ( 2 ) Jn−1 2(bβ + dα), b + dα2 /δ, dδ + bβ 2 . 0≤i,j≤n
In particular, when α = β and d = −b, then 2(bβ + dα) = 0. At this time, if n is an even number, then det0≤i,j≤n (ai,j ) = 0; if n is an odd number and n = 2k + 1, then 2 det (ai,j ) = b2k+2 γ 2k (α2 + γ)2k . (3.25) 0≤i,j≤2k+1
3.7. Only three non-zero initial values. Let a0,0 = 1, a1,0 = α, a0,1 = −β, ai,0 = a0,j = 0, i, j ≥ 2, then C0 =
n i=0
ai,0 F i = I + αF,
R0 =
n
a0,j B j = I − βB.
j=0
By (2.5), the matrix A can be rewritten as
2 A = I − αF I + αF Pα Dδ PβT + Pα Dδ PβT I − βB − Pα Dδ PβT .
(3.26)
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
−1
13 −1
Pre-multiply (I + αF)Pα−1 (I − αF)−1 and post-multiply I − βB PβT (I + βB) for (3.26), using (1.15) in the case ρ = 0, (1.16) in the case σ = 0, (1.17) in the case λ = α, and (1.18) in the case τ = −β, we have
−1 T −1 Pβ (I + βB) (I + αF)Pα−1 (I − αF)−1 · A · I − βB
2
= I + 2αF Dδ (I + βB)2 + (I + αF)2 Dδ − I + αF Dδ (I + βB)2 = Dδ + 2αFDδ − α2 β 2 F 2 Dδ B 2 − 2α2 βF 2 Dδ B ⎛ 1 δ ⎜2α ⎜ ⎜ 2α(δ − αβ) δ 2 − α2 β 2 =⎜ .. .. ⎜ . . ⎝
0
2α(δ − αβ)δ n−2
0
⎞ ⎟ ⎟ ⎟ ⎟. ⎟ ⎠
(δ 2 − α2 β 2 )δ n−2
Then
n−1 2
det A = δ (
)+1 (δ 2 − α2 β 2 )n−1 .
Therefore, we obtain Theorem 3.7. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values a0,0 = 1, a1,0 = α, a0,1 = −β, ai,0 = a0,j = 0, i, j ≥ 2. Then
n−1
(n−1)+1 . αβ + γ 2 det (ai,j ) = γ n−1 2αβ + γ
0≤i,j≤n
(3.27)
3.8. First-order geometric sequences as initial values. Let a0,0 = 0, ai,0 = i−1 j−1 ρai−1,0 + b, a0,j = σa0,j−1 + d, i, j ≥ 1, then ai,0 = b k=0 ρk , a0,j = d k=0 σ k . By (1.5) and (1.6), we get
C0 = b
n i−1
ρk F i = bF(I − F)−1 (I − ρF)−1 ,
i=1 k=0
R0 = d
j−1 n
σ k B j = dB(I − B)−1 (I − σB)−1 .
j=1 k=0
By (2.5), the matrix A can be rewritten as
A = b I − αF F(I − F)−1 (I − ρF)−1 Φ + dΦB(I − B)−1 (I − σB)−1 I − βB . (3.28)
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
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For (3.28) pre-multiply I − (1 − α)F Pα−1 (I − ρF)(I − αF)−1 and post-multiply
−1
−1 I − βB I − (1 − β)B , by using (1.9) and (1.10), we have (I − σB)PβT
I − (1 − α)F Pα−1 (I − ρF)(I − αF)−1 · A
−1
−1 · I − βB I − (1 − β)B (I − σB)PβT
= b I − (1 − α)F Pα−1 (I − F)−1 FPα Dδ (I + (β − σ)B) I − (1 − β)B
−1 + d I − (1 − α)F I + (α − ρ)F Dδ PβT B(I − B)−1 PβT I − (1 − β)B
= bFDδ I + (2β − σ − 1)B − (1 − β)(β − σ)B 2
+ d I + (2α − ρ − 1)F − (1 − α)(α − ρ)F 2 Dδ B = bFDδ + dDδ B − b(1 − β)(β − σ)FDδ B 2 − d(1 − α)(α − ρ)F 2 Dδ B
+ b(2β − σ − 1) + d(2α − ρ − 1) FDδ B ⎞ ⎛ 0 d ⎟ ⎜b ω dδ − bν ⎟ ⎜ ⎟ ⎜ .. ⎟ ⎜ bδ − dμ . ωδ =⎜ ⎟ ⎟ ⎜ . . .. .. ⎜ n−2 ⎟ (dδ − bν)δ ⎠ ⎝ (bδ − dμ)δ n−2 ωδ n−1
0
0
where ω := b(2β − σ − 1) + d(2α − ρ − 1), μ := (1 − α)(α − ρ), ν := (1 − β)(β − σ). Then 0 d b ω dδ − bν . . . bδ − dμ ωδ det A = . . .. .. n−2 (dδ − bν)δ n−2 n−1 (bδ − dμ)δ ωδ ωδ (dδ − bν)δ . . 2 (bδ − dμ)δ . ωδ = −bd .. .. n−2 . . (dδ − bν)δ (bδ − dμ)δ n−2 ωδ n−1
0
0
0
0
n = −bdδ ( 2 ) Jn−1 ω, b − dμ/δ, dδ − bν . Therefore, by Lemma 1.3, we obtain Theorem 3.8. Let {ai,j }i,j≥0 be the sequence given by the recurrence relation (1.1), and the initial values a0,0 = 0, ai,0 = ρai−1,0 + b, a0,j = σa0,j−1 + d, i, j ≥ 1. Then
n (3.29) det (ai,j ) = −bdδ ( 2 ) Jn−1 ω, b − dμ/δ, dδ − bν 0≤i,j≤n
where ω = b(2β − σ − 1) + d(2α − ρ − 1), μ = (1 − α)(α − ρ), ν = (1 − β)(β − σ). In particular, when α = β, ρ = σ and d = −b, then ω = 0 and the elements on the principal diagonal are all zero. At this time, if n is an even number, then det0≤i,j≤n (ai,j ) = 0; if n is an odd number and n = 2k + 1, then det
0≤i,j≤2k+1
2
(ai,j ) = b2k+2 (α + γ − ρ + αρ)2k (α2 + γ)2k .
(3.30)
4. Conclusions We have already seen that the initial values play an important role in calculating the determinants of the Pascal-like matrices. Changing the initial values may affect the
De-Yin Zheng: Matrix Methods for determinants of Pascal-like matrices
15
process and results of the determinant calculation. Eight different cases of initial values are considered here, and the values of the determinants of their corresponding Pascal-like matrices have been derived. Interested readers can choose other initial values to calculate the determinant of such Pascal-like matrices. Acknowledgments: The authors thank the referee for the careful reading and useful suggestions that led to an improved version of the manuscript. References [1] R. Bacher, Determinants of matrices related to the pascal triangle, Journal de Th´eorie des Nombres de Bordeaux 14 (2002), no. 1, 19-41. [2] C. Krattenthaler, Advanced determinant calculus, S´ em. Lothar. Combin., 42 (1999), pp. Art. B42q, 67 pp. (electronic). [3] C. Krattenthaler, Evaluations Of Some Determinants Of Matrices Related To The Pascal Triangle, S´ eminaire Lotharingien de Combinatoire, 1 (2002), pp. 1-19. [4] C. Krattenthaler, Advanced determinant calculus: A complement, Linear Algebra and its Applications, 411 (2005), pp. 68-166. [5] A. R. Moghaddamfar, S. Navid Salehy and S. Nima Salehy, The determinants of matrices with recursive entries, Linear Algebra Appl., 428 (2008), no. 11-12, 2468-2481. [6] A. R. Moghaddamfar and S. M. H. Pooya, Generalized pascal triangles and toeplitz matrices, Electronic Journal of Linear Algebra 18 (2009), no. 1, 564-588. [7] A. R. Moghaddamfar, S. M. H. Pooya, S. Navid Salehy and S. Nima Salehy, More on evaluating determinants, Matematicki Vesnik 64 (2012), no. 3, 211-222. [8] A. R. Moghaddamfar, S. N. Salehy and S. N. Salehy, Determinant representations of sequences: A survey, Special Matrices 2 (2014), no. 1, 46-60. [9] E. Neuwirth, Treeway Galton arrays and generalized Pascal-like determinants, Technical Report, University ofVienna, (2004), pp. 1-16. [10] M. Tan, Matrices Associated to Biindexed Linear Recurrence Relations. Ars Combinatoria., 86(2008), 305-319. [11] X. Wang, Some determinants of Pascal-like matrices, Quaestiones Mathematicae (2012), pp. 37-41. [12] H. Zakraj˘sek and M. Petkov˘sek, Pascal-like determinants are recursive, Advances in Applied Mathematics, 33 (2004), pp. 431-450. [13] Z. Z. Zhang, The Linear Algebra of Generalized Pascal Matrix, Linear Algebra Appl. 250 (1997): 51-60.