Evaluation of a family of binomial determinants.pdf

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Electronic Journal of Linear Algebra

V4% 30 Volume 30 (2015) A#% 22

2015

Evaluation of a family of binomial determinantsCharles Helou Pennsylvania State University , #722@4.%$4

James A. Sellers Pennsylvania State University , %%*@4.%$4

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ELA

EVALUATION OF A FAMILY OF BINOMIAL DETERMINANTS∗

CHARLES HELOU† AND JAMES A. SELLERS‡

Abstract. Motivated by a recent work about finite sequences where the n-th term is bounded

by n2, some classes of determinants are evaluated such as the (n− 2)× (n− 2) determinant

∆n =

xn − xk + h− 1

n− k − 1

2≤k≤n−10≤h≤n−3

, for n ≥ 3,

and more generally the n × n determinant

Dn =

xi + j

i− 1

1≤i≤n1≤j≤n

, for n ≥ 1,

where n,k,h,i,j are integers, (xk)1≤k≤n is a sequence of indeterminates over C andAB

is the usual

binomial coefficient. It is proven that

Dn = 1 and ∆n = (−1)(n−2)(n−3)

2 .

Key words. Determinants, Binomial coefficients, Row reduction.

AMS subject classifications. 11C20, 05A10, 11B65, 15B36.

1. Introduction. In a recent work [9], about finite sequences whose n-th term

does not exceed n2, there appeared the determinant

∆∗n =

n2 − k2 + h − 1

n − k − 1

2≤k≤n−10≤h≤n−3

,

with an integer n ≥ 3. One of the authors of [9], L. Haddad, conjectured after somecomputations that ∆∗

n = ±1. The authors of the present paper first proved that

∆∗n = (−1)

(n−2)(n−3)2 ,

essentially by a process of row reduction. Then, following a suggestion by G.E. An-

drews that this result should be true in a more general context, namely upon replacing

∗Received by the editors on January 30, 2015. Accepted for publication on May 19, 2015.

Handling Editor: Joao Filipe Queiro.†Department of Mathematics, Penn State Brandywine, 25 Yearsley Mill Road, Media, PA 19063,

USA ([email protected]).‡Department of Mathematics, Penn State University Park, 104 McAllister Building, University

Park, PA 16802, USA ([email protected]).312

Electronic Journal of Linear Algebra ISSN 1081-3810

A publication of the International Linear Algebra Society

Volume 30, pp. 312-321, July 2015

http://math.technion.ac.il/iic/ela



ELA

Evaluation of a Family of Binomial Determinants 313

n2 by xn and k2 by xk, where (xk)1≤k≤n is an arbitrary sequence of indeterminates

over C, the proof was extended to this general case. We then realized that the problem

can be reduced to the evaluation of a simpler, more general family of determinants,

namely

Dn =

xi + j

i − 1 1≤i≤n1≤j≤n

,

for all integers n ≥ 1. In what follows, we will establish that Dn = 1 and deduce that

∆n = (−1)(n−2)(n−3)

2 .

Results of a similar nature, involving determinants of matrices whose entries

involve binomial coefficients, can be found in [1, 2, 3, 4, 6, 10, 12]. In contrast to

these papers, we note that our determinant evaluations are strikingly simple and easy

to state. In fact, our result is a special case of the results contained in [10], but our

proof is more elementary, using only row reduction and induction.

We are thankful to L. Haddad for his conjecture and to G.E. Andrews for his

insightful suggestion.

2. The method of proof. First recall (e.g., [7] or [8]) that for an indeterminate

x over C and an integer n ≥ 0, the binomial coefficientxn

is defined by

x

n

=

x (x − 1) (x − 2) · · · (x − n + 1)

n! ,

with the convention thatx

0

= 1. It satisfies the fundamental recurrence relation

(2.1)

x

n

+

x

n + 1

=

x + 1

n + 1

, for all n ≥ 0.

Let (xk)1≤k≤n be a sequence of indeterminates over C, with n ≥ 3, and consider the

(n − 2) × (n − 2) determinant

∆n = ∆n(x1, . . . , xn) =

xn − xk + h − 1

n − k − 1

2≤k≤n−10≤h≤n−3

.

First, setting i = k − 1 and j = h + 1 allows one to rewrite ∆n as

∆n =

xn − xi+1 + j − 2

n − i − 2 1≤i≤n−2

1≤j≤n−2

.



Volume 30, pp. 312-321, July 2015




ELA

314 C. Helou and J.A. Sellers

Second, the substitution i′ = n − i − 1 transforms ∆n into

∆′n = ∆′

n(x1, . . . , xn) =

xn − xn−i′ + j − 2

i′ − 1

1≤i

′≤n−21≤j≤n−2

,

which has the same rows as ∆n but in reverse order. This order reversal consists in

respectively swapping each row of ∆n with all the rows above it. The total number

of those row swaps is

(n − 3) + (n − 4) + · · · + 2 + 1 = (n − 2)(n − 3)

2 .

Therefore,

(2.2) ∆n = (−1)(n−2)(n−3)

2 ∆′n.

The problem is thus reduced to the determination of ∆′n.

Third, setting x = xn − 2 gives

∆′

n

= x − xn−i + j

i − 1

1≤i≤n−21≤j≤n−2

.

Fourth, setting yi = x − xn−i yields

∆′n =

yi + j

i − 1

1≤i≤n−21≤j≤n−2

.

Finally, setting m = n − 2 leads to the equality

(2.3) ∆′n =

yi + j

i − 1 1≤i≤m

1≤j≤m

.

The problem is thus reduced to the evaluation of the family of determinants

Dn = Dn(x1, . . . , xn) =

xi + j

i − 1

1≤i≤n1≤j≤n

,

for n ≥ 1, where x1, x2, . . . , xn are arbitrary indeterminates.

Our primary result is now the following:

Theorem 2.1. For any positive integer n, we have

Dn = 1.



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ELA


The proof proceeds by row reduction and is presented in the next section.

Corollary 2.2. For any integer n ≥ 3, we have

∆′n = 1.

Proof . This follows from (2.3) and Theorem 2.1.

Corollary 2.3. For any integer n ≥ 3, we have

∆n = (−1)(n−2)(n−3)

2 .

Proof . This follows from (2.2) and Corollary 2.2.

Remark 2.4. An alternative method for deriving the last result is to set yk =

xn − xn−k − 2, and i = k − 1, j = h + 1. Then

∆n =yn−i−1 + j

n − i − 2

1≤i≤n−21≤j≤n−2

.

Now, reversing the order of the rows, which consists in replacing i by n − 1 − i,

transforms ∆n into

∆′n =

yi + j

i − 1

1≤i≤n−21≤j≤n−2

.

Moreover, the permutation ρ that reverses the n − 2 rows of ∆n has ⌈n−22 ⌉ orbits,

namely {1, n − 2}, {2, n − 3}, . . . It follows that (see, e.g., [11]) the sign of ρ is

ǫ (ρ) = (−1)n−22 −⌈n−2

2 ⌉ = (−1)⌊n−22 ⌋.

Hence,

∆n = (−1)⌊n−22 ⌋∆′

n = (−1)⌊n−22 ⌋,

in view of our main theorem, which yields ∆′n = 1.

3. The proof of the main theorem. We start with two results about binomial

coefficients that will be used in the proof of Theorem 2.1.

Lemma 3.1. For any integers 0 ≤ a ≤ b and n ≥ 1, and any indeterminate x

over C, we have

x + b

n − x + a

n =

b−1

h=a

x + h

n − 1

.



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ELA


Proof. By the fundamental recurrence relation for binomial coefficients (2.1),x + 1

n

−

x

n

=

x

n − 1

,

for n ≥ 1. Hence, by iterating this recurrence numerous times, we have

x + b

n

−

x + a

n

=

b−1h=a

x + h + 1

n

−

x + h

n

=

b−1h=a

x + h

n − 1

.

Lemma 3.2. For any integers 0 ≤ m ≤ n, we have

nk=m

k

m

=

n + 1

m + 1

.

Proof . For a fixed integer m ≥ 0, the proof can be completed by induction on

n ≥ m. Such an argument can be found in [5, p. 138]. This result also follows fromLemma 3.1 by taking n = m + 1, a = 0, x = m, b = n − m + 1.

We now proceed to prove Theorem 2.1 by row reduction. We start with

Dn =

(dij)1≤i≤n1≤j≤n

, where dij =

xi + j

i − 1

for 1 ≤ i, j ≤ n,

i.e.,

Dn =

1 1 1 · · · 1 · · · 1

x2+11

x2+21

x2+31

· · ·

x2+j1

· · ·

x2+n1

x3+12 x3+22 x3+32 · · · x3+j2 · · · x3+n2 x4+13

x4+23

x4+33

· · ·

x4+j3

· · ·

x4+n3

...

......

......

xi+1i−1

xi+2i−1

xi+3i−1

· · ·

xi+ji−1

· · ·

xi+ni−1

...

......

......

xn+1n−1

xn+2n−1

xn+3n−1

· · ·

xn+jn−1

· · ·

xn+nn−1

.

Denoting the i-th row of Dn by Ri, the first row reduction step consists in re-

placing Ri by Ri − di1R1 for 2 ≤ i ≤ n. This gives

D′n = d′ij1≤i≤n

1≤j≤n ,



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ELA


where

(3.1) d′ij =

dij − di1 =

xi+ji−1

−xi+1i−1

=j−1

h=1

xi+hi−2

, if 2 ≤ i ≤ n, 1 ≤ j ≤ n,

d1j = 1, if i = 1, 1 ≤ j ≤ n,

the last expression, for i ≥ 2, is obtained by using Lemma 3.1, with the usual con-

vention that an empty sum is equal to 0. Thus,

D′n =

1 1 1 · · · 1 · · · 1

0 1 2 · · · j − 1 · · · n − 1

0x3+11

2h=1

x3+h1

· · ·

j−1h=1

x3+h1

· · ·

n−1h=1

x3+h1

0

x4+12

2h=1

x4+h2

· · ·

j−1h=1

x4+h2

· · ·

n−1h=1

x4+h2

...

......

......

0xi+1i−2

2h=1

xi+hi−2

· · ·

j−1h=1

xi+hi−2

· · ·

n−1h=1

xi+hi−2

...

......

......

0xn+1n−2

2h=1

xn+hn−2

· · ·

j−1h=1

xn+hn−2

· · ·

n−1h=1

xn+hn−2

.

Moreover, we obviously have

Dn = D ′n.

Proposition 3.3. For 1 ≤ k ≤ n, let

D(k)n =

d(k)ij

1≤i≤n1≤j≤n

be the n ×n determinant obtained from Dn by applying k row reduction steps, each of

which consists in replacing the i-th row R(k−1)i of the determinant D

(k−1)n , obtained

after k − 1 such row reduction steps, by the row

R(k)i = R(k−1)

i − d(k−1)ik R(k−1)k , for k + 1 ≤ i ≤ n,

while the first k rows are unchanged, i.e.,

R(k)i = R

(k−1)i for 1 ≤ i ≤ k.

Then

d(k)ij =

j−kh=1

j−h−1k−1

xi+hi−k−1

, if k + 1 ≤ i ≤ n, 1 ≤ j ≤ n,

d(k−1)ij , if 1 ≤ i ≤ k, 1 ≤ j ≤ n,

with the convention that an empty sum (here, when j ≤ k) is equal to 0.



Volume 30, pp. 312-321, July 2015




ELA


Proof . The proof is by induction on k. The first row reduction step was applied

to Dn right before this Proposition, and it gave D′n =

d′ij1≤i≤n1≤j≤n

, which satisfies

the stated equalities for d′ij as shown in (3.1) above. So the property holds for k =

1. Assume that it holds for k − 1, where 2 ≤ k ≤ n, i.e., assume that D(k−1)n =

d(k−1)ij 1≤i≤n1≤j≤n

satisfies

d(k−1)ij =

j−(k−1)

h=1

j−h−1k−2

xi+h

i−(k−1)−1

, if k ≤ i ≤ n, 1 ≤ j ≤ n,

d(k−2)ij , if 1 ≤ i ≤ k − 1, 1 ≤ j ≤ n.

Now, as for k + 1 ≤ i ≤ n, we have R(k)i = R

(k−1)i − d

(k−1)ik R

(k−1)k , i.e.,

d(k)ij = d

(k−1)ij − d

(k−1)ik d

(k−1)kj , for 1 ≤ j ≤ n,

and by the induction assumption, there hold

d(k−1)ij =

j−k+1

h=1

j − h − 1

k − 2 xi + h

i − k ,

d(k−1)ik =

k−k+1h=1

k − h − 1

k − 2

xi + h

i − k

=

xi + 1

i − k

,

d(k−1)kj =

j−k+1h=1

j − h − 1

k − 2

xk + h

k − k

=

j−k+1h=1

j − h − 1

k − 2

.

Therefore, we get

d(k)ij =

j−k+1

h=1

j − h − 1

k − 2 xi + h

i − k −

j−k+1

h=1

j − h − 1

k − 2 xi + 1

i − k =

j−k+1h=1

j − h − 1

k − 2

xi + h

i − k

−

xi + 1

i − k

.

Moreover, by Lemma 3.1,xi + h

i − k

−

xi + 1

i − k

=

h−1r=1

xi + r

i − k − 1

,

for i > k and h ≥ 1. Hence,

d(k)ij =

j−k+1

h=1

j − h − 1

k − 2 h−1

r=1

xi + r

i − k − 1 =

j−k

r=1

xi + r

i − k − 1j−k+1

h=r+1

j − h − 1

k − 2 .



Volume 30, pp. 312-321, July 2015




ELA


Furthermore, by Lemma 3.2,

j−k+1h=r+1

j − h − 1

k − 2

=

j−r−2s=k−2

s

k − 2

=

j − r − 1

k − 1

.

Thus,

d(k)ij =

j−kr=1

j − r − 1

k − 1

xi + r

i − k − 1

,

for k + 1 ≤ i ≤ n and 1 ≤ j ≤ n.

Also, for 1 ≤ i ≤ k , since R(k)i = R

(k−1)i , we have

d(k)ij = d

(k−1)ij , for 1 ≤ i ≤ k, 1 ≤ j ≤ n.

This shows that the property holds for k , and completes the induction.

Corollary 3.4. For 1 ≤ k ≤ n, the determinant

D(k)n =

d(k)ij

1≤i≤n1≤j≤n

obtained from Dn by applying k row reduction steps as described in Proposition 3.3 is

given by

d(k)ij =

j−1i−1

, if 1 ≤ i ≤ k, 1 ≤ j ≤ n,

j−kh=1

j−h−1k−1

xi+hi−k−1

, if k + 1 ≤ i ≤ n, 1 ≤ j ≤ n,

where if 0 ≤ m < n are integers then mn = 0, and an empty sum is equal to 0.

Proof . Only the expression of d(k)ij for 1 ≤ i ≤ k and 1 ≤ j ≤ n needs to be

proved. The rest is contained in Proposition 3.3. This expression holds for k = 1

since by (3.1)

d′1j = d1j = 1 =

j − 1

1 − 1

, for 1 ≤ j ≤ n.

Assume that the expression holds for k − 1, where 2 ≤ k < n, i.e.,

d(k−1)ij =

j−1i−1

, if 1 ≤ i ≤ k − 1, 1 ≤ j ≤ n,

j−k+1

h=1j−h−1

k−2xi+h

i−k

, if k ≤ i ≤ n, 1 ≤ j ≤ n.



Volume 30, pp. 312-321, July 2015




ELA


Then, by Proposition 3.3 and Lemma 3.2, we have

d(k)ij = d

(k−1)ij =

j − 1

i − 1

, for 1 ≤ i ≤ k − 1, 1 ≤ j ≤ n,

and

d(k)kj = d

(k−1)kj =

j−k+1h=1

j − h − 1

k − 2xk + h

0

=

j−k+1h=1

j − h − 1

k − 2

=

j−2s=k−2

s

k − 2

=

j − 1

k − 1

, for 1 ≤ j ≤ n.

Hence,

d(k)ij =

j − 1

i − 1

, for 1 ≤ i ≤ k, 1 ≤ j ≤ n,

and the expression holds for k .

Remark 3.5. We have

Dn = D(k)n ,

since the determinant is invariant under the row reduction steps consisting of adding

to a row a multiple of another row. In particular,

D(n)n =

j − 1

i − 1

1≤i≤n1≤j≤n

is the determinant of an upper triangular n × n matrix whose diagonal entries are

d(n)ii = i−1i−1 = 1 for 1 ≤ i ≤ n. Therefore,

Dn = D(n)n = 1.

This concludes the proof of Theorem 2.1.

Remark 3.6. As noted in the introduction, our result is a special case of the

results contained in [10]. Indeed, in [10], Proposition 1, taking pj(x) =xj+xj−1

, which

is a polynomial of degree j −1 in x, with leading coefficient aj = 1(j−1)! , for 1 ≤ j ≤ n,

we get

xj + X i

j − 1 1≤i≤n

1≤j≤n

=

n

j=1

1

( j − 1)! · 1≤i<j≤n

(X j − X i) ,



Volume 30, pp. 312-321, July 2015




ELA


then specializing to X i = i for 1 ≤ i ≤ n, we get

xj + i

j − 1

1≤i≤n1≤j≤n

=

nj=1

1

( j − 1)! ·

1≤i<j≤n

( j − i) = 1.

Acknowledgment. We are thankful to the referee for a careful, thorough reading

of the paper, and for many helpful and interesting suggestions.

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Evaluation of a family of binomial determinants.pdf