ńņš. 1 |

C. KRATTENTHALERā

Institut fĀØr Mathematik der UniversitĀØt Wien,

u a

Strudlhofgasse 4, A-1090 Wien, Austria.

math.CO/9902004 v3 31 May 1999

E-mail: kratt@pap.univie.ac.at

WWW: http://radon.mat.univie.ac.at/People/kratt

Dedicated to the pioneer of determinant evaluations (among many other things),

George Andrews

Abstract. The purpose of this article is threefold. First, it provides the reader with

a few useful and eļ¬cient tools which should enable her/him to evaluate nontrivial de-

terminants for the case such a determinant should appear in her/his research. Second,

it lists a number of such determinants that have been already evaluated, together with

explanations which tell in which contexts they have appeared. Third, it points out

references where further such determinant evaluations can be found.

1. Introduction

Imagine, you are working on a problem. As things develop it turns out that, in

order to solve your problem, you need to evaluate a certain determinant. Maybe your

determinant is

1

det , (1.1)

i+j

1ā¤i,j,ā¤n

or

a+b

det , (1.2)

aā’i+j

1ā¤i,jā¤n

or it is possibly

Āµ+i+j

det , (1.3)

2i ā’ j

0ā¤i,jā¤nā’1

1991 Mathematics Subject Classiļ¬cation. Primary 05A19; Secondary 05A10 05A15 05A17 05A18

05A30 05E10 05E15 11B68 11B73 11C20 15A15 33C45 33D45.

Key words and phrases. Determinants, Vandermonde determinant, Cauchyā™s double alternant,

Pfaļ¬an, discrete Wronskian, Hankel determinants, orthogonal polynomials, Chebyshev polynomials,

Meixner polynomials, Meixnerā“Pollaczek polynomials, Hermite polynomials, Charlier polynomials, La-

guerre polynomials, Legendre polynomials, ultraspherical polynomials, continuous Hahn polynomials,

continued fractions, binomial coeļ¬cient, Genocchi numbers, Bernoulli numbers, Stirling numbers, Bell

numbers, Euler numbers, divided diļ¬erence, interpolation, plane partitions, tableaux, rhombus tilings,

lozenge tilings, alternating sign matrices, noncrossing partitions, perfect matchings, permutations,

inversion number, major index, descent algebra, noncommutative symmetric functions.

ā

Research partially supported by the Austrian Science Foundation FWF, grants P12094-MAT and

P13190-MAT.

1

2 C. KRATTENTHALER

or maybe

x+y+j x+y+j

ā’

det . (1.4)

x ā’ i + 2j x + i + 2j

1ā¤i,jā¤n

Honestly, which ideas would you have? (Just to tell you that I do not ask for something

impossible: Each of these four determinants can be evaluated in āclosed formā. If you

want to see the solutions immediately, plus information where these determinants come

from, then go to (2.7), (2.17)/(3.12), (2.19)/(3.30), respectively (3.47).)

Okay, let us try some row and column manipulations. Indeed, although it is not

completely trivial (actually, it is quite a challenge), that would work for the ļ¬rst two

determinants, (1.1) and (1.2), although I do not recommend that. However, I do not

recommend at all that you try this with the latter two determinants, (1.3) and (1.4). I

promise that you will fail. (The determinant (1.3) does not look much more complicated

than (1.2). Yet, it is.)

So, what should we do instead?

Of course, let us look in the literature! Excellent idea. We may have the problem

of not knowing where to start looking. Good starting points are certainly classics like

[119], [120], [121], [127] and [178]1. This will lead to the ļ¬rst success, as (1.1) does

indeed turn up there (see [119, vol. III, p. 311]). Yes, you will also ļ¬nd evaluations for

(1.2) (see e.g. [126]) and (1.3) (see [112, Theorem 7]) in the existing literature. But at

the time of the writing you will not, to the best of my knowledge, ļ¬nd an evaluation of

(1.4) in the literature.

The purpose of this article is threefold. First, I want to describe a few useful and

eļ¬cient tools which should enable you to evaluate nontrivial determinants (see Sec-

tion 2). Second, I provide a list containing a number of such determinants that have

been already evaluated, together with explanations which tell in which contexts they

have appeared (see Section 3). Third, even if you should not ļ¬nd your determinant

in this list, I point out references where further such determinant evaluations can be

found, maybe your determinant is there.

Most important of all is that I want to convince you that, today,

Evaluating determinants is not (okay: may not be) diļ¬cult!

When George Andrews, who must be rightly called the pioneer of determinant evalua-

tions, in the seventies astounded the combinatorial community by his highly nontrivial

determinant evaluations (solving diļ¬cult enumeration problems on plane partitions),

it was really diļ¬cult. His method (see Section 2.6 for a description) required a good

āguesserā and an excellent āhypergeometerā (both of which he was and is). While at

that time especially to be the latter was quite a task, in the meantime both guessing and

evaluating binomial and hypergeometric sums has been largely trivialized, as both can

be done (most of the time) completely automatically. For guessing (see Appendix A)

1

Turnbullā™s book [178] does in fact contain rather lots of very general identities satisļ¬ed by determi-

nants, than determinant āevaluationsā in the strict sense of the word. However, suitable specializations

of these general identities do also yield āgenuineā evaluations, see for example Appendix B. Since the

value of this book may not be easy to appreciate because of heavy notation, we refer the reader to

[102] for a clariļ¬cation of the notation and a clear presentation of many such identities.

ADVANCED DETERMINANT CALCULUS 3

this is due to tools like Superseeker2, gfun and Mgfun3 [152, 24], and Rate4 (which is

by far the most primitive of the three, but it is the most eļ¬ective in this context). For

āhypergeometricsā this is due to the āWZ-machineryā5 (see [130, 190, 194, 195, 196]).

And even if you should meet a case where the WZ-machinery should exhaust your com-

puterā™s capacity, then there are still computer algebra packages like HYP and HYPQ6,

or HYPERG7, which make you an expert hypergeometer, as these packages comprise

large parts of the present hypergeometric knowledge, and, thus, enable you to con-

veniently manipulate binomial and hypergeometric series (which George Andrews did

largely by hand) on the computer. Moreover, as of today, there are a few new (perhaps

just overlooked) insights which make life easier in many cases. It is these which form

large parts of Section 2.

So, if you see a determinant, donā™t be frightened, evaluate it yourself!

2. Methods for the evaluation of determinants

In this section I describe a few useful methods and theorems which (may) help you

to evaluate a determinant. As was mentioned already in the Introduction, it is always

possible that simple-minded things like doing some row and/or column operations, or

applying Laplace expansion may produce an (usually inductive) evaluation of a deter-

minant. Therefore, you are of course advised to try such things ļ¬rst. What I am

mainly addressing here, though, is the case where that ļ¬rst, āsimple-mindedā attempt

failed. (Clearly, there is no point in addressing row and column operations, or Laplace

expansion.)

Yet, we must of course start (in Section 2.1) with some standard determinants, such

as the Vandermonde determinant or Cauchyā™s double alternant. These are of course

well-known.

In Section 2.2 we continue with some general determinant evaluations that generalize

the evaluation of the Vandermonde determinant, which are however apparently not

equally well-known, although they should be. In fact, I claim that about 80 % of the

determinants that you meet in āreal life,ā and which can apparently be evaluated, are a

special case of just the very ļ¬rst of these (Lemma 3; see in particular Theorem 26 and

the subsequent remarks). Moreover, as is demonstrated in Section 2.2, it is pure routine

to check whether a determinant is a special case of one of these general determinants.

Thus, it can be really considered as a āmethodā to see if a determinant can be evaluated

by one of the theorems in Section 2.2.

2

the electronic version of the āEncyclopedia of Integer Sequencesā [162, 161], written and developed

by Neil Sloane and Simon Plouļ¬e; see http://www.research.att.com/˜njas/sequences/ol.html

3

written by Bruno Salvy and Paul Zimmermann, respectively Frederic Chyzak; available from

http://pauillac.inria.fr/algo/libraries/libraries.html

4

written in Mathematica by the author; available from http://radon.mat.univie.ac.at/People/kratt;

the Maple equivalent GUESS by FranĀøois BĀ“raud and Bruno Gauthier is available from

c e

http://www-igm.univ-mlv.fr/˜gauthier

5

Maple implementations written by Doron Zeilberger are available from

Mathematica implementations written by

http://www.math.temple.edu/˜zeilberg,

Peter Paule, Axel Riese, Markus Schorn, Kurt Wegschaider are available from

http://www.risc.uni-linz.ac.at/research/combinat/risc/software

6

written in Mathematica by the author; available from http://radon.mat.univie.ac.at/People/kratt

7

written in Maple by Bruno Ghauthier; available from http://www-igm.univ-mlv.fr/˜gauthier

4 C. KRATTENTHALER

The next method which I describe is the so-called ācondensation methodā (see Sec-

tion 2.3), a method which allows to evaluate a determinant inductively (if the method

works).

In Section 2.4, a method, which I call the āidentiļ¬cation of factorsā method, is de-

scribed. This method has been extremely successful recently. It is based on a very

simple idea, which comes from one of the standard proofs of the Vandermonde deter-

minant evaluation (which is therefore described in Section 2.1).

The subject of Section 2.5 is a method which is based on ļ¬nding one or more diļ¬eren-

tial or diļ¬erence equations for the matrix of which the determinant is to be evaluated.

Section 2.6 contains a short description of George Andrewsā™ favourite method, which

basically consists of explicitly doing the LU-factorization of the matrix of which the

determinant is to be evaluated.

The remaining subsections in this section are conceived as a complement to the pre-

ceding. In Section 2.7 a special type of determinants is addressed, Hankel determinants.

(These are determinants of the form det1ā¤i,jā¤n (ai+j ), and are sometimes also called per-

symmetric or TurĀ“nian determinants.) As is explained there, you should expect that a

a

Hankel determinant evaluation is to be found in the domain of orthogonal polynomials

and continued fractions. Eventually, in Section 2.8 a few further, possibly useful results

are exhibited.

Before we ļ¬nally move into the subject, it must be pointed out that the methods

of determinant evaluation as presented here are ordered according to the conditions a

determinant must satisfy so that the method can be applied to it, from āstringentā to

āless stringentā. I. e., ļ¬rst come the methods which require that the matrix of which

the determinant is to be taken satisļ¬es a lot of conditions (usually: it contains a lot of

parameters, at least, implicitly), and in the end comes the method (LU-factorization)

which requires nothing. In fact, this order (of methods) is also the order in which I

recommend that you try them on your determinant. That is, what I suggest is (and

this is the rule I follow):

(0) First try some simple-minded things (row and column operations, Laplace expan-

sion). Do not waste too much time. If you encounter a Hankel-determinant then

see Section 2.7.

(1) If that fails, check whether your determinant is a special case of one of the general

determinants in Sections 2.2 (and 2.1).

(2) If that fails, see if the condensation method (see Section 2.3) works. (If necessary,

try to introduce more parameters into your determinant.)

(3) If that fails, try the āidentiļ¬cation of factorsā method (see Section 2.4). Alterna-

tively, and in particular if your matrix of which you want to ļ¬nd the determinant

is the matrix deļ¬ning a system of diļ¬erential or diļ¬erence equations, try the dif-

ferential/diļ¬erence equation method of Section 2.5. (If necessary, try to introduce

a parameter into your determinant.)

(4) If that fails, try to work out the LU-factorization of your determinant (see Sec-

tion 2.6).

(5) If all that fails, then we are really in trouble. Perhaps you have to put more eļ¬orts

into determinant manipulations (see suggestion (0))? Sometimes it is worthwile

to interpret the matrix whose determinant you want to know as a linear map and

try to ļ¬nd a basis on which this map acts triangularly, or even diagonally (this

ADVANCED DETERMINANT CALCULUS 5

requires that the eigenvalues of the matrix are āniceā; see [47, 48, 84, 93, 192] for

examples where that worked). Otherwise, maybe something from Sections 2.8 or

3 helps?

A ļ¬nal remark: It was indicated that some of the methods require that your deter-

minant contains (more or less) parameters. Therefore it is always a good idea to:

Introduce more parameters into your determinant!

(We address this in more detail in the last paragraph of Section 2.1.) The more param-

eters you can play with, the more likely you will be able to carry out the determinant

evaluation. (Just to mention a few examples: The condensation method needs, at least,

two parameters. The āidentiļ¬cation of factorsā method needs, at least, one parameter,

as well as the diļ¬erential/diļ¬erence equation method in Section 2.5.)

2.1. A few standard determinants. Let us begin with a short proof of the Van-

dermonde determinant evaluation

(Xj ā’ Xi ).

Xijā’1 =

det (2.1)

1ā¤i,jā¤n

1ā¤i<jā¤n

Although the following proof is well-known, it makes still sense to quickly go through

it because, by extracting the essence of it, we will be able to build a very powerful

method out of it (see Section 2.4).

If Xi1 = Xi2 with i1 = i2, then the Vandermonde determinant (2.1) certainly vanishes

because in that case two rows of the determinant are identical. Hence, (Xi1 ā’ Xi2 )

divides the determinant as a polynomial in the Xi ā™s. But that means that the complete

product 1ā¤i<jā¤n (Xj ā’ Xi ) (which is exactly the right-hand side of (2.1)) must divide

the determinant.

On the other hand, the determinant is a polynomial in the Xi ā™s of degree at most

n

. Combined with the previous observation, this implies that the determinant equals

2

the right-hand side product times, possibly, some constant. To compute the constant,

compare coeļ¬cients of X1 X2 Ā· Ā· Ā· Xn on both sides of (2.1). This completes the proof

01 nā’1

of (2.1).

At this point, let us extract the essence of this proof as we will come back to it in

Section 2.4. The basic steps are:

1. Identiļ¬cation of factors

2. Determination of degree bound

3. Computation of the multiplicative constant.

An immediate generalization of the Vandermonde determinant evaluation is given by

the proposition below. It can be proved in just the same way as the above proof of the

Vandermonde determinant evaluation itself.

Proposition 1. Let X1 , X2 , . . . , Xn be indeterminates. If p1 , p2 , . . . , pn are polynomials

of the form pj (x) = aj xjā’1 + lower terms, then

det (pj (Xi )) = a1 a2 Ā· Ā· Ā· an (Xj ā’ Xi ). (2.2)

1ā¤i,jā¤n

1ā¤i<jā¤n

6 C. KRATTENTHALER

The following variations of the Vandermonde determinant evaluation are equally easy

to prove.

Lemma 2. The following identities hold true:

n

Xiā’j ) = (X1 Ā· Ā· Ā· Xn )ā’n

ā’ (Xi ā’ Xj )(1 ā’ Xi Xj ) (Xi2 ā’ 1),

(Xij

det

1ā¤i,jā¤n

i=1

1ā¤i<jā¤n

(2.3)

ā’(jā’1/2)

jā’1/2

ā’ Xi

det (Xi )

1ā¤i,jā¤n

n

= (X1 Ā· Ā· Ā· Xn )ā’n+1/2 (Xi ā’ Xj )(1 ā’ Xi Xj ) (Xi ā’ 1), (2.4)

1ā¤i<jā¤n i=1

ā’(jā’1)

) = 2 Ā· (X1 Ā· Ā· Ā· Xn )ā’n+1 (Xi ā’ Xj )(1 ā’ Xi Xj ) ,

det (Xijā’1 + Xi (2.5)

1ā¤i,jā¤n

1ā¤i<jā¤n

ā’(jā’1/2)

jā’1/2

det (Xi + Xi )

1ā¤i,jā¤n

n

ā’n+1/2

= (X1 Ā· Ā· Ā· Xn ) (Xi ā’ Xj )(1 ā’ Xi Xj ) (Xi + 1). (2.6)

1ā¤i<jā¤n i=1

We remark that the evaluations (2.3), (2.4), (2.5) are basically the Weyl denominator

factorizations of types C, B, D, respectively (cf. [52, Lemma 24.3, Ex. A.52, Ex. A.62,

Ex. A.66]). For that reason they may be called the āsymplecticā, the āodd orthogonalā,

and the āeven orthogonalā Vandermonde determinant evaluation, respectively.

Ī»

If you encounter generalizations of such determinants of the form det1ā¤i,jā¤n (xi j )

ā’Ī»

Ī»

or det1ā¤i,jā¤n (xi j ā’ xi j ), etc., then you should be aware that what you encounter is

basically Schur functions, characters for the symplectic groups, or characters for the

orthogonal groups (consult [52, 105, 137] for more information on these matters; see

in particular [105, Ch. I, (3.1)], [52, p. 403, (A.4)], [52, (24.18)], [52, (24.40) + ļ¬rst

paragraph on p. 411], [137, Appendix A2], [52, (24.28)]). In this context, one has to

also mention Okadaā™s general results on evaluations of determinants and Pfaļ¬ans (see

Section 2.8 for deļ¬nition) in [124, Sec. 4] and [125, Sec. 5].

Another standard determinant evaluation is the evaluation of Cauchyā™s double alter-

nant (see [119, vol. III, p. 311]),

ā’ Xj )(Yi ā’ Yj )

1ā¤i<jā¤n (Xi

1

det = . (2.7)

Xi + Yj (Xi + Yj )

1ā¤i,jā¤n

1ā¤i,jā¤n

Once you have seen the above proof of the Vandermonde determinant evaluation, you

will immediately know how to prove this determinant evaluation.

On setting Xi = i and Yi = i, i = 1, 2, . . . , n, in (2.7), we obtain the evaluation of our

ļ¬rst determinant in the Introduction, (1.1). For the evaluation of a mixture of Cauchyā™s

double alternant and Vandermondeā™s determinant see [15, Lemma 2].

ADVANCED DETERMINANT CALCULUS 7

Whether or not you tried to evaluate (1.1) directly, here is an important lesson to be

learned (it was already mentioned earlier): To evaluate (1.1) directly is quite diļ¬cult,

whereas proving its generalization (2.7) is almost completely trivial. Therefore, it is

always a good idea to try to introduce more parameters into your determinant. (That is,

in a way such that the more general determinant still evaluates nicely.) More parameters

mean that you have more objects at your disposal to play with.

The most stupid way to introduce parameters is to just write Xi instead of the row

index i, or write Yj instead of the column index j.8 For the determinant (1.1) even

both simultaneously was possible. For the determinant (1.2) either of the two (but not

both) would work. On the contrary, there seems to be no nontrivial way to introduce

more parameters in the determinant (1.4). This is an indication that the evaluation of

this determinant is in a diļ¬erent category of diļ¬culty of evaluation. (Also (1.3) belongs

to this ādiļ¬erent categoryā. It is possible to introduce one more parameter, see (3.32),

but it does not seem to be possible to introduce more.)

2.2. A general determinant lemma, plus variations and generalizations.

In this section I present an apparently not so well-known determinant evaluation that

generalizes Vandermondeā™s determinant, and some companions. As Lascoux pointed

out to me, most of these determinant evaluations can be derived from the evaluation

of a certain determinant of minors of a given matrix due to Turnbull [179, p. 505], see

Appendix B. However, this (these) determinant evaluation(s) deserve(s) to be better

known. Apart from the fact that there are numerous applications of it (them) which I

am aware of, my proof is that I meet very often people who stumble across a special

case of this (these) determinant evaluation(s), and then have a hard time to actually

do the evaluation because, usually, their special case does not show the hidden general

structure which is lurking behind. On the other hand, as I will demonstrate in a mo-

ment, if you know this (these) determinant evaluation(s) then it is a matter completely

mechanical in nature to see whether it (they) is (are) applicable to your determinant

or not. If one of them is applicable, you are immediately done.

The determinant evaluation of which I am talking is the determinant lemma from

[85, Lemma 2.2] given below. Here, and in the following, empty products (like (Xi +

An)(Xi + Anā’1 ) Ā· Ā· Ā· (Xi + Aj+1 ) for j = n) equal 1 by convention.

Lemma 3. Let X1 , . . . , Xn , A2, . . . , An, and B2 , . . . , Bn be indeterminates. Then there

holds

(Xi + An)(Xi + Anā’1 ) Ā· Ā· Ā· (Xi + Aj+1 )(Xi + Bj )(Xi + Bjā’1 ) Ā· Ā· Ā· (Xi + B2 )

det

1ā¤i,jā¤n

(Xi ā’ Xj ) (Bi ā’ Aj ). (2.8)

=

1ā¤i<jā¤n 2ā¤iā¤jā¤n

8

Other common examples of introducing more parameters are: Given that the (i, j)-entry of your

determinant is a binomial such as 2iā’j , try x+i+j (that works; see (3.30)), or even x+y+i+j (that

i+j

2iā’j y+2iā’j

x+i+j y+i+j

does not work; but see (1.2)), or 2iā’j + 2iā’j (that works; see (3.32), and consult Lemma 19

and the remarks thereafter). However, sometimes parameters have to be introduced in an unexpected

way, see (3.49). (The parameter x was introduced into a determinant of Bombieri, Hunt and van der

Poorten, which is obtained by setting x = 0 in (3.49).)

8 C. KRATTENTHALER

Once you have guessed such a formula, it is easily proved. In the proof in [85] the

determinant is reduced to a determinant of the form (2.2) by suitable column operations.

Another proof, discovered by Amdeberhan (private communication), is by condensation,

see Section 2.3. For a derivation from the above mentioned evaluation of a determinant

of minors of a given matrix, due to Turnbull, see Appendix B.

Now let us see what the value of this formula is, by checking if it is of any use in the

case of the second determinant in the Introduction, (1.2). The recipe that you should

follow is:

1. Take as many factors out of rows and/or columns of your determinant, so that all

denominators are cleared.

2. Compare your result with the determinant in (2.8). If it matches, you have found

the evaluation of your determinant.

Okay, let us do so:

n

a+b (a + b)!

det =

aā’i+j (a ā’ i + n)! (b + i ā’ 1)!

1ā¤i,jā¤n

i=1

Ć— det (a ā’ i + n)(a ā’ i + n ā’ 1) Ā· Ā· Ā· (a ā’ i + j + 1)

1ā¤i,jā¤n

Ā· (b + i ā’ j + 1)(b + i ā’ j + 2) Ā· Ā· Ā· (b + i ā’ 1)

n

(a + b)!

n

= (ā’1)( 2 )

(a ā’ i + n)! (b + i ā’ 1)!

i=1

Ć— det (i ā’ a ā’ n)(i ā’ a ā’ n + 1) Ā· Ā· Ā· (i ā’ a ā’ j ā’ 1)

1ā¤i,jā¤n

Ā· (i + b ā’ j + 1)(i + b ā’ j + 2) Ā· Ā· Ā· (i + b ā’ 1) .

Now compare with the determinant in (2.8). Indeed, the determinant in the last line is

just the special case Xi = i, Aj = ā’a ā’ j, Bj = b ā’ j + 1. Thus, by (2.8), we have a

result immediately. A particularly attractive way to write it is displayed in (2.17).

Applications of Lemma 3 are abundant, see Theorem 26 and the remarks accompa-

nying it.

In [87, Lemma 7], a determinant evaluation is given which is closely related to

Lemma 3. It was used there to establish enumeration results about shifted plane par-

titions of trapezoidal shape. It is the ļ¬rst result in the lemma below. It is ātailoredā

for the use in the context of q-enumeration. For plain enumeration, one would use the

second result. This is a limit case of the ļ¬rst (replace Xi by q Xi , Aj by ā’q ā’Aj and C

by q C in (2.9), divide both sides by (1 ā’ q)n(nā’1) , and then let q ā’ 1).

Lemma 4. Let X1 , X2 , . . . , Xn , A2, . . . , An be indeterminates. Then there hold

(C/Xi + An )(C/Xi + Anā’1 ) Ā· Ā· Ā· (C/Xi + Aj+1 )

det

1ā¤i,jā¤n

n

Ā· (Xi + An )(Xi + Anā’1 ) Ā· Ā· Ā· (Xi + Aj+1 ) = (Xi ā’ Xj )(1 ā’ C/Xi Xj ),

Aiā’1

i

i=2 1ā¤i<jā¤n

(2.9)

ADVANCED DETERMINANT CALCULUS 9

and

(Xi ā’ An ā’ C)(Xi ā’ Anā’1 ā’ C) Ā· Ā· Ā· (Xi ā’ Aj+1 ā’ C)

det

1ā¤i,jā¤n

Ā· (Xi + An )(Xi + Anā’1 ) Ā· Ā· Ā· (Xi + Aj+1 ) = (Xj ā’ Xi )(C ā’ Xi ā’ Xj ). (2.10)

1ā¤i<jā¤n

(Both evaluations are in fact special cases in disguise of (2.2). Indeed, the (i, j)-entry

of the determinant in (2.9) is a polynomial in Xi + C/Xi , while the (i, j)-entry of the

determinant in (2.10) is a polynomial in Xi ā’ C/2, both of degree n ā’ j.)

The standard application of Lemma 4 is given in Theorem 27.

In [88, Lemma 34], a common generalization of Lemmas 3 and 4 was given. In order

to have a convenient statement of this determinant evaluation, we deļ¬ne the degree

of a Laurent polynomial p(X) = N aixi , M, N ā Z, ai ā R and aN = 0, to be

i=M

deg p := N.

Lemma 5. Let X1 , X2 , . . . , Xn , A2, A3, . . . , An, C be indeterminates. If p0 , p1 , . . . , pnā’1

are Laurent polynomials with deg pj ā¤ j and pj (C/X) = pj (X) for j = 0, 1, . . . , n ā’ 1,

then

(Xi + An )(Xi + Anā’1 ) Ā· Ā· Ā· (Xi + Aj+1 )

det

1ā¤i,jā¤n

Ā· (C/Xi + An)(C/Xi + Anā’1 ) Ā· Ā· Ā· (C/Xi + Aj+1 ) Ā· pjā’1 (Xi )

n n

(Xi ā’ Xj )(1 ā’ C/Xi Xj ) Aiā’1

= piā’1 (ā’Ai) . (2.11)

i

1ā¤i<jā¤n i=1 i=1

Section 3 contains several determinant evaluations which are implied by the above

determinant lemma, see Theorems 28, 30 and 31.

Lemma 3 does indeed come out of the above Lemma 5 by setting C = 0 and

j

pj (X) = (Bk+1 + X).

k=1

Obviously, Lemma 4 is the special case pj ā” 1, j = 0, 1, . . . , n ā’ 1. It is in fact worth

stating the C = 0 case of Lemma 5 separately.

Lemma 6. Let X1 , X2 , . . . , Xn , A2, A3, . . . , An be indeterminates. If p0 , p1 , . . . , pnā’1 are

polynomials with deg pj ā¤ j for j = 0, 1, . . . , n ā’ 1, then

(Xi + An )(Xi + Anā’1 ) Ā· Ā· Ā· (Xi + Aj+1 ) Ā· pjā’1 (Xi )

det

1ā¤i,jā¤n

n

(Xi ā’ Xj )

= piā’1 (ā’Ai) . (2.12)

1ā¤i<jā¤n i=1

10 C. KRATTENTHALER

Again, Lemma 5 is tailored for applications in q-enumeration. So, also here, it may

be convenient to state the according limit case that is suitable for plain enumeration

(and perhaps other applications).

Lemma 7. Let X1 , X2 , . . . , Xn , A2, A3, . . . , An, C be indeterminates. If p0 , p1 , . . . ,

pnā’1 are polynomials with deg pj ā¤ 2j and pj (C ā’ X) = pj (X) for j = 0, 1, . . . , n ā’ 1,

then

(Xi + An )(Xi + Anā’1 ) Ā· Ā· Ā· (Xi + Aj+1 )

det

1ā¤i,jā¤n

Ā· (Xi ā’ An ā’ C)(Xi ā’ Anā’1 ā’ C) Ā· Ā· Ā· (Xi ā’ Aj+1 ā’ C) Ā· pjā’1 (Xi )

n

(Xj ā’ Xi )(C ā’ Xi ā’ Xj )

= piā’1 (ā’Ai) . (2.13)

1ā¤i<jā¤n i=1

In concluding, I want to mention that, now since more than ten years, I have a

diļ¬erent common generalization of Lemmas 3 and 4 (with some overlap with Lemma 5)

in my drawer, without ever having found use for it. Let us nevertheless state it here;

maybe it is exactly the key to the solution of a problem of yours.

Lemma 8. Let X1 , . . . , Xn , A2, . . . , An, B2 , . . . Bn , a2, . . . , an , b2 , . . . bn , and C be in-

determinates. Then there holds

ļ£«ļ£± ļ£¶

ļ£“(Xi + An ) Ā· Ā· Ā· (Xi + Aj+1 )(C/Xi + An ) Ā· Ā· Ā· (C/Xi + Aj+1 )

ļ£“

ļ£¬ļ£“(X + B ) Ā· Ā· Ā· (X + B )(C/X + B ) Ā· Ā· Ā· (C/X + B )

ļ£² j < mļ£·

ļ£¬ ļ£·

i j i 2 i j i 2

det ļ£¬ ļ£·

1ā¤i,jā¤n ļ£ļ£“(Xi + an ) Ā· Ā· Ā· (Xi + aj+1 )(C/Xi + an ) Ā· Ā· Ā· (C/Xi + aj+1 ) ļ£ø

ļ£“

ļ£“

ļ£³

(Xi + bj ) Ā· Ā· Ā· (Xi + b2 )(C/Xi + bj ) Ā· Ā· Ā· (C/Xi + b2) jā„m

(Xi ā’ Xj )(1 ā’ C/Xi Xj ) (Bi ā’ Aj )(1 ā’ C/Bi Aj )

=

1ā¤i<jā¤n 2ā¤iā¤jā¤mā’1

m n

Ć— (bi ā’ Aj )(1 ā’ C/bi Aj ) (bi ā’ aj )(1 ā’ C/bi aj )

i=2 j=m m+1ā¤iā¤jā¤n

m n mā’1 n

Ć— (Ai Ā· Ā· Ā· An ) (ai Ā· Ā· Ā· an) (B2 Ā· Ā· Ā· Bi ) (b2 Ā· Ā· Ā· bi). (2.14)

i=2 i=m+1 i=2 i=m

The limit case which goes with this determinant lemma is the following. (There is

some overlap with Lemma 7.)

ADVANCED DETERMINANT CALCULUS 11

Lemma 9. Let X1 , . . . , Xn, A2, . . . , An, B2 , . . . , Bn, a2, . . . , an , b2, . . . , bn , and C be

indeterminates. Then there holds

ļ£«ļ£± ļ£¶

(Xi + An ) Ā· Ā· Ā· (Xi + Aj+1 )(Xi ā’ An ā’ C) Ā· Ā· Ā· (Xi ā’ Aj+1 ā’ C)

ļ£“

ļ£“

ļ£¬ļ£“(X + B ) Ā· Ā· Ā· (X + B )(X ā’ B ā’ C) Ā· Ā· Ā· (X ā’ B ā’ C)

ļ£² j < mļ£·

ļ£¬ ļ£·

i j i 2 i j i 2

det ļ£¬ ļ£·

1ā¤i,jā¤n ļ£ļ£“(Xi + an ) Ā· Ā· Ā· (Xi + aj+1 )(Xi ā’ an ā’ C) Ā· Ā· Ā· (Xi ā’ aj+1 ā’ C) ļ£ø

ļ£“

ļ£“

ļ£³

(Xi + bj ) Ā· Ā· Ā· (Xi + b2 )(Xi ā’ bj ā’ C) Ā· Ā· Ā· (Xi ā’ b2 ā’ C) jā„m

(Xi ā’ Xj )(C ā’ Xi ā’ Xj ) (Bi ā’ Aj )(Bi + Aj + C)

=

1ā¤i<jā¤n 2ā¤iā¤jā¤mā’1

m n

Ć— (bi ā’ Aj )(bi + Aj + C) (bi ā’ aj )(bi + aj + C). (2.15)

i=2 j=m m+1ā¤iā¤jā¤n

If you are looking for more determinant evaluations of such a general type, then you

may want to look at [156, Lemmas A.1 and A.11] and [158, Lemma A.1].

2.3. The condensation method. This is Doron Zeilbergerā™s favourite method. It

allows (sometimes) to establish an elegant, eļ¬ortless inductive proof of a determinant

evaluation, in which the only task is to guess the result correctly.

The method is often attributed to Charles Ludwig Dodgson [38], better known as

Lewis Carroll. However, the identity on which it is based seems to be actually due to

P. Desnanot (see [119, vol. I, pp. 140ā“142]; with the ļ¬rst rigorous proof being probably

due to Jacobi, see [18, Ch. 4] and [79, Sec. 3]). This identity is the following.

Proposition 10. Let A be an n Ć— n matrix. Denote the submatrix of A in which rows

i1, i2, . . . , ik and columns j1 , j2 , . . . , jk are omitted by Aj1 ,i2 ,...,ikk . Then there holds

1 ,j2 ,...,j

i

det A Ā· det A1,n = det A1 Ā· det An ā’ det An Ā· det A1 . (2.16)

1,n 1 n 1 n

So, what is the point of this identity? Suppose you are given a family (det Mn )nā„0

of determinants, Mn being an n Ć— n matrix, n = 0, 1, . . . . Maybe Mn = Mn (a, b)

is the matrix underlying the determinant in (1.2). Suppose further that you have

already worked out a conjecture for the evaluation of det Mn (a, b) (we did in fact already

evaluate this determinant in Section 2.2, but let us ignore that for the moment),

n a b

i+j+kā’1

a+b ?

det Mn (a, b) := det = . (2.17)

aā’i+j i+j+kā’2

1ā¤i,jā¤n

i=1 j=1 k=1

Then you have already proved your conjecture, once you observe that

n

Mn (a, b) = Mnā’1 (a, b),

n

1

Mn(a, b) 1 = Mnā’1 (a, b),

1

= Mnā’1 (a + 1, b ā’ 1),

Mn (a, b) n

n

= Mnā’1 (a ā’ 1, b + 1),

Mn (a, b) 1

1,n

Mn (a, b) 1,n = Mnā’2 (a, b). (2.18)

12 C. KRATTENTHALER

For, because of (2.18), Desnanotā™s identity (2.16), with A = Mn (a, b), gives a recur-

rence which expresses det Mn (a, b) in terms of quantities of the form det Mnā’1 ( . ) and

det Mnā’2 ( . ). So, it just remains to check the conjecture (2.17) for n = 0 and n = 1, and

that the right-hand side of (2.17) satisļ¬es the same recurrence, because that completes

a perfect induction with respect to n. (What we have described here is basically the

contents of [197]. For a bijective proof of Proposition 10 see [200].)

Amdeberhan (private communication) discovered that in fact the determinant evalu-

ation (2.8) itself (which we used to evaluate the determinant (1.2) for the ļ¬rst time) can

be proved by condensation. The reader will easily ļ¬gure out the details. Furthermore,

the condensation method also proves the determinant evaluations (3.35) and (3.36).

(Also this observation is due to Amdeberhan [2].) At another place, condensation was

used by EisenkĀØlbl [41] in order to establish a conjecture by Propp [138, Problem 3]

o

about the enumeration of rhombus tilings of a hexagon where some triangles along the

border of the hexagon are missing.

The reader should observe that crucial for a successful application of the method is

the existence of (at least) two parameters (in our example these are a and b), which

help to still stay within the same family of matrices when we take minors of our original

matrix (compare (2.18)). (See the last paragraph of Section 2.1 for a few hints of how

to introduce more parameters into your determinant, in the case that you are short of

parameters.) Obviously, aside from the fact that we need at least two parameters, we

can hope for a success of condensation only if our determinant is of a special kind.

2.4. The āidentification of factorsā method. This is the method that I ļ¬nd

most convenient to work with, once you encounter a determinant that is not amenable

to an evaluation using the previous recipes. It is best to explain this method along with

an example. So, let us consider the determinant in (1.3). Here it is, together with its,

at this point, unproven evaluation,

Āµ+i+j

det

2i ā’ j

0ā¤i,jā¤nā’1

ā’Āµ ā’ 3n + i + 3

nā’1 (Āµ + i + 1) (i+1)/2

nā’1 2

2( ) i/2

= (ā’1)Ļ(nā”3 mod 4)

, (2.19)

2

(i)i

i=1

where Ļ(A) = 1 if A is true and Ļ(A) = 0 otherwise, and where the shifted factorial

(a)k is deļ¬ned by (a)k := a(a + 1) Ā· Ā· Ā· (a + k ā’ 1), k ā„ 1, and (a)0 := 1.

As was already said in the Introduction, this determinant belongs to a diļ¬erent

category of diļ¬culty of evaluation, so that nothing what was presented so far will

immediately work on that determinant.

Nevertheless, I claim that the procedure which we chose to evaluate the Vandermonde

determinant works also with the above determinant. To wit:

1. Identiļ¬cation of factors

2. Determination of degree bound

3. Computation of the multiplicative constant.

You will say: ā˜A moment please! The reason that this procedure worked so smoothly

for the Vandermonde determinant is that there are so many (to be precise: n) variables

at our disposal. On the contrary, the determinant in (2.19) has exactly one (!) variable.ā™

ADVANCED DETERMINANT CALCULUS 13

Yet ā” and this is the point that I want to make here ā” it works, in spite of having

just one variable at our disposal!.

What we want to prove in the ļ¬rst step is that the right-hand side of (2.19) divides the

determinant. For example, we would like to prove that (Āµ + n) divides the determinant

(actually, (Āµ + n) (n+1)/3 , we will come to that in a moment). Equivalently, if we set

Āµ = ā’n in the determinant, then it should vanish. How could we prove that? Well, if

it vanishes then there must be a linear combination of the columns, or of the rows, that

vanishes. So, let us ļ¬nd such a linear combination of columns or rows. Equivalently, for

Āµ = ā’n we ļ¬nd a vector in the kernel of the matrix in (2.19), respectively its transpose.

More generally (and this addresses that we actually want to prove that (Āµ + n) (n+1)/3

divides the determinant):

For proving that (Āµ + n)E divides the determinant, we ļ¬nd E linear inde-

pendent vectors in the kernel.

(For a formal justiļ¬cation that this does indeed suļ¬ce, see Section 2 of [91], and in

particular the Lemma in that section.)

Okay, how is this done in practice? You go to your computer, crank out these vectors

in the kernel, for n = 1, 2, 3, . . . , and try to make a guess what they are in general.

To see how this works, let us do it in our example. What the computer gives is the

following (we are using Mathematica here):

In[1]:= V[2]

Out[1]= {0, c[1]}

In[2]:= V[3]

Out[2]= {0, c[2], c[2]}

In[3]:= V[4]

Out[3]= {0, c[1], 2 c[1], c[1]}

In[4]:= V[5]

Out[4]= {0, c[1], 3 c[1], c[3], c[1]}

In[5]:= V[6]

Out[5]= {0, c[1], 4 c[1], 2 c[1] + c[4], c[4], c[1]}

In[6]:= V[7]

Out[6]= {0, c[1], 5 c[1], c[3], -10 c[1] + 2 c[3], -5 c[1] + c[3], c[1]}

In[7]:= V[8]

Out[7]= {0, c[1], 6 c[1], c[3], -25 c[1] + 3 c[3], c[5], -9 c[1] + c[3], c[1]}

In[8]:= V[9]

Out[8]= {0, c[1], 7 c[1], c[3], -49 c[1] + 4 c[3],

-28 c[1] + 2 c[3] + c[6], c[6], -14 c[1] + c[3], c[1]}

In[9]:= V[10]

14 C. KRATTENTHALER

Out[9]= {0, c[1], 8 c[1], c[3], -84 c[1] + 5 c[3], c[5],

196 c[1] - 10 c[3] + 2 c[5], 98 c[1] - 5 c[3] + c[5], -20 c[1] + c[3],

c[1]}

In[10]:= V[11]

Out[10]= {0, c[1], 9 c[1], c[3], -132 c[1] + 6 c[3], c[5],

648 c[1] - 25 c[3] + 3 c[5], c[7], 234 c[1] - 9 c[3] + c[5],

-27 c[1] + c[3], c[1]}

Here, V [n] is the generic vector (depending on the indeterminates c[i]) in the kernel of

the matrix in (2.19) with Āµ = ā’n. For convenience, let us denote this matrix by Mn .

You do not have to stare at these data for long to see that, in particular,

the vector (0, 1) is in the kernel of M2 ,

the vector (0, 1, 1) is in the kernel of M3,

the vector (0, 1, 2, 1) is in the kernel of M4 ,

the vector (0, 1, 3, 3, 1) is in the kernel of M5 (set c[1] = 1 and c[3] = 3),

the vector (0, 1, 4, 6, 4, 1) is in the kernel of M6 (set c[1] = 1 and c[4] = 4), etc.

Apparently,

nā’2 nā’2 nā’2 nā’2

0, , , ,..., (2.20)

0 1 2 nā’2

is in the kernel of Mn . That was easy! But we need more linear combinations. Take

a closer look, and you will see that the pattern persists (set c[1] = 0 everywhere, etc.).

It will take you no time to work out a full-ļ¬‚edged conjecture for (n + 1)/3 linear

independent vectors in the kernel of Mn.

Of course, there remains something to be proved. We need to actually prove that our

guessed vectors are indeed in the kernel. E.g., in order to prove that the vector (2.20)

is in the kernel, we need to verify that

nā’1

nā’2 ā’n + i + j

=0

jā’1 2i ā’ j

j=1

for i = 0, 1, . . . , n ā’ 1. However, verifying binomial identities is pure routine today, by

means of Zeilbergerā™s algorithm [194, 196] (see Footnote 5 in the Introduction).

Next you perform the same game with the other factors of the right-hand side product

of (2.19). This is not much more diļ¬cult. (See Section 3 of [91] for details. There,

slightly diļ¬erent vectors are used.)

Thus, we would have ļ¬nished the ļ¬rst step, āidentiļ¬cation of factors,ā of our plan: We

have proved that the right-hand side of (2.19) divides the determinant as a polynomial

in Āµ.

The second step, ādetermination of degree bound,ā consists of determining the (max-

imal) degree in Āµ of determinant and conjectured result. As is easily seen, this is n 2

in each case.

The arguments thus far show that the determinant in (2.19) must equal the right-

hand side times, possibly, some constant. To determine this constant in the third

n

step, ācomputation of the multiplicative constant,ā one compares coeļ¬cients of x( 2 ) on

ADVANCED DETERMINANT CALCULUS 15

both sides of (2.19). This is an enjoyable exercise. (Consult [91] if you do not want

to do it yourself.) Further successful applications of this procedure can be found in

[27, 30, 42, 89, 90, 92, 94, 97, 132].

Having done that, let me point out that most of the individual steps in this sort of

calculation can be done (almost) automatically. In detail, what did we do? We had to

1. Guess the result. (Indeed, without the result we could not have got started.)

2. Guess the vectors in the kernel.

3. Establish a binomial (hypergeometric) identity.

4. Determine a degree bound.

5. Compute a particular value or coeļ¬cient in order to determine the multiplicative

constant.

As I explain in Appendix A, guessing can be largely automatized. It was already

mentioned in the Introduction that proving binomial (hypergeometric) identities can

be done by the computer, thanks to the āWZ-machineryā [130, 190, 194, 195, 196] (see

Footnote 5). Computing the degree bound is (in most cases) so easy that no computer is

needed. (You may use it if you want.) It is only the determination of the multiplicative

constant (item 5 above) by means of a special evaluation of the determinant or the

n

evaluation of a special coeļ¬cient (in our example we determined the coeļ¬cient of Āµ( 2 ) )

for which I am not able to oļ¬er a recipe so that things could be carried out on a

computer.

The reader should notice that crucial for a successful application of the method

is the existence of (at least) one parameter (in our example this is Āµ) to be able to

apply the polynomiality arguments that are the āengineā of the method. If there is no

parameter (such as in the determinant in Conjecture 49, or in the determinant (3.46)

which would solve the problem of q-enumerating totally symmetric plane partitions),

then we even cannot get started. (See the last paragraph of Section 2.1 for a few hints

of how to introduce a parameter into your determinant, in the case that you are short

of a parameter.)

On the other hand, a signiļ¬cant advantage of the āidentiļ¬cation of factors methodā

is that not only is it capable of proving evaluations of the form

det(M) = CLOSED FORM,

(where CLOSED FORM means a product/quotient of āniceā factors, such as (2.19) or

(2.17)), but also of proving evaluations of the form

det(M) = (CLOSED FORM) Ć— (UGLY POLYNOMIAL), (2.21)

where, of course, M is a matrix containing (at least) one parameter, Āµ say. Exam-

ples of such determinant evaluations are (3.38), (3.39), (3.45) or (3.48). (The UGLY

POLYNOMIAL in (3.38), (3.39) and (3.48) is the respective sum on the right-hand

side, which in neither case can be simpliļ¬ed).

How would one approach the proof of such an evaluation? For one part, we already

know. āIdentiļ¬cation of factorsā enables us to show that (CLOSED FORM) divides

det(M) as a polynomial in Āµ. Then, comparison of degrees in Āµ on both sides of

(2.21) yields that (UGLY POLYNOMIAL) is a (at this point unknown) polynomial in

16 C. KRATTENTHALER

Āµ of some maximal degree, m say. How can we determine this polynomial? Nothing

āsimplerā than that: We ļ¬nd m + 1 values e such that we are able to evaluate det(M)

at Āµ = e. If we then set Āµ = e in (2.21) and solve for (UGLY POLYNOMIAL), then we

obtain evaluations of (UGLY POLYNOMIAL) at m + 1 diļ¬erent values of Āµ. Clearly,

this suļ¬ces to ļ¬nd (UGLY POLYNOMIAL), e.g., by Lagrange interpolation.

I put āsimplerā in quotes, because it is here where the crux is: We may not be able

to ļ¬nd enough such special evaluations of det(M). In fact, you may object: ā˜Why all

these complications? If we should be able to ļ¬nd m + 1 special values of Āµ for which

we are able to evaluate det(M), then what prevents us from evaluating det(M) as a

whole, for generic Āµ?ā™ When I am talking of evaluating det(M) for Āµ = e, then what I

have in mind is that the evaluation of det(M) at Āµ = e is āniceā (i.e., gives a āclosed

form,ā with no āuglyā expression involved, such as in (2.21)), which is easier to identify

(that is, to guess; see Appendix A) and in most cases easier to prove. By experience,

such evaluations are rare. Therefore, the above described procedure will only work if

the degree of (UGLY POLYNOMIAL) is not too large. (If you are just a bit short of

evaluations, then ļ¬nding other informations about (UGLY POLYNOMIAL), like the

leading coeļ¬cient, may help to overcome the problem.)

To demonstrate this procedure by going through a concrete example is beyond the

scope of this article. We refer the reader to [28, 43, 50, 51, 89, 90] for places where this

procedure was successfully used to solve diļ¬cult enumeration problems on rhombus

tilings, respectively prove a conjectured constant term identity.

2.5. A differential/difference equation method. In this section I outline a

method for the evaluation of determinants, often used by Vitaly Tarasov and Alexander

Varchenko, which, as the preceding method, also requires (at least) one parameter.

Suppose we are given a matrix M = M(z), depending on the parameter z, of which

we want to compute the determinant. Furthermore, suppose we know that M satisļ¬es

a diļ¬erential equation of the form

d

M(z) = T (z)M(z), (2.22)

dz

where T (z) is some other known matrix. Then, by elementary linear algebra, we obtain

a diļ¬erential equation for the determinant,

d

det M(z) = Tr(T (z)) Ā· det M(z), (2.23)

dz

which is usually easy to solve. (In fact, the diļ¬erential operator in (2.22) and (2.23)

could be replaced by any operator. In particular, we could replace d/dz by the diļ¬erence

operator with respect to z, in which case (2.23) is usually easy to solve as well.)

Any method is best illustrated by an example. Let us try this method on the deter-

minant (1.2). Right, we did already evaluate this determinant twice (see Sections 2.2

and 2.3), but let us pretend that we have forgotten all this.

Of course, application of the method to (1.2) itself does not seem to be extremely

promising, because that would involve the diļ¬erentiation of binomial coeļ¬cients. So,

ADVANCED DETERMINANT CALCULUS 17

let us ļ¬rst take some factors out of the determinant (as we also did in Section 2.2),

n

a+b (a + b)!

det =

aā’i+j (a ā’ i + n)! (b + i ā’ 1)!

1ā¤i,jā¤n

i=1

Ć— det (a ā’ i + n)(a ā’ i + n ā’ 1) Ā· Ā· Ā· (a ā’ i + j + 1)

1ā¤i,jā¤n

Ā· (b + i ā’ j + 1)(b + i ā’ j + 2) Ā· Ā· Ā· (b + i ā’ 1) .

Let us denote the matrix underlying the determinant on the right-hand side of this

equation by Mn (a). In order to apply the above method, we have need for a matrix

Tn (a) such that

d

Mn (a) = Tn(a)Mn(a). (2.24)

da

Similar to the procedure of Section 2.6, the best idea is to go to the computer, crank

out Tn (a) for n = 1, 2, 3, 4, . . . , and, out of the data, make a guess for Tn (a). Indeed, it

suļ¬ces that I display T5(a),

ļ£«1

ā’ 3+a+b + 4+a+b

1 1 1 4 6 6

+ 2+a+b + 3+a+b + 4+a+b

1+a+b 4+a+b

ļ£¬ 1 1 1 3

0 + 2+a+b + 3+a+b

ļ£¬ 1+a+b 3+a+b

ļ£¬ 1 1

0 0 + 2+a+b

ļ£¬ 1+a+b

ļ£ 0 0 0

0 0 0

1ļ£¶

ā’ 3+a+b + 4+a+b ā’ 1+a+b + 2+a+b ā’ 3+a+b + 4+a+b

4 8 4 1 3 3

2+a+b

ļ£·

ā’ 2+a+b + 3+a+b ā’ 2+a+b + 3+a+b

3 3 1 2 1

ļ£·

1+a+b

ļ£·

ā’ 1+a+b + 2+a+b

2 1 1

ļ£·

2+a+b

ļ£ø

1 1

1+a+b 1+a+b

0 0

(in this display, the ļ¬rst line contains columns 1, 2, 3 of T5(a), while the second line

contains the remaining columns), so that you are forced to conclude that, apparently,

it must be true that

nā’iā’1

nā’i j ā’iā’1 (ā’1)k

Tn (a) = .

jā’i a+b+nā’iā’k

k

k=0 1ā¤i,j,ā¤n

That (2.24) holds with this choice of Tn(a) is then easy to verify. Consequently, by

means of (2.23), we have

nā’1

nā’

d

det Mn (a) = det Mn (a),

da a+b+

=1

so that

nā’1

Mn(a) = constant Ā· (a + b + )nā’ . (2.25)

=1

n

The constant is found to be (ā’1)( 2 ) nā’1 !, e.g., by dividing both sides of (2.25) by

=0

n

a(2 ) , letting a tend to inļ¬nity, and applying (2.2) to the remaining determinant.

18 C. KRATTENTHALER

More sophisticated applications of this method (actually, of a version for systems of

diļ¬erence operators) can be found in [175, Proof of Theorem 5.14] and [176, Proofs of

Theorems 5.9, 5.10, 5.11], in the context of the Knizhnikā“Zamolodchikov equations.

2.6. LU-factorization. This is George Andrewsā™ favourite method. Starting point

is the well-known fact (see [53, p. 33ļ¬]) that, given a square matrix M, there exists,

under suitable, not very stringent conditions (in particular, these are satisļ¬ed if all

top-left principal minors of M are nonzero), a unique lower triangular matrix L and a

unique upper diagonal matrix U, the latter with all entries along the diagonal equal to

1, such that

M = L Ā· U. (2.26)

This unique factorization of the matrix M is known as the L(ower triangular)U(pper

triangular)-factorization of M, or as well as the GauĆ decomposition of M.

Equivalently, for a square matrix M (satisfying these conditions) there exists a unique

lower triangular matrix L and a unique upper triangular matrix U, the latter with all

entries along the diagonal equal to 1, such that

M Ā· U = L. (2.27)

Clearly, once you know L and U, the determinant of M is easily computed, as it equals

the product of the diagonal entries of L.

Now, let us suppose that we are given a family (Mn )nā„0 of matrices, where Mn is

an n Ć— n matrix, n = 0, 1, . . . , of which we want to compute the determinant. Maybe

Mn is the determinant in (1.3). By the above, we know that (normally) there exist

uniquely determined matrices Ln and Un, n = 0, 1, . . . , Ln being lower triangular, Un

being upper triangular with all diagonal entries equal to 1, such that

Mn Ā· Un = Ln . (2.28)

However, we do not know what the matrices Ln and Un are. What George Andrews

does is that he goes to his computer, cranks out Ln and Un for n = 1, 2, 3, 4, . . . (this

just amounts to solving a system of linear equations), and, out of the data, tries to

guess what the coeļ¬cients of the matrices Ln and Un are. Once he has worked out a

guess, he somehow proves that his guessed matrices Ln and Un do indeed satisfy (2.28).

This program is carried out in [10] for the family of determinants in (1.3). As it turns

out, guessing is really easy, while the underlying hypergeometric identities which are

needed for the proof of (2.28) are (from a hypergeometric viewpoint) quite interesting.

For a demonstration of the method of LU-factorization, we will content ourselves

here with trying the method on the Vandermonde determinant. That is, let Mn be the

determinant in (2.1). We go to the computer and crank out the matrices Ln and Un

for small values of n. For the purpose of guessing, it suļ¬ces that I just display the

matrices L5 and U5 . They are

ADVANCED DETERMINANT CALCULUS 19

ļ£«

1 0 0

ļ£¬1 (X2 ā’ X1 ) 0

ļ£¬

L5 = ļ£¬1 (X3 ā’ X1 ) (X3 ā’ X1 )(X3 ā’ X2 )

ļ£¬

ļ£1 (X4 ā’ X1 ) (X4 ā’ X1 )(X4 ā’ X2 )

(X5 ā’ X1 ) (X5 ā’ X1 )(X5 ā’ X2 )

1

ļ£¶

0 0

ļ£·

0 0 ļ£·

ļ£·

0 0 ļ£·

ļ£ø

(X4 ā’ X1 )(X4 ā’ X2 )(X4 ā’ X3 ) 0

(X5 ā’ X1 )(X5 ā’ X2 )(X5 ā’ X3 ) (X5 ā’ X1 )(X5 ā’ X2 )(X5 ā’ X3 )(X5 ā’ X4 )

(in this display, the ļ¬rst line contains columns 1, 2, 3 of L5 , while the second line contains

the remaining columns), and

ļ£« ļ£¶

ā’e1 (X1) ā’e3 (X1 , X2 , X3 )

1 e2 (X1 , X2 ) e4 (X1 , X2 , X3 , X4 )

ļ£¬0 ā’e3 (X1 , X2 , X3 , X4 )ļ£·

ā’e1 (X1 , X2 )

1 e2 (X1 , X2 , X3 )

ļ£¬ ļ£·

U5 = ļ£¬0 e2 (X1 , X2 , X3 , X4 )ļ£· ,

ā’e1 (X1 , X2 , X3 )

0 1

ļ£¬ ļ£·

ļ£0 ā’e1 (X1 , X2 , X3 , X4 )ļ£ø

0 0 1

0 0 0 0 1

1ā¤i1 <Ā·Ā·Ā·<im ā¤s Xi1 Ā· Ā· Ā· Xim denotes the m-th elementary

where em(X1 , X2 , . . . , Xs ) =

symmetric function.

Having seen that, it will not take you for long to guess that, apparently, Ln is given

by

jā’1

(Xi ā’ Xk )

Ln = ,

1ā¤i,jā¤n

k=1

and that Un is given by

Un = (ā’1)jā’i ejā’i (X1 , . . . , Xjā’1 ) ,

1ā¤i,jā¤n

where, of course, em (X1 , . . . ) := 0 if m < 0. That (2.28) holds with these choices of Ln

and Un is easy to verify. Thus, the Vandermonde determinant equals the product of

diagonal entries of Ln , which is exactly the product on the right-hand side of (2.1).

Applications of LU-factorization are abundant in the work of George Andrews [4,

5, 6, 7, 8, 10]. All of them concern solutions to diļ¬cult enumeration problems on

various types of plane partitions. To mention another example, Aomoto and Kato [11,

Theorem 3] computed the LU-factorization of a matrix which arose in the theory of

q-diļ¬erence equations, thus proving a conjecture by Mimachi [118].

Needless to say that this allows for variations. You may try to guess (2.26) directly

(and not its variation (2.27)), or you may try to guess the U(pper triangular)L(ower

triangular) factorization, or its variation in the style of (2.27). I am saying this because

it may be easy to guess the form of one of these variations, while it can be very diļ¬cult

to guess the form of another.

It should be observed that the way LU-factorization is used here in order to evaluate

determinants is very much in the same spirit as āidentiļ¬cation of factorsā as described in

the previous section. In both cases, the essential steps are to ļ¬rst guess something, and

then prove the guess. Therefore, the remarks from the previous section about guessing

20 C. KRATTENTHALER

and proving binomial (hypergeometric) identities apply here as well. In particular, for

guessing you are once more referred to Appendix A.

It is important to note that, as opposed to ācondensationā or āidentiļ¬cation of fac-

tors,ā LU-factorization does not require any parameter. So, in principle, it is applicable

to any determinant (which satisļ¬es the aforementioned conditions). If there are limita-

tions, then, from my experience, it is that the coeļ¬cients which have to be guessed in

LU-factorization tend to be more complicated than in āidentiļ¬cation of factorsā. That

is, guessing (2.28) (or one of its variations) may sometimes be not so easy.

2.7. Hankel determinants. A Hankel determinant is a determinant of a matrix

which has constant entries along antidiagonals, i.e., it is a determinant of the form

det (ci+j ).

1ā¤i,j,ā¤n

If you encounter a Hankel determinant, which you think evaluates nicely, then expect

the evaluation of your Hankel determinant to be found within the domain of continued

fractions and orthogonal polynomials. In this section I explain what this connection is.

To make things concrete, let us suppose that we want to evaluate

det (Bi+j+2 ), (2.29)

0ā¤i,jā¤nā’1

where Bk denotes the k-th Bernoulli number. (The Bernoulli numbers are deļ¬ned via

their generating function, ā Bk z k /k! = z/(ez ā’ 1).) You have to try hard if you

k=0

want to ļ¬nd an evaluation of (2.29) explicitly in the literature. Indeed, you can ļ¬nd

it, hidden in Appendix A.5 of [108]. However, even if you are not able to discover this

reference (which I would not have as well, unless the author of [108] would not have

drawn my attention to it), there is a rather straight-forward way to ļ¬nd an evaluation

of (2.29), which I outline below. It is based on the fact, and this is the main point of

this section, that evaluations of Hankel determinants like (2.29) are, at least implicitly,

in the literature on the theory of orthogonal polynomials and continued fractions, which

is very accessible today.

So, let us review the relevant facts about orthogonal polynomials and continued

fractions (see [76, 81, 128, 174, 186, 188] for more information on these topics).

We begin by citing the result, due to Heilermann, which makes the connection be-

tween Hankel determinants and continued fractions.

Theorem 11. (Cf. [188, Theorem 51.1] or [186, Corollaire 6, (19), on p. IV-17]). Let

ā k

(Āµk )kā„0 be a sequence of numbers with generating function k=0 Āµk x written in the

form

ā

Āµ0

Āµk xk = . (2.30)

b1 x2

1 + a0 x ā’

k=0

b2 x2

1 + a1 x ā’

1 + a2 x ā’ Ā· Ā· Ā·

Then the Hankel determinant det0ā¤i,jā¤nā’1 (Āµi+j ) equals Āµn bnā’1 bnā’2 Ā· Ā· Ā· b2 bnā’1 .

01 2 nā’2

(We remark that a continued fraction of the type as in (2.30) is called a J-fraction.)

Okay, that means we would have evaluated (2.29) once we are able to explicitly

expand the generating function ā Bk+2 xk in terms of a continued fraction of the

k=0

ADVANCED DETERMINANT CALCULUS 21

form of the right-hand side of (2.30). Using the tools explained in Appendix A, it is

easy to work out a conjecture,

ā

1/6

Bk+2 xk = , (2.31)

b1 x2

1ā’

k=0

b2 x2

1ā’

1ā’ Ā·Ā·Ā·

where bi = ā’i(i + 1)2 (i + 2)/4(2i + 1)(2i + 3), i = 1, 2, . . . . If we would ļ¬nd this

expansion in the literature then we would be done. But if not (which is the case here),

how to prove such an expansion? The key is orthogonal polynomials.

A sequence (pn (x))nā„0 of polynomials is called (formally) orthogonal if pn (x) has de-

gree n, n = 0, 1, . . . , and if there exists a linear functional L such that L(pn (x)pm (x)) =

Ī“mncn for some sequence (cn )nā„0 of nonzero numbers, with Ī“m,n denoting the Kronecker

delta (i.e., Ī“m,n = 1 if m = n and Ī“m,n = 0 otherwise).

The ļ¬rst important theorem in the theory of orthogonal polynomials is Favardā™s

Theorem, which gives an unexpected characterization for sequences of orthogonal poly-

nomials, in that it completely avoids the mention of the functional L.

Theorem 12. (Cf. [186, ThĀ“or`me 9 on p. I-4] or [188, Theorem 50.1]). Let (pn (x))nā„0

ee

be a sequence of monic polynomials, the polynomial pn (x) having degree n, n = 0, 1, . . . .

Then the sequence (pn (x)) is (formally) orthogonal if and only if there exist sequences

(an )nā„1 and (bn )nā„1 , with bn = 0 for all n ā„ 1, such that the three-term recurrence

pn+1 (x) = (an + x)pn(x) ā’ bnpnā’1 (x), for n ā„ 1, (2.32)

holds, with initial conditions p0 (x) = 1 and p1 (x) = x + a0 .

What is the connection between orthogonal polynomials and continued fractions?

This question is answered by the next theorem, the link being the generating function

of the moments.

Theorem 13. (Cf. [188, Theorem 51.1] or [186, Proposition 1, (7), on p. V-5]). Let

(pn (x))nā„0 be a sequence of monic polynomials, the polynomial pn (x) having degree n,

which is orthogonal with respect to some functional L. Let

pn+1 (x) = (an + x)pn (x) ā’ bnpnā’1 (x) (2.33)

be the corresponding three-term recurrence which is guaranteed by Favardā™s theorem.

Then the generating function ā Āµk xk for the moments Āµk = L(xk ) satisļ¬es (2.30)

k=0

with the ai ā™s and bi ā™s being the coeļ¬cients in the three-term recurrence (2.33).

Thus, what we have to do is to ļ¬nd orthogonal polynomials (pn (x))nā„0 , the three-term

recurrence of which is explicitly known, and which are orthogonal with respect to some

linear functional L whose moments L(xk ) are exactly equal to Bk+2 . So, what would

be very helpful at this point is some sort of table of orthogonal polynomials. Indeed,

there is such a table for hypergeometric and basic hypergeometric orthogonal polynomi-

als, proposed by Richard Askey (therefore called the āAskey tableā), and compiled by

Koekoek and Swarttouw [81].

Indeed, in Section 1.4 of [81], we ļ¬nd the family of orthogonal polynomials that is

of relevance here, the continuous Hahn polynomials, ļ¬rst studied by Atakishiyev and

Suslov [13] and Askey [12]. These polynomials depend on four parameters, a, b, c, d. It

22 C. KRATTENTHALER

is just the special choice a = b = c = d = 1 which is of interest to us. The theorem

below lists the relevant facts about these special polynomials.

Theorem 14. The continuous Hahn polynomials with parameters a = b = c = d = 1,

(pn (x))nā„0 , are the monic polynomials deļ¬ned by

ā

ā

ā (ā’n)k (n + 3)k (1 + x ā’1)k

2

n (n + 1)! (n + 2)!

pn (x) = ( ā’1) , (2.34)

k! (k + 1)!2

(2n + 2)! k=0

with the shifted factorial (a)k deļ¬ned as previously (see (2.19)). These polynomials

satisfy the three-term recurrence

n(n + 1)2 (n + 2)

pn+1 (x) = xpn (x) + pnā’1 (x). (2.35)

4(2n + 1)(2n + 3)

They are orthogonal with respect to the functional L which is given by

Ļā x2

L(p(x)) = p(x) dx . (2.36)

2 ā’ā sinh2 (Ļx)

Explicitly, the orthogonality relation is

n! (n + 1)!4 (n + 2)!

L(pm (x)pn (x)) = Ī“m,n . (2.37)

(2n + 2)! (2n + 3)!

In particular, L(1) = 1/6.

Now, by combining Theorems 11, 13, and 14, and by using an integral representation

of Bernoulli numbers (see [122, p. 75]),

ā

ā ā’1

1 Ļ 2

ā Ī½

BĪ½ = z dz

ā

2Ļ ā’1 sin Ļz

ā’ā ā’1

(if Ī½ = 0 or Ī½ = 1 then the path of integration is indented so that it avoids the

singularity z = 0, passing it on the negative side) we obtain without diļ¬culty the

desired determinant evaluation,

n nā’1 nā’i

i(i + 1)2 (i + 2)

1

n

(Bi+j+2 ) = (ā’1)( )

det 2

6 4(2i + 1)(2i + 3)

0ā¤i,j,ā¤nā’1

i=1

nā’1

i! (i + 1)!4 (i + 2)!

1

n

= (ā’1)( ) . (2.38)

2

6 (2i + 2)! (2i + 3)!

i=1

The general determinant evaluation which results from using continuous Hahn polyno-

mials with generic nonnegative integers a, b, c, d is worked out in [51, Sec. 5].

Let me mention that, given a Hankel determinant evaluation such as (2.38), one has

automatically proved a more general one, by means of the following simple fact (see for

example [121, p. 419]):

Lemma 15. Let x be an indeterminate. For any nonnegative integer n there holds

i+j

i+j

Ak xi+jā’k

det (Ai+j ) = det . (2.39)

k

0ā¤i,jā¤nā’1 0ā¤i,jā¤nā’1

k=0

ADVANCED DETERMINANT CALCULUS 23

The idea of using continued fractions and/or orthogonal polynomials for the evalua-

tion of Hankel determinants has been also exploited in [1, 35, 113, 114, 115, 116]. Some

of these results are exhibited in Theorem 52. See the remarks after Theorem 52 for

pointers to further Hankel determinant evaluations.

2.8. Miscellaneous. This section is a collection of various further results on deter-

minant evaluation of the general sort, which I personally like, regardless whether they

may be more or less useful.

Let me begin with a result by Strehl and Wilf [173, Sec. II], a special case of which was

already in the seventies advertised by van der Poorten [131, Sec. 4] as ā˜a determinant

evaluation that should be better knownā™. (For a generalization see [78].)

Lemma 16. Let f(x) be a formal power series. Then for any positive integer n there

holds

n

f (x) ( 2 )

iā’1

d

(aj ā’ ai ).

f(x)aj = f(x)a1 +Ā·Ā·Ā·+an

det (2.40)

dx f(x)

1ā¤i,jā¤n

1ā¤i<jā¤n

By specializing, this result allows for the quick proof of various, sometimes surprising,

determinant evaluations, see Theorems 53 and 54.

An extremely beautiful determinant evaluation is the evaluation of the determinant

of the circulant matrix.

Theorem 17. Let n by a ļ¬xed positive integer, and let a0 , a1, . . . , anā’1 be indetermi-

nates. Then

ļ£« ļ£¶

a0 a1 a2 . . . anā’1

ļ£¬anā’1 a0 a1 . . . anā’2 ļ£· nā’1

ļ£¬ ļ£·

det ļ£¬anā’2 anā’1 a0 . . . anā’3 ļ£· = (a0 + Ļi a1 + Ļ 2i a2 + Ā· Ā· Ā· + Ļ(nā’1)i anā’1 ),

ļ£¬ ļ£·

ļ£. . . . . . . . . . . . . . . . . . . . . . . . .ļ£ø i=0

a1 a2 a3 . . . a0

(2.41)

where Ļ is a primitive n-th root of unity.

Actually, the circulant determinant is just a very special case in a whole family of

determinants, called group determinants. This would bring us into the vast territory of

group representation theory, and is therefore beyond the scope of this article. It must

suļ¬ce to mention that the group determinants were in fact the cause of birth of group

representation theory (see [99] for a beautiful introduction into these matters).

The next theorem does actually not give the evaluation of a determinant, but of a

Pfaļ¬an. The Pfaļ¬an Pf(A) of a skew-symmetric (2n) Ć— (2n) matrix A is deļ¬ned by

(ā’1)c(Ļ)

Pf(A) = Aij ,

Ļ (ij)āĻ

where the sum is over all perfect matchings Ļ of the complete graph on 2n vertices,

where c(Ļ) is the crossing number of Ļ, and where the product is over all edges (ij),

i < j, in the matching Ļ (see e.g. [169, Sec. 2]). What links Pfaļ¬ans so closely to

24 C. KRATTENTHALER

determinants is (aside from similarity of deļ¬nitions) the fact that the Pfaļ¬an of a

skew-symmetric matrix is, up to sign, the square root of its determinant. That is,

det(A) = Pf(A)2 for any skew-symmetric (2n) Ć— (2n) matrix A (cf. [169, Prop. 2.2]).9

Pfaļ¬ans play an important role, for example, in the enumeration of plane partitions,

due to the results by Laksov, Thorup and Lascoux [98, Appendix, Lemma (A.11)] and

Okada [123, Theorems 3 and 4] on sums of minors of a given matrix (a combinatorial

view as enumerating nonintersecting lattice paths with varying starting and/or ending

points has been given by Stembridge [169, Theorems 3.1, 3.2, and 4.1]), and their

generalization in form of the powerful minor summation formulas due to Ishikawa and

Wakayama [69, Theorems 2 and 3].

Exactly in this context, the context of enumeration of plane partitions, Gordon [58,

implicitly in Sec. 4, 5] (see also [169, proof of Theorem 7.1]) proved two extremely useful

reductions of Pfaļ¬ans to determinants.

Lemma 18. Let (gi ) be a sequence with the property gā’i = gi , and let N be a positive

integer. Then

Pf 1ā¤i<jā¤2N gĪ± = det (giā’j + gi+jā’1 ), (2.42)

1ā¤i,jā¤N

ā’(jā’i)<Ī±ā¤jā’i

and

gĪ± j ā¤ 2N + 1

ā’(jā’i)<Ī±ā¤jā’i

= X Ā· det (giā’j ā’ gi+j ). (2.43)

Pf 1ā¤i<jā¤2N +2

X j = 2N + 2 1ā¤i,jā¤N

(In these statements only one half of the entries of the Pfaļ¬an is given, the other half

being uniquely determined by skew-symmetry).

This result looks somehow technical, but its usefulness was suļ¬ciently proved by

its applications in the enumeration of plane partitions and tableaux in [58] and [169,

Sec. 7].

Another technical, but useful result is due to Goulden and Jackson [61, Theorem 2.1].

Lemma 19. Let Fm (t), Gm (t) and Hm (t) by formal power series, with Hm (0) = 0,

m = 0, 1, . . . , n ā’ 1. Then for any positive integer n there holds

Fj (t) Fj (t)

det CT Gi (Hj (t)) = det CT Gi (0) , (2.44)

Hj (t)i Hj (t)i

0ā¤i,j,ā¤nā’1 0ā¤i,jā¤nā’1

where CT(f(t)) stands for the constant term of the Laurent series f(t).

What is the value of this theorem? In some cases, out of a given determinant eval-

uation, it immediately implies a more general one, containing (at least) one more pa-

rameter. For example, consider the determinant evaluation (3.30). Choose Fj (t) =

tj (1 + t)Āµ+j , Hj (t) = t2/(1 + t), and Gi (t) such that Gi (t2/(1 + t)) = (1 + t)k + (1 + t)ā’k

for a ļ¬xed k (such a choice does indeed exist; see [61, proof of Cor. 2.2]) in Lemma 19.

This yields

Āµā’k+i+j

Āµ+k+i+j Āµ+i+j

det + = det 2 .

2i ā’ j 2i ā’ j 2i ā’ j

0ā¤i,jā¤nā’1 0ā¤i,jā¤nā’1

9

Another point of view, beautifully set forth in [79], is that āPfaļ¬ans are more fundamental than

determinants, in the sense that determinants are merely the bipartite special case of a general sum

over matchings.ā

ADVANCED DETERMINANT CALCULUS 25

Thus, out of the validity of (3.30), this enables to establish the validity of (3.32), and

even of (3.33), by choosing Fj (t) and Hj (t) as above, but Gi (t) such that Gi (t2/(1+t)) =

(1 + t)xi + (1 + t)ā’xi , i = 0, 1, . . . , n ā’ 1.

3. A list of determinant evaluations

In this section I provide a list of determinant evaluations, some of which are very

frequently met, others maybe not so often. In any case, I believe that all of them

are useful or attractive, or even both. However, this is not intended to be, and cannot

possibly be, an exhaustive list of known determinant evaluations. The selection depends

totally on my taste. This may explain that many of these determinants arose in the

enumeration of plane partitions and rhombus tilings. On the other hand, it is exactly

this ļ¬eld (see [138, 148, 163, 165] for more information on these topics) which is a

particular rich source of nontrivial determinant evaluations. If you do not ļ¬nd āyourā

determinant here, then, at least, the many references given in this section or the general

results and methods from Section 2 may turn out to be helpful.

Throughout this section we use the standard hypergeometric and basic hypergeomet-

ric notations. To wit, for nonnegative integers k the shifted factorial (a)k is deļ¬ned (as

already before) by

(a)k := a(a + 1) Ā· Ā· Ā· (a + k ā’ 1),

so that in particular (a)0 := 1. Similarly, for nonnegative integers k the shifted q-

factorial (a; q)k is given by

(a; q)k := (1 ā’ a)(1 ā’ aq) Ā· Ā· Ā· (1 ā’ aq kā’1),

so that (a; q)0 := 1. Sometimes we make use of the notations [Ī±]q := (1 ā’ q Ī± )/(1 ā’ q),

[n]q ! := [n]q [n ā’ 1]q Ā· Ā· Ā· [1]q , [0]q ! := 1. The q-binomial coeļ¬cient is deļ¬ned by

[Ī±]q [Ī± ā’ 1]q Ā· Ā· Ā· [Ī± ā’ k + 1]q (1 ā’ q Ī±)(1 ā’ q Ī±ā’1) Ā· Ā· Ā· (1 ā’ q Ī±ā’k+1 )

Ī±

:= = .

(1 ā’ q k )(1 ā’ q kā’1 ) Ā· Ā· Ā· (1 ā’ q)

k [k]q !

q

Clearly we have limqā’1 [ Ī± ]q = Ī± .

k k

Occasionally shifted (q-)factorials will appear which contain a subscript which is a

negative integer. By convention, a shifted factorial (a)k , where k is a negative integer, is

interpreted as (a)k := 1/(a ā’ 1)(a ā’ 2) Ā· Ā· Ā· (a + k), whereas a shifted q-factorial (a; q)k ,

where k is a negative integer, is interpreted as (a; q)k := 1/(1 ā’ q aā’1)(1 ā’ q aā’2 ) Ā· Ā· Ā·

(1 ā’ q a+k ). (A uniform way to deļ¬ne the shifted factorial, for positive and negative k,

is by (a)k := Ī“(a + k)/Ī“(a), respectively by an appropriate limit in case that a or a + k

is a nonpositive integer, see [62, Sec. 5.5, p. 211f]. A uniform way to deļ¬ne the shifted

q-factorial is by means of (a; q)k := (a; q)ā/(aq k ; q)ā , see [55, (1.2.30)].)

We begin our list with two determinant evaluations which generalize the Vander-

monde determinant evaluation (2.1) in a nonstandard way. The determinants appearing

in these evaluations can be considered as āaugmentationsā of the Vandermonde deter-

minant by columns which are formed by diļ¬erentiating āVandermonde-typeā columns.

(Thus, these determinants can also be considered as certain generalized Wronskians.)

Occurences of the ļ¬rst determinant can be found e.g. in [45], [107, App. A.16], [108,

(7.1.3)], [154], [187]. (It is called āconļ¬‚uent alternantā in [107, 108].) The motivation

in [45] to study these determinants came from Hermite interpolation and the analysis

of linear recursion relations. In [107, App. A.16], special cases of these determinants

26 C. KRATTENTHALER

are used in the context of random matrices. Special cases arose also in the context of

transcendental number theory (see [131, Sec. 4]).

Theorem 20. Let n be a nonnegative integer, and let Am (X) denote the n Ć— m matrix

ļ£« ļ£¶

1 0 0 0 ... 0

ļ£¬X ļ£·

1 0 0 ... 0

ļ£¬2 ļ£·

ļ£¬X ļ£·

2X 2 0 ... 0

ļ£¬3 ļ£·,

ļ£¬X ļ£·

2

3X 6X 6 ... 0

ļ£. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ļ£ø

X nā’1 (n ā’ 1)X nā’2 (n ā’ 1)(n ā’ 2)X nā’3 . . . . . . (n ā’ 1) Ā· Ā· Ā· (n ā’ m + 1)X nā’m

i.e., any next column is formed by diļ¬erentiating the previous column with respect to

X. Given a composition of n, n = m1 + Ā· Ā· Ā· + m , there holds

mi ā’1

(Xj ā’ Xi )mi mj . (3.1)

det Am1 (X1 ) Am2 (X2 ) . . . Am (X ) = j!

1ā¤i,j,ā¤n

i=1 j=1 1ā¤i<jā¤

The paper [45] has as well an āAbel-typeā variation of this result.

Theorem 21. Let n be a nonnegative integer, and let Bm (X) denote the n Ć— m matrix

ļ£« ļ£¶

1 0 0 0 ... 0

ļ£¬X ļ£·

X X X ... X

ļ£¬2 ļ£·

ļ£¬X ļ£·

2 2 2 mā’1 2

2X 4X 8X . . . 2 X

ļ£¬3 ļ£·,

ļ£¬X ļ£·

3 3 3 mā’1 3

3X 9X 27X . . . 3 X

ļ£. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .ļ£ø

X nā’1 (n ā’ 1)X nā’1 (n ā’ 1)2 X nā’1 . . . . . . . . . . (n ā’ 1)mā’1 X nā’1

i.e., any next column is formed by applying the operator X(d/dX). Given a composition

of n, n = m1 + Ā· Ā· Ā· + m , there holds

mi ā’1

ńņš. 1 |