. 1
( 3)




Institut f¨r Mathematik der Universit¨t 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 e¬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
det , (1.1)

det , (1.2)

or it is possibly
det , (1.3)
2i ’ j

1991 Mathematics Subject Classi¬cation. Primary 05A19; Secondary 05A10 05A15 05A17 05A18
05A30 05E10 05E15 11B68 11B73 11C20 15A15 33C45 33D45.
Key words and phrases. Determinants, Vandermonde determinant, Cauchy™s double alternant,
Pfa¬an, discrete Wronskian, Hankel determinants, orthogonal polynomials, Chebyshev polynomials,
Meixner polynomials, Meixner“Pollaczek polynomials, Hermite polynomials, Charlier polynomials, La-
guerre polynomials, Legendre polynomials, ultraspherical polynomials, continuous Hahn polynomials,
continued fractions, binomial coe¬cient, Genocchi numbers, Bernoulli numbers, Stirling numbers, Bell
numbers, Euler numbers, divided di¬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.

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

or maybe
x+y+j x+y+j

det . (1.4)
x ’ i + 2j x + i + 2j

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 “closed form”. 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 ¬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 ¬rst success, as (1.1) does
indeed turn up there (see [119, vol. III, p. 311]). Yes, you will also ¬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, ¬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
e¬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 ¬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) di¬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 di¬cult enumeration problems on plane partitions),
it was really di¬cult. His method (see Section 2.6 for a description) required a good
“guesser” and an excellent “hypergeometer” (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)

Turnbull™s book [178] does in fact contain rather lots of very general identities satis¬ed by determi-
nants, than determinant “evaluations” in the strict sense of the word. However, suitable specializations
of these general identities do also yield “genuine” 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 clari¬cation of the notation and a clear presentation of many such identities.

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 e¬ective in this context). For
“hypergeometrics” this is due to the “WZ-machinery”5 (see [130, 190, 194, 195, 196]).
And even if you should meet a case where the WZ-machinery should exhaust your com-
puter™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, don™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 ¬rst. What I am
mainly addressing here, though, is the case where that ¬rst, “simple-minded” attempt
failed. (Clearly, there is no point in addressing row and column operations, or Laplace
Yet, we must of course start (in Section 2.1) with some standard determinants, such
as the Vandermonde determinant or Cauchy™s double alternant. These are of course
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 “real life,” and which can apparently be evaluated, are a
special case of just the very ¬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 “method” to see if a determinant can be evaluated
by one of the theorems in Section 2.2.
the electronic version of the “Encyclopedia of Integer Sequences” [162, 161], written and developed
by Neil Sloane and Simon Plou¬e; see http://www.research.att.com/˜njas/sequences/ol.html
written by Bruno Salvy and Paul Zimmermann, respectively Frederic Chyzak; available from
written in Mathematica by the author; available from http://radon.mat.univie.ac.at/People/kratt;
the Maple equivalent GUESS by Fran¸ois B´raud and Bruno Gauthier is available from
c e
Maple implementations written by Doron Zeilberger are available from
Mathematica implementations written by
Peter Paule, Axel Riese, Markus Schorn, Kurt Wegschaider are available from
written in Mathematica by the author; available from http://radon.mat.univie.ac.at/People/kratt
written in Maple by Bruno Ghauthier; available from http://www-igm.univ-mlv.fr/˜gauthier

The next method which I describe is the so-called “condensation method” (see Sec-
tion 2.3), a method which allows to evaluate a determinant inductively (if the method
In Section 2.4, a method, which I call the “identi¬cation of factors” 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 ¬nding one or more di¬eren-
tial or di¬erence equations for the matrix of which the determinant is to be evaluated.
Section 2.6 contains a short description of George Andrews™ 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 det1¤i,j¤n (ai+j ), and are sometimes also called per-
symmetric or Tur´nian determinants.) As is explained there, you should expect that 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 ¬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 “stringent” to
“less stringent”. I. e., ¬rst come the methods which require that the matrix of which
the determinant is to be taken satis¬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 “identi¬cation of factors” method (see Section 2.4). Alterna-
tively, and in particular if your matrix of which you want to ¬nd the determinant
is the matrix de¬ning a system of di¬erential or di¬erence equations, try the dif-
ferential/di¬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 e¬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 ¬nd a basis on which this map acts triangularly, or even diagonally (this

requires that the eigenvalues of the matrix are “nice”; see [47, 48, 84, 93, 192] for
examples where that worked). Otherwise, maybe something from Sections 2.8 or
3 helps?
A ¬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 “identi¬cation of factors” method needs, at least, one parameter,
as well as the di¬erential/di¬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 ’ Xi ).
Xij’1 =
det (2.1)

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 ’ Xi2 )
divides the determinant as a polynomial in the Xi ™s. But that means that the complete
product 1¤i<j¤n (Xj ’ 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 ™s of degree at most
. Combined with the previous observation, this implies that the determinant equals
the right-hand side product times, possibly, some constant. To compute the constant,
compare coe¬cients of X1 X2 · · · Xn on both sides of (2.1). This completes the proof
01 n’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. Identi¬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 xj’1 + lower terms, then

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

The following variations of the Vandermonde determinant evaluation are equally easy
to prove.
Lemma 2. The following identities hold true:
Xi’j ) = (X1 · · · Xn )’n
’ (Xi ’ Xj )(1 ’ Xi Xj ) (Xi2 ’ 1),

’ Xi
det (Xi )
= (X1 · · · Xn )’n+1/2 (Xi ’ Xj )(1 ’ Xi Xj ) (Xi ’ 1), (2.4)
1¤i<j¤n i=1

) = 2 · (X1 · · · Xn )’n+1 (Xi ’ Xj )(1 ’ Xi Xj ) ,
det (Xij’1 + Xi (2.5)

det (Xi + Xi )
= (X1 · · · Xn ) (Xi ’ Xj )(1 ’ Xi Xj ) (Xi + 1). (2.6)
1¤i<j¤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 “symplectic”, the “odd orthogonal”,
and the “even orthogonal” Vandermonde determinant evaluation, respectively.
If you encounter generalizations of such determinants of the form det1¤i,j¤n (xi j )
or det1¤i,j¤n (xi j ’ 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) + ¬rst
paragraph on p. 411], [137, Appendix A2], [52, (24.28)]). In this context, one has to
also mention Okada™s general results on evaluations of determinants and Pfa¬ans (see
Section 2.8 for de¬nition) in [124, Sec. 4] and [125, Sec. 5].
Another standard determinant evaluation is the evaluation of Cauchy™s double alter-
nant (see [119, vol. III, p. 311]),
’ Xj )(Yi ’ Yj )
1¤i<j¤n (Xi
det = . (2.7)
Xi + Yj (Xi + Yj )

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
¬rst determinant in the Introduction, (1.1). For the evaluation of a mixture of Cauchy™s
double alternant and Vandermonde™s determinant see [15, Lemma 2].

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 di¬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 di¬erent category of di¬culty of evaluation. (Also (1.3) belongs
to this “di¬erent category”. 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 Vandermonde™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 + An’1 ) · · · (Xi + Aj+1 ) for j = n) equal 1 by convention.
Lemma 3. Let X1 , . . . , Xn , A2, . . . , An, and B2 , . . . , Bn be indeterminates. Then there

(Xi + An)(Xi + An’1 ) · · · (Xi + Aj+1 )(Xi + Bj )(Xi + Bj’1 ) · · · (Xi + B2 )

(Xi ’ Xj ) (Bi ’ Aj ). (2.8)
1¤i<j¤n 2¤i¤j¤n

Other common examples of introducing more parameters are: Given that the (i, j)-entry of your
determinant is a binomial such as 2i’j , try x+i+j (that works; see (3.30)), or even x+y+i+j (that
2i’j y+2i’j
x+i+j y+i+j
does not work; but see (1.2)), or 2i’j + 2i’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).)

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:
a+b (a + b)!
det =
a’i+j (a ’ i + n)! (b + i ’ 1)!
— det (a ’ i + n)(a ’ i + n ’ 1) · · · (a ’ i + j + 1)

· (b + i ’ j + 1)(b + i ’ j + 2) · · · (b + i ’ 1)
(a + b)!
= (’1)( 2 )
(a ’ i + n)! (b + i ’ 1)!
— det (i ’ a ’ n)(i ’ a ’ n + 1) · · · (i ’ a ’ j ’ 1)

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

Now compare with the determinant in (2.8). Indeed, the determinant in the last line is
just the special case Xi = i, Aj = ’a ’ j, Bj = b ’ 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 ¬rst result in the lemma below. It is “tailored”
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 ¬rst (replace Xi by q Xi , Aj by ’q ’Aj and C
by q C in (2.9), divide both sides by (1 ’ q)n(n’1) , and then let q ’ 1).
Lemma 4. Let X1 , X2 , . . . , Xn , A2, . . . , An be indeterminates. Then there hold

(C/Xi + An )(C/Xi + An’1 ) · · · (C/Xi + Aj+1 )
· (Xi + An )(Xi + An’1 ) · · · (Xi + Aj+1 ) = (Xi ’ Xj )(1 ’ C/Xi Xj ),
i=2 1¤i<j¤n


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

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

(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 ’ C/2, both of degree n ’ 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 de¬ne the degree
of a Laurent polynomial p(X) = N aixi , M, N ∈ Z, ai ∈ R and aN = 0, to be
deg p := N.
Lemma 5. Let X1 , X2 , . . . , Xn , A2, A3, . . . , An, C be indeterminates. If p0 , p1 , . . . , pn’1
are Laurent polynomials with deg pj ¤ j and pj (C/X) = pj (X) for j = 0, 1, . . . , n ’ 1,

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

· (C/Xi + An)(C/Xi + An’1 ) · · · (C/Xi + Aj+1 ) · pj’1 (Xi )
n n
(Xi ’ Xj )(1 ’ C/Xi Xj ) Ai’1
= pi’1 (’Ai) . (2.11)
1¤i<j¤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
pj (X) = (Bk+1 + X).

Obviously, Lemma 4 is the special case pj ≡ 1, j = 0, 1, . . . , n ’ 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 , . . . , pn’1 are
polynomials with deg pj ¤ j for j = 0, 1, . . . , n ’ 1, then

(Xi + An )(Xi + An’1 ) · · · (Xi + Aj+1 ) · pj’1 (Xi )
(Xi ’ Xj )
= pi’1 (’Ai) . (2.12)
1¤i<j¤n i=1

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

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

· (Xi ’ An ’ C)(Xi ’ An’1 ’ C) · · · (Xi ’ Aj+1 ’ C) · pj’1 (Xi )
(Xj ’ Xi )(C ’ Xi ’ Xj )
= pi’1 (’Ai) . (2.13)
1¤i<j¤n i=1

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

«± 
(Xi + An ) · · · (Xi + Aj+1 )(C/Xi + An ) · · · (C/Xi + Aj+1 )

¬(X + B ) · · · (X + B )(C/X + B ) · · · (C/X + B )
 j < m·
¬ ·
i j i 2 i j i 2
det ¬ ·
1¤i,j¤n (Xi + an ) · · · (Xi + aj+1 )(C/Xi + an ) · · · (C/Xi + aj+1 ) 

(Xi + bj ) · · · (Xi + b2 )(C/Xi + bj ) · · · (C/Xi + b2) j≥m

(Xi ’ Xj )(1 ’ C/Xi Xj ) (Bi ’ Aj )(1 ’ C/Bi Aj )
1¤i<j¤n 2¤i¤j¤m’1
m n
— (bi ’ Aj )(1 ’ C/bi Aj ) (bi ’ aj )(1 ’ C/bi aj )
i=2 j=m m+1¤i¤j¤n
m n m’1 n
— (Ai · · · An ) (ai · · · an) (B2 · · · Bi ) (b2 · · · 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.)

Lemma 9. Let X1 , . . . , Xn, A2, . . . , An, B2 , . . . , Bn, a2, . . . , an , b2, . . . , bn , and C be
indeterminates. Then there holds
«± 
(Xi + An ) · · · (Xi + Aj+1 )(Xi ’ An ’ C) · · · (Xi ’ Aj+1 ’ C)

¬(X + B ) · · · (X + B )(X ’ B ’ C) · · · (X ’ B ’ C)
 j < m·
¬ ·
i j i 2 i j i 2
det ¬ ·
1¤i,j¤n (Xi + an ) · · · (Xi + aj+1 )(Xi ’ an ’ C) · · · (Xi ’ aj+1 ’ C) 

(Xi + bj ) · · · (Xi + b2 )(Xi ’ bj ’ C) · · · (Xi ’ b2 ’ C) j≥m

(Xi ’ Xj )(C ’ Xi ’ Xj ) (Bi ’ Aj )(Bi + Aj + C)
1¤i<j¤n 2¤i¤j¤m’1
m n
— (bi ’ Aj )(bi + Aj + C) (bi ’ aj )(bi + aj + C). (2.15)
i=2 j=m m+1¤i¤j¤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 Zeilberger™s favourite method. It
allows (sometimes) to establish an elegant, e¬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. 140“142]; with the ¬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 — 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

det A · det A1,n = det A1 · det An ’ det An · 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 )n≥0
of determinants, Mn being an n — 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
a+b ?
det Mn (a, b) := det = . (2.17)
a’i+j i+j+k’2
i=1 j=1 k=1

Then you have already proved your conjecture, once you observe that
Mn (a, b) = Mn’1 (a, b),
Mn(a, b) 1 = Mn’1 (a, b),
= Mn’1 (a + 1, b ’ 1),
Mn (a, b) n
= Mn’1 (a ’ 1, b + 1),
Mn (a, b) 1
Mn (a, b) 1,n = Mn’2 (a, b). (2.18)

For, because of (2.18), Desnanot™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 Mn’1 ( . ) and
det Mn’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) satis¬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 ¬rst time) can
be proved by condensation. The reader will easily ¬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 Eisenk¨lbl [41] in order to establish a conjecture by Propp [138, Problem 3]
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 “identification of factors” method. This is the method that I ¬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,
2i ’ j

’µ ’ 3n + i + 3
n’1 (µ + i + 1) (i+1)/2
n’1 2
2( ) i/2
= (’1)χ(n≡3 mod 4)
, (2.19)

where χ(A) = 1 if A is true and χ(A) = 0 otherwise, and where the shifted factorial
(a)k is de¬ned by (a)k := a(a + 1) · · · (a + k ’ 1), k ≥ 1, and (a)0 := 1.
As was already said in the Introduction, this determinant belongs to a di¬erent
category of di¬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. Identi¬cation of factors
2. Determination of degree bound
3. Computation of the multiplicative constant.
You will say: ˜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.™

Yet ” and this is the point that I want to make here ” it works, in spite of having
just one variable at our disposal!.
What we want to prove in the ¬rst step is that the right-hand side of (2.19) divides the
determinant. For example, we would like to prove that (µ + n) divides the determinant
(actually, (µ + n) (n+1)/3 , we will come to that in a moment). Equivalently, if we set
µ = ’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 ¬nd such a linear combination of columns or rows. Equivalently, for
µ = ’n we ¬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 (µ + n) (n+1)/3
divides the determinant):

For proving that (µ + n)E divides the determinant, we ¬nd E linear inde-
pendent vectors in the kernel.

(For a formal justi¬cation that this does indeed su¬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]

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],

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 µ = ’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.
n’2 n’2 n’2 n’2
0, , , ,..., (2.20)
0 1 2 n’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-¬‚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
n’2 ’n + i + j
j’1 2i ’ j

for i = 0, 1, . . . , n ’ 1. However, verifying binomial identities is pure routine today, by
means of Zeilberger™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 di¬cult. (See Section 3 of [91] for details. There,
slightly di¬erent vectors are used.)
Thus, we would have ¬nished the ¬rst step, “identi¬cation of factors,” of our plan: We
have proved that the right-hand side of (2.19) divides the determinant as a polynomial
in µ.
The second step, “determination of degree bound,” consists of determining the (max-
imal) degree in µ 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
step, “computation of the multiplicative constant,” one compares coe¬cients of x( 2 ) on

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 coe¬cient in order to determine the multiplicative
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 “WZ-machinery” [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
evaluation of a special coe¬cient (in our example we determined the coe¬cient of µ( 2 ) )
for which I am not able to o¬er a recipe so that things could be carried out on a
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 µ) to be able to
apply the polynomiality arguments that are the “engine” 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 signi¬cant advantage of the “identi¬cation of factors method”
is that not only is it capable of proving evaluations of the form
(where CLOSED FORM means a product/quotient of “nice” factors, such as (2.19) or
(2.17)), but also of proving evaluations of the form
det(M) = (CLOSED FORM) — (UGLY POLYNOMIAL), (2.21)
where, of course, M is a matrix containing (at least) one parameter, µ 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 simpli¬ed).
How would one approach the proof of such an evaluation? For one part, we already
know. “Identi¬cation of factors” enables us to show that (CLOSED FORM) divides
det(M) as a polynomial in µ. Then, comparison of degrees in µ on both sides of
(2.21) yields that (UGLY POLYNOMIAL) is a (at this point unknown) polynomial in

µ of some maximal degree, m say. How can we determine this polynomial? Nothing
“simpler” than that: We ¬nd m + 1 values e such that we are able to evaluate det(M)
at µ = e. If we then set µ = e in (2.21) and solve for (UGLY POLYNOMIAL), then we
obtain evaluations of (UGLY POLYNOMIAL) at m + 1 di¬erent values of µ. Clearly,
this su¬ces to ¬nd (UGLY POLYNOMIAL), e.g., by Lagrange interpolation.
I put “simpler” in quotes, because it is here where the crux is: We may not be able
to ¬nd enough such special evaluations of det(M). In fact, you may object: ˜Why all
these complications? If we should be able to ¬nd m + 1 special values of µ for which
we are able to evaluate det(M), then what prevents us from evaluating det(M) as a
whole, for generic µ?™ When I am talking of evaluating det(M) for µ = e, then what I
have in mind is that the evaluation of det(M) at µ = e is “nice” (i.e., gives a “closed
form,” with no “ugly” 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 ¬nding other informations about (UGLY POLYNOMIAL), like the
leading coe¬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 di¬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 satis¬es
a di¬erential equation of the form

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

where T (z) is some other known matrix. Then, by elementary linear algebra, we obtain
a di¬erential equation for the determinant,

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

which is usually easy to solve. (In fact, the di¬erential operator in (2.22) and (2.23)
could be replaced by any operator. In particular, we could replace d/dz by the di¬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 di¬erentiation of binomial coe¬cients. So,

let us ¬rst take some factors out of the determinant (as we also did in Section 2.2),
a+b (a + b)!
det =
a’i+j (a ’ i + n)! (b + i ’ 1)!
— det (a ’ i + n)(a ’ i + n ’ 1) · · · (a ’ i + j + 1)

· (b + i ’ j + 1)(b + i ’ j + 2) · · · (b + i ’ 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
Mn (a) = Tn(a)Mn(a). (2.24)
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
su¬ces that I display T5(a),
’ 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
¬ 1 1 1 3
0 + 2+a+b + 3+a+b
¬ 1+a+b 3+a+b
¬ 1 1
0 0 + 2+a+b
¬ 1+a+b
 0 0 0
0 0 0
’ 3+a+b + 4+a+b ’ 1+a+b + 2+a+b ’ 3+a+b + 4+a+b
4 8 4 1 3 3
’ 2+a+b + 3+a+b ’ 2+a+b + 3+a+b
3 3 1 2 1
’ 1+a+b + 2+a+b
2 1 1

1 1
1+a+b 1+a+b
0 0
(in this display, the ¬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
n’i j ’i’1 (’1)k
Tn (a) = .
j’i a+b+n’i’k
k=0 1¤i,j,¤n

That (2.24) holds with this choice of Tn(a) is then easy to verify. Consequently, by
means of (2.23), we have
det Mn (a) = det Mn (a),
da a+b+

so that
Mn(a) = constant · (a + b + )n’ . (2.25)
The constant is found to be (’1)( 2 ) n’1 !, e.g., by dividing both sides of (2.25) by
a(2 ) , letting a tend to in¬nity, and applying (2.2) to the remaining determinant.

More sophisticated applications of this method (actually, of a version for systems of
di¬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 Knizhnik“Zamolodchikov equations.

2.6. LU-factorization. This is George Andrews™ favourite method. Starting point
is the well-known fact (see [53, p. 33¬]) that, given a square matrix M, there exists,
under suitable, not very stringent conditions (in particular, these are satis¬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 · 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 Gauß 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 · 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 )n≥0 of matrices, where Mn is
an n — 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 · 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 coe¬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 su¬ces that I just display the
matrices L5 and U5 . They are

1 0 0
¬1 (X2 ’ X1 ) 0
L5 = ¬1 (X3 ’ X1 ) (X3 ’ X1 )(X3 ’ X2 )
1 (X4 ’ X1 ) (X4 ’ X1 )(X4 ’ X2 )
(X5 ’ X1 ) (X5 ’ X1 )(X5 ’ X2 )

0 0
0 0 ·
0 0 ·

(X4 ’ X1 )(X4 ’ X2 )(X4 ’ X3 ) 0
(X5 ’ X1 )(X5 ’ X2 )(X5 ’ X3 ) (X5 ’ X1 )(X5 ’ X2 )(X5 ’ X3 )(X5 ’ X4 )

(in this display, the ¬rst line contains columns 1, 2, 3 of L5 , while the second line contains
the remaining columns), and
« 
’e1 (X1) ’e3 (X1 , X2 , X3 )
1 e2 (X1 , X2 ) e4 (X1 , X2 , X3 , X4 )
¬0 ’e3 (X1 , X2 , X3 , X4 )·
’e1 (X1 , X2 )
1 e2 (X1 , X2 , X3 )
¬ ·
U5 = ¬0 e2 (X1 , X2 , X3 , X4 )· ,
’e1 (X1 , X2 , X3 )
0 1
¬ ·
0 ’e1 (X1 , X2 , X3 , X4 )
0 0 1
0 0 0 0 1

1¤i1 <···<im ¤s Xi1 · · · 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
(Xi ’ Xk )
Ln = ,
and that Un is given by
Un = (’1)j’i ej’i (X1 , . . . , Xj’1 ) ,

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 di¬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-di¬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 di¬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 “identi¬cation of factors” as described in
the previous section. In both cases, the essential steps are to ¬rst guess something, and
then prove the guess. Therefore, the remarks from the previous section about guessing

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 “condensation” or “identi¬cation of fac-
tors,” LU-factorization does not require any parameter. So, in principle, it is applicable
to any determinant (which satis¬es the aforementioned conditions). If there are limita-
tions, then, from my experience, it is that the coe¬cients which have to be guessed in
LU-factorization tend to be more complicated than in “identi¬cation of factors”. 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 ).

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)

where Bk denotes the k-th Bernoulli number. (The Bernoulli numbers are de¬ned via
their generating function, ∞ Bk z k /k! = z/(ez ’ 1).) You have to try hard if you
want to ¬nd an evaluation of (2.29) explicitly in the literature. Indeed, you can ¬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 ¬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
∞ k
(µk )k≥0 be a sequence of numbers with generating function k=0 µk x written in the

µk xk = . (2.30)
b1 x2
1 + a0 x ’
b2 x2
1 + a1 x ’
1 + a2 x ’ · · ·
Then the Hankel determinant det0¤i,j¤n’1 (µi+j ) equals µn bn’1 bn’2 · · · b2 bn’1 .
01 2 n’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 ∞ Bk+2 xk in terms of a continued fraction of the

form of the right-hand side of (2.30). Using the tools explained in Appendix A, it is
easy to work out a conjecture,

Bk+2 xk = , (2.31)
b1 x2
b2 x2
1’ ···
where bi = ’i(i + 1)2 (i + 2)/4(2i + 1)(2i + 3), i = 1, 2, . . . . If we would ¬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))n≥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)) =
δmncn for some sequence (cn )n≥0 of nonzero numbers, with δm,n denoting the Kronecker
delta (i.e., δm,n = 1 if m = n and δm,n = 0 otherwise).
The ¬rst important theorem in the theory of orthogonal polynomials is Favard™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, Th´or`me 9 on p. I-4] or [188, Theorem 50.1]). Let (pn (x))n≥0
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 )n≥1 and (bn )n≥1 , with bn = 0 for all n ≥ 1, such that the three-term recurrence
pn+1 (x) = (an + x)pn(x) ’ bnpn’1 (x), for n ≥ 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))n≥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) ’ bnpn’1 (x) (2.33)
be the corresponding three-term recurrence which is guaranteed by Favard™s theorem.
Then the generating function ∞ µk xk for the moments µk = L(xk ) satis¬es (2.30)
with the ai ™s and bi ™s being the coe¬cients in the three-term recurrence (2.33).
Thus, what we have to do is to ¬nd orthogonal polynomials (pn (x))n≥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 “Askey table”), and compiled by
Koekoek and Swarttouw [81].
Indeed, in Section 1.4 of [81], we ¬nd the family of orthogonal polynomials that is
of relevance here, the continuous Hahn polynomials, ¬rst studied by Atakishiyev and
Suslov [13] and Askey [12]. These polynomials depend on four parameters, a, b, c, d. It

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))n≥0 , are the monic polynomials de¬ned by

√ (’n)k (n + 3)k (1 + x ’1)k
n (n + 1)! (n + 2)!
pn (x) = ( ’1) , (2.34)
k! (k + 1)!2
(2n + 2)! k=0

with the shifted factorial (a)k de¬ned as previously (see (2.19)). These polynomials
satisfy the three-term recurrence
n(n + 1)2 (n + 2)
pn+1 (x) = xpn (x) + pn’1 (x). (2.35)
4(2n + 1)(2n + 3)
They are orthogonal with respect to the functional L which is given by
π∞ x2
L(p(x)) = p(x) dx . (2.36)
2 ’∞ sinh2 (πx)
Explicitly, the orthogonality relation is
n! (n + 1)!4 (n + 2)!
L(pm (x)pn (x)) = δm,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]),

∞ ’1
1 π 2
√ ν
Bν = z dz

2π ’1 sin πz
’∞ ’1

(if ν = 0 or ν = 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 di¬culty the
desired determinant evaluation,
n n’1 n’i
i(i + 1)2 (i + 2)
(Bi+j+2 ) = (’1)( )
det 2
6 4(2i + 1)(2i + 3)
i! (i + 1)!4 (i + 2)!
= (’1)( ) . (2.38)
6 (2i + 2)! (2i + 3)!
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
Ak xi+j’k
det (Ai+j ) = det . (2.39)
0¤i,j¤n’1 0¤i,j¤n’1

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 ˜a determinant
evaluation that should be better known™. (For a generalization see [78].)
Lemma 16. Let f(x) be a formal power series. Then for any positive integer n there
f (x) ( 2 )
(aj ’ ai ).
f(x)aj = f(x)a1 +···+an
det (2.40)
dx f(x)

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 ¬xed positive integer, and let a0 , a1, . . . , an’1 be indetermi-
nates. Then
« 
a0 a1 a2 . . . an’1
¬an’1 a0 a1 . . . an’2 · n’1
¬ ·
det ¬an’2 an’1 a0 . . . an’3 · = (a0 + ωi a1 + ω 2i a2 + · · · + ω(n’1)i an’1 ),
¬ ·
. . . . . . . . . . . . . . . . . . . . . . . . . i=0
a1 a2 a3 . . . a0
where ω 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
su¬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
Pfa¬an. The Pfa¬an Pf(A) of a skew-symmetric (2n) — (2n) matrix A is de¬ned by

Pf(A) = Aij ,
π (ij)∈π

where the sum is over all perfect matchings π of the complete graph on 2n vertices,
where c(π) is the crossing number of π, and where the product is over all edges (ij),
i < j, in the matching π (see e.g. [169, Sec. 2]). What links Pfa¬ans so closely to

determinants is (aside from similarity of de¬nitions) the fact that the Pfa¬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) — (2n) matrix A (cf. [169, Prop. 2.2]).9
Pfa¬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 Pfa¬ans to determinants.
Lemma 18. Let (gi ) be a sequence with the property g’i = gi , and let N be a positive
integer. Then

Pf 1¤i<j¤2N g± = det (gi’j + gi+j’1 ), (2.42)

g± j ¤ 2N + 1
= X · det (gi’j ’ gi+j ). (2.43)
Pf 1¤i<j¤2N +2
X j = 2N + 2 1¤i,j¤N

(In these statements only one half of the entries of the Pfa¬an is given, the other half
being uniquely determined by skew-symmetry).
This result looks somehow technical, but its usefulness was su¬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 ’ 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
0¤i,j,¤n’1 0¤i,j¤n’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)µ+j , Hj (t) = t2/(1 + t), and Gi (t) such that Gi (t2/(1 + t)) = (1 + t)k + (1 + t)’k
for a ¬xed k (such a choice does indeed exist; see [61, proof of Cor. 2.2]) in Lemma 19.
This yields
µ+k+i+j µ+i+j
det + = det 2 .
2i ’ j 2i ’ j 2i ’ j
0¤i,j¤n’1 0¤i,j¤n’1

Another point of view, beautifully set forth in [79], is that “Pfa¬ans are more fundamental than
determinants, in the sense that determinants are merely the bipartite special case of a general sum
over matchings.”

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)’xi , i = 0, 1, . . . , n ’ 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 ¬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 ¬nd “your”
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 de¬ned (as
already before) by
(a)k := a(a + 1) · · · (a + k ’ 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 ’ a)(1 ’ aq) · · · (1 ’ aq k’1),
so that (a; q)0 := 1. Sometimes we make use of the notations [±]q := (1 ’ q ± )/(1 ’ q),
[n]q ! := [n]q [n ’ 1]q · · · [1]q , [0]q ! := 1. The q-binomial coe¬cient is de¬ned by
[±]q [± ’ 1]q · · · [± ’ k + 1]q (1 ’ q ±)(1 ’ q ±’1) · · · (1 ’ q ±’k+1 )
:= = .
(1 ’ q k )(1 ’ q k’1 ) · · · (1 ’ q)
k [k]q !

Clearly we have limq’1 [ ± ]q = ± .
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 ’ 1)(a ’ 2) · · · (a + k), whereas a shifted q-factorial (a; q)k ,
where k is a negative integer, is interpreted as (a; q)k := 1/(1 ’ q a’1)(1 ’ q a’2 ) · · ·
(1 ’ q a+k ). (A uniform way to de¬ne the shifted factorial, for positive and negative k,
is by (a)k := “(a + k)/“(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 de¬ne the shifted
q-factorial is by means of (a; q)k := (a; q)∞/(aq k ; q)∞ , 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 “augmentations” of the Vandermonde deter-
minant by columns which are formed by di¬erentiating “Vandermonde-type” columns.
(Thus, these determinants can also be considered as certain generalized Wronskians.)
Occurences of the ¬rst determinant can be found e.g. in [45], [107, App. A.16], [108,
(7.1.3)], [154], [187]. (It is called “con¬‚uent alternant” 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

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 — m matrix
« 
1 0 0 0 ... 0
¬X ·
1 0 0 ... 0
¬2 ·
¬X ·
2X 2 0 ... 0
¬3 ·,
¬X ·
3X 6X 6 ... 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 
X n’1 (n ’ 1)X n’2 (n ’ 1)(n ’ 2)X n’3 . . . . . . (n ’ 1) · · · (n ’ m + 1)X n’m
i.e., any next column is formed by di¬erentiating the previous column with respect to
X. Given a composition of n, n = m1 + · · · + m , there holds
mi ’1
(Xj ’ Xi )mi mj . (3.1)
det Am1 (X1 ) Am2 (X2 ) . . . Am (X ) = j!
i=1 j=1 1¤i<j¤

The paper [45] has as well an “Abel-type” variation of this result.
Theorem 21. Let n be a nonnegative integer, and let Bm (X) denote the n — m matrix
« 
1 0 0 0 ... 0
¬X ·
X X X ... X
¬2 ·
¬X ·
2 2 2 m’1 2
2X 4X 8X . . . 2 X
¬3 ·,
¬X ·
3 3 3 m’1 3
3X 9X 27X . . . 3 X
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
X n’1 (n ’ 1)X n’1 (n ’ 1)2 X n’1 . . . . . . . . . . (n ’ 1)m’1 X n’1
i.e., any next column is formed by applying the operator X(d/dX). Given a composition
of n, n = m1 + · · · + m , there holds
mi ’1

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