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\def\al{\alpha}
\def\be{\beta}
\def\ga{\gamma}
\def\de{\delta}
\def\ep{\varepsilon}
\def\ze{\zeta}
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\def\th{\theta}
\def\vt{\vartheta}
\def\io{\iota}
\def\ka{\kappa}
\def\la{\lambda}
\def\rh{\rho}
\def\si{\sigma}
\def\ta{\tau}
\def\ph{\varphi}
\def\ch{\chi}
\def\ps{\psi}
\def\om{\omega}
\def\Ga{\Gamma}
\def\De{\Delta}
\def\Th{\Theta}
\def\La{\Lambda}
\def\Si{\Sigma}
\def\Ph{\Phi}
\def\Ps{\Psi}
\def\Om{\Omega}
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\begin{document}

\newbox\Adr
\setbox\Adr\vbox{
\centerline{\sc J. M. Brunat$^{1*}$,
C.~Krattenthaler$^{2\dagger\ddagger}$, A.~Lascoux$^{3\dagger}$ and 
A. Montes$^{1*}$}
\vskip18pt
\centerline{$^1$Departament de Matem\`atica Aplicada II,}
\centerline{Universitat Polit\`ecnica de Catalunya,}
\centerline{Jordi Girona 1--3,   08034 Barcelona,   Spain}
\centerline{WWW: \footnotesize\tt http://www-ma2.upc.edu/\~{}montes}
\vskip18pt
\centerline{$^2$Institut Camille Jordan, Universit\'e Claude Bernard
Lyon-I,}
\centerline{21, avenue Claude Bernard, F-69622 Villeurbanne Cedex,
France}
\centerline{WWW: \footnotesize\tt http://igd.univ-lyon1.fr/\~{}kratt}
\vskip18pt
\centerline{$^3$Institut Gaspard Monge, Universit\'e de
Marne-la-Vall\'ee,} 
\centerline{F-77454 Marne-la-Vall\'ee Cedex~2,
France}
\centerline{WWW: \footnotesize\tt http://www-igm.univ-mlv.fr/\~{}al}
}

\title{Some composition determinants}

\author[J. M. Brunat, C.~Krattenthaler, A.~Lascoux and A. Montes]{\box\Adr}

\address{Departament de Matem\`atica Aplicada II,
Universitat Polit\`ecnica de Catalunya,
Jordi Girona 1--3,   08034 Barcelona,   Spain.}

\address{Institut Camille Jordan, Universit\'e Claude Bernard Lyon-I, 
21, avenue Claude Bernard, F-69622 Villeurbanne Cedex, France.}

\address{Institut Gaspard Monge, Universit\'e de Marne-la-Vall\'ee,
F-77454 Marne-la-Vall\'ee Cedex~2, France.}

\thanks{$^*$Work partially supported by the Ministerio de Ciencia y
Tecnolog\'\i a under projects BFM2003-00368 and MTM2004-01728 and
by the Generalitat de Catalunya under project 2001 SGR 00224.\newline
 \indent$^\dagger$Research partially supported by EC's IHRP Programme,
grant HPRN-CT-2001-00272, ``Algebraic Combinatorics in Europe."\newline
 \indent$^\ddagger$Current address: Fakult\"at f\"ur Mathematik, Universit\"at Wien,
 Nordbergstra{\ss}e~15, A-1090 Vienna, Austria}

\subjclass[2000]{Primary 05A19;
 Secondary 05A10 11C20 15A15}
\keywords{Binomial determinants, power determinants, 
compositions, Chu--Van\-der\-monde summation.}

\begin{abstract}
We compute several parametric determinants in which rows and columns are indexed by
compositions, where the entries are either products of
binomial coefficients or products of
powers. These results generalize previous determinant evaluations due
to the first and fourth author [{\it SIAM J. Matrix Anal.\ Appl.} {\bf 23}
(2001), 459--471] and [``A polynomial generalization of the 
power-compositions determinant," {\it Linear Multilinear Algebra} 
(to appear)], and they prove two conjectures of the second author
[``Advanced determinant calculus: a complement," preliminary version].
\end{abstract}

\maketitle

\section{Introduction}

A composition of a non-negative integer $n$ 
is a vector $(\al_1,\al_2,\dots,\al_k)$ of non-negative integers such that
$\al_1+\al_2+\dots+\al_k=n$, for some $k$. For a fixed $k$, 
let $\mathcal C(n,k)$ denote the corresponding set of compositions of $n$.
While working on a problem in global optimisation, 
two of the authors \cite{BrMoAA}
discovered the following surprising determinant evaluation.
It allowed them to show how to explicitly express a multivariable
polynomial as a {\it difference of convex functions}.
In the statement, we use standard multi-index notation: 
if $\boldsymbol\al=(\al_1,\al_2,\dots,\al_k)$ and
$\boldsymbol \be=(\be_1,\be_2,\dots,\be_k)$ are two compositions, we
let
$$\boldsymbol\al^{\boldsymbol \be}:=\al_1^{\be_1}\al_2^{\be_2}\cdots
\al_k^{\be_k},$$
where $0^0$ is interpreted as $1$.

\begin{Theorem} \label{thm:Brunat}
For any positive integers $n$ and $k$, we have
\begin{equation} \label{eq:Brunat} 
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left(\boldsymbol\al^{\boldsymbol \be}\right)=
n^{{\binom{ n + k-1} k}+k-1}\,
 \prod_{i = 1}^{ n-1}
      i^{( n -i+1) {\binom {n + k-i-1} { k-2}}}.
\end{equation}
\end{Theorem}

In the preliminary version \cite{KratBZa} of \cite{KratBZ}, 
the second author observed empirically that there seemed to be
a polynomial generalisation of this theorem.

\begin{Conjecture}[{\cite[Conjecture~57]{KratBZa}}] \label{conj:Brunat1}
For any positive integers $n$ and $k$, we have
\begin{equation} \label{eq:Brunat1} 
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left((x+\boldsymbol\al)^{\boldsymbol \be}\right)=
(kx+n)^{{\binom{ n + k-1} k}}n^{k-1}\,
 \prod_{i = 1}^{ n-1}
      i^{( n -i+1) {\binom {n + k-i-1} { k-2}}},
\end{equation}
where $x$ is a variable, and $x+\boldsymbol\al$ is short for
$(x+\al_1,x+\al_2,\dots,x+\al_k)$.
\end{Conjecture}

Morover, he also worked out a binomial version of this conjecture.
Extending our
multi-index notation, let
$$\binom{\boldsymbol\al}{\boldsymbol \be}:=
\binom{\al_1}{\be_1}\binom{\al_2}{\be_2}\cdots
\binom{\al_k}{\be_k}.$$

\begin{Conjecture}[{\cite[Conjecture~58]{KratBZa}}] \label{conj:Brunat2}
For any positive integers $n$ and $k$, we have
\begin{equation} \label{eq:Brunat2a} 
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left(\binom{x+\boldsymbol\al+\boldsymbol\be}{\boldsymbol \be}\right)=
 \frac{\displaystyle\prod_{i = 0}^{n-1}{( kx+n+k+i) }^
       {{\binom{ k+i-1} { k-1}}}}{\displaystyle\prod_{i = 1}^{n}
      i^{{\binom{n+k-i-1} {k-1}}}},
\end{equation}
where $x$ is a variable, and $x+\boldsymbol\al+\boldsymbol\be$ 
is short for
$(x+\al_1+\be_1,x+\al_2+\be_2,\dots,x+\al_k+\be_k)$.
\end{Conjecture}

In the recent paper \cite{BrMoAB}, the first and fourth author 
succeeded to prove
Conjecture~\ref{conj:Brunat1}. In fact, they established 
the following multivariable generalisation.

\begin{Theorem} \label{thm:Brunat3}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ be a vector of indeterminates.
Then, for any positive integers $n$ and $k$, we have
\begin{equation} \label{eq:Brunat3} 
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left((\mathbf x+\boldsymbol\al)^
    {\boldsymbol \be}\right)=
 \left({ \vert \mathbf x\vert+ n }\right)
      ^{\binom{n+k-1} {k}}
\prod _{i=1} ^{n}i^{(k-1)\binom {n+k-i-1}{k-1}},
\end{equation}
where $\mathbf x+\boldsymbol\al$ is short for
$(x_1+\al_1,x_2+\al_2,\dots,x_k+\al_k)$, and
where $\vert \mathbf x\vert=x_1+x_2+\dots+x_k$. 
\end{Theorem}

The purpose of this paper is to, in some sense, explain the
miraculous existence of all these formulae. We do this by introducing
further variables, $\la_1,\la_2,\dots,\la_k$, 
in the binomial determinant in
\eqref{eq:Brunat2a}, and by proving an evaluation theorem for the
resulting determinant. All the afore-mentioned determinant evaluations
are then special cases, respectively limit cases, of this new theorem.
To be precise, the main result of this paper is the following
determinant evaluation.

\begin{Theorem} \label{thm:Brunat4}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ and
$\boldsymbol\la=(\la_1,\la_2,\dots,\la_k)$ be vectors of
indeterminates.
Then, for any positive integers $n$ and $k$, we have
\begin{equation} \label{eq:Brunat4} 
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left(\binom{\mathbf x+\boldsymbol\la\boldsymbol\al}
    {\boldsymbol \be}\right)=\left(\prod _{j=1}
^{k}\la_j\right)^{\binom {n+k-1}k}
\frac {\displaystyle\underset{\ep_1+\dots+\ep_k<n}
{\prod _{\ep_1,\dots,\ep_k\ge0} ^{}} \left(
n+ \sum _{j=1} ^{k}\left(\frac {x_j} {\la_j}-\frac {\ep_j}
{\la_j}\right) \right)}
{\displaystyle\prod _{i=1} ^{n}i^{\binom{n+k-i-1} {k-1}}},
\end{equation}
where $\mathbf x+\boldsymbol\la\boldsymbol\al$ is short for
$(x_1+\la_1\al_1,x_2+\la_2\al_2,\dots,x_k+\la_k\al_k)$.
\end{Theorem}

Using the elementary property $\binom Xm=(-1)^m\binom {-X+m-1}m$
of binomial coefficients, we show
that an equivalent way to write the same result is as follows.

\begin{Theorem} \label{thm:Brunat4b}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ and
$\boldsymbol\la=(\la_1,\la_2,\dots,\la_k)$ be vectors of
indeterminates.
Then, with notation as in Theorem~{\em\ref{thm:Brunat4}}, 
for any positive integers $n$ and $k$ we have
\begin{equation} \label{eq:Brunat4b} 
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left(\binom{\mathbf x+\boldsymbol\la\boldsymbol\al+\boldsymbol\be}
    {\boldsymbol \be}\right)=\left(\prod _{j=1}
^{k}\la_j\right)^{\binom {n+k-1}k}
\frac {\displaystyle\underset{\ep_1+\dots+\ep_k<n}
{\prod _{\ep_1,\dots,\ep_k\ge0} ^{}} \left(n+
 \sum _{j=1} ^{k}\left(\frac {x_j+1} {\la_j}+\frac {\ep_j}
{\la_j}\right) \right)}
{\displaystyle\prod _{i=1} ^{n}i^{\binom{n+k-i-1} {k-1}}}.
\end{equation}
\end{Theorem}

In order to see how Conjecture~\ref{conj:Brunat2} is implied by these
results, it is convenient to first state separately the special cases of
Theorems~\ref{thm:Brunat4} and \ref{thm:Brunat4b} where all the
$\la_i$'s are identical.
In that case, the product in the numerator on the right-hand sides of
\eqref{eq:Brunat4} and \eqref{eq:Brunat4b} can be rearranged by
grouping together the factors corresponding to compositions
$\boldsymbol \varepsilon=(\ep_1,\ep_2,\dots,\ep_k)$ of $n-i$, which become
identical. Taking into account that the number of such compositions
is $\binom {n-i+k-1}{k-1}$, this yields the following two
corollaries.

\begin{Corollary} \label{thm:Brunat2}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ be a vector of indeterminates,
and let $\la$ be an indeterminate.
Then, for any positive integers $n$ and $k$, we have
\begin{equation} \label{eq:Brunat2} 
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left(\binom{\mathbf x+\la\boldsymbol\al}
    {\boldsymbol \be}\right)=\la^{(k-1)\binom {n+k-1}k}
\prod _{i=1} ^{n} \left(\frac{{ \vert \mathbf x\vert+(\la-1)n+i }}
 i\right)
      ^{\binom{n+k-i-1} {k-1}},
\end{equation}
where $\mathbf x+\la\boldsymbol\al$ is short for
$(x_1+\la\al_1,x_2+\la\al_2,\dots,x_k+\la\al_k)$, and
where $\vert \mathbf x\vert=x_1+x_2+\dots+x_k$, as before. 
\end{Corollary}

\begin{Corollary} \label{thm:Brunat2b}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ be a vector of indeterminates,
and let $\la$ be an indeterminate.
Then, with notation as in Corollary~{\em\ref{thm:Brunat2}}, 
for any positive integers $n$ and $k$ we have
\begin{equation} \label{eq:Brunat2b} 
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left(\binom{\mathbf x+\la\boldsymbol\al+\boldsymbol\be}
    {\boldsymbol \be}\right)=\la^{(k-1)\binom {n+k-1}k}
\prod _{i=1} ^{n} \left(\frac{{ \vert \mathbf x\vert+(\la +1)n+k-i }}
 i\right)
      ^{\binom{n+k-i-1} {k-1}}.
\end{equation}
\end{Corollary}

Clearly, Conjecture~\ref{conj:Brunat2} is the special case of the above
corollary where $\la=1$ and $x_i=x$ for all $i$. 

Theorem~\ref{thm:Brunat3} is also implied by
Theorem~\ref{thm:Brunat4}. To see this, we shall show that, by 
extracting the highest homogeneous component in \eqref{eq:Brunat4}
(this could also be realised by an appropriate limit),
we obtain the following corollary.

\begin{Corollary} \label{thm:Brunat5a}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ and
$\boldsymbol\la=(\la_1,\la_2,\dots,\la_k)$ be vectors of
indeterminates.
Then, with notation as in Theorem~{\em\ref{thm:Brunat4}}, 
for any positive integers $n$ and $k$ we have
\begin{equation} \label{eq:Brunat5a} 
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left((\mathbf x+\boldsymbol\la\boldsymbol\al)^
    {\boldsymbol \be}\right)=\left(\prod _{j=1}
^{k}\la_j\right)^{\binom {n+k-1}k}
\left(n+ \sum _{j=1} ^{k}\frac {x_j} {\la_j}\right)^{\binom {n+k-1}k}
\prod _{i=1} ^{n}i^{(k-1)\binom {n+k-i-1}{k-1}}.
\end{equation}
\end{Corollary}

The corresponding special case where all the $\la_i$'s are identical
is the following.

\begin{Corollary} \label{thm:Brunat1a}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ be a vector of indeterminates,
and let $\la$ be an indeterminate.
Then, with notation as in Corollary~{\em\ref{thm:Brunat2}}, 
for any positive integers $n$ and $k$ we have
\begin{equation} \label{eq:Brunat1a} 
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left((\mathbf x+\la\boldsymbol\al)^
    {\boldsymbol \be}\right)=\la^{(k-1)\binom {n+k-1}k}
 \left({ \vert \mathbf x\vert+\la n }\right)
      ^{\binom{n+k-1} {k}}
\prod _{i=1} ^{n}i^{(k-1)\binom {n+k-i-1}{k-1}}.
\end{equation}
\end{Corollary}

Clearly, Theorem~\ref{thm:Brunat3} is the special case $\la=1$ of this
corollary.

In the next section, we give proofs of Theorems~\ref{thm:Brunat4}
and \ref{thm:Brunat4b}, and of
Corollary~\ref{thm:Brunat5a} (and, thus, of
Corollaries~\ref{thm:Brunat2}, \ref{thm:Brunat2b} and
\ref{thm:Brunat1a} also). 
%We provide in fact two proofs
%of our main theorem, Theorem~\ref{thm:Brunat4}. 
%Our first proof is based on the ``identification of factors" technique
%(see \cite[Sec.~2.4]{KratBN}). In contrast, our second proof
%extends the inductive procedure from
%\cite{BrMoAA,BrMoAB} that was used in the original proofs of the special cases
%in Theorems~\ref{thm:Brunat} and \ref{thm:Brunat3}.
In contrast to the inductive procedure in
\cite{BrMoAA,BrMoAB} that was used in the original proofs of Theorems~\ref{thm:Brunat} and \ref{thm:Brunat3},
our proof is based on the ``identification of factors" technique
(see \cite[Sec.~2.4]{KratBN}).
As it turns out, the crucial identity
in our proof is the multivariate version of the Chu--Vandermonde
summation formula (see Lemma~\ref{lem:ChuVand}).
Finally, in the last section, we derive analogues of
Theorems~\ref{thm:Brunat4} and \ref{thm:Brunat4b}, and
of Corollary~\ref{thm:Brunat5a} for the
subdeterminants in which we restrict the rows and columns to 
compositions of $n$ with exactly
$k$ {\it positive} summands.

\section{The proofs}

\begin{Lemma} \label{lem:ChuVand}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ be a vector of indeterminates, and
let $n$ and $k$ be non-negative integers. Then
$$\sum _{\boldsymbol\de\in\mathcal C(n,k)} ^{}
\binom {\mathbf x}{\boldsymbol\de} =
\sum _{\de_1+\dots+\de_k=n} ^{}\binom {\mathbf x}{\boldsymbol\de} =
\binom {\vert \mathbf x\vert}n.$$
\end{Lemma}

\begin{proof}
The Chu--Vandermonde summation formula (see e.g.\ \cite[Sec.~5.1,
(5.27)]{GrKPAA}) reads
$$
\sum _{r=0} ^{s}\binom Mr\binom {N}{s-r}=\binom {M+N}s.$$
On the basis of this formula, the assertion of the lemma is
easily proved by induction on $k$.
\end{proof}

\begin{proof}%[First proof of Theorem~\ref{thm:Brunat4}]
[Proof of Theorem~\ref{thm:Brunat4}]
We prove the theorem by the identification of factors method
described in \cite[Sec.~2.4]{KratBN}.
For convenience, let us write $M(n,k)$ for the matrix of which we want
to compute the determinant, that is,
$$M(n,k)=\left(\binom{\mathbf x+\boldsymbol\la\boldsymbol\al}
    {\boldsymbol \be}\right)_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}.$$

\medskip
{\it Step~1. The term
$$T(n,k,\mathbf x,\boldsymbol\la,\boldsymbol\ep):=n\prod _{j=1} ^{k}\la_j+
\sum _{t=1} ^{k} {x_t}\underset{j\ne t}{\prod _{j=1} ^{k}}\la_j
-\sum _{t=1} ^{k}  {\ep_t} \underset{j\ne t}{\prod _{j=1} ^{k}}\la_j$$
divides $\det M(n,k)$ for any composition $\boldsymbol\ep\in\mathcal
C(n-i,k)$, $1\le i\le n$.} (It should be noted that
$T(n,k,\mathbf x,\boldsymbol\la,\boldsymbol\ep)$ is the factor
corresponding to $\ep_1,\ep_2,\dots,\ep_k$ in the product 
in the numerator on the
right-hand side of \eqref{eq:Brunat4}, up to multiplication by $
\prod _{j=1} ^{k}\la_j$.) To prove this assertion, we
find a vector in the kernel of
\begin{equation} \label{eq:Mnk}
M(n,k)\Big\vert_{T(n,k,\mathbf x,\boldsymbol\la,\boldsymbol\ep)=0}.
\end{equation}
This kernel lives in the free vector space generated by the
compositions in $\mathcal C(n,k)$. Given  $\boldsymbol \de\in\mathcal
C(n,k)$, let us denote the corresponding
element (``unit vector") in this vector space by $e_{\boldsymbol\de}$.
Then we claim that the vector
\begin{equation} \label{eq:v} 
v_{\boldsymbol\ep}:=
\sum _{\boldsymbol\de\in\mathcal C(i,k)} ^{}
\left(\sum _{t=1} ^{k}\frac {\de_t} {\la_t}\right)
\binom {\boldsymbol \de+\boldsymbol \ep}{\boldsymbol \de}
e_{\boldsymbol\de+\boldsymbol\ep}
\end{equation}
is in the kernel of the matrix \eqref{eq:Mnk}. To
see this, we calculate, using Lemma~\ref{lem:ChuVand} and the
notation $\mathbf u_t=(0,\dots,0,1,0,\dots,0)$ (with the $1$ in
position $t$),
\begin{align*}
\text{coefficient of $e_\al$ in }M(n,k)\cdot v_{\boldsymbol\ep}&=\sum
_{\boldsymbol\de\in\mathcal C(i,k)} ^{} 
\binom{\mathbf x+\boldsymbol\la\boldsymbol\al}
    {\boldsymbol \de+\boldsymbol \ep}
\left(\sum _{t=1} ^{k}\frac {\de_t} {\la_t}\right)
\binom{\boldsymbol\de+\boldsymbol\ep}
    {\boldsymbol \de}\\
&\kern-3cm
=\binom {\mathbf x+\boldsymbol\la\boldsymbol\al}
{\boldsymbol \ep}
\sum _{t=1} ^{k}\sum
_{\vert\boldsymbol\de\vert=i} ^{} \frac {\de_t} {\la_t}
\binom{\mathbf x+\boldsymbol\la\boldsymbol\al-\boldsymbol \ep}
    {\boldsymbol \de}\\
&\kern-3cm
=\binom {\mathbf x+\boldsymbol\la\boldsymbol\al}
{\boldsymbol \ep}
\sum _{t=1} ^{k}\frac {1} {\la_t}(x_t+\la_t\al_t-\ep_t)
\sum
_{\vert\boldsymbol\de\vert=i} ^{} 
\binom{\mathbf x+\boldsymbol\la\boldsymbol\al-\boldsymbol \ep-\mathbf u_t}
    {\boldsymbol \de-\mathbf u_t}\\
&\kern-3cm
=\binom {\mathbf x+\boldsymbol\la\boldsymbol\al}
{\boldsymbol \ep}
\left(n+\sum _{t=1} ^{k}\left(\frac {x_t} {\la_t}-
\frac {\ep_t} {\la_t}\right)\right)
\binom{\vert\mathbf x+\boldsymbol\la\boldsymbol\al\vert-(n-i)-1}
    {i-1}.
\end{align*}
Since $i\ge1$, the occurrence of the factor in the middle implies
$$M(n,k)\Big\vert_{T(n,k,\mathbf x,\boldsymbol\la,\boldsymbol\ep)=0}\cdot
v_{\boldsymbol\ep}=0.$$

%Next we show that the vectors $v_{\boldsymbol\ep}$, $\boldsymbol\ep\in\mathcal
%C(n-i,k)$, are linearly independent. Suppose that
%\begin{equation} \label{eq:lin}
%\sum _{\boldsymbol\ep\in\mathcal C(n-i,k)} ^{}\mu_{\boldsymbol\ep}
%v_{\boldsymbol\ep}=0,
%\end{equation}
%for some scalars $\mu_{\boldsymbol\ep}$. In order to see that all 
%$\mu_{\boldsymbol\ep}$'s must vanish, we consider the ``support" of
%\eqref{eq:lin}, 
%\begin{equation} \label{eq:S}
%S:=\{\boldsymbol\ep\in\mathcal C(n-i,k):\mu_{\boldsymbol\ep}\ne0\}. 
%\end{equation}
%Let $\boldsymbol\et$ be an element in the support $S$ such that
%$$\Vert\boldsymbol\et\Vert_\infty:=\max\{\et_j:1\le j\le k\}$$
%is maximal among all elements of $S$. Then, in \eqref{eq:lin}, we
%consider all $v_{\boldsymbol\ep}$ with a non-zero coefficient of
%\begin{equation} \label{eq:eta}
%e_{\boldsymbol\et+(0,\dots,0,i,0,\dots,0)}, 
%\end{equation}
%the $i$ sitting in the $r$-th place, where $r$ is the place of a
%maximum among all components of $\boldsymbol\et$, that is,
%$$\et_r=\Vert\boldsymbol\et\Vert_\infty.$$
%Going back to the definition \eqref{eq:v} of $v_{\boldsymbol\ep}$, a
%non-zero coefficient of \eqref{eq:eta} in $v_{\boldsymbol\ep}$ can
%only occur if there is a $\boldsymbol\de$ such that
%$$\boldsymbol\de+\boldsymbol\ep=\boldsymbol\et+(0,\dots,0,i,0,\dots,0).$$
%This, in its turn, implies that $\de_j+\ep_j=\et_j$ for $j\ne r$. In
%particular, $\ep_r\le\et_j$ for $j\ne r$. But then we have
%$$\ep_r=(n-i)-\underset{j\ne r}{
%\sum _{j=1} ^{k}}\ep_j=\et_r+\underset{j\ne r}{
%\sum _{j=1} ^{k}}(\et_j-\ep_j)\ge\et_r.
%$$
%Thus, if $\boldsymbol\ep\ne\boldsymbol\et$, then $\ep_r>\et_r$, that is,
%$$\Vert\boldsymbol\ep\Vert_\infty>\Vert\boldsymbol\et\Vert_\infty.$$
%But then $\boldsymbol\et$ was not maximal in $S$ with respect to
%$\Vert.\Vert_\infty$, a contradiction. Hence, the only
%$\boldsymbol\ep$ in $S$ such that $v_{\boldsymbol\ep}$ contains the
%vector \eqref{eq:eta} with a  non-zero coefficient is $\boldsymbol\et$
%itself. But since the vector \eqref{eq:eta} is one of the basis
%vectors of our vector space, it must be annihilated in
%\eqref{eq:lin}. This is only possible if $\mu_{\boldsymbol\et}=0$, a
%contradiction to our assumption that $\boldsymbol\et$ was an element
%of the support $S$. Thus, $S$ was in fact empty, proving the
%independence of the vectors $v_{\boldsymbol\ep}$,
%$\boldsymbol\ep\in\mathcal C(n-i,k)$.

\medskip
{\it Step~2. Comparison of degrees.}
By inspection,
the (total) degree in the $x_i$'s and $\la_i$'s 
of the determinant on the left-hand side of
\eqref{eq:Brunat4} is at most $n\cdot\vert \mathcal C(n,k)\vert=n\binom
{n+k-1}n$.
%\begin{align*}
%\sum _{\boldsymbol\be\in\mathcal C(n,k)} ^{}\be_1&=
%\sum _{i=0} ^{n}i\cdot\vert\mathcal C(n-i,k-1)\vert\\&=
%\sum _{i=0} ^{n}i\binom {n-i+k-2}{k-2}\\
%&=-\sum _{i=0} ^{n}(n-i)\binom {n-i+k-2}{k-2}+
%\sum _{i=0} ^{n}n\binom {n-i+k-2}{k-2}\\
%&=-(k-1)\sum _{i=0} ^{n}\binom {n-i+k-2}{k-1}+
%n\sum _{i=0} ^{n}\binom {n-i+k-2}{k-2}\\
%&=-(k-1)\binom {n+k-1}{k}+n\binom {n+k-1}{k-1}\\
%&=\binom {n+k-1}{k}.
%\end{align*}
%Here, we used special instances of the Chu--Vandermonde summation to
%evaluate the sums over $i$.
On the other hand, the degree in the $x_i$'s and $\la_i$'s 
of the right-hand side of
\eqref{eq:Brunat4} is equal to $k\binom {n+k-1}k=n\binom {n+k-1}n$.
%\begin{equation} \label{eq:vand1} 
%\sum _{i=1} ^{n}\binom {n+k-i-1}{k-1}=\binom {n+k-1}k.
%\end{equation}
%Again, we used a special instance of the Chu--Vandermonde summation to
%evaluate the sum over $i$.

Since the degree bound on the determinant is the same as the degree
of the right-hand side of \eqref{eq:Brunat4}, 
the determinant must be equal to the
right-hand side up to a possible multiplicative constant
which is independent of the $x_i$'s and $\la_i$'s. 
%Moreover, since, by symmetry, the same
%is also true for $x_i$, $i=2,3,\dots,k$, 
We conclude that
\vbox{
\begin{multline} \label{eq:const}
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left(\binom{\mathbf x+\boldsymbol\la\boldsymbol\al}
    {\boldsymbol \be}\right)=\text{const}(n,k)
\left(\prod _{j=1}
^{k}\la_j\right)^{\binom {n+k-1}k}\\
\times
{\displaystyle\underset{\ep_1+\dots+\ep_k<n}
{\prod _{\ep_1,\dots,\ep_k\ge0} ^{}} \left(
n+ \sum _{j=1} ^{k}\left(\frac {x_j} {\la_j}-\frac {\ep_j}
{\la_j}\right) \right)},
\end{multline}}

\noindent
where $\text{const}(n,k)$ is independent of $\mathbf x$ and $\boldsymbol\la$. 

\medskip
{\it Step~3. Computation of the multiplicative constant.}
If we set $x_i=0$ and $\la_i=1$ for all $i$, then the determinant on the
left-hand side of \eqref{eq:Brunat4} becomes triangular with $1$s on
the diagonal. Thus, we obtain that
$$1=\text{const}(n,k)\underset{\ep_1+\dots+\ep_k<n}
{\prod _{\ep_1,\dots,\ep_k\ge0} ^{}} \left(
n - \sum _{j=1} ^{k}{\ep_j}\right) =
\text{const}(n,k)\prod _{i=1} ^{n} i^{\vert\mathcal C(n-i,k)\vert}.
$$
This implies that
$$\text{const}(n,k)=\prod _{i=1} ^{n} i^{-\binom {n+k-i-1}{k-1}},$$
%In order to compute the multiplicative constant
%$C(n,k,\la)$ in \eqref{eq:const}, we proceed by induction on $n+k$.
%
%To start the induction, we consider the case where $n$ is
%arbitrary and $k=1$, in which case
%a straight-forward computation from \eqref{eq:const} yields that 
%$C(n,1,\la)=1$, and the case where $n=1$ and $k$ is arbitrary, in
%which case it is not difficult to see from \eqref{eq:const} that
%$C(1,k,\la)=\la^{k-1}$. 
%
%For carrying out the induction step, we put $x_1=0$ in \eqref{eq:const}. 
%For this specialization, the matrix $M(n,k)$ decomposes in a block
%form if we put the rows and columns corresponding to compositions in
%$\mathcal C(n,k)$ with first summand equal to zero first. More
%precisely, for $x_1=0$ we have
%$$M(n,k)=\begin{pmatrix} A&0\\ *&B\end{pmatrix},$$
%where $A=M(n,k-1)\big\vert_{x_1=0}$, and 
%$$B=\left(\binom{\la\al_1}
%    {\be_1}
%\prod _{i=2} ^{k}\binom{x_i+\la\al_i}
%    {\be_i}\right)_{\boldsymbol\al,\boldsymbol\be\in\mathcal
%C'(n,k)},$$
%where $\mathcal C'(n,k)$ denotes the set of compositions of $n$ into
%$k$ non-negative summands out of which the first one is positive.
%If we now factor $\la\al_1$ out of the $\boldsymbol \al$-th row of $B$
%for all $\boldsymbol \al\in\mathcal C'(n,k)$,
%and if we factor $\be_1$ out of the $\boldsymbol\be$-th column of $B$
%for all $\boldsymbol \be\in\mathcal C'(n,k)$, then the remaining
%matrix is $M(n-1,k)\big\vert_{x_1=\la-1}$. Thus,
%\begin{equation} \label{eq:Gl} 
%\det M(n,k)\Big\vert_{x_1=0}=\la^{\vert C'(n,k)\vert}
%\det M(n,k-1)\Big\vert_{x_1=0}\cdot
%\det M(n-1,k)\Big\vert_{x_1=\la-1}.
%\end{equation}
%However, the cardinality of $C'(n,k)$ is the same as the cardinality
%of $C(n-1,k)$, namely $\binom {n+k-2}{k-1}$. If we then substitute
%\eqref{eq:const} in \eqref{eq:Gl} and cancel factors which are common
%on both sides, then we arrive at
%$$C(n,k,\la)=\la^{\binom {n+k-2}{k-1}}C(n,k-1,\la)C(n-1,k,\la).$$
%The unique solution to this recursion with initial conditions
%$C(n,1,\la)=1$ and\break $C(1,k,\la)=\la^{k-1}$ is 
%$$C(n,k,\la)=\la^{(k-1)\binom {n+k-1}k},$$
completing the proof of the theorem.
\end{proof}

%\begin{proof}[Second proof of Theorem~\ref{thm:Brunat4}]
%Since this proof is closely modelled after those in
%\cite{BrMoAA,BrMoAB}, we shall be somewhat brief here.
%
%Let $M(n,k,\mathbf{x},\boldsymbol\lambda)$ be the matrix with rows and columns
%labelled by the $k$-compositions of $n$ in reverse lexicographic order
%(that is, we first compare the {\it last\/} summands of two compositions
%and we put the composition with smaller last summand first,
%if the last summands should be equal, then we compare the next-to-last
%summands, etc.) and with the
%$(\bm{\alpha},\bm{\beta})$-entry  equal to
%$\binom {\mathbf{x}+\boldsymbol\lambda\bm{\alpha}}{\bm{\beta}}$. Moreover, we write
%$\Delta(n,k,\mathbf{x},\boldsymbol\lambda)$ for the determinant of this matrix, that is,
%$$\Delta(n,k,\mathbf{x},\boldsymbol\lambda)=
%\det \left(\binom {\mathbf{x}+\boldsymbol\lambda\bm{\alpha}}{\bm{\beta}}\right)
%_{\bfalpha,\bfbeta\in\mathcal C(n,k)}.$$
%
%We prove \eqref{eq:Brunat4} by induction on $k$.
%For $k=1$ the result is trivial. Consider now $k=2$. Put
%$\mathbf{x}=(x_1,x_2)$ and  consider the principal minor of order
%$r+1$ of $\Delta(n,2,(x_1,x_2),(\la_1,\la_2))$,
%$$
%D_r(n,x_1,x_2,\la_1,\la_2)
%=\det_{0\le i,j\le r}\left(\binom {x_1+\lambda_1(n-i)}{n-j}
%\binom {x_2+\lambda_2 i}j\right).
%$$
%Taking out factors which are common to rows, respectively to columns,
%it is easy to see that this determinant can be evaluated using
%Lemma~3 from \cite{KratBN} with $n$ replaced by $r+1$, $X_i$ replaced
%by $\la i$, $A_j$ replaced by $(-x_1-\la_1(n+1)+n-j+1)/\la_1$, 
%and $B_j$ replaced by $(x_2-\la_2-j+2)/\la_2$. After simplification, we obtain
%\begin{multline} \label{Dr}
%D_r(n,x_1,x_2,\la_1,\la_2)=
%(\la_1\la_2)^{\binom {r+1} 2}
% \left(\prod_{i=0}^r\frac {(n-r)!} {(n-i)!} \binom
%{x_1+\la_1(n-i)}{n-r}\right)\\
%\times
%\left(\prod_{1\le i\le j\le r} \left(\frac {x_1-n+j} {\la_1}+\frac 
%{x_2-i+1} {\la_2}+n\right) \right).
%\end{multline}
%
%Next, let $k\ge 2$ (note that the following discussion also includes the case
%$k=2$). 
%Setting $\bar{\mathbf{u}}=(u_1,\dots,u_{k-1})$, an entry
%$\binom {\mathbf{x}+\lambda\bm{\alpha}}{\bm{\beta}}$ in
%$M(n,k,\mathbf{x},\boldsymbol\lambda)$ can be written in the form
%$\binom {\bar{\mathbf{x}}+\lambda\bar{\bm{\alpha}}}{\bar{\bm{\beta}}}
%\binom {x_k+\lambda\alpha_k)}{\beta_k}$. For $0\le i,j\le n$, let
%$S_{ij}$ be the matrix with entries
%$\binom {\bar{\mathbf{x}}+\lambda\bar{\bm{\alpha}}}{\bar{\bm{\beta}}}$
%where $\bm{\alpha}$ and $\bm{\beta}$ satisfy $\alpha_k=i$ and
%$\beta_k=j$. Thus, the submatrix of $M(n,k,\mathbf{x},\boldsymbol\lambda)$
%formed by the entries labelled $(\bm{\alpha},\bm{\beta})$ with
%$\alpha_k=i$ and $\beta_k=j$ can be written as
%$\left(S_{ij}\binom {x_k+\lambda i}j\right)$.  Note that
%$$
%S_{rr}=M(n-r,k-1,\bar{\mathbf{x}},\boldsymbol\lambda).
%$$
%Define $f_0(i,j)=\binom {x_k+\lambda i}j$. Then
%$M(n,k,\mathbf{x},\boldsymbol\lambda)$ admits the block
%decomposition
%$$
%M(n,k,\mathbf{x},\boldsymbol\lambda)=(S_{ij}f_0(i,j))_{0\le i,j\le n}.
%$$
%The idea is to put $M(n,k,\mathbf{x},\boldsymbol\lambda)$ in block triangular form
%in such a way that at each step only (multiples of) earlier columns 
%are added to later columns in the reverse lexicographic order.
%
%By induction,
%assume that $\Delta(n,k,\mathbf{x},\lambda)=\det (S_{ij}f_r(i,j))$ where
%$S_{ij}=\left(\binom {\bar{\mathbf{x}}+\lambda\bar{\bm{\alpha}}}
%{\bar{\bm{\beta}}}\right)$,
%with $\alpha_k=i$, $\beta_k=j$, and $f_r(i,j)=0$ for $i<r$ and $j>i$.
%
%Fix a column $\bm{\beta}$ with $\beta_k=j>r$.  For each
%$\bm{\gamma}\in C(n,k)$ with $\gamma_k=r$ and $\gamma_\ell\ge \beta_\ell$
%for $1\le \ell\le k-1$, add to the column $\bm{\beta}$ the column
%$\bm{\gamma}$ multiplied by
%$$
%-\frac{\binom {\bar{\bfgamma}}{\bar{\bfbeta}}}
%{\binom {| \bar{\mathbf{x}} |+\lambda(n-r)-n+j}{j-r}}
%\frac{f_r(r,j)}{f_r(r,r)}.
%$$
%Then, using Lemma~\ref{lem:ChuVand},
%it can be seen that, for $i\ge r$ and $j>r$, each block
%$S_{i,j}f_r(i,j)$ is modified to $S_{i,j}f_{r+1}(i,j)$ where
%\begin{align}
%f_0(i,j)
%&=
%\binom {x_k+\lambda i}j, \nonumber\\
%f_{r+1}(i,j) &= f_r(i,j)- \frac{\binom {| \bar{\mathbf{x}} |
%+\lambda(n-i)-(n-j)}{j-r}} {\binom {| \bar{\mathbf{x}} |
%+\lambda(n-r)-(n-j)}{j-r}} \cdot
%\frac{f_r(r,j)}{f_r(r,r)}f_r(i,r),
%\label{recurrence}
%\end{align}
%and $f_{r+1}(i,j)=f_r(i,j)$ for $j\le r$ and $f_{r+1}(r,j)=0$ for
%$j>r$. After $n$ iterations, we obtain a block-triangular matrix
%where the diagonal blocks are $S_{rr}f_r(r,r)$. Therefore
%\begin{equation}
%\label{rfinal}
%\Delta(n,k,\mathbf{x},\lambda)
%=\prod_{r=0}^n \left(\Delta(n-r,k-1,\mathbf{x},\lambda)
%f_r(r,r)^{\binom {n-r+k-2} { k-2}}\right).
%\end{equation}
%
%The values $\Delta(n-r,k-1,\mathbf{x},\lambda)$ are given by the induction
%hypothesis. To calculate $f_r(r,r)$ falls back to the case $k=2$. Apply the
%above procedure to calculate $\Delta(n,2,(y,z),\lambda)$. In this case the
%recurrence \eqref{recurrence} becomes 
%\begin{align}
%f_0(i,j)
%&=
%\binom {z+\lambda i}j, \nonumber \\
%f_{r+1}(i,j)
%&=
%f_r(i,j)-
%\frac{\binom {y+\lambda(n-i)-(n-j)}{j-r}}{\binom {y+\lambda(n-r)-(n-j)}{j-r}}
%\cdot \frac{f_r(r,j)}{f_r(r,r)}f_r(i,r).
%\label{r2}
%\end{align}
%In the end, the block-triangular matrix that we obtain is actually a
%{\it triangular} matrix. Since we never permuted rows or columns, and
%since we always added multiples of columns of smaller index to columns
%of larger index, the manipulations that we performed left also
%the value of each {\it principal minor} unchanged. 
%Recall that we denoted the principal minor of order $r+1$
%of $\Delta(n,2,(y,z),\lambda)$ by $D_r(n,y,z,\lambda)$.
%In view of the above remarks, the triangular form
%at the end of the elimination procedure yields the identity
%$$
%\frac{D_r(n,y,z,\lambda)}{D_{r-1}(n,y,z,\lambda)}
%=\binom {y+\lambda(n-r)}{n-r}f_r(r,r).
%$$
%However, the left-hand side can be calculated using~(\ref{Dr}). Thus,
%we arrive at
%$$
%f_r(r,r)={\lambda^r}
%\frac{\prod_{i=1}^r(y+z+(\lambda-1)n+i)}
%{\prod_{i=1}^{r-1}(y+(\lambda-1)n-\lambda i+r)}.
%$$
%The recurrence~(\ref{recurrence}) corresponds to
%recurrence~(\ref{r2}) for $y=| \bar{\mathbf{x}} |$ and $z=x_k$.
%Thus, the value of $f_r(r,r)$ for any $k$ is
%$$
%f_r(r,r)=\lambda^r \frac{\prod_{i=1}^r(| \mathbf{x} |
%+(\lambda-1)n+i)} {\prod_{i=1}^{r-1} (| \bar{\mathbf{x}} |
%+(\lambda-1)n-\lambda i+r)}.
%$$
%Substituting this in~(\ref{rfinal}) and using the induction hypothesis,
%one obtains (\ref{eq:Brunat4}).
%\end{proof}

\begin{proof}[Proof of the equivalence of Theorem~\ref{thm:Brunat4}
and \ref{thm:Brunat4b}]
If we replace $\la_i$ by $-\la_i$ and $x_i$ by $-x_i-1$ for all $i$ in
\eqref{eq:Brunat4}, and then use the
identity $\binom Xm=(-1)^m\binom {-X+m-1}m$, then we obtain
\begin{align}\notag 
&\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left((-1)^n
\binom{\mathbf x+\boldsymbol\la\boldsymbol\al+\boldsymbol\be}
    {\boldsymbol \be}\right)=\\
\label{eq:-1}
&
=(-1)^{k\binom {n+k-1}k}
\left(\prod _{j=1}
^{k}\la_j\right)^{\binom {n+k-1}k}
\frac {\displaystyle\underset{\ep_1+\dots+\ep_k<n}
{\prod _{\ep_1,\dots,\ep_k\ge0} ^{}} \left(
n+ \sum _{j=1} ^{k}\left(\frac {x_j+1} {\la_j}+\frac {\ep_j}
{\la_j}\right) \right)}
{\displaystyle\prod _{i=1} ^{n}i^{\binom{n+k-i-1} {k-1}}}.
\end{align}
Except for the signs, this is exactly \eqref{eq:Brunat4b}. However,
we have
\begin{multline*}
 n\cdot\vert \mathcal C(n,k)\vert+k\binom {n+k-1} k
= n\binom {n+k-1} { k-1}+k\binom {n+k-1} k
=2\frac{(n+k-1)!}{(k-1)!\,(n-1)!}.
\end{multline*}
Since this is an even number, the signs in \eqref{eq:-1} do indeed cancel.
\end{proof}


\begin{proof}[Proof of Corollary~\ref{thm:Brunat5a}]
The right-hand and left-hand sides of \eqref{eq:Brunat4} are both
polynomials in the $\la_i$'s and the $x_i$'s. As we already observed in
Step~2 of the first proof of Theorem~\ref{thm:Brunat4},
the (total) degree in the $\la_i$'s and the
$x_i$'s of the determinant on the left-hand side is equal to
$$n\cdot\vert\mathcal C(n,k)\vert=n\binom {n+k-1}{k-1}=
k\binom {n+k-1}k,$$ 
and the degree in the $\la_i$'s and the
$x_i$'s of the expression on the right-hand side is exactly the same
value.
Therefore, if we extract the homogeneous parts in the $\la_i$'s and the
$x_i$'s of degree $k\binom {n+k-1}k$ in \eqref{eq:Brunat4}, we obtain
\begin{equation*}
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left(\frac {(\mathbf x+\boldsymbol\la\boldsymbol\al)^{\boldsymbol\be}}
    {\boldsymbol \be!}\right)=\left(\prod _{j=1}
^{k}\la_j\right)^{\binom {n+k-1}k}
\frac {\displaystyle\underset{\ep_1+\dots+\ep_k<n}
{\prod _{\ep_1,\dots,\ep_k\ge0} ^{}} \left(
n+ \sum _{j=1} ^{k}\frac {x_j} {\la_j} \right)}
{\displaystyle\prod _{i=1} ^{n}i^{\binom{n+k-i-1} {k-1}}},
\end{equation*}
where $\boldsymbol \be!=
\prod _{i=1} ^{k}\be_i!$, or, equivalently,
\begin{equation*}
\det_{\boldsymbol\al,\boldsymbol\be\in\mathcal
C(n,k)}\left( {(\mathbf x+\boldsymbol\la\boldsymbol\al)^{\boldsymbol\be}}
\right)=\left(\prod _{j=1}
^{k}\la_j\right)^{\binom {n+k-1}k}
\frac{\left(n+ \sum _{j=1} ^{k}\frac {x_j} {\la_j}\right)^{\sum _{i=0}
^{n-1}\vert\mathcal C(i,k)\vert}}
{\prod _{i=1} ^{n}i^{\binom{n+k-i-1} {k-1}}}
\Bigg(
\prod _{\boldsymbol\be\in\mathcal
C(n,k)} ^{}\boldsymbol \be!\Bigg).
\end{equation*}
Since 
$$\sum _{i=0} ^{n-1}\vert\mathcal C(i,k)\vert=
\sum _{i=0} ^{n-1}\binom {i+k-1}{k-1}=\binom {n+k-1}k,$$
the only missing piece for the proof of the corollary is the
verification of the identity
\begin{equation} \label{eq:beta}
 \prod _{\boldsymbol\be\in\mathcal
C(n,k)} ^{}\boldsymbol \be!=
\prod _{\be_1+\dots+\be_k=n} ^{}\be_1!\cdots\be_k!=
\prod _{i=1} ^{n} i
      ^{k\binom{n+k-i-1} {k-1}}.
\end{equation}
This can, for example, be done by induction on $n+k$, by using the
obvious recurrence
$$\Pi(n,k)=
\prod _{i=0} ^{n}\left(i!^{\vert\mathcal C(n-i,k-1)\vert}\Pi(n-i,k-1)\right),$$
where
$$\Pi(n,k)=\prod _{\be_1+\dots+\be_k=n} ^{}\be_1!\cdots\be_k!.$$
\end{proof}


\section{Determinants for compositions with only positive parts}

Let $\mathcal C^*(n,k)$ denote the set of all compositions of $n$ with
exactly $k$ positive summands.
Then we have the following theorem.

\begin{Theorem}
\label{DP}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ and
$\boldsymbol\la=(\la_1,\la_2,\dots,\la_k)$ be vectors of
indeterminates.
Then, for any positive integers $n$ and $k$, $n\ge k$, we have
\begin{multline} \label{eq:DP}
\det_{\bfalpha,\bfbeta\in \mathcal C^*(n,k)}
\left(\binom{\mathbf{x}+\boldsymbol\lambda\bfalpha}{\bfbeta} \right)=
\left(\prod _{j=1}
^{k}\la_j\right)^{\binom {n-1}k}
%
\left(
\prod_{i=1}^{k}\prod_{j=1}^{n-k+1}
\left(\frac { x_i+\lambda_i j}j\right)^{\binom {n-j-1} { k-2}}
\right)\\
\times
\frac {\displaystyle\underset{\ep_1+\dots+\ep_k<n-k}
{\prod _{\ep_1,\dots,\ep_k\ge0} ^{}} \left(
n+ \sum _{j=1} ^{k}\left(\frac {x_j-1} {\la_j}-\frac {\ep_j}
{\la_j}\right) \right)}
{\displaystyle\prod _{i=1} ^{n}i^{\binom{n-i-1} {k-1}}}.
\end{multline}
\end{Theorem}
\begin{proof}
Let $\mathbf{1}$ be the $k$-vector with all entries equal to 1.
The mapping $\mathcal C^*(n,k)\rightarrow \mathcal C(n-k,k)$ defined by 
$\boldsymbol{\alpha}=(\alpha_1,\ldots,\alpha_k)
\mapsto \boldsymbol{\alpha}-\mathbf{1}
=(\alpha_1-1,\ldots, \alpha_k-1)$
is bijective. Thus, we have
\begin{align*}
\det_{\boldsymbol{\alpha},\boldsymbol{\beta}\in \mathcal C^*(n,k)}
&\left(\binom {\mathbf{x}+\boldsymbol\lambda\boldsymbol{\alpha}}
{\boldsymbol{\beta}}\right)  = 
\det_{\boldsymbol{\alpha},\boldsymbol{\beta}\in \mathcal C^*(n,k)}
\left( 
\binom {\mathbf{x}+\boldsymbol\lambda+\boldsymbol\lambda(\boldsymbol{\alpha}-\mathbf{1})}
{\boldsymbol{\beta}-\mathbf{1}+\mathbf{1}}
\right) \\
&= 
\det_{\boldsymbol{\alpha},\boldsymbol{\beta}\in \mathcal C(n-k,k)}
\left(
\binom {\mathbf{x}+\boldsymbol\lambda+\boldsymbol\lambda\boldsymbol{\alpha}}
{\boldsymbol{\beta}+\mathbf{1}}
\right)  \\
&=  
\det_{\boldsymbol{\alpha},\boldsymbol{\beta}\in \mathcal C(n-k,k)}
\left(
 \binom{\mathbf{x}+\boldsymbol\lambda-\mathbf{1}+\boldsymbol\lambda\bfalpha}{\bfbeta}
\prod _{i=1} ^{k}\frac {
x_i+\lambda_i+\lambda_i\alpha_i}{\beta_i+1}
\right)  \\
&=
\left(
\prod_{\boldsymbol{\alpha}\in \mathcal C(n-k,k)}
\prod _{i=1} ^{k}\frac {
x_i+\lambda_i(\alpha_i+1)}{\alpha_i+1}
\right)\det_{\boldsymbol{\alpha},\boldsymbol{\beta}\in \mathcal C(n-k,k)}
\left(
 \binom{\mathbf{x}+\boldsymbol\lambda-\mathbf{1}+\boldsymbol\lambda\bfalpha}{\bfbeta}
\right).
\end{align*}
In the product, a ratio $(x_i+\lambda_i (j+1))/(j+1)$ 
appears as many times as there are
compositions in $\mathcal C(n-k-j,k-1)$, i.e., ${\binom {n-j-2} { k-2}}$ times.
Thus,
\begin{align*}
\prod_{\boldsymbol{\alpha}\in \mathcal C(n-k,k)}
\prod _{i=1} ^{k}\frac {
x_i+\lambda_i(\alpha_i+1)}{\alpha_i+1}
&=\prod_{i=1}^k\prod_{j=0}^{n-k} 
\left(\frac {x_i+\lambda_i(j+1)} {j+1}\right)^{\binom {n-j-2} { k-2}}\\
&=
\prod_{i=1}^k\prod_{j=1}^{n-k+1} \left(\frac 
{x_i+\lambda_i j}j\right)^{\binom {n-j-1} { k-2}}.
\end{align*}
Substituting this and using Theorem~\ref{thm:Brunat4} with $n$ replaced 
by $n-k$ and $\mathbf x$ replaced by $\mathbf x+\boldsymbol\la-\mathbf1$ 
to evaluate the determinant, we obtain the desired formula. 
\end{proof}  

By replacing $\la_i$ by $-\la_i$ and $x_i$ by $-x_i-1$ for all $i$ in
\eqref{eq:DP}, and then using the
identity $\binom Xm=(-1)^m\binom {-X+m-1}m$, we obtain the following
equivalent form of Theorem~\ref{DP}.

\begin{Theorem}
\label{DP2}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ and
$\boldsymbol\la=(\la_1,\la_2,\dots,\la_k)$ be vectors of
indeterminates.
Then, for any positive integers $n$ and $k$, $n\ge k$, we have
\begin{multline} \label{eq:DP2}
\det_{\bfalpha,\bfbeta\in \mathcal C^*(n,k)}
\left(\binom{\mathbf{x}+\boldsymbol\lambda\bfalpha+\bfbeta}{\bfbeta} \right)=
\left(\prod _{j=1}
^{k}\la_j\right)^{\binom {n-1}k}
%
\left(
\prod_{i=1}^{k}\prod_{j=1}^{n-k+1}
\left(\frac { x_i+\lambda_i j+1}j\right)^{\binom {n-j-1} { k-2}}
\right)\\
\times
\frac {\displaystyle\underset{\ep_1+\dots+\ep_k<n-k}
{\prod _{\ep_1,\dots,\ep_k\ge0} ^{}} ^{} \left(
n+ \sum _{j=1} ^{k}\left(\frac {x_j+2} {\la_j}+\frac {\ep_j}
{\la_j}\right) \right)}
{\displaystyle\prod _{i=1} ^{n}i^{\binom{n-i-1} {k-1}}}.
\end{multline}
\end{Theorem}


Extracting the highest homogeneous component in Theorem~\ref{DP}, we obtain
the following analogue of Corollary~\ref{thm:Brunat1a}.

\begin{Corollary} \label{thm:Brunat1b}
Let $\mathbf x=(x_1,x_2,\dots,x_k)$ and
$\boldsymbol\la=(\la_1,\la_2,\dots,\la_k)$ be vectors of
indeterminates.
Then, for any positive integers $n$ and $k$, $n\ge k$, we have
\begin{multline}
\det_{\bfalpha,\bfbeta\in \mathcal C^*(n,k)}
\left((\mathbf{x}+\boldsymbol\lambda\bfalpha)^{\bfbeta} \right)=
\left(\prod _{j=1}
^{k}\la_j\right)^{\binom {n-1}k}
%
\left(
\prod_{i=1}^{k}\prod_{j=1}^{n-k+1}
\left( { x_i+\lambda_i j}\right)^{\binom {n-j-1} { k-2}}
\right)\\
\times
\left(
{\displaystyle\underset{\ep_1+\dots+\ep_k<n-k}
{\prod _{\ep_1,\dots,\ep_k\ge0} ^{}} \left(
n+ \sum _{j=1} ^{k}\frac {x_j} {\la_j}\right)}\right)
\prod _{i=1} ^{n-k}i^{(k-1)\binom {n-i-1}{k-1}}.
\end{multline}
\end{Corollary}



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\end{document}

