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\ihead{A determinant identity for symmetric matrices, version \today}
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\begin{document}
\section*{A determinant identity for symmetric matrices}

\textit{Darij Grinberg,
%TCIMACRO{\TeXButton{TeX field}{\today}}%
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\today
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}

\subsection{The formula}

A strange asymmetry is haunting linear algebra: Alternating matrices (i.e.,
square matrices $A$ that satisfy $A^{T}=-A$ and have a diagonal full of
zeroes) are known to have a rich and beautiful algebraic theory, while little
has been said algebraically about symmetric matrices (i.e., square matrices
$A$ that satisfy $A^{T}=A$). It is as if symmetric matrices \textquotedblleft
belong to\textquotedblright\ the realm of analysis (where the spectral theorem
and various decompositions help manage them over $\mathbb{R}$), whereas
alternating matrices have their home in algebra and algebraic combinatorics
(as seen most elegantly in the theory of Pfaffians \cite[\S 12.12]{Loehr-BC},
\cite{DreWen95}, \cite{Knuth95}).

In the 1970s, Procesi discovered a remarkable determinantal identity for
symmetric matrices that disrupts this pattern. He used this identity to
classify the invariants of the orthogonal group. In this note, we shall give
it a new and simpler proof.

To state this identity, let us first introduce some notations.

If $a$ and $b$ are two integers, then $\left[  a,b\right]  $ shall denote the
set $\left\{  a,a+1,a+2,\ldots,b\right\}  $ (this is empty if $a>b$). We let
$\left[  n\right]  $ denote the set $\left[  1,n\right]  =\left\{
1,2,\ldots,n\right\}  $, where $n$ is any nonnegative integer. Given any set
$U$, we let $S_{U}$ denote the group of all permutations of $U$. In
particular, $S_{\left[  n\right]  }$ is the well-known $n$-th symmetric group
$S_{n}$. When the set $U$ is finite, we will denote the sign of a permutation
$\sigma\in S_{U}$ by $\left(  -1\right)  ^{\sigma}$.

Let $\mathbb{K}$ be a commutative ring. Fix an integer $k\geq0$. Let
$A=\left(  a_{i,j}\right)  _{1\leq i,j\leq2k}\in\mathbb{K}^{2k\times2k}$ be a
symmetric $2k\times2k$-matrix. Given any two $k$-tuples $\left(  i_{1}%
,i_{2},\ldots,i_{k}\right)  $ and $\left(  j_{1},j_{2},\ldots,j_{k}\right)  $
of elements of $\left[  2k\right]  $, we define $A_{i_{1},i_{2},\ldots,i_{k}%
}^{j_{1},j_{2},\ldots,j_{k}}$ to be the $k\times k$-matrix $\left(
a_{i_{x},j_{y}}\right)  _{1\leq x,y\leq k}$. If $i_{1}<i_{2}<\cdots<i_{k}$ and
$j_{1}<j_{2}<\cdots<j_{k}$, then this matrix is a submatrix of $A$; otherwise,
it is a \textquotedblleft submatrix in a wider sense\textquotedblright,
allowing for repeated and permuted rows/columns.

The identity we shall be concerned with is the following:

\begin{theorem}
\label{thm.symm-procesi}Let $p\in\left[  k\right]  $. Then,%
\[
\sum_{\sigma\in S_{\left[  p,k+p\right]  }}\left(  -1\right)  ^{\sigma}%
\det\left(  A_{1,2,\ldots,p-1,\sigma\left(  p\right)  ,\sigma\left(
p+1\right)  ,\ldots,\sigma\left(  k\right)  }^{\sigma\left(  k+1\right)
,\sigma\left(  k+2\right)  ,\ldots,\sigma\left(  k+p\right)
,k+p+1,k+p+2,\ldots,2k}\right)  =0.
\]

\end{theorem}

\begin{example}
\label{exa.thm.symm-procesi.1}Let $k=2$ and $p=2$. Then, Theorem
\ref{thm.symm-procesi} says that
\[
\sum_{\sigma\in S_{\left[  2,4\right]  }}\left(  -1\right)  ^{\sigma}%
\det\left(  A_{1,\sigma\left(  2\right)  }^{\sigma\left(  3\right)
,\sigma\left(  4\right)  }\right)  =0.
\]
Expanding this sum, we can rewrite this as%
\begin{align}
&  \det\left(  A_{1,2}^{3,4}\right)  -\det\left(  A_{1,3}^{2,4}\right)
-\det\left(  A_{1,4}^{3,2}\right)  -\det\left(  A_{1,2}^{4,3}\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\det\left(  A_{1,3}^{4,2}\right)  +\det\left(
A_{1,4}^{2,3}\right)  =0. \label{eq.eq.thm.symm-procesi.1.2}%
\end{align}
The six addends on the left hand side actually come in three pairs of equal
addends, since a determinant changes sign when two of its columns are swapped:%
\begin{align*}
\det\left(  A_{1,2}^{4,3}\right)   &  =-\det\left(  A_{1,2}^{3,4}\right)  ;\\
\det\left(  A_{1,3}^{4,2}\right)   &  =-\det\left(  A_{1,3}^{2,4}\right)  ;\\
\det\left(  A_{1,4}^{3,2}\right)   &  =-\det\left(  A_{1,4}^{2,3}\right)  .
\end{align*}
Collecting these equal addends together, we can rewrite the equality
(\ref{eq.eq.thm.symm-procesi.1.2}) as%
\[
2\left(  \det\left(  A_{1,2}^{3,4}\right)  -\det\left(  A_{1,3}^{2,4}\right)
+\det\left(  A_{1,4}^{2,3}\right)  \right)  =0.
\]
Since this is a polynomial identity in the entries of $A$, we can divide it by
$2$ (working in a polynomial ring over $\mathbb{Z}$ instead of $\mathbb{K}$)
and obtain
\begin{equation}
\det\left(  A_{1,2}^{3,4}\right)  -\det\left(  A_{1,3}^{2,4}\right)
+\det\left(  A_{1,4}^{2,3}\right)  =0. \label{eq.eq.thm.symm-procesi.1.4}%
\end{equation}

\end{example}

More generally, proceeding as in Example \ref{exa.thm.symm-procesi.1}, we can
reduce Theorem \ref{thm.symm-procesi} as follows:

\begin{theorem}
\label{thm.symm-procesi-red}Let $p\in\left[  k\right]  $. Let $T_{\left[
p,k+p\right]  }$ be the subgroup of $S_{\left[  p,k+p\right]  }$ consisting of
those permutations $\sigma$ that fix the subset $\left[  p,k\right]  $ (that
is, satisfy $\sigma\left(  \left[  p,k\right]  \right)  =\left[  p,k\right]
$). Let $L$ be a left transversal (i.e., a system of distinct representatives
for the left cosets) of $T_{\left[  p,k+p\right]  }$ in $S_{\left[
p,k+p\right]  }$. Then,%
\[
\sum_{\sigma\in L}\left(  -1\right)  ^{\sigma}\det\left(  A_{1,2,\ldots
,p-1,\sigma\left(  p\right)  ,\sigma\left(  p+1\right)  ,\ldots,\sigma\left(
k\right)  }^{\sigma\left(  k+1\right)  ,\sigma\left(  k+2\right)
,\ldots,\sigma\left(  k+p\right)  ,k+p+1,k+p+2,\ldots,2k}\right)  =0.
\]

\end{theorem}

Theorem \ref{thm.symm-procesi-red} appears in \cite[\S 13.8.3, Lemma]%
{Proces06} (and, in a more inchoate form, in \cite[Lemma 5.2]{DeCPro76}),
where it is used to identify the invariants of the orthogonal group.

As in Example \ref{exa.thm.symm-procesi.1}, we can easily derive Theorem
\ref{thm.symm-procesi-red} from Theorem \ref{thm.symm-procesi}: Just observe
that the determinant $\det\left(  A_{1,2,\ldots,p-1,\sigma\left(  p\right)
,\sigma\left(  p+1\right)  ,\ldots,\sigma\left(  k\right)  }^{\sigma\left(
k+1\right)  ,\sigma\left(  k+2\right)  ,\ldots,\sigma\left(  k+p\right)
,k+p+1,k+p+2,\ldots,2k}\right)  $ gets multiplied by the sign $\left(
-1\right)  ^{\tau}$ when we multiply the permutation $\sigma\in S_{\left[
p,k+p\right]  }$ by a permutation $\tau\in T_{\left[  p,k+p\right]  }$ from
the right (since this merely permutes the rows and the columns of the matrix,
and the signs of the relevant permutations multiply to $\left(  -1\right)
^{\tau}$). This observation lets us bunch equal addends in Theorem
\ref{thm.symm-procesi-red} together into blocks, thus yielding%
\[
\left\vert T_{\left[  p,k+p\right]  }\right\vert \cdot\sum_{\sigma\in
L}\left(  -1\right)  ^{\sigma}\det\left(  A_{1,2,\ldots,p-1,\sigma\left(
p\right)  ,\sigma\left(  p+1\right)  ,\ldots,\sigma\left(  k\right)  }%
^{\sigma\left(  k+1\right)  ,\sigma\left(  k+2\right)  ,\ldots,\sigma\left(
k+p\right)  ,k+p+1,k+p+2,\ldots,2k}\right)  =0.
\]
Finally, we can divide by the positive integer $\left\vert T_{\left[
p,k+p\right]  }\right\vert $ after making the WLOG assumption that the entries
of $A$ are indeterminates in a polynomial ring over $\mathbb{Z}$.

Hence, it remains to prove Theorem \ref{thm.symm-procesi}. This was done in
\cite[\S 13.8.3, Lemma]{Proces06} and \cite[Lemma 5.2]{DeCPro76} by a
complicated induction. We shall instead give a simple proof that uses nothing
but the definition of a determinant.

\subsection{The proof}

\begin{proof}
[Proof of Theorem \ref{thm.symm-procesi}.]We extend each permutation
$\sigma\in S_{\left[  p,k+p\right]  }$ to a permutation $\widehat{\sigma}\in
S_{\left[  2k\right]  }$ by setting
\[
\widehat{\sigma}\left(  i\right)  :=%
\begin{cases}
\sigma\left(  i\right)  , & \text{if }i\in\left[  p,k+p\right]  ;\\
i, & \text{if }i\notin\left[  p,k+p\right]
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for all }i\in\left[  2k\right]  .
\]
This extended permutation $\widehat{\sigma}$ agrees with the original $\sigma$
on the inputs $p,p+1,\ldots,k+p$. Thus, it shall lead to no confusion if we
simply write $\sigma\left(  i\right)  $ for $\widehat{\sigma}\left(  i\right)
$ whenever $i\in\left[  2k\right]  $. We will therefore do so, for brevity's
sake. Thus, we can rewrite the determinant%
\[
\det\left(  A_{1,2,\ldots,p-1,\sigma\left(  p\right)  ,\sigma\left(
p+1\right)  ,\ldots,\sigma\left(  k\right)  }^{\sigma\left(  k+1\right)
,\sigma\left(  k+2\right)  ,\ldots,\sigma\left(  k+p\right)
,k+p+1,k+p+2,\ldots,2k}\right)
\]
in Theorem \ref{thm.symm-procesi} in the nicer form%
\[
\det\left(  A_{\sigma\left(  1\right)  ,\sigma\left(  2\right)  ,\ldots
,\sigma\left(  k\right)  }^{\sigma\left(  k+1\right)  ,\sigma\left(
k+2\right)  ,\ldots,\sigma\left(  2k\right)  }\right)  .
\]
Thus, for each $\sigma\in S_{\left[  p,k+p\right]  }$, we have%
\begin{align*}
&  \det\left(  A_{1,2,\ldots,p-1,\sigma\left(  p\right)  ,\sigma\left(
p+1\right)  ,\ldots,\sigma\left(  k\right)  }^{\sigma\left(  k+1\right)
,\sigma\left(  k+2\right)  ,\ldots,\sigma\left(  k+p\right)
,k+p+1,k+p+2,\ldots,2k}\right) \\
&  =\det\left(  A_{\sigma\left(  1\right)  ,\sigma\left(  2\right)
,\ldots,\sigma\left(  k\right)  }^{\sigma\left(  k+1\right)  ,\sigma\left(
k+2\right)  ,\ldots,\sigma\left(  2k\right)  }\right) \\
&  =\det\left(  a_{\sigma\left(  i\right)  ,\ \sigma\left(  k+j\right)
}\right)  _{1\leq i,j\leq k}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }A_{\sigma\left(  1\right)  ,\sigma\left(  2\right)
,\ldots,\sigma\left(  k\right)  }^{\sigma\left(  k+1\right)  ,\sigma\left(
k+2\right)  ,\ldots,\sigma\left(  2k\right)  }\right) \\
&  =\sum_{\zeta\in S_{k}}\left(  -1\right)  ^{\zeta}\prod_{i=1}^{k}%
a_{\sigma\left(  i\right)  ,\ \sigma\left(  k+\zeta\left(  i\right)  \right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of a determinant}\right)
.
\end{align*}
Multiplying this with $\left(  -1\right)  ^{\sigma}$ and summing the result
over all $\sigma\in S_{\left[  p,k+p\right]  }$, we obtain%
\begin{align}
&  \sum_{\sigma\in S_{\left[  p,k+p\right]  }}\left(  -1\right)  ^{\sigma}%
\det\left(  A_{1,2,\ldots,p-1,\sigma\left(  p\right)  ,\sigma\left(
p+1\right)  ,\ldots,\sigma\left(  k\right)  }^{\sigma\left(  k+1\right)
,\sigma\left(  k+2\right)  ,\ldots,\sigma\left(  k+p\right)
,k+p+1,k+p+2,\ldots,2k}\right) \nonumber\\
&  =\sum_{\sigma\in S_{\left[  p,k+p\right]  }}\left(  -1\right)  ^{\sigma
}\sum_{\zeta\in S_{k}}\left(  -1\right)  ^{\zeta}\prod_{i=1}^{k}%
a_{\sigma\left(  i\right)  ,\ \sigma\left(  k+\zeta\left(  i\right)  \right)
}\nonumber\\
&  =\sum_{\zeta\in S_{k}}\left(  -1\right)  ^{\zeta}\sum_{\sigma\in S_{\left[
p,k+p\right]  }}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{k}a_{\sigma\left(
i\right)  ,\ \sigma\left(  k+\zeta\left(  i\right)  \right)  }.
\label{pf.thm.symm-procesi.5}%
\end{align}


Now, let $\zeta\in S_{k}$ be arbitrary. We shall show that
\begin{equation}
\sum_{\sigma\in S_{\left[  p,k+p\right]  }}\left(  -1\right)  ^{\sigma}%
\prod_{i=1}^{k}a_{\sigma\left(  i\right)  ,\ \sigma\left(  k+\zeta\left(
i\right)  \right)  }=0. \label{pf.thm.symm-procesi.subsum}%
\end{equation}


Indeed, the set $\left[  p,k+p\right]  $ has size $\left\vert \left[
p,k+p\right]  \right\vert =k+1>k$. Hence, by the pigeonhole principle, there
exists some $j\in\left[  k\right]  $ such that both numbers $j$ and
$k+\zeta\left(  j\right)  $ belong to the set $\left[  p,k+p\right]
$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $\left\vert
\left\{  j,k+\zeta\left(  j\right)  \right\}  \cap\left[  p,k+p\right]
\right\vert \leq1$ for each $j\in\left[  k\right]  $. Hence,%
\[
\left\vert \bigcup_{j=1}^{k}\left(  \left\{  j,k+\zeta\left(  j\right)
\right\}  \cap\left[  p,k+p\right]  \right)  \right\vert \leq\sum_{j=1}%
^{k}\underbrace{\left\vert \left\{  j,k+\zeta\left(  j\right)  \right\}
\cap\left[  p,k+p\right]  \right\vert }_{\leq1}\leq\sum_{j=1}^{k}1=k.
\]
In view of
\begin{align*}
\bigcup_{j=1}^{k}\left(  \left\{  j,k+\zeta\left(  j\right)  \right\}
\cap\left[  p,k+p\right]  \right)   &  =\underbrace{\left(  \bigcup_{j=1}%
^{k}\left\{  j,k+\zeta\left(  j\right)  \right\}  \right)  }%
_{\substack{=\left\{  1,2,\ldots,k,k+\zeta\left(  1\right)  ,k+\zeta\left(
2\right)  ,\ldots,k+\zeta\left(  k\right)  \right\}  \\=\left[  2k\right]
\\\text{(since }\zeta\in S_{k}\text{ ensures that the numbers }k+\zeta\left(
1\right)  ,k+\zeta\left(  2\right)  ,\ldots,k+\zeta\left(  k\right)
\\\text{are precisely }k+1,k+2,\ldots,2k\text{)}}}\cap\left[  p,k+p\right] \\
&  =\left[  2k\right]  \cap\left[  p,k+p\right]  =\left[  p,k+p\right]  ,
\end{align*}
this rewrites as $\left\vert \left[  p,k+p\right]  \right\vert \leq k$. But
this contradicts $\left\vert \left[  p,k+p\right]  \right\vert >k$.}. Consider
this $j$. The two numbers $j$ and $k+\zeta\left(  j\right)  $ are furthermore
distinct (since the former is $\leq k$ while the latter is $>k$); thus, the
transposition $\tau:=t_{j,k+\zeta\left(  j\right)  }\in S_{\left[
p,k+p\right]  }$ exists (since $j$ and $k+\zeta\left(  j\right)  $ belong to
$\left[  p,k+p\right]  $) and has sign $\left(  -1\right)  ^{\tau}=-1$ (since
any transposition has sign $-1$). Since $S_{\left[  p,k+p\right]  }$ is a
group, we have $\sigma\tau\in S_{\left[  p,k+p\right]  }$ for each $\sigma\in
S_{\left[  p,k+p\right]  }$ (because $\tau\in S_{\left[  p,k+p\right]  }$).

When we replace a permutation $\sigma\in S_{\left[  p,k+p\right]  }$ by the
composition $\sigma\tau$, the sign $\left(  -1\right)  ^{\sigma}$ flips (since
$\left(  -1\right)  ^{\sigma\tau}=\left(  -1\right)  ^{\sigma}%
\underbrace{\left(  -1\right)  ^{\tau}}_{=-1}=-\left(  -1\right)  ^{\sigma}$),
but the product $\prod_{i=1}^{k}a_{\sigma\left(  i\right)  ,\ \sigma\left(
k+\zeta\left(  i\right)  \right)  }$ remains unchanged (since the only factor
that sees any change is the factor for $i=j$, which changes from
$a_{\sigma\left(  j\right)  ,\ \sigma\left(  k+\zeta\left(  j\right)  \right)
}$ to
\begin{align*}
a_{\left(  \sigma\tau\right)  \left(  j\right)  ,\ \left(  \sigma\tau\right)
\left(  k+\zeta\left(  j\right)  \right)  }  &  =a_{\sigma\left(
k+\zeta\left(  j\right)  \right)  ,\ \sigma\left(  j\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\tau\text{ swaps }j\text{ with
}k+\zeta\left(  j\right)  \right) \\
&  =a_{\sigma\left(  j\right)  ,\ \sigma\left(  k+\zeta\left(  j\right)
\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since the matrix }A\text{ is
symmetric}\right)  ,
\end{align*}
which of course just means that it stays the same). Hence, by pairing up each
even permutation $\sigma\in S_{\left[  p,k+p\right]  }$ with the odd
permutation $\sigma\tau\in S_{\left[  p,k+p\right]  }$ in the sum
(\ref{pf.thm.symm-procesi.subsum}), we cause all addends to cancel with their
respective partners. Thus, the whole sum simplifies to $0$. This proves
(\ref{pf.thm.symm-procesi.subsum}).

Now, forget that we fixed $\zeta$. We thus have proved
(\ref{pf.thm.symm-procesi.subsum}) for each $\zeta\in S_{k}$. Now,
(\ref{pf.thm.symm-procesi.5}) becomes%
\begin{align*}
&  \sum_{\sigma\in S_{\left[  p,k+p\right]  }}\left(  -1\right)  ^{\sigma}%
\det\left(  A_{1,2,\ldots,p-1,\sigma\left(  p\right)  ,\sigma\left(
p+1\right)  ,\ldots,\sigma\left(  k\right)  }^{\sigma\left(  k+1\right)
,\sigma\left(  k+2\right)  ,\ldots,\sigma\left(  k+p\right)
,k+p+1,k+p+2,\ldots,2k}\right) \\
&  =\sum_{\zeta\in S_{k}}\left(  -1\right)  ^{\zeta}\underbrace{\sum
_{\sigma\in S_{\left[  p,k+p\right]  }}\left(  -1\right)  ^{\sigma}\prod
_{i=1}^{k}a_{\sigma\left(  i\right)  ,\ \sigma\left(  k+\zeta\left(  i\right)
\right)  }}_{\substack{=0\\\text{(by (\ref{pf.thm.symm-procesi.subsum}))}}}=0.
\end{align*}
This proves Theorem \ref{thm.symm-procesi}.
\end{proof}

\begin{remark}
\label{rmk.symm-procesi.partial-sym}In our above proof, we did not use the
symmetry of $A$ in full; we only used the requirement%
\begin{equation}
a_{u,v}=a_{v,u}\ \ \ \ \ \ \ \ \ \ \text{for all }u,v\in\left[  p,k+p\right]
. \label{eq.rmk.symm-procesi.partial-sym.sym}%
\end{equation}
Indeed, when we argued that $a_{\sigma\left(  k+\zeta\left(  j\right)
\right)  ,\ \sigma\left(  j\right)  }=a_{\sigma\left(  j\right)
,\ \sigma\left(  k+\zeta\left(  j\right)  \right)  }$, we could easily observe
that $\sigma\left(  k+\zeta\left(  j\right)  \right)  \in\left[  p,k+p\right]
$ (since the choice of $j$ ensures that $k+\zeta\left(  j\right)  \in\left[
p,k+p\right]  $, and since $\sigma$ preserves the set $\left[  p,k+p\right]
$) and $\sigma\left(  j\right)  \in\left[  p,k+p\right]  $ (since the choice
of $j$ ensures that $j\in\left[  p,k+p\right]  $, and since $\sigma$ preserves
the set $\left[  p,k+p\right]  $). Thus, we obtain slightly more general
versions of Theorem \ref{thm.symm-procesi} and Theorem
\ref{thm.symm-procesi-red}.
\end{remark}

\subsection{Generalizations and variants}

Our above proof of Theorem \ref{thm.symm-procesi} offers several opportunities
for variation. Among other things:

\begin{itemize}
\item The signs $\left(  -1\right)  ^{\zeta}$ in the definition of a
determinant can be removed or replaced by other conjugation-invariant
functions of $\zeta$; this causes the determinants to be replaced by
permanents or immanants \cite{GouJac92}.

\item The signs $\left(  -1\right)  ^{\sigma}$ in Theorem
\ref{thm.symm-procesi} can be removed, but then the matrix $A$ must be assumed
to be skew-symmetric rather than symmetric. Or, to be more precise, we only
need to assume that $a_{u,v}=-a_{v,u}$ for any two distinct(!) elements $u$
and $v$ of $\left[  p,k+p\right]  $.
\end{itemize}

Some of these variants, however, are trivial. Indeed, if $k>1$, then the
variant of Theorem \ref{thm.symm-procesi} with determinants replaced by
permanents falls prey to a trivial cancellation argument (since the permanent
of a matrix does not change if we swap some columns or some rows). The same
happens with the version of Theorem \ref{thm.symm-procesi} for skew-symmetric
matrices and with no $\left(  -1\right)  ^{\sigma}$ signs. However, if we make
both changes at once (i.e., remove the $\left(  -1\right)  ^{\sigma}$ signs,
require $A$ to be skew-symmetric, and replace determinants by permanents),
then we obtain the following nontrivial result:

\begin{theorem}
Let $p\in\left[  k\right]  $. Instead of assuming that $A$ be symmetric, let
us assume that $a_{u,v}=-a_{v,u}$ for any two distinct elements $u$ and $v$ of
$\left[  p,k+p\right]  $. Then,%
\[
\sum_{\sigma\in S_{\left[  p,k+p\right]  }}\operatorname*{per}\left(
A_{1,2,\ldots,p-1,\sigma\left(  p\right)  ,\sigma\left(  p+1\right)
,\ldots,\sigma\left(  k\right)  }^{\sigma\left(  k+1\right)  ,\sigma\left(
k+2\right)  ,\ldots,\sigma\left(  k+p\right)  ,k+p+1,k+p+2,\ldots,2k}\right)
=0,
\]
where $\operatorname*{per}B$ denotes the permanent of a matrix $B$.
\end{theorem}

\begin{question}
Can we prove Theorem \ref{thm.symm-procesi-red} directly by sign-reversing
involution, without having to divide by a positive integer?
\end{question}

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\end{document}