Consider a generic $n \times n$ matrix $M$.


Define the $(n-1) \times n$ matrix $M\_q$ to be $M$ with the $q$th row omitted, and assume that $M\_q$ possesses a right inverse, $R\_q$:


$$R\_q = M\_q^T (M\_q M\_q^T)^{-1}$$


The components of $R\_q$ will be *rational* functions of the components of $M$, with the determinant $\det(M\_q M\_q^T)$ as their denominator.


Now, consider the $n \times n$ matrix $C$ of cofactors of $M$. That is, $C\_{i j}$ is $(-1)^{i+j}$ times the determinant of the submatrix of $M$ obtained by omitting the $i$th row and the $j$th column.


I have found by direct computation for low values of $n$ that the antisymmetric products of the $q$th row of $C$ and the components of $R\_q$:


$$(A\_q)\_{i j k} = C\_{q i} (R\_q)\_{j k} - C\_{q j} (R\_q)\_{i k}$$


are *polynomials* of degree $n-2$ in the components of $M$, with the determinant that one might expect to be present in the denominator, inherited from $R\_q$, factoring out.


To give an example, for $n=3$:


$$M\_1=
\left(
\begin{array}{ccc}
 m\_{2,1} & m\_{2,2} & m\_{2,3} \\
 m\_{3,1} & m\_{3,2} & m\_{3,3} \\
\end{array}
\right)$$


$$C=
\left(
\begin{array}{ccc}
 m\_{2,2} m\_{3,3}-m\_{2,3} m\_{3,2} & m\_{2,3} m\_{3,1}-m\_{2,1} m\_{3,3} & m\_{2,1} m\_{3,2}-m\_{2,2} m\_{3,1} \\
 m\_{1,3} m\_{3,2}-m\_{1,2} m\_{3,3} & m\_{1,1} m\_{3,3}-m\_{1,3} m\_{3,1} & m\_{1,2} m\_{3,1}-m\_{1,1} m\_{3,2} \\
 m\_{1,2} m\_{2,3}-m\_{1,3} m\_{2,2} & m\_{1,3} m\_{2,1}-m\_{1,1} m\_{2,3} & m\_{1,1} m\_{2,2}-m\_{1,2} m\_{2,1} \\
\end{array}
\right)$$


$$
R\_1 = M\_1^T (M\_1 M\_1^T)^{-1} = \\
{\scriptsize
\frac{
\left(
\begin{array}{cc}
 m\_{2,1} \left(m\_{3,2}^2+m\_{3,3}^2\right)-m\_{3,1} \left(m\_{2,2} m\_{3,2}+m\_{2,3} m\_{3,3}\right) & m\_{3,1} \left(m\_{2,2}^2+m\_{2,3}^2\right)
 -m\_{2,1} \left(m\_{2,2} m\_{3,2}+m\_{2,3} m\_{3,3}\right) \\
 m\_{2,2} \left(m\_{3,1}^2+m\_{3,3}^2\right)-m\_{3,2} \left(m\_{2,1} m\_{3,1}+m\_{2,3} m\_{3,3}\right) & m\_{3,2} \left(m\_{2,1}^2+m\_{2,3}^2\right)
 -m\_{2,2} \left(m\_{2,1} m\_{3,1}+m\_{2,3} m\_{3,3}\right) \\
 m\_{2,3} \left(m\_{3,1}^2+m\_{3,2}^2\right)-m\_{3,3} \left(m\_{2,1} m\_{3,1}+m\_{2,2} m\_{3,2}\right) & m\_{3,3} \left(m\_{2,1}^2+m\_{2,2}^2\right)
 -m\_{2,3} \left(m\_{2,1} m\_{3,1}+m\_{2,2} m\_{3,2}\right) \\
\end{array}
\right)}
{\left(m\_{3,2}^2+m\_{3,3}^2\right) m\_{2,1}^2-2 m\_{2,3} m\_{3,1} m\_{3,3} m\_{2,1}+m\_{2,3}^2 \left(m\_{3,1}^2+m\_{3,2}^2\right)-2 m\_{2,2} m\_{3,2}
 \left(m\_{2,1} m\_{3,1}+m\_{2,3} m\_{3,3}\right)+m\_{2,2}^2 \left(m\_{3,1}^2+m\_{3,3}^2\right)}
}
$$


$$(A\_1)\_{i j k}= C\_{1 i} (R\_1)\_{j k} - C\_{1 j} (R\_1)\_{i k} = 
\left(
\begin{array}{ccc}
 \left(
\begin{array}{c}
 0 \\
 0 \\
\end{array}
\right) & \left(
\begin{array}{c}
 m\_{3,3} \\
 -m\_{2,3} \\
\end{array}
\right) & \left(
\begin{array}{c}
 -m\_{3,2} \\
 m\_{2,2} \\
\end{array}
\right) \\
 \left(
\begin{array}{c}
 -m\_{3,3} \\
 m\_{2,3} \\
\end{array}
\right) & \left(
\begin{array}{c}
 0 \\
 0 \\
\end{array}
\right) & \left(
\begin{array}{c}
 m\_{3,1} \\
 -m\_{2,1} \\
\end{array}
\right) \\
 \left(
\begin{array}{c}
 m\_{3,2} \\
 -m\_{2,2} \\
\end{array}
\right) & \left(
\begin{array}{c}
 -m\_{3,1} \\
 m\_{2,1} \\
\end{array}
\right) & \left(
\begin{array}{c}
 0 \\
 0 \\
\end{array}
\right) \\
\end{array}
\right)$$


My question is: why are the components of $A\_q$ polynomials rather than rational functions? Can this be proved for general $n$? And can $A\_q$ be reduced, for general $n$, to a simpler expression that makes no reference to $R\_q$?


(Edited to change question from sum to individual components).



Why does this antisymmetric product factor out a determinant?
$\let\sumnonlimits\sum
\let\prodnonlimits\prod
\renewcommand{\sum}{\sumnonlimits\limits}
\renewcommand{\prod}{\prodnonlimits\limits}
$ I will write $R$ for $R\_{q}$, because $q$ is constant. In the following, I am
going to assume that $R$ is an *arbitrary* right inverse of $M\_{q}$, rather
than the specific right inverse $M\_{q}^{T}\left( M\_{q}M\_{q}^{T}\right)
^{-1}$ which you have suggested. This makes the statement a tad more general
and rids us of a red herring.


I shall prove that for any
$i\in\left\{ 1,2,...,n\right\} $, $j\in\left\{ 1,2,...,n\right\} $ and
$k\in\left\{ 1,2,...,n-1\right\} $, the value $C\_{qi}R\_{jk}-C\_{qj}R\_{ik}$ is
a homogeneous polynomial of degree $n-2$ in the entries of the matrix $M\_{q}$
(independently, and independent, of the choice of right inverse $R$).


For every $\ell\in\left\{ 1,2,...,n\right\} $, we abbreviate $C\_{q\ell}$ by
$v\_{q}$. Thus,


$v\_{\ell}=C\_{q\ell}=\left( -1\right) ^{q+\ell}\det\left(
\underbrace{M\text{ without row }q}\_{=M\_{q}}\text{ and column }\ell\right) $


**(1)** $=\left( -1\right) ^{q+\ell}\det\left( M\_{q}\text{ without column
}\ell\right) $.


From this it easy to obtain


**(2)** $\sum\_{\ell\in\left\{ 1,2,...,n\right\} }\left( M\_{q}\right)
\_{j\ell}v\_{\ell}=0$ for every $j \in \left\{1,2,...,n-1\right\}$.


[*Proof of **(2)**:* Let $j \in \left\{1,2,...,n-1\right\}$. Let $G$ be the result of inserting a copy of row $j$ of the matrix $M\_{q}$ between the rows $q-1$ and $q$ of this matrix. Then, $G$ is an $n\times n$-matrix with two equal rows, and hence has determinant $\det G=0$. But Laplace expansion of $\det G$ along the row we have inserted yields


$\det G=\sum\_{\ell\in\left\{ 1,2,...,n\right\} }\left( M\_{q}\right)
\_{j\ell}\left( -1\right) ^{q+\ell}\det\left( M\_{q}\text{ without column
}\ell\right) $,


since the entries of this row are $\left( M\_{q}\right) \_{j\ell}$ and the
cofactors corresponding to this row are $\left( -1\right) ^{q+\ell}
\det\left( M\_{q}\text{ without column }\ell\right) $. Now, **(1)** yields


$\sum\_{\ell\in\left\{ 1,2,...,n\right\} }\left( M\_{q}\right) \_{j\ell
}v\_{\ell}$


$=\sum\_{\ell\in\left\{ 1,2,...,n\right\} }\left( M\_{q}\right) \_{j\ell
}\left( -1\right) ^{q+\ell}\det\left( M\_{q}\text{ without column }
\ell\right) $


$=\det G=0$,


so that **(2)** is proven.]


Since $M$ is generic, its cofactors $v\_{1}$, $v\_{2}$, $...$, $v\_{n}$ are nonzero.


Fix $i\in\left\{ 1,2,...,n\right\} $. Let $N\_{i}$ denote the matrix $M\_{q}$
without column $i$. Thus, $\det\left( N\_{i}\right) =\det\left( M\_{q}\text{
without column }i\right) =\left( -1\right) ^{q+i}v\_{i}$ (because **(1)**
yields $v\_{i}=\left( -1\right) ^{q+i}\det\left( M\_{q}\text{ without column
}i\right) $).


Let $S\_{i}$ denote the matrix $\left( R\_{jk}-\dfrac{v\_{j}}{v\_{i}}
R\_{ik}\right) \_{j\in\left\{ 1,2,...,n\right\} ;\ k\in\left\{
1,2,...,n-1\right\} }$. The $i$-th row of this matrix $S\_{i}$ is zero; let
$S\_{i}^{\prime}$ denote the matrix $S\_{i}$ without row $i$.


Now, we claim that $N\_{i}S\_{i}^{\prime}=I\_{n-1}$ (the $\left( n-1\right)
\times\left( n-1\right) $ identity matrix). Indeed, for every $\left(
j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$, the $\left( j,k\right)
$-th entry of $N\_{i}S\_{i}^{\prime}$ is


$\left( N\_{i}S\_{i}^{\prime}\right) \_{jk}=\sum\_{u\in\left\{
1,2,...,n-1\right\} }\left( N\_{i}\right) \_{ju}\left( S\_{i}^{\prime
}\right) \_{uk}$


$=\sum\_{\ell\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\}
}\left( M\_{q}\right) \_{j\ell}\underbrace{\left( S\_{i}\right) \_{\ell k}
}\_{=R\_{\ell k}-\dfrac{v\_{\ell}}{v\_{i}}R\_{ik}}$


(since $N\_{i}$ is the matrix $M\_{q}$ without column $i$, while $S\_{i}^{\prime
}$ is the matrix $S\_{i}$ without row $i$)


$=\sum\_{\ell\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\}
}\left( M\_{q}\right) \_{j\ell}\left( R\_{\ell k}-\dfrac{v\_{\ell}}{v\_{i}
}R\_{ik}\right) $


$=\sum\_{\ell\in\left\{ 1,2,...,n\right\} }\left( M\_{q}\right) \_{j\ell
}\left( R\_{\ell k}-\dfrac{v\_{\ell}}{v\_{i}}R\_{ik}\right) $


(here, we added an $\ell=i$ addend to the sum; this did not change the sum
because this addend is $0$)


$=\underbrace{\sum\_{\ell\in\left\{ 1,2,...,n\right\} }\left( M\_{q}\right)
\_{j\ell}R\_{\ell k}}\_{\substack{=\left( M\_{q}R\right) \_{jk}=\delta
\_{jk}\\\text{(since }R\text{ is a}\\\text{right inverse of }M\_{q}\text{)}
}}-\underbrace{\sum\_{\ell\in\left\{ 1,2,...,n\right\} }\left( M\_{q}\right)
\_{j\ell}\dfrac{v\_{\ell}}{v\_{i}}R\_{ik}}\_{=\dfrac{1}{v\_{i}}R\_{ik}\sum\_{\ell
\in\left\{ 1,2,...,n\right\} }\left( M\_{q}\right) \_{j\ell}v\_{\ell}}$


$=\delta\_{jk}-\dfrac{1}{v\_{i}}R\_{ik}\underbrace{\sum\_{\ell\in\left\{
1,2,...,n\right\} }\left( M\_{q}\right) \_{j\ell}v\_{\ell}}
\_{\substack{=0\\\text{(by \textbf{(2)})}}}=\delta\_{jk}$.


This shows that $N\_{i}S\_{i}^{\prime}=I\_{n-1}$, and thus $S\_{i}^{\prime}
=N\_{i}^{-1}$ (since $N\_{i}$ and $S\_{i}^{\prime}$ are $\left( n-1\right)
\times\left( n-1\right) $-matrices).


Now, we recall Cramer's rule for the inverse of a matrix. It essentially says
that the inverse of a square matrix is obtained by dividing its adjoint by its
determinant. In other words, if $X$ is an invertible $p\times p$-matrix for
some $p\in\mathbb{N}$, and $j$ and $k$ are elements of $\left\{
1,2,...,p\right\} $, then


**(3)** $\left( X^{-1}\right) \_{jk}=\dfrac{1}{\det X}\left( -1\right)
^{j+k}\det\left( X\text{ without row }k\text{ and column }j\right) $.


Now, recall that $S\_{i}^{\prime}=N\_{i}^{-1}$. Hence,


$\left( S\_{i}^{\prime}\right) \_{jk}=\left( N\_{i}^{-1}\right) \_{jk}
=\dfrac{1}{\det\left( N\_{i}\right) }\left( -1\right) ^{j+k}\det\left(
N\_{i}\text{ without row }k\text{ and column }j\right) $ (by **(3)**)


$=\dfrac{1}{\left( -1\right) ^{q+i}v\_{i}}\left( -1\right) ^{j+k}
\det\left( N\_{i}\text{ without row }k\text{ and column }j\right) $ (since
$\det\left( N\_{i}\right) =\left( -1\right) ^{q+i}v\_{i}$)


$=\dfrac{1}{v\_{i}}\left( -1\right) ^{i+j+q+k}\det\left( N\_{i}\text{ without
row }k\text{ and column }j\right) $


for all $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$. In other words,


**(4)** $v\_{i}\left( S\_{i}^{\prime}\right) \_{jk}=\left( -1\right)
^{i+j+q+k}\det\left( N\_{i}\text{ without row }k\text{ and column }j\right) $


for all $\left( j,k\right) \in\left\{ 1,2,...,n-1\right\} ^{2}$.


Let $\mathbf{B}$ be the (unique) increasing bijection from $\left\{
1,2,...,n\right\} \setminus\left\{ i\right\} $ to $\left\{
1,2,...,n-1\right\} $. Then, for all $j\in\left\{ 1,2,...,n\right\}
\setminus\left\{ i\right\} $ and $k\in\left\{ 1,2,...,n-1\right\} $, we have


**(5)** $v\_{i}\left( S\_{i}\right) \_{jk}=\left( -1\right) ^{i+\mathbf{B}
\left( j\right) +q+k}\det\left( M\_{q}\text{ without row }k\text{ and
columns }i\text{ and }j\right) $.


(Indeed, this follows from **(4)**, applied to $\mathbf{B}\left( j\right) $
instead of $j$, because $\left( S\_{i}^{\prime}\right) \_{\mathbf{B}\left(
j\right) ,k}=\left( S\_{i}\right) \_{jk}$ (since $S\_{i}^{\prime}$ is the
matrix $S\_{i}$ without row $i$) and because column $j$ of $M\_{q}$ is column
$\mathbf{B}\left( j\right) $ of $N\_{i}$ (since $N\_{i}$ is the matrix $M\_{q}$
without column $i$).)


Now, fix $j\in\left\{ 1,2,...,n\right\} $ and $k\in\left\{
1,2,...,n-1\right\} $. We need to show that $C\_{qi}R\_{jk}-C\_{qj}R\_{ik}$ is a
homogeneous polynomial of degree $n-2$ in the entries of the matrix $M\_{q}$.
We WLOG assume that $j\neq i$ (since otherwise, $C\_{qi}R\_{jk}-C\_{qj}R\_{ik}
=0$). Thus, $j\in\left\{ 1,2,...,n\right\} \setminus\left\{ i\right\} $.
Since $C\_{qi}=v\_{i}$ and $C\_{qj}=v\_{j}$ (by the definitions of $v\_{i}$ and
$v\_{j}$), we have


$\underbrace{C\_{qi}}\_{=v\_{i}}R\_{jk}-\underbrace{C\_{qj}}\_{=v\_{j}}R\_{ik}
=v\_{i}R\_{jk}-v\_{j}R\_{ik}=v\_{i}\underbrace{\left( R\_{jk}-\dfrac{v\_{j}}{v\_{i}
}R\_{ik}\right) }\_{\substack{=\left( S\_{i}\right) \_{jk}\\\text{(by the
definition of }S\_{i}\text{)}}}$


**(6)** $=v\_{i}\left( S\_{i}\right) \_{jk}=\left( -1\right) ^{i+\mathbf{B}\left(
j\right) +q+k}\det\left( M\_{q}\text{ without row }k\text{ and columns
}i\text{ and }j\right) $


(by **(5)**),


which is obviously a homogeneous polynomial of degree $n-2$ in the entries of
the matrix $M\_{q}$ (and independent of the choice of $R$), qed.


Thanks for a very nice question, and sorry for this mess of an answer...


**PS.** I believe the genericity of $M$ is not required for **(6)** to hold. Does anyone see a nice proof of this? I don't.


**PPS.** Here is a more general statement which, I think, is true. Let $A$ be a commutative ring. Let $N$ be an $\left(n-1\right) \times n$-matrix over $A$. Let $s \in A^n$ be a vector. For every $\ell \in \left\{1,2,...,n\right\}$, let $p\_\ell$ denote the scalar $\left( -1\right) ^{\ell}\det\left( N \text{ without column
}\ell\right)$. If $w$ is a vector and $i$ is an integer, we denote by $w\_i$ the $i$-th coordinate of $w$ (whenever this makes sense). Then, every two distinct $i \in \left\{1,2,...,n\right\}$ and $j \in \left\{1,2,...,n\right\}$ satisfy


**(11)** $p\_i s\_j - p\_j s\_i = \sum\_{k=1}^{n-1} \pm \left(Ns\right)\_k \det\left(N \text{ without row } k \text{ and columns } i \text{ and } j\right)$,


where $\pm$ is something like $\left(-1\right)^{i+j+\left[i<j\right]+k}$ using the Iverson bracket.


If this is proven (and this shouldn't be too hard -- it's a polynomial identity, so you can assume as much genericity as you wish), the original result is obtained by setting $N = M\_q$ and $s = \left(k\text{-th column of } R\right)$.


**PPPS.** Indeed, **(11)** is not hard to prove. Let $e\_1, e\_2, ..., e\_n$ be the $n$ standard basis vectors of $A^n$. Let $P$ be the $n \times \left(n-1\right)$-matrix whose columns (from left to right) are $e\_1, e\_2, ..., \widehat{e\_i}, ..., \widehat{e\_j}, ..., e\_n, s$ (where $\widehat{\text{something}}$ means omission, and the order of $i$ and $j$ is not necessarily the one we have shown). Then, $NP$ is the $\left(n-1\right)\times \left(n-1\right)$-matrix whose columns (from left to right) are the columns $1, 2, ..., \widehat{i}, ..., \widehat{j}, ..., n$ of $N$ and the column-vector $Ns$. The right hand side of **(11)** is the determinant of this matrix $NP$, computed by Laplace expansion along its last column. The left hand side of **(11)** is the same determinant, computed using the [Cauchy-Binet formula](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula) (which takes a rather simple form here because if we remove the $\ell$-th row from $P$ for some $\ell \notin \left\{i, j\right\}$, then the resulting matrix has two rows with only their rightmost entries nonzero, and so has determinant $0$).


**PPPPS.** Here is a different way to formulate the above proof of **(11)**,
using exterior algebra instead of the Cauchy-Binet formula and giving some
more detail.


Let $e\_{1}$, $e\_{2}$, $...$, $e\_{n}$ be the $n$ standard basis vectors of
$A^{n}$. Let $f\_{1}$, $f\_{2}$, $...$, $f\_{n-1}$ be the $n-1$ standard basis
vectors of $A^{n-1}$. We identify the matrix $N\in A^{\left( n-1\right)
\times n}$ with the $A$-linear map $A^{n}\rightarrow A^{n-1}$ it represents.
As usual, we use the notation ''$\widehat{\text{some term}}$'' for omission of a term in a product or list (for example,
$\left( 1,2,...,\widehat{5},...,8\right) =\left( 1,2,3,4,6,7,8\right) $).
We also use the Iverson bracket notation, i.e., whenever $\mathcal{A}$ is a
logical statement, we write $\left[ \mathcal{A}\right] $ for the integer
$\left\{
\begin{array}
[c]{c}
1,\text{ if }\mathcal{A}\text{ is true;}\\
0,\text{ if }\mathcal{A}\text{ is false}
\end{array}
\right. $.


Let $\mathbf{f}$ be the element $f\_{1}\wedge f\_{2}\wedge...\wedge f\_{n-1}$ of
$\wedge^{n-1}\left( A^{n-1}\right) $. It is known that $\left( \mathbf{f}\right) $
is an $A$-module basis of $\wedge^{n-1}\left( A^{n-1}\right) $.


Let $i$ and $j$ be two distinct elements of $\left\{ 1,2,...,n\right\} $. We
have $s=\sum\_{k=1}^{n}s\_{k}e\_{k}$, so that


$e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{i}}\wedge...\wedge\widehat{e\_{j}
}\wedge...\wedge e\_{n}\wedge s$


$=e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{i}}\wedge...\wedge
\widehat{e\_{j}}\wedge...\wedge e\_{n}\wedge\left( \sum\_{k=1}^{n}s\_{k}
e\_{k}\right) $


**(12)** $=\sum\_{k=1}^{n}s\_{k}e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{i}
}\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge e\_{n}\wedge e\_{k}$


(where the notation $e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{i}}
\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge e\_{n}$ means that both $e\_{i}$
and $e\_{j}$ are omitted, but does not necessarily imply that $i<j$). Of all
the addends of the sum on the right hand side of **(12)**, only those for
$k=i$ and for $k=j$ have a chance to be nonzero (every other summand contains
a wedge product with two equal factors, and thus vanishes). Hence, **(12)**
simplifies to


$e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{i}}\wedge...\wedge\widehat{e\_{j}
}\wedge...\wedge e\_{n}\wedge s$


$=s\_{i}\underbrace{e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{i}}
\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge e\_{n}\wedge e\_{i}}\_{=\left(
-1\right) ^{i+\left[ i<j\right] }e\_{1}\wedge e\_{2}\wedge...\wedge
\widehat{e\_{j}}\wedge...\wedge e\_{n}}+s\_{j}\underbrace{e\_{1}\wedge e\_{2}
\wedge...\wedge\widehat{e\_{i}}\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge
e\_{n}\wedge e\_{j}}\_{=\left( -1\right) ^{j+1-\left[ i<j\right] }e\_{1}\wedge
e\_{2}\wedge...\wedge\widehat{e\_{i}}\wedge...\wedge e\_{n}}$


$=s\_{i}\left( -1\right) ^{i+\left[ i<j\right] }e\_{1}\wedge e\_{2}
\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge e\_{n}+s\_{j}\left( -1\right)
^{j+1-\left[ i<j\right] }e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{i}
}\wedge...\wedge e\_{n}$


$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left( s\_{j}\left(
-1\right) ^{i}e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{i}}\wedge...\wedge
e\_{n}-s\_{i}\left( -1\right) ^{j}e\_{1}\wedge e\_{2}\wedge...\wedge
\widehat{e\_{j}}\wedge...\wedge e\_{n}\right) $.


Applying the linear map $\wedge^{n-1}N$ to both sides of this equality results in


$\left( \wedge^{n-1}N\right) \left( e\_{1}\wedge e\_{2}\wedge...\wedge
\widehat{e\_{i}}\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge e\_{n}\wedge
s\right) $


$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left( s\_{j}\left(
-1\right) ^{i}\underbrace{\left( \wedge^{n-1}N\right) \left( e\_{1}\wedge
e\_{2}\wedge...\wedge\widehat{e\_{i}}\wedge...\wedge e\_{n}\right) }
\_{=\det\left( N\text{ without column }i\right) \mathbf{f}}\right. $


$\left. -s\_{i}\left( -1\right) ^{j}\underbrace{\left( \wedge
^{n-1}N\right) \left( e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{j}}
\wedge...\wedge e\_{n}\right) }\_{=\det\left( N\text{ without column
}j\right) \mathbf{f}}\right) $


$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left( s\_{j}
\underbrace{\left( -1\right) ^{i}\det\left( N\text{ without column
}i\right) }\_{=p\_{i}} \mathbf{f} -s\_{i}\underbrace{\left( -1\right) ^{j}
\det\left( N\text{ without column }j\right) }\_{=p\_{j}} \mathbf{f} \right) $


$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left( s\_{j}p\_{i}
\mathbf{f}-s\_{i}p\_{j}\mathbf{f}\right) =\left( -1\right) ^{\left[
i<j\right] +i+j-1}\left( p\_{i}s\_{j}-p\_{j}s\_{i}\right) \mathbf{f}$,


so that


$\left( p\_{i}s\_{j}-p\_{j}s\_{i}\right) \mathbf{f}$


**(13)** $=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\left(
\wedge^{n-1}N\right) \left( e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{i}
}\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge e\_{n}\wedge s\right) $.


On the other hand, using $\wedge$ to denote the multiplication in the exterior
algebra $\wedge\left( A^{n-1}\right) $, we see


$\left( \wedge^{n-1}N\right) \left( e\_{1}\wedge e\_{2}\wedge...\wedge
\widehat{e\_{i}}\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge e\_{n}\wedge
s\right) $


$=\underbrace{\left( \wedge^{n-2}N\right) \left( e\_{1}\wedge e\_{2}
\wedge...\wedge\widehat{e\_{i}}\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge
e\_{n}\right) }\_{=\sum\_{\ell=1}^{n-1}\det\left( N\text{ without columns
}i\text{ and }j\text{ and row }\ell\right) f\_{1}\wedge f\_{2}\wedge
...\wedge\widehat{f\_{\ell}}\wedge...\wedge f\_{n-1}}\wedge\underbrace{\left(
Ns\right) }\_{=\sum\_{k=1}^{n-1}\left( Ns\right) \_{k}f\_{k}}$


$=\left( \sum\_{\ell=1}^{n-1}\det\left( N\text{ without columns }i\text{ and
}j\text{ and row }\ell\right) f\_{1}\wedge f\_{2}\wedge...\wedge
\widehat{f\_{\ell}}\wedge...\wedge f\_{n-1}\right) \wedge\left( \sum
\_{k=1}^{n-1}\left( Ns\right) \_{k}f\_{k}\right) $


**(14)** $=\sum\_{\ell=1}^{n-1}\sum\_{k=1}^{n-1}\left( Ns\right) \_{k}
\det\left( N\text{ without columns }i\text{ and }j\text{ and row }
\ell\right) f\_{1}\wedge f\_{2}\wedge...\wedge\widehat{f\_{\ell}}\wedge...\wedge
f\_{n-1}\wedge f\_{k}$.


Of all the addends of the inner sum on the right hand side of **(14)**, only
those for $k=\ell$ have a chance to be nonzero (because all other addends have
the form $f\_{1}\wedge f\_{2}\wedge...\wedge\widehat{f\_{\ell}}\wedge...\wedge
f\_{n-1}\wedge f\_{k}$ for $k\neq\ell$, which is a wedge product with two equal
factors and thus equals $0$). Hence, **(14)** simplifies to


$\left( \wedge^{n-1}N\right) \left( e\_{1}\wedge e\_{2}\wedge...\wedge
\widehat{e\_{i}}\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge e\_{n}\wedge
s\right) $


$=\sum\_{\ell=1}^{n-1}\left( Ns\right) \_{\ell}\det\left( N\text{ without
columns }i\text{ and }j\text{ and row }\ell\right) \underbrace{f\_{1}\wedge
f\_{2}\wedge...\wedge\widehat{f\_{\ell}}\wedge...\wedge f\_{n-1}\wedge f\_{\ell}
}\_{=\left( -1\right) ^{n-1-\ell}f\_{1}\wedge f\_{2}\wedge...\wedge f\_{n-1}}$


$=\sum\_{\ell=1}^{n-1}\left( Ns\right) \_{\ell}\det\left( N\text{ without
columns }i\text{ and }j\text{ and row }\ell\right) \left( -1\right)
^{n-1-\ell}\underbrace{f\_{1}\wedge f\_{2}\wedge...\wedge f\_{n-1}}\_{=\mathbf{f}
}$


$=\sum\_{\ell=1}^{n-1}\left( Ns\right) \_{\ell}\det\left( N\text{ without
columns }i\text{ and }j\text{ and row }\ell\right) \left( -1\right)
^{n-1-\ell}\mathbf{f}$.


Thus, **(13)** becomes


$\left( p\_{i}s\_{j}-p\_{j}s\_{i}\right) \mathbf{f}$


$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\underbrace{\left(
\wedge^{n-1}N\right) \left( e\_{1}\wedge e\_{2}\wedge...\wedge\widehat{e\_{i}
}\wedge...\wedge\widehat{e\_{j}}\wedge...\wedge e\_{n}\wedge s\right) }
\_{=\sum\_{\ell=1}^{n-1}\left( Ns\right) \_{\ell}\det\left( N\text{ without
columns }i\text{ and }j\text{ and row }\ell\right) \left( -1\right)
^{n-1-\ell}\mathbf{f}}$


$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\sum\_{\ell=1}^{n-1}\left(
Ns\right) \_{\ell}\det\left( N\text{ without columns }i\text{ and }j\text{
and row }\ell\right) \left( -1\right) ^{n-1-\ell}\mathbf{f}$.


Since $\left( \mathbf{f}\right) $ is a basis of $\wedge^{n-1}\left(
A^{n-1}\right) $, we can compare coefficients before $\mathbf{f}$ in this
equality, and obtain


$p\_{i}s\_{j}-p\_{j}s\_{i}$


$=\left( -1\right) ^{\left[ i<j\right] +i+j-1}\sum\_{\ell=1}^{n-1}\left(
Ns\right) \_{\ell}\det\left( N\text{ without columns }i\text{ and }j\text{
and row }\ell\right) \left( -1\right) ^{n-1-\ell}$


$=\sum\_{\ell=1}^{n-1}\left( -1\right) ^{\left[ i<j\right] +i+j+n-\ell
}\left( Ns\right) \_{\ell}\det\left( N\text{ without columns }i\text{ and
}j\text{ and row }\ell\right) $


$=\sum\_{k=1}^{n-1}\left( -1\right) ^{\left[ i<j\right] +i+j+n-k}\left(
Ns\right) \_{k}\det\left( N\text{ without columns }i\text{ and }j\text{ and
row }k\right) $.


This proves **(11)**, up to all the sign errors I surely have made.


As a consequence, we obtain a new proof of your statement, and it does no longer rely on genericity of $M$.


