# Why does this antisymmetric product factor out a determinant?

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).

• Shouldn't the sum in the definition of $A_q$ end at $k = n-1$ ? – darij grinberg Jul 28 '14 at 12:31
• You're right about the bound on the sum. Sorry! I've fixed it. – Greg Egan Jul 28 '14 at 13:00
• Am I being stupid or is the thing polynomial even if you don't sum over $k$ ? And (without summing) it should be an $\left(n-2\right)\times\left(n-2\right)$-minor of $M$ (up to sign). (I am not 100% sure of my proof so far.) – darij grinberg Jul 28 '14 at 13:44
• You're absolutely right, there's no need to sum. The summation arose in the problem I was tackling and I assumed it was crucial. I should have checked! The separate results do indeed look like higher-order minors (though it's not yet clear to me why). – Greg Egan Jul 28 '14 at 14:07

$\let\sumnonlimits\sum \let\prodnonlimits\prod \renewcommand{\sum}{\sumnonlimits\limits} \renewcommand{\prod}{\prodnonlimits\limits}$ 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 (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 $M$.

• Thanks! It might take me some time to work through this and be sure I understand it. – Greg Egan Jul 28 '14 at 15:16
• You're welcome! Let me know if something is wrong. – darij grinberg Jul 28 '14 at 15:24
• This looks perfect to me. I'll have to think about non-generic $M$; so long as $M_q$ still has a right inverse maybe we can argue by continuity from the generic case. – Greg Egan Jul 28 '14 at 23:51
• That's a wonderful generalisation! I had tried to use the Cauchy-Binet formula on the original problem, but I couldn't see how to make it fit. – Greg Egan Jul 29 '14 at 23:13