Let $A$ be an $n\times n$ skew-symmetric matrix of rank $r$.
Given subsets $X$ and $Y$ of row and column indices respectively, let $A\_{X,Y}$ denote the submatrix of $A$ obtained by only keeping rows with indices in $X$ and columns with indices in $Y$.


Prove that for any subsets $X, Y\subseteq \{1, 2, \ldots, n\}$ each of size $r$, we have


$$\det A\_{X,X} \cdot \det A\_{Y,Y} = (-1)^r (\det A\_{X,Y})^2.$$


I've heard that this theorem is due to Frobenius, but have not been able to track down a reference that proves this result.



Minors of low rank skew-symmetric matrix
Since no one else has posted a complete answer so far, let me give one.
Note that it is only complete in the sense of answering the OP's question;
several other questions arise that I cannot easily address.


In Section 1, I will prove the main result (Theorem 1), which is more general
than the OP's equality. In Section 2, I will derive the latter from the
former. In Sections 3 and 4, I will generalize the statement and ask further questions.


1. On size-$r$ minors of a rank-$r$ matrix
==========================================


We fix a field $\mathbb{K}$. (In Section 3, we will generalize this to a
commutative ring.)


Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $. For each $n\in\mathbb{N}$,
let $\left[ n\right] $ denote the set $\left\{ 1,2,\ldots,n\right\} $.


If $A=\left( a\_{i,j}\right) \_{1\leq i\leq n,\ 1\leq j\leq m}\in
\mathbb{K}^{n\times m}$ is an $n\times m$-matrix (for some $n,m\in\mathbb{N}
$), and if $I\subseteq\left[ n\right] $ and $J\subseteq\left[ m\right] $
are arbitrary subsets, then $A\_{I,J}$ will denote the submatrix $\left(
a\_{i\_{x},j\_{y}}\right) \_{1\leq x\leq p,\ 1\leq y\leq q}$ of $A$, where the
subsets $I$ and $J$ have been written as $I=\left\{ i\_{1}<i\_{2}<\cdots
<i\_{p}\right\} $ and $J=\left\{ j\_{1}<j\_{2}<\cdots<j\_{q}\right\} $. (Thus,
$A\_{I,J}$ is the matrix obtained from $A$ upon removing all rows other than
the rows indexed by elements of $I$ and removing all columns other than the
columns indexed by elements of $J$.)


The crucial result is the following:



> 
> **Theorem 1.** Let $n,m,r\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times m}$ be a
> matrix such that $\operatorname\*{rank}A\leq r$. Let $X$ and $Y$ be subsets of
> $\left[ n\right] $, and let $U$ and $V$ be subsets of $\left[ m\right] $;
> assume that $\left\vert X\right\vert =\left\vert Y\right\vert =\left\vert
> U\right\vert =\left\vert V\right\vert =r$. Then,
> \begin{align\*}
> \det\left( A\_{X,U}\right) \cdot\det\left( A\_{Y,V}\right) =\det\left(
> A\_{X,V}\right) \cdot\det\left( A\_{Y,U}\right) .
> \end{align\*}
> 
> 
> 


Theorem 1 is surprisingly easy to prove using the following two basic facts:



> 
> **Lemma 2.** Let $n,m,r\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times m}$ be a
> matrix such that $\operatorname\*{rank}A\leq r$. Then, there exist two matrices
> $B\in\mathbb{K}^{n\times r}$ and $C\in\mathbb{K}^{r\times m}$ such that $A=BC$.
> 
> 
> 


*Proof of Lemma 2.* The particular case of Lemma 2 when $\operatorname\*{rank}
A=r$ is a well-known result in elementary linear algebra, but since the
general case is rarely stated, let me sketch the proof (for the general case):


Let $\operatorname\*{Col}A$ denote the span of the columns of $A$. This is a
$\mathbb{K}$-vector subspace of $\mathbb{K}^{n\times1}$. Its dimension is
$\dim\left( \operatorname\*{Col}A\right) =\operatorname\*{rank}A\leq r$. In
other words, the $\mathbb{K}$-vector space $\operatorname\*{Col}A$ has
dimension $\leq r$. Thus, this $\mathbb{K}$-vector space has a basis with at
most $r$ elements. Therefore, this vector space can be generated by exactly
$r$ elements (just take a basis with at most $r$ elements, and insert the zero
vector enough times to get exactly $r$ generators). In other words, there
exist $r$ vectors $v\_{1},v\_{2},\ldots,v\_{r}$ that span the vector space
$\operatorname\*{Col}A$. Consider these $v\_{1},v\_{2},\ldots,v\_{r}$.


Now, let $B\in\mathbb{K}^{n\times r}$ be the matrix whose $r$ columns are
$v\_{1},v\_{2},\ldots,v\_{r}$. For each $i\in\left\{ 1,2,\ldots,m\right\} $, we
have
\begin{align\*}
\left( \text{the }i\text{-th column of }A\right) & \in\operatorname\*{Col}
A\qquad\left( \text{by the definition of }\operatorname\*{Col}A\right) \\
& =\operatorname\*{span}\left( v\_{1},v\_{2},\ldots,v\_{r}\right)
\end{align\*}
(since $v\_{1},v\_{2},\ldots,v\_{r}$ span $\operatorname\*{Col}A$); thus, there
exist scalars $c\_{i,1},c\_{i,2},\ldots,c\_{i,r}\in\mathbb{K}$ such that
\begin{align\*}
\left( \text{the }i\text{-th column of }A\right) =c\_{i,1}v\_{1}+c\_{i,2}
v\_{2}+\cdots+c\_{i,r}v\_{r}.
\end{align\*}
Consider these scalars $c\_{i,1},c\_{i,2},\ldots,c\_{i,r}$. Define the matrix
$C\in\mathbb{K}^{r\times m}$ by $C=\left( c\_{j,i}\right) \_{1\leq i\leq
r,\ 1\leq j\leq m}$. Then, it is easy to see that $A=BC$. This proves Lemma 2.
$\blacksquare$



> 
> **Lemma 3.** Let $n,m,r\in\mathbb{N}$. Let $B\in\mathbb{K}^{n\times r}$ and
> $C\in\mathbb{K}^{r\times m}$ be two matrices. Let $X\subseteq\left[ n\right]
> $ and $U\subseteq\left[ m\right] $ be two subsets. Then,
> \begin{align\*}
> \left( BC\right) \_{X,U}=B\_{X,\left[ r\right] }C\_{\left[ r\right] ,U}.
> \end{align\*}
> 
> 
> 


*Proof of Lemma 3.* Straightforward entry-by-entry verification (using the
definition of matrix multiplication). $\blacksquare$


*Proof of Theorem 1.* Lemma 2 yields that there exist two matrices
$B\in\mathbb{K}^{n\times r}$ and $C\in\mathbb{K}^{r\times m}$ such that
$A=BC$. Consider these $B$ and $C$. Now, Lemma 3 yields $\left( BC\right)
\_{X,U}=B\_{X,\left[ r\right] }C\_{\left[ r\right] ,U}$. In view of $A=BC$,
this rewrites as $A\_{X,U}=B\_{X,\left[ r\right] }C\_{\left[ r\right] ,U}$.
But the matrix is $B\_{X,\left[ r\right] }$ is square (since $\left\vert
X\right\vert =r=\left\vert \left[ r\right] \right\vert $), and so is the
matrix $C\_{\left[ r\right] ,U}$ (since $\left\vert U\right\vert
=r=\left\vert \left[ r\right] \right\vert $); hence, $\det\left(
B\_{X,\left[ r\right] }C\_{\left[ r\right] ,U}\right) =\det\left(
B\_{X,\left[ r\right] }\right) \cdot\det\left( C\_{\left[ r\right]
,U}\right) $. In view of $A\_{X,U}=B\_{X,\left[ r\right] }C\_{\left[
r\right] ,U}$, this rewrites as
\begin{equation}
\det\left( A\_{X,U}\right) =\det\left( B\_{X,\left[ r\right] }\right)
\cdot\det\left( C\_{\left[ r\right] ,U}\right) .
\label{eq.darij1.pf.t1.AXU}
\tag{1}
\end{equation}
Similar reasoning shows that
\begin{align}
\det\left( A\_{X,V}\right) & =\det\left( B\_{X,\left[ r\right] }\right)
\cdot\det\left( C\_{\left[ r\right] ,V}\right) ;
\label{eq.darij1.pf.t1.AXV}
\tag{2}
\\
\det\left( A\_{Y,U}\right) & =\det\left( B\_{Y,\left[ r\right] }\right)
\cdot\det\left( C\_{\left[ r\right] ,U}\right) ;
\label{eq.darij1.pf.t1.AYU}
\tag{3}
\\
\det\left( A\_{Y,V}\right) & =\det\left( B\_{Y,\left[ r\right] }\right)
\cdot\det\left( C\_{\left[ r\right] ,V}\right) .
\label{eq.darij1.pf.t1.AYV}
\tag{4}
\end{align}


Multiplying the equalities \eqref{eq.darij1.pf.t1.AXU} and
\eqref{eq.darij1.pf.t1.AYV}, we obtain
\begin{align\*}
\det\left( A\_{X,U}\right) \cdot\det\left( A\_{Y,V}\right) & =\det\left(
B\_{X,\left[ r\right] }\right) \cdot\det\left( C\_{\left[ r\right]
,U}\right) \cdot\det\left( B\_{Y,\left[ r\right] }\right) \cdot\det\left(
C\_{\left[ r\right] ,V}\right) \\
& =\underbrace{\det\left( B\_{X,\left[ r\right] }\right) \cdot\det\left(
C\_{\left[ r\right] ,V}\right) }\_{\substack{=\det\left( A\_{X,V}\right)
\\\text{(by \eqref{eq.darij1.pf.t1.AXV})}}}\cdot\underbrace{\det\left(
B\_{Y,\left[ r\right] }\right) \cdot\det\left( C\_{\left[ r\right]
,U}\right) }\_{\substack{=\det\left( A\_{Y,U}\right) \\\text{(by
\eqref{eq.darij1.pf.t1.AYU})}}}\\
& =\det\left( A\_{X,V}\right) \cdot\det\left( A\_{Y,U}\right) .
\end{align\*}
This proves Theorem 1. $\blacksquare$


2. The skew-symmetric case
==========================


Recall that a square matrix $A\in\mathbb{K}^{n\times n}$ is said to be
*skew-symmetric* if it satisfies $A^{T}=-A$. Now you claim:



> 
> **Theorem 4.** Let $A\in\mathbb{K}^{n\times n}$ be a skew-symmetric matrix.
> Let $r\in\mathbb{N}$ be such that $\operatorname\*{rank}A\leq r$. Let $X$ and
> $Y$ be subsets of $\left[ n\right] $ such that $\left\vert X\right\vert
> =\left\vert Y\right\vert =r$. Then,
> \begin{align\*}
> \det\left( A\_{X,X}\right) \cdot\det\left( A\_{Y,Y}\right) =\left(
> -1\right) ^{r}\cdot\left( \det\left( A\_{X,Y}\right) \right) ^{2}.
> \end{align\*}
> 
> 
> 


*Proof of Theorem 4.* The matrix $A$ is skew-symmetric; thus, $A^{T}=-A$.
Now,
\begin{align\*}
\left( A\_{X,Y}\right) ^{T} & =\left( A^{T}\right) \_{Y,X}=\left(
-A\right) \_{Y,X}\qquad\left( \text{since }A^{T}=-A\right) \\
& =-A\_{Y,X}.
\end{align\*}
Therefore, $\det\left( \left( A\_{X,Y}\right) ^{T}\right) =\det\left(
-A\_{Y,X}\right) =\left( -1\right) ^{r}\det\left( A\_{Y,X}\right) $ (since
$A\_{Y,X}$ is an $r\times r$-matrix). But the determinant of a matrix does not
change when the matrix is transposed. Hence,
\begin{align\*}
\det\left( A\_{X,Y}\right) =\det\left( \left( A\_{X,Y}\right) ^{T}\right)
=\left( -1\right) ^{r}\det\left( A\_{Y,X}\right) .
\end{align\*}


Now, Theorem 1 (applied to $U=X$ and $V=Y$) yields
\begin{align\*}
\det\left( A\_{X,X}\right) \cdot\det\left( A\_{Y,Y}\right) =\det\left(
A\_{X,Y}\right) \cdot\det\left( A\_{Y,X}\right) .
\end{align\*}
Comparing this with
\begin{align\*}
& \left( -1\right) ^{r}\cdot\left( \det\left( A\_{X,Y}\right) \right)
^{2}\\
& =\left( -1\right) ^{r}\cdot\det\left( A\_{X,Y}\right) \cdot
\underbrace{\det\left( A\_{X,Y}\right) }\_{=\left( -1\right) ^{r}\det\left(
A\_{Y,X}\right) }\\
& =\left( -1\right) ^{r}\cdot\det\left( A\_{X,Y}\right) \cdot\left(
-1\right) ^{r}\det\left( A\_{Y,X}\right) \\
& =\underbrace{\left( -1\right) ^{r}\cdot\left( -1\right) ^{r}}\_{=\left(
-1\right) ^{2r}=1}\cdot\det\left( A\_{X,Y}\right) \cdot\det\left(
A\_{Y,X}\right) =\det\left( A\_{X,Y}\right) \cdot\det\left( A\_{Y,X}\right)
,
\end{align\*}
we obtain
\begin{align\*}
\det\left( A\_{X,X}\right) \cdot\det\left( A\_{Y,Y}\right) =\left(
-1\right) ^{r}\cdot\left( \det\left( A\_{X,Y}\right) \right) ^{2}.
\end{align\*}
This proves Theorem 4. $\blacksquare$


Note that if $\mathbb{K}$ is a field of characteristic $\neq2$, then the rank
of any skew-symmetric matrix $A\in\mathbb{K}^{n\times n}$ is even. This is a
known fact, implicit in [Keith Conrad, *Bilinear
Forms*](https://kconrad.math.uconn.edu/blurbs/linmultialg/bilinearform.pdf)
(combine Theorem 1.6 and formula (5.3) in this document).


3. Generalizing to commutative rings
====================================


Now, let us generalize our situation. Instead of requiring $\mathbb{K}$ to be
a field, we now merely assume that $\mathbb{K}$ be a commutative ring. Does
Theorem 1 still hold? The answer will depend on how we define
$\operatorname\*{rank}A$. There is no "obviously right" definition of "rank" of
a matrix over a commutative ring, but there are several contenders. We look
specifically for a notion that generalizes the statement
"$\operatorname\*{rank}A\leq r$". Two candidates are the following:



> 
> We say that a matrix $A\in\mathbb{K}^{n\times m}$ has *strong rank $\leq r$*
> (for some $r\in\mathbb{N}$) if there exist two matrices $B\in\mathbb{K}
> ^{n\times r}$ and $C\in\mathbb{K}^{r\times m}$ such that $A=BC$.
> 
> 
> 



> 
> We say that a matrix $A\in\mathbb{K}^{n\times m}$ has *weak rank $\leq r$*
> (for some $r\in\mathbb{N}$) if every two $\left( r+1\right) $-element
> subsets $X\subseteq\left[ n\right] $ and $Y\subseteq\left[ m\right] $
> satisfy $\det\left( A\_{X,Y}\right) =0$.
> 
> 
> 


It is easy to see (using the Cauchy--Binet theorem) that every matrix that has
strong rank $\leq r$ must also have weak rank $\leq r$. The converse is not
true (in fact, for $n=2$ and $r=1$, it boils down to [the pre-pre-Schreier condition](https://mathoverflow.net/questions/14555/first-order-ufd-factorial-ring-condition-pre-schreier-rings)).


Our above proof of Theorem 1 obviously proves the following generalization:



> 
> **Theorem 5.** Let $n,m,r\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times m}$ be a
> matrix that has strong rank $\leq r$. Let $X$ and $Y$ be subsets of $\left[
> n\right] $, and let $U$ and $V$ be subsets of $\left[ m\right] $; assume
> that $\left\vert X\right\vert =\left\vert Y\right\vert =\left\vert
> U\right\vert =\left\vert V\right\vert =r$. Then,
> \begin{align\*}
> \det\left( A\_{X,U}\right) \cdot\det\left( A\_{Y,V}\right) =\det\left(
> A\_{X,V}\right) \cdot\det\left( A\_{Y,U}\right) .
> \end{align\*}
> 
> 
> 


But we also have the following more general fact:



> 
> **Theorem 6.** Let $n,m,r\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times m}$ be a
> matrix that has weak rank $\leq r$. Let $X$ and $Y$ be subsets of $\left[
> n\right] $, and let $U$ and $V$ be subsets of $\left[ m\right] $; assume
> that $\left\vert X\right\vert =\left\vert Y\right\vert =\left\vert
> U\right\vert =\left\vert V\right\vert =r$. Then,
> \begin{align\*}
> \det\left( A\_{X,U}\right) \cdot\det\left( A\_{Y,V}\right) =\det\left(
> A\_{X,V}\right) \cdot\det\left( A\_{Y,U}\right) .
> \end{align\*}
> 
> 
> 


The only way I can currently prove Theorem 6 is by reducing it to Theorem 5
using the "meta-theorem" saying that every polynomial identity in the entries
of a matrix that holds (universally, i.e., over every $\mathbb{K}$) for all
matrices having strong rank $\leq r$ must also hold for all matrices having
weak rank $\leq r$. This "meta-theorem" follows from some standard monomial
theory, specifically the result that the quotient of the coordinate ring of
$\mathbb{K}^{n\times m}$ modulo the ideal generated by all $\left(
r+1\right) \times\left( r+1\right) $-minors can be embedded into the
coordinate ring of $\mathbb{K}^{n\times r}\times\mathbb{K}^{r\times m}$ via
the $A=BC$ substitution (with $B\in\mathbb{K}^{n\times r}$ and $C\in
\mathbb{K}^{r\times m}$). The latter fact is proved implicitly in Section 5 of
[Richard G. Swan, *On the straightening law for minors of a matrix*, https://arxiv.org/abs/1605.06696v1](https://arxiv.org/abs/1605.06696v1).



> 
> **Question 7.** Is there a "natural" proof of Theorem 6?
> 
> 
> 


4. Pfaffians
============


If we are already looking at skew-symmetric matrices, it isn't a long stretch
to move on to alternating matrices any more.


Recall that a square matrix $A\in\mathbb{K}^{n\times n}$ is said to be
*alternating* if it is skew-symmetric (i.e., satisfies $A^{T}=-A$) and all its
diagonal entries are $0$. If the field $\mathbb{K}$ has characteristic $\neq
2$, then every alternating matrix is skew-symmetric; but this is not true in
characteristic $2$. Any alternating matrix $A\in\mathbb{K}^{n\times n}$ has a
well-defined [Pfaffian](https://en.wikipedia.org/wiki/Pfaffian)
$\operatorname\*{Pf}A\in\mathbb{K}$ satisfying $\det A=\left(
\operatorname\*{Pf}A\right) ^{2}$. Note that $\operatorname\*{Pf}A=0$ if $n$ is
odd. Properties of Pfaffians come up in the literature all the time; there is
[a MathOverflow question collecting references on
them](https://mathoverflow.net/questions/280362/detailed-modern-references-for-basic-properties-of-pfaffians-over-commutative-ri).
(It is probably worth adding the recent preprint [arXiv:2008.04247v1](https://arxiv.org/abs/2008.04247v1).)


It is now not too hard to show the following:



> 
> **Theorem 8.** Let $\mathbb{K}$ be a field. Let $n,r\in\mathbb{N}$. Let
> $A\in\mathbb{K}^{n\times n}$ be an alternating matrix such that
> $\operatorname\*{rank}A\leq r$. Let $X$ and $Y$ be subsets of $\left[
> n\right] $ such that $\left\vert X\right\vert =\left\vert Y\right\vert =r$.
> Then,
> \begin{align\*}
> \operatorname\*{Pf}\left( A\_{X,X}\right) \cdot\operatorname\*{Pf}\left(
> A\_{Y,Y}\right) =\det\left( A\_{X,Y}\right) .
> \end{align\*}
> 
> 
> 


Indeed, this relies on the following analogue of Lemma 2:



> 
> **Lemma 9.** Let $\mathbb{K}$ be a field. Let $n,r\in\mathbb{N}$. Let
> $A\in\mathbb{K}^{n\times n}$ be an alternating matrix such that
> $\operatorname\*{rank}A\leq r$. Then, there exist a matrix $S\in\mathbb{K}
> ^{r\times n}$ and an alternating matrix $B\in\mathbb{K}^{r\times r}$ such that
> $A=S^{T}BS$.
> 
> 
> 


*Proof of Lemma 9 (rough outline).* We will use the results of the
(expository) paper [Keith Conrad, *Bilinear
Forms*](https://kconrad.math.uconn.edu/blurbs/linmultialg/bilinearform.pdf).
We translate the claim into the language of alternating bilinear forms. Set
$V=\mathbb{K}^{n\times1}$, and let $f:V\times V\rightarrow\mathbb{K}$ be the
alternating bilinear form corresponding to $A$. (Thus, explicitly, it is the
form $\left( x,y\right) \mapsto x^{T}Ay$.) Let $R$ be the subspace $\left\{
v\in V\ \mid\ f\left( v,w\right) =0\text{ for all }w\in V\right\}
=\operatorname\*{Ker}A$ of $V$. (This is often called the *radical* of $f$.)
Then, $f$ descends to an alternating bilinear form $\overline{f}$ on the
quotient space $V/R$. Choose any basis $\left( w\_{1},w\_{2},\ldots
,w\_{g}\right) $ for the latter quotient space; then,
\begin{align\*}
g=\dim\left( V/R\right) =\underbrace{\dim V}\_{=n}-\dim\underbrace{R}
\_{=\operatorname\*{Ker}A}=n-\dim\left( \operatorname\*{Ker}A\right)
=\operatorname\*{rank}A\leq r.
\end{align\*}
Let $B\_{0}\in\mathbb{K}^{g\times g}$ be the alternating matrix that represents
the bilinear form $\overline{f}$ on this basis $\left( w\_{1},w\_{2}
,\ldots,w\_{g}\right) $. Let $S\_{0}\in\mathbb{K}^{g\times n}$ be the matrix
that represents the projection map $V\rightarrow V/R$ with respect to the
standard basis of $V=\mathbb{K}^{n\times1}$ and the basis $\left( w\_{1}
,w\_{2},\ldots,w\_{g}\right) $ of $V/R$. Then, it is easy to see that
$A=S\_{0}^{T}B\_{0}S\_{0}$. By adding $r-g$ zero rows and columns to $B\_{0}$ and
$r-g$ zero rows to $S\_{0}$, we obtain two matrices $B\in\mathbb{K}^{r\times
r}$ and $S\in\mathbb{K}^{r\times n}$ such that $B$ is alternating and such
that $A=S^{T}BS$. This proves Lemma 9. $\blacksquare$


[*Remark:* Lemma 9 can be made a lot stronger if we assume
$\operatorname\*{rank}A=r$. Indeed, in this case, we can say that $r$ is even,
and that $A$ can be written in the form $A=S^{T}JS$, where $S\in
\mathbb{K}^{r\times n}$ is some matrix, and where $J$ is the "standard"
skew-symmetric $r\times r$-matrix (i.e., the block matrix $\left(
\begin{array}
[c]{cc}
0 & I\_{r/2}\\
-I\_{r/2} & 0
\end{array}
\right) $). This follows from the fact that the alternating bilinear form
$\overline{f}$ in the above proof of Lemma 9 is non-degenerate and thus has a
symplectic basis. (And this basis has size $g=r$, because
$\operatorname\*{rank}A=r$.)]



> 
> **Lemma 10.** Let $r\in\mathbb{N}$. Let $B\in\mathbb{K}^{r\times r}$ be an
> alternating matrix. Let $K\in\mathbb{K}^{r\times r}$ be any matrix. Then,
> \begin{align\*}
> \operatorname\*{Pf}\left( K^{T}BK\right) =\det K\cdot\operatorname\*{Pf}B.
> \end{align\*}
> 
> 
> 


For proofs of Lemma 10, see (e.g.) Teorema 8.6.3 (1) in [Marco Manetti,
*Algebra lineare, per matematici*,
2020-09-13](https://www1.mat.uniroma1.it/people/manetti/dispense/algebralineare.pdf).
A more general fact -- which relates to Lemma 10 in the same way as the
Cauchy-Binet formula relates to the classical $\det\left( UV\right) =\det
U\cdot\det V$ -- is Ishikawa/Wakayama's "Minor Summation Formula" as stated in
[Masao Ishikawa, Masato Wakayama, *Minor Summation Formula of Pfaffians*,
Linear and Multilinear Algebra **39** (1995), pp. 285--305](https://doi.org/10.1080/03081089508818403).


*Proof of Theorem 8 (rough outline).* Lemma 9 yields that there exist a matrix
$S\in\mathbb{K}^{r\times n}$ and an alternating matrix $B\in\mathbb{K}
^{r\times r}$ such that $A=S^{T}BS$. Consider these $S$ and $B$.


From $A=S^{T}BS$, it is easy to see that
\begin{equation}
A\_{X,Y}=\underbrace{\left( S^{T}\right) \_{X,\left[ r\right] }}\_{=\left(
S\_{\left[ r\right] ,X}\right) ^{T}}BS\_{\left[ r\right] ,Y}=\left(
S\_{\left[ r\right] ,X}\right) ^{T}BS\_{\left[ r\right] ,Y}.
\label{eq.darij1.pf.t8.4}
\tag{5}
\end{equation}
Hence,
\begin{align}
\det\left( A\_{X,Y}\right) & =\det\left( \left( S\_{\left[ r\right]
,X}\right) ^{T}BS\_{\left[ r\right] ,Y}\right) =\underbrace{\det\left(
\left( S\_{\left[ r\right] ,X}\right) ^{T}\right) }\_{=\det\left(
S\_{\left[ r\right] ,X}\right) }\cdot\det B\cdot\det\left( S\_{\left[
r\right] ,Y}\right) \nonumber\\
& =\det\left( S\_{\left[ r\right] ,X}\right) \cdot\det B\cdot\det\left(
S\_{\left[ r\right] ,Y}\right) .
\label{eq.darij1.pf.t8.5}
\tag{6}
\end{align}
Similarly to \eqref{eq.darij1.pf.t8.4}, we also obtain
\begin{align\*}
A\_{X,X}=\left( S\_{\left[ r\right] ,X}\right) ^{T}BS\_{\left[ r\right]
,X},
\end{align\*}
so that
\begin{align\*}
\operatorname\*{Pf}\left( A\_{X,X}\right) =\operatorname\*{Pf}\left( \left(
S\_{\left[ r\right] ,X}\right) ^{T}BS\_{\left[ r\right] ,X}\right)
=\det\left( S\_{\left[ r\right] ,X}\right) \cdot\operatorname\*{Pf}B
\end{align\*}
(by Lemma 10). Likewise,
\begin{align\*}
\operatorname\*{Pf}\left( A\_{Y,Y}\right) =\det\left( S\_{\left[ r\right]
,Y}\right) \cdot\operatorname\*{Pf}B.
\end{align\*}
Multiplying the preceding two equalities and recalling that $\left(
\operatorname\*{Pf}B\right) ^{2}=\det B$, we obtain


\begin{align\*}
\operatorname\*{Pf}\left( A\_{X,X}\right) \cdot\operatorname\*{Pf}\left(
A\_{Y,Y}\right) =\det\left( S\_{\left[ r\right] ,X}\right) \cdot\det\left(
S\_{\left[ r\right] ,Y}\right) \cdot\det B=\det\left( A\_{X,Y}\right)
\end{align\*}
(by \eqref{eq.darij1.pf.t8.5}). This proves Theorem 8. $\blacksquare$


Note that squaring the equality in Theorem 8 yields the claim of Theorem 4
(albeit only in the case when $A$ is alternating).



> 
> **Question 9.** Can Theorem 8 be generalized to commutative rings $\mathbb{K}$
> ? This is trickier than one might expect, as the notions of "weak rank" and
> "strong rank" probably need to be updated.
> 
> 
> 


