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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.

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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 (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).

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.

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 $\operatorname*{Pf}A\in\mathbb{K}$ 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. (It is probably worth adding the recent preprint arXiv: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. 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 8.6.4 (1) in Marco Manetti, Algebra lineare, per matematici, 2018-09-13. 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.

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.

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A partial solution: Assume $X$ and $Y$ partition $\{1,\dots,n\}$. Hence $A$ is in the block form $$ A=\begin{bmatrix} B_{r\times r}&C_{r\times r}\\ -^{\rm{T}}C_{r\times r}&D_{r\times r} \end{bmatrix} $$ where $B$ and $D$ are skew-symmetric. One needs to show that if the rank of the matrix above is $r$, then $\det(B)\det(D)=(-1)^r\left(\det(C)\right)^2$. The dimension of the null space must be $r$. A column vector $\begin{bmatrix} v_{r\times 1}\\ w_{r\times 1} \end{bmatrix}$ is killed by $A$ iff \begin{equation*} \begin{cases} Bv+Cw=\mathbf{0}\\ -^{\rm{T}}Cv+Dw=\mathbf{0} \end{cases}. \end{equation*} If $B$ is invertible, we can solve the first equation for $v$ to obtain $v=-B^{-1}Cw$. Substituting in the second equation yields $(^{\rm{T}}CB^{-1}C+D)w=0$. Thus the dimension of the null space of $A$ is the same as that of the $r\times r$ matrix $^{\rm{T}}CB^{-1}C+D$. Hence this matrix must be zero. We conclude that $^{\rm{T}}CB^{-1}C=-D$. Taking determinants of both sides we obtain $\det(^{\rm{T}}CB^{-1}C)=(-1)^r\det(D)$, and hence $\det(B)\det(D)=(-1)^r\left(\det(C)\right)^2$.

If $B$ is singular, it suffices to argue that $C$ is singular as well. Aiming for a contradiction, if $C$ is invertible one may solve the first equation for $v$ to obtain $w=-C^{-1}Bv$. Substituting in the second equation, a similar argument shows $^{\rm{T}}C+DC^{-1}B=\mathbf{O}_{r\times r}$. This is a contradiction since $^{\rm{T}}C$ is invertible while $B$ and hence $DC^{-1}B$ are singular.

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