# Minors of low rank skew-symmetric matrix

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.

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 $$I=\left\{ i_{1} and $$J=\left\{ j_{1}. (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 $$$$\det\left( A_{X,U}\right) =\det\left( B_{X,\left[ r\right] }\right) \cdot\det\left( C_{\left[ r\right] ,U}\right) . \label{eq.darij1.pf.t1.AXU} \tag{1}$$$$ 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 $$$$A_{X,Y}=\underbrace{\left( S^{T}\right) _{X,\left[ r\right] }}_{=\left( S_{\left[ r\right] ,X}\right) ^{T}}BS_{\left[ r\right] ,Y}=\left( S_{\left[ r\right] ,X}\right) ^{T}BS_{\left[ r\right] ,Y}. \label{eq.darij1.pf.t8.4} \tag{5}$$$$ Hence, \begin{align} \det\left( A_{X,Y}\right) & =\det\left( \left( S_{\left[ r\right] ,X}\right) ^{T}BS_{\left[ r\right] ,Y}\right) =\underbrace{\det\left( \left( S_{\left[ r\right] ,X}\right) ^{T}\right) }_{=\det\left( S_{\left[ r\right] ,X}\right) }\cdot\det B\cdot\det\left( S_{\left[ r\right] ,Y}\right) \nonumber\\ & =\det\left( S_{\left[ r\right] ,X}\right) \cdot\det B\cdot\det\left( S_{\left[ r\right] ,Y}\right) . \label{eq.darij1.pf.t8.5} \tag{6} \end{align} Similarly to \eqref{eq.darij1.pf.t8.4}, we also obtain \begin{align*} A_{X,X}=\left( S_{\left[ r\right] ,X}\right) ^{T}BS_{\left[ r\right] ,X}, \end{align*} so that \begin{align*} \operatorname*{Pf}\left( A_{X,X}\right) =\operatorname*{Pf}\left( \left( S_{\left[ r\right] ,X}\right) ^{T}BS_{\left[ r\right] ,X}\right) =\det\left( S_{\left[ r\right] ,X}\right) \cdot\operatorname*{Pf}B \end{align*} (by Lemma 10). Likewise, \begin{align*} \operatorname*{Pf}\left( A_{Y,Y}\right) =\det\left( S_{\left[ r\right] ,Y}\right) \cdot\operatorname*{Pf}B. \end{align*} Multiplying the preceding two equalities and recalling that $$\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.

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.