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.
 A: 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.   
