It turns out that a special case of something I'm working on gives, as a corollary, a rather 19th-century-looking elementary statement about the rank of a certain symmetric matrix. I thought I would post it here in the hope that someone might recognize it as something familiar.


Let $A$ be a $2\times n$ matrix over a field $\mathbb k$, where $n\geq 3$. Let $|A\_{ij}|$ denote its minors, for $1\leq i,j\leq n$, taking the convention that $|A\_{ii}|=0$. Let $B$ be the $n\times n$ symmetric matrix with $B\_{ij}=|A\_{ij}|^2$, for $1\leq i,j\leq n$. It can be shown that, if at least three columns of $A$ are pairwise linearly independent, then $B$ has rank exactly $3$.


In other words, over the polynomial ring $R:={\mathbb k}[x\_{ij}\colon 1\leq i<j\leq n]$, the matrix $C$ with $C\_{ij}=x\_{ij}^2$ for $i\neq j$ and diagonal entries zero has the property that its ideal of $k\times k$ minors is contained in the ideal of Plücker relations (defining the Grassmannian of $2$-planes in ${\mathbb k}^n$).


My question then: is there an easy way to see why this is true or equivalent to something classical? Thanks!



A symmetric bilinear form and a Plücker identity
$\newcommand{\NN}{\mathbb{N}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\tup}[1]{\left( #1 \right)}
\newcommand{\ive}[1]{\left[ #1 \right]}
\newcommand{\rank}{\operatorname{rank}}$ 
I let $\NN$ be the set $\set{0, 1, 2, \ldots}$. For each $n \in \NN$, let $\ive{n}$ denote the set $\set{1, 2, \ldots, n}$.


Fix a field $k$. We can more generally allow $k$ to be any commutative ring, as long as we define the rank of a matrix $A$ over $k$ correctly -- namely, as the largest $i \in \NN$ such that $A$ has at least one nonzero minor of size $i$. In the following, I shall state all results only for fields $k$, but (when necessary) I will comment on the case of a commutative ring in the proofs.


Let $n \in \NN$. Let $A$ be a $2 \times n$-matrix over $k$. Let $A\_1, A\_2, \ldots, A\_n$ be the $n$ columns of $A$; these are $n$ column vectors of size $2$. For each $i, j \in \ive{n}$, let $A\_{i, j}$ be the $2 \times 2$-matrix over $k$ whose two columns are $A\_i$ and $A\_j$.


Note that any $i, j \in \ive{n}$ satisfy $\det\tup{A\_{i,j}} = - \det\tup{A\_{j,i}}$ (since the matrices $A\_{i,j}$ and $A\_{j,i}$ differ only in the order of their columns) and thus $\tup{\det\tup{A\_{i,j}}}^2 = \tup{\det\tup{A\_{j,i}}}^2$. Also, any $i \in \ive{n}$ satisfies $\det\tup{A\_{i,i}} = 0$ (since the matrix $A\_{i,i}$ has two equal columns).


Let $B$ be the $n \times n$-matrix over $k$ whose $\tup{i, j}$-th entry is $\tup{\det\tup{A\_{i,j}}}^2$ for all $i, j \in \ive{n}$. Thus,
\begin{equation}
B =
\begin{pmatrix}
 \tup{\det\tup{A\_{1,1}}}^2 & \tup{\det\tup{A\_{1,2}}}^2 & \cdots & \tup{\det\tup{A\_{1,n}}}^2 \\
 \tup{\det\tup{A\_{2,1}}}^2 & \tup{\det\tup{A\_{2,2}}}^2 & \cdots & \tup{\det\tup{A\_{2,n}}}^2 \\
 \vdots & \vdots & \ddots & \vdots \\
 \tup{\det\tup{A\_{n,1}}}^2 & \tup{\det\tup{A\_{n,2}}}^2 & \cdots & \tup{\det\tup{A\_{n,n}}}^2
\end{pmatrix}
.
\end{equation}
Note that this matrix $B$ is symmetric (since any $i, j \in \ive{n}$ satisfy $\tup{\det\tup{A\_{i,j}}}^2 = \tup{\det\tup{A\_{j,i}}}^2$), and its diagonal entries are $0$ (since each $i \in \ive{n}$ satisfies $\det\tup{A\_{i,i}} = 0$, because the two columns of the matrix $A\_{i, i}$ are equal).



> 
> **Proposition 1.** We have $\rank B \leq 3$.
> 
> 
> 


To prove this proposition, we need the following lemmas:



> 
> **Lemma 2.** Let $p\_1, p\_2, \ldots, p\_n$ be $n$ elements of $k$, and let $q\_1, q\_2, \ldots, q\_n$ be $n$ elements of $k$. Let $X$ be the $n \times n$-matrix over $k$ whose $\tup{i, j}$-th entry is $p\_i q\_j$ for all $i, j \in \ive{n}$. Then, $\rank X \leq 1$.
> 
> 
> 


*Proof of Lemma 2.* Let $P$ be the column vector $\tup{p\_1, p\_2, \ldots, p\_n}^T$, and let $Q$ be the row vector $\tup{q\_1, q\_2, \ldots, q\_n}$. Then, $X = PQ$, so that $\rank X = \rank\tup{PQ} \leq \rank P \leq 1$ (since $P$ is a $n \times 1$-matrix). (We have here relied on the fact that $\rank\tup{PQ} \leq \rank P$, which is well-known when $k$ is a field. When $k$ is a commutative ring, it is still true, because the Cauchy-Binet theorem shows that any size-$i$ minor of $PQ$ is a $k$-linear combination of size-$i$ minors of $P$.) Thus, Lemma 2 is proven. $\blacksquare$



> 
> **Lemma 3.** Let $C$ and $D$ be two matrices of the same size. Then, $\rank\tup{C + D} \leq \rank C + \rank D$.
> 
> 
> 


*Proof of Lemma 3.* This is well-known when $k$ is a field. In the more general case when $k$ is a commutative ring, we can prove it as follows: There is a well-known formula that expresses $\det\tup{A + B}$ (where $A$ and $B$ are two $p \times p$-matrices) as a sum of terms of the form $\pm \tup{\text{a size-$i$ minor of } A } \tup{\text{a size-$j$ minor of } B }$ with $i + j = p$. (See, e.g., Theorem 6.160 in [my *Notes on the combinatorial fundamentals of algebra*, 10 January 2019](https://github.com/darijgr/detnotes/releases/tag/2019-01-10) for this formula. Note that $i$ and $j$ can be $0$, in which case it should be kept in mind that size-$0$ minors equal $1$.) Using this formula, we see that any minor of $C + D$ having size $> \rank C + \rank D$ equals $0$ (because if we let $A$ be the corresponding minor of $C$ and $B$ be the corresponding minor of $D$, then all the terms in the sum will be $0$). Thus, $\rank\tup{C + D} \leq \rank C + \rank D$. $\blacksquare$



> 
> **Lemma 4.** Let $C\_1, C\_2, \ldots, C\_p$ be any $p$ matrices of a given size. Then, $\rank\tup{C\_1 + C\_2 + \cdots + C\_p} \leq \rank\tup{C\_1} + \rank\tup{C\_2} + \cdots + \rank\tup{C\_p}$.
> 
> 
> 


*Proof of Lemma 4.* This follows by induction on $p$, using Lemma 3. $\blacksquare$


*Proof of Proposition 1.* Let $a\_1, a\_2, \ldots, a\_n$ be the $n$ entries of the first row of $A$. Let $b\_1, b\_2, \ldots, b\_n$ be the $n$ entries of the second row of $A$. Then, the column vector $A\_i$ can be written as $A\_i = \tup{a\_i, b\_i}^T$ for each $i \in \ive{n}$. Hence, for any $i, j \in \ive{n}$, we have
$A\_{i, j} = \begin{pmatrix} a\_i & a\_j \\ b\_i & b\_j \end{pmatrix}$ and thus
\begin{align}
 \tup{\det\tup{A\_{i,j}}}^2
 &= \tup{\det \begin{pmatrix} a\_i & a\_j \\ b\_i & b\_j \end{pmatrix}}^2
 = \tup{a\_i b\_j - b\_i a\_j}^2 \\
 &= a\_i^2 b\_j^2 - 2 a\_i b\_j b\_i a\_j + b\_i^2 a\_j^2
 = a\_i^2 b\_j^2 - 2 a\_i b\_i a\_j b\_j + b\_i^2 a\_j^2 .
\end{align}


Now, define three $n \times n$-matrices $C$, $D$ and $E$ over $k$ as follows:


* Let $C$ be the $n \times n$-matrix whose $\tup{i, j}$-th entry is $a\_i^2 b\_j^2$ for all $i, j \in \ive{n}$.
* Let $D$ be the $n \times n$-matrix whose $\tup{i, j}$-th entry is $- 2 a\_i b\_i a\_j b\_j$ for all $i, j \in \ive{n}$.
* Let $E$ be the $n \times n$-matrix whose $\tup{i, j}$-th entry is $b\_i^2 a\_j^2$ for all $i, j \in \ive{n}$.


Then, for all $i, j \in \ive{n}$, the $\tup{i, j}$-th entry of the matrix $C + D + E$ is
$a\_i^2 b\_j^2 - 2 a\_i b\_i a\_j b\_j + b\_i^2 a\_j^2 = \tup{\det\tup{A\_{i,j}}}^2$;
but this is the same as the $\tup{i, j}$-th entry of the matrix $B$.
Thus, $C + D + E = B$.


Lemma 2 (applied to $p\_i = a\_i^2$, $q\_j = b\_j^2$ and $X = C$) yields $\rank C \leq 1$.
Lemma 2 (applied to $p\_i = - 2 a\_i b\_i$, $q\_j = a\_j b\_j$ and $X = D$) yields $\rank D \leq 1$.
Lemma 2 (applied to $p\_i = b\_i^2$, $q\_j = a\_j^2$ and $X = E$) yields $\rank E \leq 1$.
Now, Lemma 4 (applied to $p = 3$, $C\_1 = C$, $C\_2 = D$ and $C\_3 = E$) yields
\begin{equation}
 \rank\tup{C + D + E} \leq \underbrace{\rank C}\_{\leq 1} + \underbrace{\rank D}\_{\leq 1} + \underbrace{\rank E}\_{\leq 1} \leq 1 + 1 + 1 = 3.
\end{equation}
In view of $C + D + E = B$, this rewrites as $\rank B \leq 3$. This proves Proposition 1. $\blacksquare$


Next, we claim:



> 
> **Proposition 5.** Let $p, q, r$ be three elements of $\ive{n}$ such that $p < q < r$. Let $\overline{B}$ be the submatrix of $B$ formed by removing all rows except for the $p$-th, $q$-th and $r$-th rows and all columns except for the $p$-th, $q$-th and $r$-th columns. (This is a $3 \times 3$-matrix.) Then,
>  \begin{equation}
>  \det \overline{B} = 2 \tup{\det\tup{A\_{q, r}}}^2 \tup{\det\tup{A\_{p, r}}}^2 \tup{\det\tup{A\_{p, q}}}^2 .
> \end{equation}
> 
> 
> 


This will follow from the following lemma, which is easily checked by hand:



> 
> **Lemma 6.** Let $u, v, w \in k$. Then,
>  \begin{equation}
>  \det \begin{pmatrix}
>  0 & w & v \\ w & 0 & u \\ v & u & 0
>  \end{pmatrix}
>  = 2 u v w .
> \end{equation}
> 
> 
> 


*Proof of Proposition 5.* The definition of $\overline{B}$ yields
\begin{equation}
\overline{B} =
\begin{pmatrix}
 \tup{\det\tup{A\_{p,p}}}^2 & \tup{\det\tup{A\_{p,q}}}^2 & \tup{\det\tup{A\_{p,r}}}^2 \\
 \tup{\det\tup{A\_{q,p}}}^2 & \tup{\det\tup{A\_{q,q}}}^2 & \tup{\det\tup{A\_{q,r}}}^2 \\
 \tup{\det\tup{A\_{r,p}}}^2 & \tup{\det\tup{A\_{r,q}}}^2 & \tup{\det\tup{A\_{r,r}}}^2
\end{pmatrix}
=
\begin{pmatrix}
 0 & \tup{\det\tup{A\_{p,q}}}^2 & \tup{\det\tup{A\_{p,r}}}^2 \\
 \tup{\det\tup{A\_{p,q}}}^2 & 0 & \tup{\det\tup{A\_{q,r}}}^2 \\
 \tup{\det\tup{A\_{p,r}}}^2 & \tup{\det\tup{A\_{q,r}}}^2 & 0
\end{pmatrix}
\end{equation}
(since any $i, j \in \ive{n}$ satisfy $\tup{\det\tup{A\_{i,j}}}^2 = \tup{\det\tup{A\_{j,i}}}^2$, and since each $i \in \ive{n}$ satisfies $\det\tup{A\_{i,i}} = 0$). Hence,
\begin{equation}
 \det \overline{B}
 = \det
 \begin{pmatrix}
 0 & \tup{\det\tup{A\_{p,q}}}^2 & \tup{\det\tup{A\_{p,r}}}^2 \\
 \tup{\det\tup{A\_{p,q}}}^2 & 0 & \tup{\det\tup{A\_{q,r}}}^2 \\
 \tup{\det\tup{A\_{p,r}}}^2 & \tup{\det\tup{A\_{q,r}}}^2 & 0
 \end{pmatrix}
 = 2 \tup{\det\tup{A\_{q, r}}}^2 \tup{\det\tup{A\_{p, r}}}^2 \tup{\det\tup{A\_{p, q}}}^2
\end{equation}
(by Lemma 6, applied to $u = \tup{\det\tup{A\_{q, r}}}^2$, $v = \tup{\det\tup{A\_{p, r}}}^2$ and $w = \tup{\det\tup{A\_{p, q}}}^2$). This proves Proposition 5. $\blacksquare$



> 
> **Proposition 7.** Assume that there exist $p, q, r \in \ive{n}$ such that $p < q < r$ and $2 \tup{\det\tup{A\_{q, r}}}^2 \tup{\det\tup{A\_{p, r}}}^2 \tup{\det\tup{A\_{p, q}}}^2 \neq 0$. Then, $\rank B = 3$.
> 
> 
> 


*Proof of Proposition 7.* Consider these $p, q, r$ whose existence we have assumed. Consider the $3 \times 3$-matrix $\overline{B}$ defined in Proposition 5. Then, Proposition 5 yields
\begin{equation}
 \det \overline{B} = 2 \tup{\det\tup{A\_{q, r}}}^2 \tup{\det\tup{A\_{p, r}}}^2 \tup{\det\tup{A\_{p, q}}}^2 \neq 0.
\end{equation}
But $\overline{B}$ is a submatrix of $B$, and thus $\det \overline{B}$ is a minor of $B$ of size $3$. Hence, there exists a nonzero minor of $B$ of size $3$ (since $\det \overline{B} \neq 0$). Thus, $\rank B \geq 3$. Combining this with $\rank B \leq 3$ (which follows from Proposition 1), we obtain $\rank B = 3$. This proves Proposition 7. $\blacksquare$



> 
> **Proposition 8.** Assume that $k$ has characteristic $2$. (If $k$ is just a commutative ring, then we should instead assume that $2 = 0$ in $k$.) Then, $\rank B \leq 2$.
> 
> 
> 


*Proof of Proposition 8.* This is analogous to the proof of Proposition 1, but we additionally need to observe that $\rank D \leq 0$ (instead of $\rank D \leq 1$), because all entries of $D$ are $0$ (since they contain the factor $2$). $\blacksquare$


