$\newcommand{\NN}{\mathbb{N}}
\newcommand{\ZZ}{\mathbb{Z}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\tup}[1]{\left( #1 \right)}
\newcommand{\ive}[1]{\left[ #1 \right]}
\newcommand{\rank}{\operatorname{rank}}$
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 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 $2$). $\blacksquare$