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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!

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    $\begingroup$ So $A_{ij}$ stands for the $2\times 2$-matrix formed by the $i$-th and $j$-th columns of $A$, right? $\endgroup$ – darij grinberg Nov 6 '18 at 18:35
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    $\begingroup$ In that case, let me denote the entries of the $1$-st row of $A$ by $a_1, a_2, \ldots, a_n$, and the entries of the $2$-nd row of $A$ by $b_1, b_2, \ldots, b_n$. Then, $\left|A_{ij}\right| = a_i b_j - a_j b_i$ for all $i$ and $j$ (including the case when $i = j$). Thus, $B = C + D + E$, where $C, D, E$ are three $n \times n$-matrices whose $\left(i, j\right)$-th entries are $a_i^2 b_j^2, 2 a_i a_j b_i b_j, a_j^2 b_i^2$, respectively. All three matrices $C, D, E$ have rank $1$, since each of them can be immediately written in the form $uv$ for a column vector $u$ and ... $\endgroup$ – darij grinberg Nov 6 '18 at 18:39
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    $\begingroup$ ... a row vector $v$. Thus, their sum $B = C + D + E$ has rank $\leq 1 + 1 + 1 = 3$. To prove that it has rank exactly $3$, it is probably necessary to do some amount of computation, namely computing its determinant in the $n = 3$ case. But I'm not sure what the exact condition here would be, because the one you're giving (three columns of $A$ being linearly independent) is never satisfied! $\endgroup$ – darij grinberg Nov 6 '18 at 18:41
  • $\begingroup$ Actually, Sage tells me that $\det B = 2 \prod\limits_{i<j} \left|A_{ij}\right|^2$ when $n = 3$ (note that this is also trivial to compute by hand). Thus, the correct condition is "at least two columns of $A$ are linearly independent". $\endgroup$ – darij grinberg Nov 6 '18 at 18:46
  • $\begingroup$ That should tell us that each of the $2\times 2$ minors should be nonzero, though. $\endgroup$ – Graham Denham Nov 6 '18 at 19:48
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$\newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\set}[1]{\left\{ #1 \right\}} \newcommand{\abs}[1]{\left| #1 \right|} \newcommand{\tup}[1]{\left( #1 \right)} \newcommand{\ive}[1]{\left[ #1 \right]} \newcommand{\rank}{\operatorname{rank}}$ 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 $2$). $\blacksquare$

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  • $\begingroup$ Well done! I'm going to leave the "accepted answer" box unchecked for a few hours, just to bait a few more readers. I'm curious to hear if anyone has met "B" before out in nature. $\endgroup$ – Graham Denham Nov 6 '18 at 21:25
  • $\begingroup$ @GrahamDenham: I have met a matrix rather similar to $B$ (but much more complicated): See Theorem 2.1 in Darij Grinberg, On a double Sylvester determinant. $\endgroup$ – darij grinberg Nov 6 '18 at 21:45

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