# A symmetric bilinear form and a Plücker identity

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 $$R:={\mathbb k}[x_{ij}\colon 1\leq i, the matrix $$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!

• So $A_{ij}$ stands for the $2\times 2$-matrix formed by the $i$-th and $j$-th columns of $A$, right? – darij grinberg Nov 6 '18 at 18:35
• 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 ... – darij grinberg Nov 6 '18 at 18:39
• ... 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! – darij grinberg Nov 6 '18 at 18:41
• 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". – darij grinberg Nov 6 '18 at 18:46
• That should tell us that each of the $2\times 2$ minors should be nonzero, though. – Graham Denham Nov 6 '18 at 19:48


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, $$$$B = \begin{pmatrix} \tup{\det\tup{A_{1,1}}}^2 & \tup{\det\tup{A_{1,2}}}^2 & \cdots & \tup{\det\tup{A_{1,n}}}^2 \\ \tup{\det\tup{A_{2,1}}}^2 & \tup{\det\tup{A_{2,2}}}^2 & \cdots & \tup{\det\tup{A_{2,n}}}^2 \\ \vdots & \vdots & \ddots & \vdots \\ \tup{\det\tup{A_{n,1}}}^2 & \tup{\det\tup{A_{n,2}}}^2 & \cdots & \tup{\det\tup{A_{n,n}}}^2 \end{pmatrix} .$$$$ Note that this matrix $$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 $$$$\rank\tup{C + D + E} \leq \underbrace{\rank C}_{\leq 1} + \underbrace{\rank D}_{\leq 1} + \underbrace{\rank E}_{\leq 1} \leq 1 + 1 + 1 = 3.$$$$ In view of $$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, $$$$\det \overline{B} = 2 \tup{\det\tup{A_{q, r}}}^2 \tup{\det\tup{A_{p, r}}}^2 \tup{\det\tup{A_{p, q}}}^2 .$$$$

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

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

Proof of Proposition 5. The definition of $$\overline{B}$$ yields $$$$\overline{B} = \begin{pmatrix} \tup{\det\tup{A_{p,p}}}^2 & \tup{\det\tup{A_{p,q}}}^2 & \tup{\det\tup{A_{p,r}}}^2 \\ \tup{\det\tup{A_{q,p}}}^2 & \tup{\det\tup{A_{q,q}}}^2 & \tup{\det\tup{A_{q,r}}}^2 \\ \tup{\det\tup{A_{r,p}}}^2 & \tup{\det\tup{A_{r,q}}}^2 & \tup{\det\tup{A_{r,r}}}^2 \end{pmatrix} = \begin{pmatrix} 0 & \tup{\det\tup{A_{p,q}}}^2 & \tup{\det\tup{A_{p,r}}}^2 \\ \tup{\det\tup{A_{p,q}}}^2 & 0 & \tup{\det\tup{A_{q,r}}}^2 \\ \tup{\det\tup{A_{p,r}}}^2 & \tup{\det\tup{A_{q,r}}}^2 & 0 \end{pmatrix}$$$$ (since any $$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$$). Hence, $$$$\det \overline{B} = \det \begin{pmatrix} 0 & \tup{\det\tup{A_{p,q}}}^2 & \tup{\det\tup{A_{p,r}}}^2 \\ \tup{\det\tup{A_{p,q}}}^2 & 0 & \tup{\det\tup{A_{q,r}}}^2 \\ \tup{\det\tup{A_{p,r}}}^2 & \tup{\det\tup{A_{q,r}}}^2 & 0 \end{pmatrix} = 2 \tup{\det\tup{A_{q, r}}}^2 \tup{\det\tup{A_{p, r}}}^2 \tup{\det\tup{A_{p, q}}}^2$$$$ (by Lemma 6, applied to $$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 $$$$\det \overline{B} = 2 \tup{\det\tup{A_{q, r}}}^2 \tup{\det\tup{A_{p, r}}}^2 \tup{\det\tup{A_{p, q}}}^2 \neq 0.$$$$ But $$\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 $$2$$). $$\blacksquare$$

• 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. – Graham Denham Nov 6 '18 at 21:25
• @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. – darij grinberg Nov 6 '18 at 21:45