do you know this determinant (basic commutative algebra)?

Let $\ell_1,\dots,\ell_n$ be $d+1$-variate linear forms over complex numbers in variables $X=(X_0,\dots,X_d)$. Consider the $(n-d)$-fold products $$\ell_{i_1}(X)\ell_{i_2}(X)\dots\ell_{i_{n-d}}(X)=\sum_{|I|=n-d}a_{I,J}X^I,\ J=(i_1,\dots,i_{n-d}),\\ 1\leq i_1<i_2<\dots < i_{n-d}\leq n.$$

Define $\binom{n}{d}\times\binom{n}{d}$-matrix $A$ with entries $A_{IJ}=a_{I,J}$. Then it appears that $$\det A=C\prod_{K}L_K,\quad K=(k_1,\dots,k_{d+1}),\ 1\leq k_1< k_2<\dots<k_{d+1},$$ where each $L_K$ is a $(d+1)\times (d+1)$-minor of the $(d+1)\times n$-matrix $L$ of the coefficients of $\ell_1,\dots,\ell_n$, and $C\neq 0$.

For instance, let $n=4$, $d=2$, and $$L=\left(\begin{array}{rrrr} a_{0} & a_{1} & a_{2} & a_{3} \\ b_{0} & b_{1} & b_{2} & b_{3} \\ c_{0} & c_{1} & c_{2} & c_{3} \end{array}\right)$$

Then $$A = {\scriptsize \left(\begin{array}{rrrrrr} a_{0} a_{1} & a_{0} a_{2} & a_{0} a_{3} & a_{1} a_{2} & a_{1} a_{3} & a_{2} a_{3} \\ a_{1} b_{0} + a_{0} b_{1} & a_{2} b_{0} + a_{0} b_{2} & a_{3} b_{0} + a_{0} b_{3} & a_{2} b_{1} + a_{1} b_{2} & a_{3} b_{1} + a_{1} b_{3} & a_{3} b_{2} + a_{2} b_{3} \\ a_{1} c_{0} + a_{0} c_{1} & a_{2} c_{0} + a_{0} c_{2} & a_{3} c_{0} + a_{0} c_{3} & a_{2} c_{1} + a_{1} c_{2} & a_{3} c_{1} + a_{1} c_{3} & a_{3} c_{2} + a_{2} c_{3} \\ b_{0} b_{1} & b_{0} b_{2} & b_{0} b_{3} & b_{1} b_{2} & b_{1} b_{3} & b_{2} b_{3} \\ b_{1} c_{0} + b_{0} c_{1} & b_{2} c_{0} + b_{0} c_{2} & b_{3} c_{0} + b_{0} c_{3} & b_{2} c_{1} + b_{1} c_{2} & b_{3} c_{1} + b_{1} c_{3} & b_{3} c_{2} + b_{2} c_{3} \\ c_{0} c_{1} & c_{0} c_{2} & c_{0} c_{3} & c_{1} c_{2} & c_{1} c_{3} & c_{2} c_{3} \end{array}\right)}$$

and $$\det A=(-a_{3}b_{2}c_{1}+a_{2}b_{3}c_{1}+a_{3}b_{1}c_{2}-a_{1}b_{3}c_{2}-a_{2}b_{1}c_{3}+ a_{1}b_{2}c_{3}) \\ \times (a_{3} b_{2}c_{0} - a_{2} b_{3} c_{0} - a_{3} b_{0} c_{2} + a_{0} b_{3} c_{2} + a_{2} b_{0} c_{3} - a_{0} b_{2} c_{3})\\ \times (- a_{3} b_{1} c_{0}+a_{1} b_{3} c_{0} + a_{3} b_{0}c_{1}-a_{0} b_{3} c_{1}-a_{1}b_{0} c_{3} + a_{0} b_{1} c_{3})\\ \times (a_{2} b_{1} c_{0} - a_{1} b_{2}c_{0} - a_{2} b_{0} c_{1} + a_{0} b_{2} c_{1} + a_{1} b_{0} c_{2} - a_{0} b_{1} c_{2})\\ =L_{(234)}L_{(123)}L_{(124)}L_{(134)}.$$ This (and more - namely we would like to know how $A^{-1}$ looks like) must be well-known, but we cannot find relevant references.

Update II. If one instead takes $n-d-1$-fold products of $\ell_i$, then one gets, in the same way, a $\binom{n-1}{d}\times\binom{n}{d+1}$-matrix, with determinants of $\binom{n-1}{d}\times\binom{n-1}{d}$-minors factoring into products of $L_K$ as above. More precisely, if a $\binom{n-1}{d}\times\binom{n-1}{d}$-minor $M$ misses one $\ell_i$ then one arrives to the situation outlined above, which we know how to deal with, thanks to David's answer. Otherwise, we still can see that $\det M$ is divisible by $L_K$, where $K$ is one of $d+1$-subsets of $(1,\dots,n)$ distinct from the complement of $J=(i_1,\dots,i_{n-d-1})$ in $(1,\dots,n)$, where is $J$ corresponding to a column of $M$; there are $\binom{n-1}{d+1}=\binom{n}{d+1}-\binom{n-1}{d}$ such $K$. If $L_K$ vanishes then the forms $\ell_t$ comprising its columns have a common zero $z$, and as $K\cap J\neq\emptyset$, the vector $(z^I)$ is in the left kernel of $M$. Degree count now shows that $\det M$ factors into the product of $L_K$. (Some $\det M$ vanish identically, and this apparently has to do with the homology of a simplicial complex related to the index sets $J$ of its columns).

• @David: it's my typo, I fixed it now; it should be $|J|=n-d$, not $d$. Thanks for spotting it! – Dima Pasechnik Apr 11 '12 at 18:30
• This reminds me of determinants I've seen in discussions of resultants. I don't know much, but perhaps "resultant" is a sufficient buzzword for you to find out more? – Theo Johnson-Freyd Apr 11 '12 at 18:39
• I don't understand why some entries of your example matrix $A$ have one addend while others have two. Shouldn't each one have two? Also, my bets are on your IMHO to be an IIRC ;) – darij grinberg Apr 16 '12 at 4:30
• @darij $(a_0 x+b_0 y+c_0 z) ( a_1 x+b_1 y + c_1 z) = (a_0 a_1) x^2 + (a_0 b_1 + a_1 b_0) xy + \cdots + (c_0 c_1) z^2$. The entries in the first column of $A$ are the coefficients of the RHS. – David E Speyer Apr 16 '12 at 6:21
• it looks as if vanishing of a minor is implied by nontrivial top homology of a simplicial complex constructed from the index set corresponding to the columns of the minor: each column corresponds to a subset of forms, so one needs to take the complement of this subset in the set of forms as a facet of the complex (thus one gets a facet for each column). – Dima Pasechnik Apr 16 '12 at 18:34

Here's a proof; I don't know any references. I will show that, if $L_K=0$, then $\det A=0$. Since the $L_K$ are distinct irreducible polynomials, this shows that $\prod L_K$ divides $\det A$, and the two sides have the same degree.
Suppose that $L_K=0$. Then the linear forms $\ell_{k_1}$, ..., $\ell_{k_{d+1}}$ have a common zero; call it $(z_0, z_1, \ldots, z_n)$. Every $I$ has nonempty intersection with $K$, since $|I|+|K|=(n-d)+(d+1) = n+1$. So every product $\ell_{i_1} \cdots \ell_{i_{n-d}}$ vanishes at $(z_0, z_1, \ldots, z_d)$.
That means that the vector of monomials $(z^J)$ in in the left kernel of $A$, so $\det A=0$.