Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).

share|improve this question
    
@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
1  
@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 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
show 1 more comment

1 Answer

up vote 12 down vote accepted

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 $\det A=0$.

share|improve this answer
    
Thank you. We totally overlooked this... –  Dima Pasechnik Apr 12 '12 at 1:22
    
I added an update to the question to pose a more general question. –  Dima Pasechnik Apr 16 '12 at 3:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.