# Why does this matrix have zero determinant?

This curious identity arose from studying reductions of the maximal ideal in certain monomial algebra. It can be proved "by hand", (i.e, using Macaulay 2), but I am seeking a more conceptual understanding and related references if they exist.

Let $$R$$ be a commutative ring. For two vectors $$v=(a,b,c,d), w=(A,B,C,D)\in R^4$$, we define $$v\star w:= (aA,aB+bA,bB, cC,cD+dC,dD)\in R^6$$. Given any 3 vectors $$v_1,v_2,v_3\in R^4$$, we can form a $$6\times 6$$ matrix $$M$$ whose rows are $$v_i\star v_j$$, $$1\leq i,j\leq 3$$. Then: $$\det(M)=0$$

It is not clear to me how to explain this. The kernel of $$M$$ is a column of degree $$6$$ polynomials, so the relations are quite complicated.

Question: Is there a way to conceptually explain the vanishing of $$\det(M)$$? Have you seen similar identities?

Three vectors $$v_1,v_2,v_3$$ lie in a hyperplane $$H:\alpha x+\beta y+\gamma z+\delta t=0$$, in this plane we have $$Q(v,v):=(\alpha x+\beta y)^2-(\gamma z+\delta t)^2=0,\forall v\in H$$. Thus by polarization $$Q(v,w)=\frac 14 (Q(v+w,v+w)-Q(v-w,v-w))=0$$ for all $$v,w\in H$$ that yields a relation between columns of your matrix: if $$v=(a,b,c,d), w=(A,B,C,D)$$, then $$Q(v,w)=\alpha^2 aA+\beta^2 bB+\alpha\beta(aB+bA)-\gamma^2 cC-\delta^2 dD-\gamma\delta(cD+dC)$$.
• So in other words you are saying that the vector $(\alpha^2,\alpha\beta,\beta^2,-\gamma^2,-\gamma\delta,-\delta^2)$ is in the kernel of the matrix. Since the $\alpha,\beta,\gamma,\delta$ are cubic in the vectors $v_1,v_2,v_3$, this explains the degree $6$ polynomials. – Zach Teitler Oct 27 '19 at 8:40