Rank-four principal-minor conjecture for scaling matrices

Let $C$ be a complex symmetric $d\times d$ scaling matrix of rank $4$, so $C_{ii}=0$ and $C_{ij}\neq0$ for $i\neq j$. Choose a symmetric rank factorization

$$C=XQX^{\mathsf T},$$

where $X\in\mathbb C^{d\times4}$ has full column rank and $Q\in\mathbb C^{4\times4}$ is symmetric and invertible. For every four-element subset $I$,

$$\det C[I,I]=\det(Q)\det(X[I,:])^2. \tag{1}$$

Hence the nonvanishing principal four-minors are exactly the bases of the rank-four row matroid of $X$.

For every triple $\{i,j,k\}$,

$$\det C[\{i,j,k\},\{i,j,k\}] =2C_{ij}C_{ik}C_{jk}\neq0.$$

Thus every three rows of $X$ are independent, and the represented matroid is paving.

We use the paving-matroid base theorem proved by Chayim Lowen: a rank-$r$ paving matroid on $d$ elements has at least

$$\binom{d-1}{r-1}$$

bases, with equality exactly for $U_{r-1,d-1}\oplus U_{1,1}$. See the proof at https://mathoverflow.net/questions/412516/lower-bounds-for-the-number-of-bases-of-a-paving-matroid.

Applying this theorem with $r=4$ and using (1) gives at least

$$\binom{d-1}{3}$$

nonzero principal four-minors. Equality holds exactly when some index $\alpha$ is a coloop and the remaining $d-1$ indices form $U_{3,d-1}$, or equivalently exactly when the nonvanishing principal four-minors are those indexed by four-subsets containing $\alpha$.

All principal two- and three-minors are nonzero, all principal one-minors vanish, and every principal minor of order at least five vanishes. Consequently the equality-case delta-matroid is precisely

$$\mathcal F_\alpha =\{\varnothing\}\cup\binom{[d]}2\cup\binom{[d]}3 \cup\{I\in\binom{[d]}4:\alpha\in I\}.$$

Finally, equality is attained for every $d\geq4$. Put $m=d-1$, let

D_{ij}=(i-j)^2\quad(1\leq i,j\leq m),\qquad u=D\mathbf1,\qquad C=\begin{pmatrix}D&u\u^{\mathsf T}&0\end{pmatrix}.

Every off-diagonal entry of $C$ is strictly positive. Writing $t=(1,\ldots,m)^{\mathsf T}$, we have

$$D=t^{\circ2}\mathbf1^{\mathsf T} -2tt^{\mathsf T} +\mathbf1(t^{\circ2})^{\mathsf T},$$

so $\operatorname{rank}D\leq3$. Every three-element principal minor equals

$$2(i-j)^2(i-k)^2(j-k)^2\neq0,$$

and therefore $\operatorname{rank}D=3$. Because $u=D\mathbf1$, block congruence gives

$$C\sim\operatorname{diag} \left(D,-\mathbf1^{\mathsf T}D\mathbf1\right),$$

and $\mathbf1^{\mathsf T}D\mathbf1>0$; hence $\operatorname{rank}C=4$. The first $m$ indices have matroid rank three, every triple among them is independent, and the final index is a coloop. Thus exactly

$$\binom{d-1}{3}$$

principal four-minors are nonzero. This proves both the sharp bound and its complete equality characterization.