Sharp Self-Chollet Inequalities for Rank-Two Correlation Matrices of Order Five

1. Introduction

For an $n\times n$ matrix $A=\left(a_{ij}\right)$, write

$$perA=\sum _{\sigma \in S_n}{\displaystyle \prod _{i=1}^n}{a}_{i,\sigma \left(i\right)}.$$

Chollet asked in 1982 whether

$$per(A\circ B)\le per\left(A\right)per\left(B\right)$$

(1)

holds for positive semidefinite Hermitian matrices A and B [1]. Here ∘ denotes the Hadamard product. In the real symmetric setting it is enough to prove the self-conjugate inequality

$$per(A\circ A)\le per{\left(A\right)}^2.$$

(2)

The conjecture was proved in orders at most three by Gregorac and Hentzel [2]. Hutchinson proved the real order-four case [3], and Rodtes subsequently proved the full complex order-four case [4]. Recent work treats several annular and order-five ridge classes [5], but the unrestricted order-five problem remains open. Zhang’s analytic study of maximizing correlation matrices provides additional structural background [6].

Rank-two correlation matrices are a natural first boundary stratum. Pate determined their sharp permanent lower bound [7]; Drury later recorded the related question whether $per(C\circ C)\le per\left(C\right)$ holds for every real rank-two correlation matrix [8]. The present paper concerns instead the self-Chollet ratio

$$\mathcal{R}\left(C\right)={\displaystyle \frac{per(C\circ C)}{per{\left(C\right)}^2}}$$

and determines its exact maximum in order five.

Our main result is the following.

Theorem 1. 

Let $C\in {\mathbb{R}}^{5\times 5}$ be a correlation matrix with $rankC\le 2$. Then

$$per(C\circ C)\le {\displaystyle \frac{137}{1440}}per{\left(C\right)}^2.$$

(3)

The constant is best possible. Equality holds if and only if, after permuting the indices and changing signs of Gram vectors, the five unoriented Gram lines form a regular pentagon in ${\mathbb{RP}}^1$. Equivalently, an equality representative is

$$C_{jk}=cos{\displaystyle \frac{(j-k)\pi }{5}},0\le j,k\le 4.$$

(4)

Since $137/1440<1$, Theorem 1 proves a strict strengthening of (2) on this stratum. It also applies, after diagonal normalization, to arbitrary real positive semidefinite $5\times 5$ matrices of rank at most two; zero diagonal entries give zero rows and are trivial.

Our second result removes a full-dimensional neighborhood of the identity from any order-five counterexample search.

Theorem 2. 

Let $C\in {\mathbb{R}}^{5\times 5}$ be a correlation matrix. If ${\lambda }_{min}\left(C\right)\ge 3/4$, then

$$per(C\circ C)\le per{\left(C\right)}^2.$$

Equality in this spectral region holds only for $C=I$.

The paper is organized as follows. Section 2 records the normalization, a symmetric-tensor identity, and a two-phase minimization lemma. Section 3, Section 4 and Section 5 prove Theorem 1. Section 6 proves Theorem 2. The exact finite checks used to audit the displayed expansions are described in Section 7.

2. Preliminaries

2.1. Reduction to Correlation Matrices

Suppose first that $A⪰0$ has positive diagonal, and put

$$D=diag(a_{11}^{-1/2},\dots ,a_{55}^{-1/2}),C=DAD.$$

Then C is a correlation matrix. For a diagonal matrix E,

$$per\left(EME\right)=\left(\prod _i{e}_i^2\right)perM.$$

(5)

Applying this once to A and once to $A\circ A$ shows that $\mathcal{R}$ is invariant under positive diagonal congruence. If $a_{ii}=0$, positive semidefiniteness forces the ith row and column to vanish. This proves the claimed reduction, including all boundary cases.

2.2. A Symmetric-Tensor Permanent Identity

Let $v_1,\dots ,v_n\in {\mathbb{C}}^d$, and write ${\ell }_j\left(z\right)={\sum }_{r=1}^d{v}_{jr}{z}_r$. For a multi-index $\alpha =({\alpha }_1,\dots ,{\alpha }_d)$ with $\left|\alpha \right|=n$, set $\left({\displaystyle \genfrac{}{}{0pt}{}{n}{\alpha }}\right)=n!/({\alpha }_1!\dots {\alpha }_d!)$.

Lemma 1. 

If $G={\left(\langle v_i,v_j\rangle \right)}_{i,j=1}^n$, then

$$perG=n!\sum _{\left|\alpha \right|=n}{\displaystyle \frac{{\left|\left[z^{\alpha }\right]{\prod }_{j=1}^n{\ell }_j\left(z\right)\right|}^2}{\left(\genfrac{}{}{0pt}{}{n}{\alpha }\right)}}.$$

(6)

Proof. 

Let ${\mathsf{P}}_{\mathrm{sym}}$ be the orthogonal projection from ${\left({\mathbb{C}}^d\right)}^{\otimes n}$ onto its symmetric subspace. Directly from the definition,

$$perG=n!{∥{\mathsf{P}}_{\mathrm{sym}}(v_1\otimes \dots \otimes v_n)∥}^2.$$

Group the standard tensor basis by multiplicity vector $\alpha$. The normalized symmetrization of a word of type $\alpha$ has norm one, while the coefficient of that vector is the coefficient of $z^{\alpha }$ in ${\prod }_j{\ell }_j\left(z\right)$ divided by $\sqrt{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{\alpha }}\right)}$. Summing the squared coordinates gives (6). □

2.3. A Two-Phase Minimum

Lemma 2. 

For $A,B,C\ge 0$, the minimum of

$$Acos\varphi +Bcos\psi +Ccos(\varphi -\psi )$$

over $(\varphi ,\psi )\in {\mathbb{R}}^2$ is one of

$$B-A-C,-B-|A-C|,$$

(7)

or, when

$$|A-C|\le {\displaystyle \frac{AC}{B}}\le A+C,$$

(8)

the interior value

$$-{\displaystyle \frac{A^2{B}^2+B^2{C}^2+C^2{A}^2}{2ABC}}.$$

(9)

The assertion at a zero coefficient is interpreted by continuity.

Proof. 

Fix $\psi$. Minimizing in $\varphi$ gives

$$Bcos\psi -\sqrt{A^2+C^2+2ACcos\psi }.$$

Thus it remains to minimize the convex function

$$h\left(z\right)=Bz-\sqrt{A^2+C^2+2ACz},-1\le z\le 1.$$

The two endpoint values are (7). An interior critical point satisfies

$$\sqrt{A^2+C^2+2ACz}={\displaystyle \frac{AC}{B}}.$$

It exists precisely under (8), and substitution gives (9). □

3. The Rank-Two Reduction

Choose unit Gram vectors

$$x_j=(cos{\theta }_j,sin{\theta }_j),1\le j\le 5,$$

and put

$$\begin{array}{cccc}\hfill q_j& =e^{2i{\theta }_j},\hfill & \hfill a& =e_1(q_1,\dots ,q_5),\hfill \\ \hfill b& =e_2(q_1,\dots ,q_5),\hfill & \hfill \epsilon & =e_5(q_1,\dots ,q_5).\hfill \end{array}$$

Here $e_k$ denotes the kth elementary symmetric function. Let

$$u={\left|a\right|}^2,v={\left|b\right|}^2,S=10+2u+v.$$

In the complex orthonormal basis of ${\mathbb{C}}^2$ adapted to rotations, the coordinate polynomial of $x_j$ is, up to an irrelevant common convention,

$$2^{-1/2}(q_j^{-1/2}X+q_j^{1/2}Z).$$

Lemma 1, together with $|e_4|=|e_1|$ and $|e_3|=|e_2|$, gives

$$P:=perC={\displaystyle \frac{120}{32}}{\displaystyle \sum _{k=0}^5}{\displaystyle \frac{|e_k{\left(q\right)|}^2}{\left(\genfrac{}{}{0pt}{}{5}{k}\right)}}={\displaystyle \frac{3}{4}}(10+2u+v)={\displaystyle \frac{3}{4}}S.$$

(10)

For the Hadamard square, use the Gram vectors $x_j\otimes x_j$ in ${Sym}^2\left({\mathbb{C}}^2\right)$. Their coordinate polynomials in an orthonormal basis are

$$L_j(X,Y,Z)={\displaystyle \frac{q_j^{-1}}{2}}X+{\displaystyle \frac{1}{\sqrt{2}}}Y+{\displaystyle \frac{q_j}{2}}Z.$$

Consequently Lemma 1 gives the directly checkable finite sum

$$Q:=per(C\circ C)=120\sum _{r+s+t=5}{\displaystyle \frac{{\left|\left[X^r{Y}^s{Z}^t\right]{\prod }_{j=1}^5{L}_j(X,Y,Z)\right|}^2}{5!/(r!s!t!)}}.$$

(11)

Using

$$\overline{e_1}={\displaystyle \frac{e_4}{\epsilon }},\overline{e_2}={\displaystyle \frac{e_3}{\epsilon }},$$

and collecting the twenty-one terms in (11) yields

$$Q={\displaystyle \frac{N}{128}},$$

(12)

where

$$\begin{array}{cc}\hfill N=& 685+94u+97v+21u^2+13uv+4v^2\hfill \\ \hfill & -80Re\left(a^2\overline{b}\right)-6Re\left(\overline{\epsilon }a{b}^2\right)-12Re\left(\epsilon a{\overline{b}}^{,3}\right).\hfill \end{array}$$

(13)

Equations (11) and (13) provide two independent descriptions of the same finite expansion.

By (10)–(12), inequality (3) is equivalent to

$$G:=137S^2-20N\ge 0.$$

(14)

Rotate all $q_j$ by a common unimodular factor so that $\epsilon =1$. Write

$$a=x{e}^{i\alpha },b=y{e}^{i\beta },x,y\ge 0,$$

and set

$$\varphi =2\alpha -\beta ,\psi =\alpha +2\beta .$$

Since $\varphi -\psi =\alpha -3\beta$, expansion of (14) gives

$$\begin{array}{cc}\hfill G=& 3600x^2+800y^2+128x^4+288x^2{y}^2+57y^4\hfill \\ \hfill & +1600x^{2}ycos\varphi +120x{y}^{2}cos\psi +240x{y}^{3}cos(\varphi -\psi ).\hfill \end{array}$$

(15)

4. Positivity of the Phase Minimum

Apply Lemma 2 to

$$A=1600x^{2}y,B=120x{y}^2,C=240x{y}^3,$$

(16)

and introduce the nonnegative variables

$$X={\displaystyle \frac{x}{5}},Y={\displaystyle \frac{y}{10}}.$$

(17)

We verify nonnegativity on each of the three possible branches.

4.1. The Endpoint $B-A-C$

After division by 10000, expression (15) becomes

$$\begin{array}{cc}\hfill H_1=& 8X^4+72X^2{Y}^2-40X^{2}Y+9X^2-120X{Y}^3\hfill \\ \hfill & +6X{Y}^2+57Y^4+8Y^2.\hfill \end{array}$$

The identity

$$\begin{array}{cc}\hfill H_1=& 8{(X^2-Y)}^2+{\displaystyle \frac{6}{19}}{(-2XY-3Y^2+5X)}^2\hfill \\ \hfill & +{\displaystyle \frac{21}{19}}{(-8XY+7Y^2+X)}^2\hfill \end{array}$$

(18)

proves $H_1\ge 0$.

4.2. The Endpoint $-B-|A-C|$

For $A\ge C$ and $C\ge A$, respectively, the two polynomials obtained after division by 10000 are $z^{\mathsf{T}}{M}_{-}z$ and $z^{\mathsf{T}}{M}_{+}z$, where

$$z={(X^2,XY,Y^2,X,Y)}^{\mathsf{T}},$$

$$M_{-}=\left(\begin{array}{ccccc}8& 0& -6& 0& -2\\ 0& 84& 60& -18& 6\\ -6& 60& 57& -9& 0\\ 0& -18& -9& 9& 0\\ -2& 6& 0& 0& 8\end{array}\right),$$

(19)

and

$$M_{+}=\left(\begin{array}{ccccc}8& 0& -11& 0& -3\\ 0& 94& -60& 23& 11\\ -11& -60& 57& -14& 0\\ 0& 23& -14& 9& 0\\ -3& 11& 0& 0& 8\end{array}\right).$$

(20)

The successive leading principal minors are

$$(8,672,6480,23328,0),(8,752,2690,8723,0).$$

(21)

For either matrix the leading $4\times 4$ block is therefore positive definite, and the determinant-zero Schur complement is zero. Hence both matrices are positive semidefinite. Their kernels are, respectively,

$$ker{M}_{-}=span{(1,-1,1,-1,1)}^{\mathsf{T}},ker{M}_{+}=span{(-1,-1,-1,1,1)}^{\mathsf{T}}.$$

(22)

4.3. The Interior Phase Minimum

The phase-dependent contribution in (15) becomes

$$-400x^2-1600x^2{y}^2-9y^4.$$

After division by 80000, it remains to prove

$$H_0=X^4-41X^2{Y}^2+6Y^4+X^2+Y^2\ge 0.$$

(23)

The upper inequality in (8) is, after (16)–(17), precisely

$$20XY\le X+3Y^2.$$

(24)

The exact remainder identity

$$\begin{array}{cc}\hfill H_0=& X^4+Y^2+{\displaystyle \frac{1}{400}}\left(359X^2-246X{Y}^2+2031Y^4\right)\hfill \\ \hfill & -{\displaystyle \frac{41}{400}}(20XY-X-3Y^2)(20XY+X+3Y^2)\hfill \end{array}$$

(25)

shows that the last line is nonnegative under (24). The quadratic form in $(X,Y^2)$ on the first line is positive definite because

$$359·2031-{123}^2=714000>0.$$

(26)

Thus $H_0\ge 0$. This completes the proof of (14) and hence the inequality in Theorem 1.

5. Equality

We now track equality in the preceding certificates. In the interior branch, (25) and (26) force $X=Y=0$. On the two $-B-|A-C|$ branches, (22) and the nonnegativity of all five entries of z again force $X=Y=0$.

For the remaining endpoint, the three squares in (18) vanish simultaneously. The first gives $Y=X^2$. The second then reduces to

$$X(3X^3+2X^2-5)=X(X-1)(3X^2+5X+5)=0.$$

Thus the only zeros in the nonnegative quadrant are $(0,0)$ and $(1,1)$. The latter would require $\left|a\right|=5$, $\left|b\right|=10$ and $arg\left(a^2\overline{b}\right)=\pi$. But $\left|a\right|=5$ forces equality in the triangle inequality for $a={\sum }_j{q}_j$, hence all $q_j$ coincide. Then $a=5q$, $b=10q^2$, and $a^2\overline{b}=250>0$, a contradiction.

Therefore equality is equivalent to

$$a=b=0.$$

(27)

Since the $q_j$ are unimodular, $e_4=\epsilon \overline{e_1}$ and $e_3=\epsilon \overline{e_2}$. Hence (27) gives

$${\displaystyle \prod _{j=1}^5}(z-q_j)=z^5-\epsilon .$$

The $q_j$ are therefore a rotated regular pentagon. Since $q_j=e^{2i{\theta }_j}$ records the unoriented Gram line, this is exactly the equality classification in Theorem 1. Conversely, for the regular configuration,

$$perC={\displaystyle \frac{15}{2}},per(C\circ C)={\displaystyle \frac{685}{128}},\mathcal{R}\left(C\right)={\displaystyle \frac{137}{1440}}.$$

(28)

The proof of Theorem 1 is complete.

6. A Uniform Spectral Shell

We prove Theorem 2. If $C\ne I$, write

$$C=(1-s)I+sG,s=1-{\lambda }_{min}\left(C\right)\in (0,1/4],$$

(29)

where $G=\left(g_{ij}\right)$ is a singular correlation matrix. Put $H=G-I$ and

$$T=\sum _{1\le i<j\le 5}{g}_{ij}^2.$$

(30)

The cycle-cover expansion is

$$per(I+sH)=1+p_2{s}^2+p_3{s}^3+p_4{s}^4+p_5{s}^5.$$

(31)

There are 10 transpositions, 20 oriented 3-cycles, 15 products of two transpositions, 30 oriented 4-cycles, 20 covers of type $(2,3)$, and 24 oriented 5-cycles. Applying AM–GM to the absolute edge weights, and using $|g_{ij}|\le 1$, gives

$$p_2=T,|p_3|\le 2T,|p_4|\le {\displaystyle \frac{9}{2}}T,\left|p_5\right|\le {\displaystyle \frac{22}{5}}T.$$

(32)

For completeness, the constants can be read edgewise. Every edge lies in three triangles. Among unoriented 4-cycles it appears six times, and among pairs of disjoint transpositions it appears three times. In the $(2,3)$ covers it appears once as the doubled transposition edge and three times as a triangle edge; in the twelve unoriented Hamilton cycles it appears six times. These incidences give exactly (32).

The off-diagonal entries of $C\circ C$ are $s^2{g}_{ij}^2$. Thus

$$per(C\circ C)=1+q_2{s}^4+q_3{s}^6+q_4{s}^8+q_5{s}^{10},$$

(33)

where all $q_j\ge 0$. Applying the same estimates to $g_{ij}^2$ and using ${\sum }_{i<j}{g}_{ij}^4\le T$ gives

$$q_2\le T,q_3\le 2T,q_4\le {\displaystyle \frac{9}{2}}T,q_5\le {\displaystyle \frac{22}{5}}T.$$

(34)

Let $E=p_2{s}^2+p_3{s}^3+p_4{s}^4+p_5{s}^5$. Since $E^2\ge 0$, equations (32)–(34) yield

$$\begin{array}{cc}\hfill per{\left(C\right)}^2-per(C\circ C)& =2E+E^2-(q_2{s}^4+q_3{s}^6+q_4{s}^8+q_5{s}^{10})\hfill \\ \hfill & \ge T{s}^{2}f\left(s\right),\hfill \end{array}$$

(35)

where

$$f\left(s\right)=2-4s-10s^2-{\displaystyle \frac{44}{5}}{s}^3-2s^4-{\displaystyle \frac{9}{2}}{s}^6-{\displaystyle \frac{22}{5}}{s}^8.$$

(36)

The polynomial f is strictly decreasing on $[0,\infty )$, and

$$f(1/4)={\displaystyle \frac{37441}{163840}}>0.$$

(37)

If $C\ne I$, then $T>0$; hence (35) is strict. The identity matrix gives equality, proving Theorem 2.

7. Computational Reproducibility

The proofs above are analytic; no numerical optimization is used in either theorem. A short exact-arithmetic script performs the following independent finite checks:

• it recomputes (10) directly from the $5!$ permanent terms and recomputes (13) from (11);
• it derives the three phase-branch polynomials from (15), verifies the sum-of-squares identity (18), and checks the exact $LD{L}^{\mathsf{T}}$ data and kernels for (19)–(20);
• it verifies the remainder identity (25), the cycle counts in Section 6, and the rational value (37).

The scripts use only symbolic integers and rational numbers. Source code, tests, search manifests, and file hashes accompany the manuscript.