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  pdftitle={A linear gap for the maximal cp-rank},
  pdfauthor={Yair Lavi}
}

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\title{A linear gap for the maximal cp-rank}
\author{Yair Lavi}
\date{28 September 2026}

\begin{document}
\maketitle

\begin{abstract}

We prove that the maximal cp-rank $p_n$ satisfies $p_n\ge n(n-3)/2$ for odd $n\ge5$ and $p_n\ge n(n-3)/2-1$ for even $n\ge6$. Together with the known upper bound $p_n\le\binom{n+1}{2}-4$, this gives $p_n=n^2/2+O(n)$.
\end{abstract}

\noindent\textbf{Keywords.} Completely positive matrix; cp-rank; copositive matrix; exposed face; cyclic incidence matrix.

\noindent\textbf{2020 Mathematics Subject Classification.} 15B48, 15A23, 05B20.

\section{Introduction}

A real symmetric $n\times n$ matrix $A$ is \textbf{completely positive} if
$A=BB^{\mathsf T}$ for an entrywise nonnegative real matrix $B$ with finitely
many columns. Its \textbf{cp-rank} $\operatorname{cpr}(A)$ is the least possible
number of columns of a matrix $B$ in such a representation; set
$\operatorname{cpr}(0)=0$. Write $\operatorname{CP}_n$ for the cone of these
matrices and

\begin{equation}
p_n=\max\{\operatorname{cpr}(A):A\in\operatorname{CP}_n\}.
\end{equation}

Write $\mathbb R_+^n=\{x\in\mathbb R^n:x_i\ge0\text{ for every }i\}$;
thus $x\ge0$ means that every coordinate of $x$ is nonnegative. A
real symmetric $n\times n$ matrix $S$ is \textbf{copositive} if
$x^{\mathsf T}Sx\ge0$ for every $x\in\mathbb R_+^n$.

Drew, Johnson, and Loewy conjectured $p_n=\lfloor n^2/4\rfloor$
for $n\ge4$ \cite{ref1}. Bomze, Schachinger, and Ullrich disproved this at
orders $7$--$11$ \cite{ref5} and at every order $n\ge12$ \cite{ref3}. The general
upper bound $p_n\le\binom{n+1}{2}-4$ for $n\ge5$ is \cite[Corollary~5.2]{ref2}.
Relative to this upper bound, the lower bounds of \cite{ref3} and
\cite[Theorem~7.4]{ref4} leave gaps of $O(n^{3/2})$ and
$O(n\log\log n)$, respectively.

We follow \cite[Sections~2--3]{ref5} and \cite[Lemmas~2.1--2.2 and Example~3.1]{ref3}
in using copositive matrices with nonzero vectors $u\ge0$ satisfying
$u^{\mathsf T}Su=0$ and in deleting the same coordinates from rows and
columns. Let $\widehat S$ be the order-seven matrix of \cite[Example~1]{ref5}.
With zero-based indices modulo $7$, the choice
$t=4\pi-7\arccos(-1/6)$ in our family gives
$S_{ij}=\widehat S_{2i,2j}/162$. Here we give
a uniform family in every odd order, prove its copositivity, classify
all such vectors up to positive scaling, and obtain explicit lower
bounds in both parities without finite-order computation.

\begin{theorem}
For every integer $n\ge5$, put
$\varepsilon_n=0$ if $n$ is odd and $\varepsilon_n=1$ if $n$ is even. Then

\begin{equation}
\boxed{\frac{n(n-3)}2-\varepsilon_n
\ \le\ p_n\ \le\ \binom{n+1}{2}-4.}
\end{equation}
\end{theorem}

Thus $p_n=\tfrac12n^2+O(n)$. The difference between the upper and lower
expressions in (1.2) is $2n-4+\varepsilon_n$. The construction gives
the lower expression on a particular subset of $\operatorname{CP}_n$;
(1.2) does not assert an exact value of $p_n$.

At $n=5$, our bound $5$ compares with $p_5=6$ \cite[Remark~5.1]{ref2}.
At $n=6$, our bound $8$ compares with $9\le p_6\le15$
\cite[Table~2]{ref3} and \cite[Theorem~6.1]{ref2}. At $n=7,13$ our bounds
match published lower bounds $14,65$ \cite[Example~1]{ref5} and \cite[Table~1]{ref3};
at $n=14,15$ they give $76,90$, below $80,95$ \cite[Table~1]{ref3}.
For $n=8,\ldots,12$, our values $19,27,34,44,53$ exceed the
published $18,26,30,37,50$ \cite[Examples~2--3]{ref5}, \cite[Table~1]{ref3}, and \cite[Table~1]{ref4}.

The proof classifies the finitely many nonnegative vectors $u\ne0$ with
$u^{\mathsf T}Su=0$ for a suitable copositive $S$, up to positive scaling.
Every nonzero factor column of a completely positive matrix orthogonal
to $S$ is a positive multiple of one of them. Lemma 2.1 shows that the
dimension of the span of their outer products is the maximum cp-rank
among these matrices; Sections 3--5 compute it in odd and even order.

\section{Finite zero sets and cp-rank}

Let $S$ be any real symmetric $n\times n$ matrix. A nonzero vector
$u\in\mathbb R_+^n$ with $u^{\mathsf T}Su=0$ is a \textbf{nonnegative zero} of
$S$. Its \textbf{support} is $\operatorname{supp}(u)=\{i:u_i>0\}$. The ray
generated by a nonnegative zero $u$ is $\{cu:c\ge0\}$; two nonnegative
zeros determine the same ray exactly when one is a positive multiple of
the other. We call such a ray a \textbf{zero ray} of $S$.

On the space of real symmetric $n\times n$ matrices, we use the
\textbf{Frobenius inner product}
$\langle S,A\rangle=\operatorname{tr}(SA)$. An \textbf{exposed face} of a cone
$K$ is its intersection with the zero set of a linear functional that
is nonnegative on $K$. If $S$ is copositive, then for every
$A=BB^{\mathsf T}\in\operatorname{CP}_n$, writing the columns of $B$ as
$b_1,\ldots,b_s$ gives
$\langle S,A\rangle=\sum_{j=1}^s b_j^{\mathsf T}Sb_j\ge0$.
Hence the following set is an exposed face of the
completely positive cone $\operatorname{CP}_n$:

\begin{equation}
F(S)=\{A\in\operatorname{CP}_n:\langle S,A\rangle=0\}.
\end{equation}

The following finite-zero principle appears in
\cite[Lemma~2.1]{ref5} and \cite[Lemmas~2.1--2.2]{ref3}.

\begin{lemma}[finite-zero face]
Suppose a copositive $S$ has exactly
$q<\infty$ nonnegative zero rays, represented by
$u^{(1)},\ldots,u^{(q)}$. If

\begin{equation}
h=\dim\operatorname{span}\{u^{(a)}u^{(a)\mathsf T}:1\le a\le q\},
\end{equation}

then the maximum cp-rank of matrices in $F(S)$ is $h$.

\end{lemma}

\begin{proof}
If $q=0$, then $F(S)=\{0\}$ and $h=0$. Otherwise, choose a
factorization $A=BB^{\mathsf T}$ for $A\in F(S)$. Copositivity makes
every summand's pairing with $S$ nonnegative, so each nonzero column of
$B$ lies on a listed zero ray. Thus $F(S)$ is generated by their outer
products. The
conic form of Carath\'eodory's theorem says that any point of a cone
spanning an $h$-dimensional space is a nonnegative combination of at
most $h$ generators, so cp-rank is at most $h$. A point of $F(S)$
outside the finitely many spans of fewer than $h$ generators has
cp-rank $h$. Such a point exists because a finite union of proper
linear subspaces cannot cover an $h$-dimensional cone.

\end{proof}

\section{A copositive family in odd order}

Let $n=2m+1\ge5$ and index coordinates by $\mathbb Z/n\mathbb Z$ in
cyclic order. We also regard index $i$ as the vertex at polar angle
$2\pi i/n$ of the unit regular $n$-gon. The \textbf{cyclic distance} $d(i,j)$
is the smaller number of steps along the polygon from $i$ to $j$, so
$1\le d(i,j)\le m$ for $i\ne j$. A \textbf{cyclic interval of $L$ steps} is
$\{i,i+1,\ldots,i+L\}$ modulo $n$, where $0\le L<n$. Fix

\begin{equation}
0<t<\frac{2\pi}{2n-3},\qquad
a=\frac{2\pi+3t}{n},\qquad
\phi_k=ak-t\quad(1\le k\le m),
\end{equation}

and define a real symmetric matrix $S$ by

\begin{equation}
S_{ii}=1,\qquad S_{ij}=\cos\phi_{d(i,j)}\quad(i\ne j).
\end{equation}

At $n=5$, the matrices in (3.2) belong, up to a permutation of
coordinates, to Hildebrand's order-five copositive family \cite{ref6}.

This matrix is \textbf{circulant}: its entries satisfy
$S_{i+h,j+h}=S_{ij}$ for every $h\in\mathbb Z/n\mathbb Z$.

The interval in (3.1) gives $0<\phi_1<\cdots<\phi_m<\pi$.
Indeed $\phi_1=(2\pi-(n-3)t)/n>0$ and
$\phi_m=\pi-\pi/n+(n-3)t/(2n)<\pi$.
The \textbf{pair angle} of $\{i,j\}$ is $\phi_{d(i,j)}$. For an index set
$I$, the \textbf{principal submatrix} on $I$ is
$S_I=(S_{ij})_{i,j\in I}$.

For three distinct vertices $v_1,v_2,v_3$ in cyclic order, choose
representatives $0\le v_1<v_2<v_3<n$. Their \textbf{cyclic gaps} are
$k=v_2-v_1$, $\ell=v_3-v_2$, and $r=n+v_1-v_3$: the positive numbers
of steps from $v_1$ to $v_2$, from $v_2$ to $v_3$, and from $v_3$ back
to $v_1$. They satisfy $k+\ell+r=n$; starting the cyclic listing at a
different vertex only permutes them cyclically. We call the triple
\textbf{surrounding} if $k,\ell,r\le m$. Then each gap is the cyclic
distance of its endpoint pair, and the three pair angles sum to
$a(k+\ell+r)-3t=an-3t=2\pi$.

For example, let $n=7$, $m=3$, $t=\pi/9$ (which satisfies (3.1)), and
therefore $a=\pi/3$.
The triple $\{0,2,4\}$ has gaps $(k,\ell,r)=(2,2,3)$, so it is
surrounding. Its pair distances are $d(0,2)=d(2,4)=2$ and $d(4,0)=3$;
the pairs $\{0,2\}$ and $\{2,4\}$ each have angle $\phi_2=5\pi/9$,
while $\{0,4\}$ has angle $\phi_3=8\pi/9$. These three angles sum to
$2\pi$.

\begin{lemma}[triangle zeros]
Let $T=\{v_1,v_2,v_3\}$ be a
surrounding triple, with its vertices listed in cyclic order and gaps
$k,\ell,r$ as above. Up to multiplication by a positive scalar, there
is exactly one nonnegative zero $u$ of $S$ with
$\operatorname{supp}(u)=T$. A representative, with all other
coordinates zero, is

\begin{equation}
u_{v_1}=\sin\phi_\ell,\qquad
u_{v_2}=\sin\phi_r,\qquad
u_{v_3}=\sin\phi_k,
\end{equation}

\end{lemma}

\begin{proof}
For the vertices in cyclic order, take planar unit vectors
$w_1,w_2,w_3$ at angles $0,\phi_k,\phi_k+\phi_\ell$.
Their \textbf{Gram matrix}, the matrix $(w_a\cdot w_b)_{a,b=1}^3$ of pairwise
inner products, is $S_T$, and
$\sin\phi_\ell w_1+\sin\phi_r w_2+\sin\phi_k w_3=0$:
the vertical coordinate is
$\sin\phi_k(\sin\phi_r+\sin(\phi_k+\phi_\ell))=0$,
and the horizontal coordinate is
$\sin\phi_\ell+\sin(\phi_k+\phi_r)=0$, using
$\phi_k+\phi_\ell+\phi_r=2\pi$.
Thus (3.3) is the unique nonnegative kernel ray of the rank-two
positive semidefinite matrix $S_T$.
\end{proof}

In the seven-vertex example, (3.3) gives
$u_0=u_4=\sin(5\pi/9)$ and $u_2=\sin(8\pi/9)$, with all other
coordinates zero.

We now prove that $S$ is copositive and has no other zeros. An \textbf{open
semicircle} means an open arc of angular length $\pi$ on the unit
circle. The next lemma concerns the regular $n$-gon defined above.

\begin{lemma}[four-vertex geometry]
A vertex set containing no
surrounding triple lies in an open semicircle, hence in a cyclic interval
of at most $m$ steps. A four-vertex set containing a surrounding triple
contains exactly two such triples. They share a pair $\{i,j\}$.

\end{lemma}

\begin{proof}
A surrounding triple is precisely a triple of polygon vertices
whose convex hull contains the origin: three points on a circle enclose
its center exactly when they do not fit in an open semicircle, equivalently
when none of their cyclic gaps exceeds $m$. If the origin were in the convex
hull of a vertex set, planar Carath\'eodory's theorem would put it in the
convex hull of at most three of those vertices. One vertex cannot
contain the origin, and a segment between two vertices could contain
it only if they were antipodal, impossible for odd $n$. Hence a set
with no surrounding triple has a convex hull omitting the origin.
A linear functional strictly positive on this convex hull places all
its vertices in an open half-plane through the origin, hence in an open
semicircle spanning at most $m$ cyclic steps.

For four vertices let $g_1,g_2,g_3,g_4$ be their cyclic gaps, summing
to $2m+1$. If any $g_i>m$, no triple is surrounding. Otherwise the
triple obtained by omitting a vertex is surrounding exactly when the
two gaps merged by that omission sum to at most $m$. The sums
$g_1+g_2$ and $g_3+g_4$ add to $2m+1$, so exactly one is at most $m$;
the same holds for $g_2+g_3$ and $g_4+g_1$. Hence exactly two triples
are surrounding. They share a pair.
\end{proof}

\begin{lemma}[overlapping triangle zeros]
If $u,v$ represent the rays of the two
surrounding triples in a four-set, then $u^{\mathsf T}Sv>0$.

\end{lemma}

\begin{proof}
Because $S$ depends only on cyclic distance, rotations and
reflections preserve it. Relabel the common pair as $\{0,k\}$ with
$1\le k\le m$. A third vertex of a surrounding triple through this
pair lies in $\{m+1,\ldots,m+k\}$. Write the third vertices of
$u,v$ as $p,q$, respectively, and set
$d=d(p,q)=|p-q|$, so $1\le d\le m-1$.
Since $1\le p-k,n-p\le m$, we have
$ap-2t=\phi_k+\phi_{p-k}=2\pi-\phi_{n-p}$, and likewise for $q$.
Choose planar unit vectors at angles
$0,\phi_k,ap-2t,aq-2t$, indexed by $0,k,p,q$, and let $G$ be their
Gram matrix. Since $an=2\pi+3t$, its entries agree with $S$ on each
surrounding triple, while $G_{pq}=\cos(ad)$. Each of $u,v$ is a
dependence among its triple's planar vectors, so $u^{\mathsf T}Gv=0$.
The only entry at which $S$ and $G$ can differ in $u^{\mathsf T}Sv$
is the pair $\{p,q\}$:

\begin{equation}
u^{\mathsf T}Sv=u_pv_q[\cos(ad-t)-\cos(ad)]
=2u_pv_q\sin(ad-t/2)\sin(t/2)>0.
\end{equation}

Here $ad-t/2>0$ follows from $a>t$, and $ad<\pi$ follows from
$d\le m-1$ and $t<2\pi/(n-3)$, a weaker consequence of (3.1).
\end{proof}

\begin{proposition}[complete zero list]
The matrix $S$ is copositive,
and its nonnegative zero rays are exactly those in Lemma 3.1.

\end{proposition}

\begin{proof}
Minimize $x^{\mathsf T}Sx$ on the simplex
$\Delta=\{x\ge0:\sum_i x_i=1\}$, and let $I$ be the support of a
minimizer. If $I$ contains no surrounding triple, Lemma 3.2 places it
in a cyclic interval $\{s,s+1,\ldots,s+L\}$ modulo $n$, with $L\le m$.
For each $i\in I$, choose the unique $r_i\in\{0,\ldots,L\}$ with
$i\equiv s+r_i\pmod n$, and let
$G_{ij}=\cos(a(r_i-r_j))$, a planar Gram matrix. For $i\ne j$ in
the interval, $d=|r_i-r_j|\le m$ and

\begin{equation}
S_{ij}-G_{ij}=2\sin(ad-t/2)\sin(t/2)>0.
\end{equation}

Indeed $a>t$ and
$am-t/2=\pi-\pi/n+(2n-3)t/(2n)<\pi$. The diagonals agree.
Since $G$ is positive semidefinite, the off-diagonal correction gives
$x^{\mathsf T}Sx>0$ for supports of size at least two;
singleton supports have value one.

If $I$ is a surrounding triple, $S_I$ is positive semidefinite by
Lemma 3.1. If $I$ contains a surrounding triple and at least one more
vertex, choose a four-set inside $I$ containing it. Let $u,v$ be its two
triangle rays, each normalized to have coordinate sum one. Then
$y=u-v$ has coordinate sum zero, and Lemma 3.3 gives
$y^{\mathsf T}Sy=-2u^{\mathsf T}Sv<0$.
Both $x+\epsilon y$ and $x-\epsilon y$ remain in $\Delta$ for small
$\epsilon>0$, but the average of their quadratic values is
$x^{\mathsf T}Sx+\epsilon^2 y^{\mathsf T}Sy<x^{\mathsf T}Sx$,
contradicting minimality. Hence the minimum is nonnegative and $S$ is
copositive.

A normalized zero minimizes on $\Delta$, so the cases force its
support to be a surrounding triple; $S_I$ then forces (3.3).
\end{proof}

\section{The odd-order face dimension}

Index the rows of $M$ and $H$ by surrounding triples $T$. Index the
columns of the zero--one matrix $M$ by unordered pairs
$e=\{i,j\}$ with $0\le i<j<n$, and set
$M_{T,e}=1$ if $e\subset T$ and $M_{T,e}=0$ otherwise. For each $T$,
let $u^{(T)}$ be the representative in (3.3), with
$u_i^{(T)}=0$ for $i\notin T$. Index the columns of $H$ by
$(i,j)$ with $0\le i\le j<n$ and define

\[
H_{T,(i,j)}=u_i^{(T)}u_j^{(T)}\qquad(0\le i\le j<n).
\]

Thus $(i,i)$ labels a diagonal coordinate, while for $i<j$ the column
$(i,j)$ of $H$ corresponds to the column $\{i,j\}$ of $M$.
The rank of $H$ is the dimension $h$ in (2.2).

\begin{lemma}[rank transfer]
$\operatorname{rank}H=\operatorname{rank}M$.

\end{lemma}

\begin{proof}
For a triple $T=\{i,j,k\}$, let
$g_T=\sin\phi_{d(i,j)}\sin\phi_{d(i,k)}\sin\phi_{d(j,k)}>0$.
For a pair $e=\{a,b\}$ write $d(e)=d(a,b)$, and let $H_{\mathrm{off}}$
be the block of $H$ with only off-diagonal columns. If $e\subset T$,
equation (3.3) gives $u_au_b\sin\phi_{d(e)}=g_T$. Thus

\begin{equation}
H_{\mathrm{off}}=\operatorname{diag}(g_T)\,M\,
\operatorname{diag}\bigl(1/\sin\phi_{d(e)}\bigr),
\end{equation}

Both diagonal factors are invertible, so this block has rank
$\operatorname{rank}M$. For every zero $u$ and vertex $i$ with
$u_i>0$, varying that coordinate slightly in both directions preserves
nonnegativity. Copositivity and $u^{\mathsf T}Su=0$ therefore give
$(Su)_i=0$. If $u_i=0$, the following equation is trivial. Hence

\begin{equation}
u_i^2=-\sum_{j\ne i}S_{ij}u_iu_j.
\end{equation}

Each diagonal column of $H$ is a fixed linear combination of its
off-diagonal columns. It adds no rank.
\end{proof}

\begin{lemma}[cyclic incidence rank]
For every odd $n=2m+1\ge5$,

\begin{equation}
\operatorname{rank}M=\binom n2-n=\frac{n(n-3)}2.
\end{equation}

\end{lemma}

\begin{proof}
Because $n$ is odd, every unordered pair has a unique form
$\{i,i+k\}$ with $1\le k\le m$. Label it by $(i,k)$ and write
$x_{i,k}$ for its coordinate.
For a surrounding triple with directed cyclic gaps $k,\ell,r\le m$
and $k+\ell+r=n$, the equation $Mx=0$ is

\begin{equation}
x_{i,k}+x_{i+k,\ell}+x_{i+k+\ell,r}=0.
\end{equation}

Over $\mathbb C$, the functions $i\mapsto z^i$ for $z^n=1$ form a
basis of functions on $\mathbb Z/n\mathbb Z$. Thus every array $x$ has
the Fourier expansion

\[
x_{i,k}=\sum_{z^n=1}z^if_k(z),\qquad
f_k(z)=\frac1n\sum_{i=0}^{n-1}z^{-i}x_{i,k}.
\]

Substitution into (4.4) and comparison of the independent functions
$i\mapsto z^i$ show that the complex kernel splits into \textbf{modes}
$x_{i,k}=z^if_k$, one for each $z$. The equations for a mode are

\begin{equation}
f_k+z^kf_\ell+z^{k+\ell}f_r=0
\qquad(k+\ell+r=n;\ 1\le k,\ell,r\le m).
\end{equation}

For $z\ne1$, subtract the equations for $(k,m+1-k,m)$ and
$(k,m,m+1-k)$; the result is (4.6).

\begin{equation}
(1-z^m)f_{m+1-k}=(1-z^{m+1-k})f_m.
\end{equation}

Since $\gcd(m,n)=1$, $z^m\ne1$. As $k$ varies, (4.6) fixes every
$f_j$ as a multiple of $1-z^j$. Conversely these values satisfy
(4.5), whose left side telescopes to $1-z^n=0$. Thus the kernel in
each nontrivial mode has dimension one.

For $z=1$, compare the equations for
$(k,m+1-k,m)$ and $(k+1,m+1-k,m-1)$,
$1\le k\le m-1$. They give
$f_{k+1}-f_k=f_m-f_{m-1}$, so $f_k=Ak+B$. Every equation (4.5) then
reduces to $nA+3B=0$. This mode also has dimension one. There are $n$
values of $z$, so $\dim_{\mathbb C}\ker M=n$. The matrix $M$ is
real, hence its real nullity is $n$ as well. With $\binom n2$ columns,
(4.3) follows.
\end{proof}

By Proposition 3.4, Lemmas 2.1, 4.1, and 4.2, the exposed face
$F(S)\subseteq\operatorname{CP}_n$ has
maximum cp-rank exactly $n(n-3)/2$. This proves the odd case of the
lower bound in Theorem 1.1.

\section{One-coordinate restriction and even order}

Let $N=2m\ge6$, let $n=N+1=2m+1$, and form the odd-order matrix
$S$ in (3.2). Delete coordinate $0$ to obtain an $N\times N$
principal submatrix $S'$. It is copositive because a nonnegative
vector on the surviving coordinates extends by zero. Such a vector is
a zero of $S'$ exactly when its zero extension is a zero of $S$.
Proposition 3.4 therefore says that the zero rays of $S'$ are precisely
the surrounding-triple rays avoiding $0$.

Let $M'$ be the submatrix of $M$ with rows for surrounding triples
avoiding $0$ and columns for pairs of surviving vertices.
The rank transfer in Lemma 4.1 applies to this
sublist: equation (4.1) restricts to its rows and surviving pairs, and
(4.2) uses only surviving coordinates. We calculate $\operatorname{rank}M'$.

\begin{proposition}[restriction rank]
For every $m\ge3$,

\begin{equation}
\operatorname{rank}M'=\binom{2m}{2}-2m-1
=\frac{N(N-3)}2-1.
\end{equation}

\end{proposition}

\begin{proof}
Divide the columns of the full odd-order matrix $M$ into
$P_0$, the pairs avoiding $0$, and $P_\bullet$, the pairs through $0$.
Divide its rows into those avoiding $0$, whose $P_0$ block is $M'$,
and the rows $R_0$ of surrounding triples through $0$. Write $U$ for
the row space of $M'$
padded by zeros on $P_\bullet$, and $W$ for the row space of $R_0$.
Since $\operatorname{rowspace}M=U+W$,

\begin{equation}
\operatorname{rank}M'=\operatorname{rank}M-\operatorname{rank}R_0
+\dim(U\cap W).
\end{equation}

A triple $\{0,j,k\}$ with $1\le j<k\le2m$ is surrounding exactly when
$j\le m<k$ and $k-j\le m$. Encode these triples by the edges of a
bipartite graph $B$ with left vertices $1,\ldots,m$ and right
vertices $m+1,\ldots,2m$, where $j$ is adjacent to
$m+1,\ldots,j+m$. Write $E(B)$ for its edge set. The graph is
connected and has

\begin{equation}
|E(B)|=\sum_{j=1}^m j=\frac{m(m+1)}2.
\end{equation}

Each row of $R_0$ contains the $P_0$ coordinate indexed by its pair
$\{j,k\}$, which occurs in no other row of $R_0$. Thus its rows are
independent and $\operatorname{rank}R_0=|E(B)|$.
For a pair $\{a,b\}$, let $b_{ab}$ be its unit coordinate vector in
the space indexed by all unordered pairs. An element of $W$ is
$\sum_{jk\in E(B)}\lambda_{jk}(b_{0j}+b_{0k}+b_{jk})$. Its
$P_\bullet$ coordinates vanish precisely when
$\sum_{f\ni v}\lambda_f=0$ at each vertex $v$ of $B$. Define
$Z(B)=\{\lambda\in\mathbb R^{E(B)}:\sum_{f\ni v}\lambda_f=0
\text{ for every vertex }v\}$, and use the same definition for
subgraphs $G$ of $B$. For an edge weighting $\lambda\in\mathbb R^{E(G)}$,
its support is $\{e\in E(G):\lambda_e\ne0\}$. Orient each edge of $G$
from left to right, and let $D_G$ be its vertex-edge matrix, with
$-1$ at the left endpoint, $+1$ at the right endpoint, and zero
elsewhere. The equations $D_G\lambda=0$ differ from the defining
vertex-sum equations only by signs, so $Z(G)=\ker D_G$ is the real
cycle space of $G$. The map
$\pi(\lambda)=\sum_{jk\in E(B)}\lambda_{jk}b_{jk}$ is injective, so
$\dim(U\cap W)$ equals the dimension of the $\lambda\in Z(B)$ for
which $\pi(\lambda)\in U$.

The edge $e_*=\{m,m+1\}$ belongs to $B$ but its column in $M'$ is
zero. Indeed no surrounding triple avoiding $0$ contains both
$m$ and $m+1$: a third vertex on either complementary arc makes one
cyclic gap exceed $m$. Hence $\pi(\lambda)\in U$ implies
$\lambda_{e_*}=0$. Let $B-e_*$ be the graph obtained by deleting
this edge, and regard $Z(B-e_*)$ as a subspace of $\mathbb R^{E(B)}$
by assigning zero to its $e_*$ coordinate. This gives the upper
inclusion

\begin{equation}
\{\lambda\in Z(B):\pi(\lambda)\in U\}\subseteq Z(B-e_*),
\end{equation}

We prove the reverse inclusion by four-cycles. For any cycle in a
bipartite graph, its \textbf{cycle vector} assigns alternating $+1$ and
$-1$ to successive edges and zero to all other edges. A four-cycle in $B$
has vertices $j<j'\le m<k<k'$ with $k'-j\le m$ and alternating
pair-coordinate vector, the image of its cycle vector under $\pi$,
$v=b_{jk}-b_{j'k}+b_{j'k'}-b_{jk'}$.
It avoids $e_*$ exactly when $j'\le m-1$ or $k\ge m+2$. In the first
case set $\ell=j'+m+1>k'$; in the second set
$\ell=k-m-1<j$. The defining inequalities of $B$ show in either
case that all four triples $\{j,k,\ell\}$,
$\{j',k,\ell\}$, $\{j',k',\ell\}$, and
$\{j,k',\ell\}$ are surrounding and avoid $0$. Indeed, for
$x\in\{j,j'\}$ and $y\in\{k,k'\}$ the cyclic gaps of
$\{x,y,\ell\}$ are, in the first case,
$y-x\le m$, $\ell-y\le m$, $n-\ell+x=m+x-j'\le m$;
in the second case they are $x-\ell\le m$, $y-x\le m$,
$n-y+\ell=m+k-y\le m$. If $T_{xyz}$
denotes the incidence row of $\{x,y,z\}$, cancellation of the pairs through
$\ell$ gives

\begin{equation}
T_{jk\ell}-T_{j'k\ell}+T_{j'k'\ell}-T_{jk'\ell}=v.
\end{equation}

Thus every four-cycle vector of $B-e_*$ maps into $U$.

These four-cycles span $Z(B-e_*)$. For any nonzero $\lambda$ in this
space, a vertex incident with exactly one edge in its support would
violate the vertex-sum equation. Thus every vertex incident with a
support edge has degree at least two there, and the finite support
subgraph contains a cycle. Subtract a multiple of that
cycle's vector to eliminate one support edge, without adding edges to
the support; finitely many repetitions show that cycle vectors span
the space. Now consider a cycle $C$ of length at least six.
Choose its largest left vertex $j_*$ other
than $m$. Every right vertex $k$ on $C$ has a cycle-neighbor
$j_a\ne m$, so $k\le j_a+m\le j_*+m$. Thus $j_*$ is adjacent in
$B-e_*$ to every right vertex of $C$; this edge cannot be $e_*$, since
$j_*\ne m$. There are at least three right vertices but only two
cycle-neighbors of $j_*$, so $C$ has a chord, an edge joining two
nonconsecutive vertices of $C$. It divides $C$ into two shorter cycles.
Choose the signs of their alternating cycle vectors so their chord
entries cancel; their sum is the cycle vector of $C$. Repeating this
split expresses that vector as a sum of four-cycle vectors. Also,
$B-e_*$ remains connected when $m\ge3$: $m$ meets
every right vertex except $m+1$, the latter meets every left vertex
except $m$, and $m,m+1$ are joined by the path
$m,(m+2),(m-1),(m+1)$. For $G=B-e_*$, the equation
$D_G^{\mathsf T}c=0$ makes the vertex labels $c$ equal across every
edge. Connectedness makes $c$ constant, so $\operatorname{rank}D_G=2m-1$
and

\begin{equation}
\dim Z(B-e_*)=\bigl(|E(B)|-1\bigr)-2m+1
=\frac{m(m-3)}2.
\end{equation}

Equations (5.4)--(5.6) show that $\dim(U\cap W)=m(m-3)/2$.
Substitute this, (5.3), and
$\operatorname{rank}M=\binom{2m+1}{2}-(2m+1)$ from Lemma 4.2 into
(5.2). The result is
$\operatorname{rank}M'=2m^2-3m-1$, which is (5.1).
\end{proof}

\begin{proof}[Proof of Theorem 1.1]
The odd lower bound follows after Lemma 4.2.
For even $n=N\ge6$, the rank transfer of Lemma 4.1 applies to the
surviving zero rays, so Proposition 5.1 and Lemma 2.1 give a matrix
in the exposed face $F(S')\subseteq\operatorname{CP}_N$ of cp-rank
$N(N-3)/2-1$. The upper bound is
\cite[Corollary~5.2]{ref2}. Subtracting the two expressions in (1.2)
gives $2n-4+\varepsilon_n$.
\end{proof}

\section*{Acknowledgments}

The proofs presented in this paper were carried out by GPT-5.6-sol,
GPT-6 Astra, Claude Fable 5, and Claude Fable 5.1, under the guidance
of the author. The author has reviewed the resulting proof arguments.
Responsibility for the final text rests with the author.


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