Rational factorizations of completely positive matrices

TL;DR

This paper proves that any rational matrix inside the cone of completely positive matrices can be factored into rational matrices, ensuring rational cp-factorizations exist within the interior of this cone.

Contribution

It establishes that all rational matrices in the interior of the completely positive cone admit rational cp-factorizations, filling a gap in the understanding of matrix factorizations.

Findings

Rational matrices in the interior have rational cp-factorizations

The result applies to all matrices within the interior of the cone

Provides a constructive proof for rational factorizations

Abstract

In this note it is proved that every rational matrix which lies in the interior of the cone of completely positive matrices also has a rational cp-factorization.

Equations

$${\mathcal{C}\mathcal{P}}_n=\mathrm{cone}\{x{x}^{\mathsf{T}}:x\in {\mathbb{R}}_{\ge 0}^n\}.$$

$$A=\sum _{i=1}^m{\alpha }_i{x}_i{x}_i^{\mathsf{T}}\text{with }{\alpha }_i\ge 0\text{ and }{x}_i\in {\mathbb{R}}_{\ge 0}^n,\text{for }i=1,\dots ,m.$$

$$\left|{\alpha }_i-\frac{p_i}{q}\right|\le \frac{\epsilon }{q}\text{for all }i=1,\dots ,n.$$

$$\mathrm{int}(K)=\{x\in E:⟨x,y⟩>0\text{ for all }y\in K^{\ast }\setminus \{0\}\},$$

$$K^{\ast }=(\mathrm{cl}(K){)}^{\ast },$$

$${\mathcal{C}\mathcal{O}\mathcal{P}}_n={\mathcal{C}\mathcal{P}}_n^{\ast }=\{B\in {\mathcal{S}}^n:⟨A,B⟩\ge 0\text{ for all }A\in \mathcal{C}\mathcal{P}\}.$$

$$\mathrm{int}({\mathcal{C}\mathcal{O}\mathcal{P}}_n)=\{B\in {\mathcal{S}}^n:⟨B,x{x}^{\mathsf{T}}⟩>0\text{ for all }x\in {\mathbb{R}}_{\ge 0}^n\setminus \{0\}\}.$$

$${\widetilde{\mathcal{C}\mathcal{P}}}_n=\mathrm{cone}\{v{v}^{\mathsf{T}}:v\in {\mathbb{Z}}_{\ge 0}^n\}.$$

$$\mathcal{R}=\{B\in {\mathcal{S}}^n:⟨B,v{v}^{\mathsf{T}}⟩\ge 1\text{ for all }v\in {\mathbb{Z}}_{\ge 0}^n\setminus \{0\}\},$$

$$0\le \frac{1}{\epsilon }(x+\epsilon y{)}^{\mathsf{T}}B(x+\epsilon y)=2x^{\mathsf{T}}By+\epsilon y^{\mathsf{T}}By$$

$$0\le \frac{1}{\epsilon }(x-\epsilon y{)}^{\mathsf{T}}B(x-\epsilon y)=-2x^{\mathsf{T}}By+\epsilon y^{\mathsf{T}}By$$

$$(\alpha x+y{)}^{\mathsf{T}}B(\alpha x+y)=y^{\mathsf{T}}By\text{for }\alpha \in \mathbb{R}\text{ and }y\in {\mathbb{R}}^n.$$

$$y=qx-p\text{where}\Vert y{\Vert }_{\infty }\le \epsilon .$$

$$1\le ⟨B,p{p}^{\mathsf{T}}⟩=(qx-y{)}^{\mathsf{T}}B(qx-y)=y^{\mathsf{T}}By\le C\Vert y{\Vert }_{\infty }^2\le C{\epsilon }^2,$$

$$\mathcal{P}(A,\lambda )=\{B\in {\mathcal{S}}^n:⟨A,B⟩\le \lambda ,⟨B,v{v}^{\mathsf{T}}⟩\ge 1\text{ for all }v\in {\mathbb{Z}}_{\ge 0}^n\setminus \{0\}\}$$

$${\widetilde{\mathcal{C}\mathcal{P}}}_n^{\ast }=(\mathrm{cl}({\widetilde{\mathcal{C}\mathcal{P}}}_n){)}^{\ast }={\mathcal{C}\mathcal{P}}_n^{\ast }={\mathcal{C}\mathcal{O}\mathcal{P}}_n,$$

$$1=⟨B_k,v_k{v}_k^{\mathsf{T}}⟩=\Vert v_k{\Vert }^2⟨B_k,u_k{u}_k^{\mathsf{T}}⟩.$$

$$A\in \mathrm{cone}\{v_i{v}_i^{\mathsf{T}}:i=1,\dots ,m\}.$$

$$⟨C,A⟩<0\text{and}⟨C,v_i{v}_i^{\mathsf{T}}⟩\ge 0\text{ for all }i=1,\dots ,m.$$

$$B^{\ast }+\mu C\in \mathcal{P}(A,\lambda )\text{but}⟨B^{\ast }+\mu C,A⟩<⟨B^{\ast },A⟩,$$

Full text

1. Introduction

The cone of completely positive matrices is central to copositive programming, see [3] and also to several topics in matrix theory, see [1]. However, so far, this cone is quite mysterious, many basic questions about it are open. In [2] Berman, Dür, and Shaked-Monderer ask: Given a matrix $A\in\mathcal{CP}_{n}$ all of whose entries are integral, does $A$ always have a rational cp-factorization?

The cone of completely positive matrices is defined as the convex cone spanned by symmetric rank-$1$-matrices $xx^{\sf T}$ where $x$ lies in the nonnegative orthant $\mathbb{R}^{n}_{\geq 0}$:

[TABLE]

A cp-factorization of a matrix $A$ is a factorization of the form

[TABLE]

We talk about a rational cp-factorization when the $\alpha_{i}$’s are rational numbers and when the $x_{i}$’s are rational vectors. Of course, in a rational cp-factorization we can assume that the $x_{i}$’s are integral vectors.

In this note we prove the following theorem:

Theorem 1.1**.**

Every rational matrix which lies in the interior of the cone of completely positive matrices has a rational cp-factorization.

So to fully answer the question of Berman, Dür, and Shaked-Monderer, it remains to consider the boundary of $\mathcal{CP}_{n}$.

2. Proof of Theorem 1.1

For the proof we will need a classical result from simultaneous Diophantine approximation, a theorem of Dirichlet, which we state here. One can find a proof of Dirichlet’s theorem for example in the book [4, Theorem 5.2.1] of Grötschel, Lovász, and Schrijver.

Theorem 2.1**.**

Let $\alpha_{1},\ldots,\alpha_{n}$ be real numbers and let $\varepsilon$ be a real number with $0<\varepsilon<1$. Then there exist integers $p_{1},\ldots,p_{n}$ and a natural number $q$ with $1\leq q\leq\varepsilon^{-n}$ such that

[TABLE]

The next lemma collects standard, easy-to-prove facts about convex cones. Let $E$ be a Euclidean space with inner product $\langle\cdot,\cdot\rangle$. Let $K\subseteq E$ be a proper convex cone, which means that $K$ is closed, has a nonempty interior, and satisfies $K\cap(-K)=\{0\}$. Its dual cone is defined as $K^{*}=\{y\in E:\langle x,y\rangle\geq 0\text{ for all }x\in K\}$.

Lemma 2.2**.**

Let $K\subseteq E$ be a proper convex cone. Then,

[TABLE]

where $\operatorname{int}(K)$ is the topological interior of $K$, and

[TABLE]

where $\operatorname{cl}(K)$ is the topological closure of $K$.

We need some more notation: With $\mathcal{S}^{n}$ we denote the vector space of symmetric matrices with $n$ rows and $n$ columns which is a Euclidean space with inner product $\langle A,B\rangle=\operatorname{Trace}(AB)=\sum_{i,j=1}^{n}A_{ij}B_{ij}$. The cone of copositive matrices is the dual cone of $\mathcal{CP}_{n}$:

[TABLE]

Its interior equals

[TABLE]

We also define the following rational subcone of $\mathcal{CP}_{n}$:

[TABLE]

We prepare the proof of the paper’s main result by two lemmata which might be useful facts themselves.

Lemma 2.3**.**

The set

[TABLE]

is contained in the interior of the cone of copositive matrices $\mathcal{COP}_{n}$.

Proof.

Since the set of nonnegative rational vectors $\mathbb{Q}^{n}_{\geq 0}$ lies dense in the nonnegative orthant $\mathbb{R}_{\geq 0}^{n}$, we have the inclusion $\mathcal{R}\subseteq\mathcal{COP}_{n}$. Suppose for contradiction that the set on the left is not contained in $\operatorname{int}(\mathcal{COP}_{n})$: There is a matrix $B$ with $\langle B,vv^{\sf T}\rangle\geq 1$ for all $v\in\mathbb{Z}^{n}_{\geq 0}\setminus\{0\}$ and there is a nonzero vector $x\in\mathbb{R}^{n}_{\geq 0}$ with $\langle B,xx^{\sf T}\rangle=0$.

By induction on $n$ (and reordering if necessary) we may assume that all entries of $x$ are strictly positive, $x_{i}>0$ for all $i=1,\ldots,n$, since otherwise, we can reduce the situation to the case of smaller dimension by considering a suitable submatrix of $B$.

Hence, the vector $x$ lies in the interior of the nonnegative orthant. Therefore, and because $B\in\mathcal{COP}_{n}$, we have for every vector $y\in\mathbb{R}^{n}$ and $\varepsilon>0$ sufficiently small the inequality

[TABLE]

and similarly

[TABLE]

From this, equality $x^{\mathsf{T}}B=0$ follows. From this, we also see that $B$ is positive semidefinite. This implies that

[TABLE]

We apply Dirichlet’s approximation theorem, Theorem 2.1 to the vector $x$ and to $\varepsilon\in(0,1)$. We obtain a vector $p=(p_{1},\ldots,p_{n})$ and a natural number $q$. Since $x_{i}>0$ we may without loss of generality assume that $p_{i}\geq 0$. Thus, by the assumption $B\in\mathcal{R}$, we have $\langle B,pp^{\sf T}\rangle\geq 1$.

Define

[TABLE]

Since $B$ is positive semidefinite, there is a constant $C$ such that $y^{\sf T}By\leq C\|y\|_{\infty}^{2}$ for all $y\in\mathbb{R}^{n}$. Putting everything together we get

[TABLE]

which yields a contradiction for small enough values of $\varepsilon$. ∎

Lemma 2.4**.**

Let $A$ be a completely positive matrix which lies in the interior of $\mathcal{CP}_{n}$ and let $\lambda$ be a sufficiently large positive real number. Then the set

[TABLE]

is a full-dimensional polytope.

Proof.

For sufficiently large $\lambda$ a sufficiently small ball around a suitable multiple of $A$ is contained in $P(A,\lambda)$, which shows that $P(A,\lambda)$ has full dimension.

By the theorem of Minkowski and Weyl, see for example [5, Corollary 7.1c], polytopes are exactly bounded polyhedra. So it suffices to show that the set $\mathcal{P}(A,\lambda)$ is a bounded polyhedron.

First we show that $\mathcal{P}(A,\lambda)$ is bounded: For suppose not. Then there is $B_{0}\in\mathcal{P}(A,\lambda)$ and $B_{1}\in\mathcal{S}^{n}$, with $B_{1}\neq 0$, so that the ray $B_{0}+\alpha B_{1}$, with $\alpha\geq 0$, lies completely in $\mathcal{P}(A,\lambda)$. In particular $\langle B_{1},vv^{\sf T}\rangle\geq 0$ for all $v\in\mathbb{Z}^{n}_{\geq 0}$. Hence, $B_{1}$ lies in the dual cone of $\tilde{\mathcal{CP}}_{n}$. On the other hand $\langle A,B_{1}\rangle\leq 0$. Hence, by Lemma 2.2 (1), $B_{1}\not\in\mathcal{COP}_{n}\setminus\{0\}$, but by Lemma 2.2 (2),

[TABLE]

so $B_{1}=0$, yielding a contradiction.

Now we show that $\mathcal{P}(A,\lambda)$ is a polyhedron: For suppose not. Then there is a sequence $v_{i}\in\mathbb{Z}^{n}_{\geq 0}\setminus\{0\}$ of infinitely many pairwise different nonzero lattice vectors so that there are $B_{i}\in\mathcal{P}(A,\lambda)$ with $\langle B_{i},v_{i}v_{i}^{\sf T}\rangle=1$. Since $\mathcal{P}(A,\lambda)$ is compact, there exists a subsequence $B_{i_{j}}$ which converges to $B^{*}\in\mathcal{P}(A,\lambda)$. Define the sequence $u_{i_{j}}=v_{i_{j}}/\|v_{i_{j}}\|$ which lies in the compact set $\mathbb{R}^{n}_{\geq 0}\cap S^{n-1}$ where $S^{n-1}$ denotes the unit sphere. Hence there is a subsequence converging to $u^{*}\in S^{n-1}$, in particular $u^{*}\neq 0$. Denote the indices of this subsequence with $k$, then

[TABLE]

When $k$ tends to infinity, the squared norms $\|v_{k}\|^{2}$ tend to infinity as well, since we use infinitely many pairwise different lattice vectors and there exist only finitely many lattice vectors up to some given norm. So $\langle B_{k},u_{k}u_{k}^{\sf T}\rangle$ tends to $\langle B^{*},u^{*}(u^{*})^{\sf T}\rangle=0$, and by Lemma 2.3 we obtain a contradiction. ∎

Now we prove the main result and finish the paper.

Proof of Theorem 1.1.

Let $A$ be matrix having rational entries only and lying in the interior of the cone of completely positive matrices. Then $\mathcal{P}(A,\lambda)$ is a polytope according to the previous lemma. We minimize the linear functional $B\mapsto\langle A,B\rangle$ over $\mathcal{P}(A,\lambda)$. The minimum is attained at one of the polytopes’ vertices, $B^{*}\in\mathcal{P}(A,\lambda)$. Then we choose those lattice vectors $v_{i}\in\mathbb{Z}_{\geq 0}^{n}$, with $i=1,\ldots,m$ for which equality $\langle B^{*},v_{i}v_{i}^{\sf T}\rangle=1$ holds. Because of the minimality of $\langle A,B^{*}\rangle$ it follows

[TABLE]

Otherwise, see for example [5, Theorem 7.1], we find a separating linear hyperplane orthogonal to $C$ separating $A$ and $\operatorname{cone}\{v_{i}v_{i}^{\sf T}:i=1,\ldots,m\}$:

[TABLE]

Then for sufficiently small $\mu>0$ we would have

[TABLE]

which contradicts the minimality of $\langle A,B^{*}\rangle$.

We apply Carathéodory’s theorem (see for example [5, Corollary 7.1i]) to (3) and choose a subset $I\subseteq\{1,\ldots,m\}$ so that $v_{i}v_{i}^{\sf T}$ are linearly independent and so that $A$ lies in $\operatorname{cone}\{v_{i}v_{i}^{\sf T}:i\in I\}$. Since $A$ is a rational matrix and since the $v_{i}v_{i}^{\sf T}$’s are linearly independent rational matrices, there is a unique choice of rational numbers $\alpha_{i}\in\mathbb{Q}_{\geq 0}$, with $i\in I$, so that $A=\sum_{i\in I}\alpha_{i}v_{i}v_{i}^{\sf T}$ holds, which gives a desired rational cp-factorization. ∎

Acknowledgements

We thank Naomi Shaked-Monderer for comments on the manuscript.