An Optimality Gap Test for a Semidefinite Relaxation of a Quadratic Program with Two Quadratic Constraints

TL;DR

This paper introduces a necessary and sufficient test to determine when solutions to a quadratic program with two quadratic constraints can be obtained from its semidefinite relaxation, even in nonconvex cases, and when strong duality holds.

Contribution

It generalizes existing methods by providing a test applicable to nonconvex quadratic programs with two constraints, without requiring strict convexity.

Findings

Test confirms no optimality gap in one example

Test identifies an optimality gap in another example

Method extends applicability beyond prior assumptions

Abstract

We propose a necessary and sufficient test to determine whether a solution for a general quadratic program with two quadratic constraints (QC2QP) can be computed from that of a specific convex semidefinite relaxation, in which case we say that there is no optimality gap. Originally intended to solve a nonconvex optimal control problem, we consider the case in which the cost and both constraints of the QC2QP may be nonconvex. We obtained our test, which also ascertains when strong duality holds, by generalizing a closely-related method by Ai and Zhang. An extension was necessary because, while the method proposed by Ai and Zhang also allows for two quadratic constraints, it requires that at least one is strictly convex. In order to illustrate the usefulness of our test, we applied it to two examples that do not satisfy the assumptions required by prior methods. Our test guarantees that there is no optimality gap for the first example -- a solution is also computed from the relaxation -- and we used it to establish that an optimality gap exists in the second.

Equations

$$\underset{\mathbf{z}\in {\mathbb{R}}^n}{\mathrm{minimize}}$$

$$q_1(\mathbf{z})={\mathbf{z}}^T{Q}_1\mathbf{z}+2b_1^T\mathbf{z}+c_1\le 0,$$

$$q_2(\mathbf{z})={\mathbf{z}}^T{Q}_2\mathbf{z}+2b_2^T\mathbf{z}+c_2\le 0,$$

$$\underset{\mathbf{x}\in \mathcal{X}}{\mathrm{minimize}}$$

$$\mathbf{x}\in \mathcal{C}.$$

$$\underset{\mathbf{z}\in {\mathbb{R}}^n,\mathbf{t}\in \mathbb{R}}{\mathrm{minimize}}$$

$$M(q_i)\text{vbox}\text{scalebox}0.5\bullet \left[\begin{array}{c}\mathbf{t}\\ \mathbf{z}\end{array}\right]{\left[\begin{array}{c}\mathbf{t}\\ \mathbf{z}\end{array}\right]}^T={\mathbf{z}}^T{Q}_i\mathbf{z}+2\mathbf{t}{b}_i^T\mathbf{z}+{\mathbf{t}}^2{c}_i\le 0,i\in \{1,2\},$$

$${\mathbf{t}}^2=1,$$

$$M(q_0)=\left[\begin{array}{cc}0& b_0^T\\ b_0& Q_0\end{array}\right],M(q_i)=\left[\begin{array}{cc}{c}_i& b_i^T\\ b_i& Q_i\end{array}\right],i\in \{1,2\}.$$

$$x=\left[\begin{array}{c}t\\ z\end{array}\right].$$

$$\underset{\mathbf{X}\in {\mathcal{S}}^{n+1}}{\mathrm{minimize}}$$

$$M(q_i)\text{vbox}\text{scalebox}0.5\bullet \mathbf{X}\le 0,i\in \{1,2\},$$

$$I_{00}\text{vbox}\text{scalebox}0.5\bullet \mathbf{X}=1,$$

$$\mathbf{X}⪰0,$$

$$\underset{\mathbf{Z}\in {\mathcal{S}}^{n+1},{\mathbf{y}}_0,{\mathbf{y}}_1,{\mathbf{y}}_2\in \mathbb{R}}{\mathrm{maximize}}$$

$${\mathbf{y}}_0{I}_{00}-{\mathbf{y}}_{1}M(q_1)-{\mathbf{y}}_{2}M(q_2)+\mathbf{Z}=M(q_0),$$

$${\mathbf{y}}_i\ge 0,i\in \{1,2\},$$

$$\mathbf{Z}⪰0.$$

$$M(q_i)\text{vbox}\text{scalebox}0.5\bullet X$$

$$I_{00}\text{vbox}\text{scalebox}0.5\bullet X$$

$$z^T{Q}_{i}z+2b_i^{T}z+c_i<0,i\in \{1,2\}.$$

$$M(q_0)-y_0{I}_{00}+y_{1}M(q_1)+y_{2}M(q_2)\succ 0.$$

$$Q_0+y_1{Q}_1+y_2{Q}_2$$

$$\xi (y_1,y_2{)}^T(Q_0+y_1{Q}_1+y_2{Q}_2{)}^{-1}\xi (y_1,y_2)$$

$$XZ$$

$$y_{1}M(q_1)\text{vbox}\text{scalebox}0.5\bullet X$$

$$y_{2}M(q_2)\text{vbox}\text{scalebox}0.5\bullet X$$

$$M(q_1)\text{vbox}\text{scalebox}0.5\bullet {\hat{x}}_i{\hat{x}}_i^T$$

$$(M(q_2)\text{vbox}\text{scalebox}0.5\bullet {\hat{x}}_1{\hat{x}}_1^T)(M(q_2)\text{vbox}\text{scalebox}0.5\bullet {\hat{x}}_2{\hat{x}}_2^T)$$

$$M(q_1)\text{vbox}\text{scalebox}0.5\bullet {\hat{x}}_1{\hat{x}}_2^T\ne 0.$$

$$\hat{X}=P_1^T{\Lambda }_1{P}_1,$$

$$\hat{Z}=P_2^T{\Lambda }_2{P}_2,$$

$${\lambda }_{ij}^{\ast }:=\{\begin{array}{rl}{\lambda }_{ij},& {\lambda }_{ij}\ge {ϵ}_2\\ 0,& {\lambda }_{ij}<{ϵ}_2\end{array},i\in \{1,2\},j\in \{1,2,\dots ,n+1\}.$$

$$X^{\ast }$$

$$Z^{\ast }$$

$$y_i^{\ast }$$

$$\mathrm{rank}(\hat{X},{ϵ}_2)$$

$$\mathrm{rank}(\hat{Z},{ϵ}_2)$$

$$|M(q_1)\text{vbox}\text{scalebox}0.5\bullet x_i^{\ast }(x_i^{\ast }{)}^T|$$

$$\displaystyle\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\left(M(q_{2})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{$\displaystyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{$\textstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{$\scriptstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{$\scriptscriptstyle\bullet$}}}}}x_{1}^{*}(x_{1}^{*})^{T}\right)\left(M(q_{2})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{$\displaystyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{$\textstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{$\scriptstyle\bullet$}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{$\scriptscriptstyle\bullet$}}}}}x_{2}^{*}(x_{2}^{*})^{T}\right)$$

$$|M(q_1)\text{vbox}\text{scalebox}0.5\bullet x_1^{\ast }(x_2^{\ast }{)}^T|$$

Full text

**Keywords. ** quadratically constrained quadratic program, semidefinite relaxation, strong duality, nonconvex optimization

**AMS subject classifications. ** 90C26, 90C46

1 Introduction

We consider the following real-valued quadratic program with two quadratic constraints (QC2QP):

[TABLE]

where $Q_{0}$, $Q_{1}$, and $Q_{2}$ are $n\times n$-dimensional real symmetric matrices; $b_{0}$, $b_{1}$ and $b_{2}$ are $n$-dimensional real vectors; and $c_{1}$ and $c_{2}$ are real constants.

Main problem: We seek to solve (QP0) without positive semidefiniteness restrictions on $Q_{0},Q_{1}$, and $Q_{2}$, which, generally, makes the problem nonconvex.

1.1 Brief overview of existing related work

Existing work explored two distinct approaches to obtain a globally optimal solution to (QP0). The first approach, which we adopt to develop our method, uses a semidefinite relaxation of (QP0) whose (Lagrange) dual is convex and identical to that of (QP0). The second approach seeks to exploit the structure of the QC2QP, possibly subject to additional restrictions, to characterize globally optimal solutions in a way that numerically tractable methods can be used. Subsequently, we proceed to describe previous work on both approaches.

Following the first approach, Ai and Zhang [1] introduce a necessary and sufficient condition for strong duality for the Celis-Dennis-Tapia (CDT) subproblem of minimizing a nonconvex quadratic cost over the intersection of an ellipsoid and an elliptical cylinder [4] (corresponding to $Q_{1}$ being positive definite and $Q_{2}$ being positive semidefinite, respectively, in (QP0)), which is a special case of QC2QP used in the extended trust region method [20]. Their result shows that strong duality holds, and a primal optimal solution can be obtained from a semidefinite relaxation, if and only if optimal solutions of the dual and the relaxation violate the so-called Property I comprising three algebraic conditions. Subsequent work by Yuan et al. in [21] shows that adding second-order cone (SOC) constraints to a CDT subproblem for which Property I holds may narrow or even eliminate the duality gap. In the latter case, a globally optimal solution to the original problem can be computed from a solution of the semidefinite relaxation with an SOC reformulation.

Another relaxation technique is to solve the QC2QP in the complex domain. In [2], Beck and Eldar use such a methodology to introduce a necessary and sufficient condition for strong duality, using the classical extended S-Lemma of Fradkov and Yakubovich [7]. If strong duality holds, a globally optimal solution to the original problem can be obtained by solving a quadratic feasibility problem. Using the necessary and sufficient condition and the convexity of a quadratic mapping, they subsequently prove a sufficient condition for strong duality for the real-valued QC2QP. Huang and Zhang [12] propose a sufficient condition for strong duality in the complex-valued problem in which a globally optimal solution to the original problem can be obtained from a semidefinite relaxation if strong duality holds. Their result is derived using a matrix rank-one decomposition for complex Hermitian matrices.

Following the second approach, Peng and Yuan [14] prove a necessary condition for global optimality in QC2QP. Specifically, the number of negative eigenvalues of the Hessian of the Lagrangian is characterized at a globally optimal solution. For the CDT subproblem, Bomze and Overton [3] prove necessary and sufficient conditions for global and local optimality using copositivity.

1.2 Our main contribution

We seek to use a specific semidefinite relaxation to find a solution for (QP0) for the case in which there are no positive semidefiniteness restrictions on $Q_{0},Q_{1}$, and $Q_{2}$. When a solution for (QP0) can be determined from that of the relaxation we say that there is no optimality gap. The relaxation is cast as a convex semidefinite program (SDP) for which an optimal solution can be determined efficiently using existing software. The dual of the semidefinite relaxation is convex and is also the dual of (QP0). This motivates the analysis in section 3, where we propose the so-called Property I+ defined by four algebraic conditions that determine, based on solutions of the relaxation and its dual, when an optimality gap exists. Our main result is Theorem 3.2, which states precisely a necessary and sufficient condition for the existence of an optimality gap based on Property I+. As we discuss in detail in section 5, Theorem 3.2 extends the closely-related result of [1] in the following ways:

• •

The assumption in [1] that either $Q_{1}$ or $Q_{2}$ must be positive definite is replaced in our work by the weaker requirement that the dual of (QP0) satisfies Slater’s condition.

• •

In the particular case when $Q_{1}$, or $Q_{2}$, is positive definite the above-mentioned Property I+ is equivalent to Property I used in [1] to determine when there is an optimality gap. Hence, our work presents no advantage relative to [1] when $Q_{1}$, or $Q_{2}$, is positive definite.

A nonconvex optimal control problem studied by Cheng and Martins in [6] motivated the unexampled QC2QP considered here, in which neither $Q_{1}$ nor $Q_{2}$ is assumed positive definite.

1.3 Structure of the article

We start with reviewing in section 2 the key results of [1]. In doing so, we also present the essential concepts used in [1], which include the semidefinite relaxation used here. We define Property I+ and subsequently state our main result (Theorem 3.2) in section 3. In section 4, we describe an algorithm to implement the test of Theorem 3.2, and we also discuss relevant numerical considerations. In addition, in section 4, we apply our algorithm to two QC2QP examples that do not satisfy the assumptions required by previous methods. More specifically, we compute the optimal solution for the first example after we establish that it has no optimality gap. In contrast, we establish that there is an optimality gap for the second. In section 5, we explain why Theorem 3.2 extends the closely-related result of [1]. In section 6, we summarize the paper and discuss future research directions that could leverage our results, in concert with related recent work, to obtain methods that would not only be numerically more robust, but also more general. In section 7, we present a detailed proof of Theorem 3.2. The necessity portion of the proof embeds the description of a method to construct a solution to (QP0) from that of its convex relaxation, for the case in which the test in Theorem 3.2 guarantees that there is no optimality gap.

1.4 Notation and conventions

Throughout the paper, we adopt the following notation, which is mostly borrowed from [1]: We denote the set of real numbers with $\mathbb{R}$. We use $\mathcal{S}^{n}$ to denote the set of symmetric matrices in $\mathbb{R}^{n\times n}$. We use the dot notation to denote the matrix inner product, that is, A\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}B:=\operatorname{Tr}(AB^{T}) for $A,B\in\mathbb{R}^{n\times n}$, where $\operatorname{Tr}(AB^{T})$ denotes the trace of $AB^{T}$. We use $\det(C)$ to denote the determinant of a square matrix $C$. We use $\operatorname{rank}(D)$ and $\operatorname{rank}(D,\epsilon)$ to denote the rank and the numerical rank with tolerance $\epsilon$, respectively, of a matrix $D$. A positive (semi)definite matrix $M$ is denoted by $M\succ(\succeq)0$. We use $0_{n\times m}$ to denote a matrix in $\mathbb{R}^{n\times m}$ with all entries being [math] and $I_{n}$ to denote an $n\times n$-dimensional identity matrix. A diagonal matrix is denoted by $\text{diag}(a_{1},a_{2},\dots,a_{n})$, where $a_{1},a_{2},\dots,a_{n}\in\mathbb{R}$ are the diagonal entries. The null space of a linear mapping $L:\mathbb{V}\rightarrow\mathbb{W}$ between two vector spaces $\mathbb{V}$ and $\mathbb{W}$ is denoted by $\mathcal{N}(L)$. We use $|a|$ to denote the absolute value of a real-valued constant or variable $a$. We use the term polynomial time, which is defined in [20], to indicate that the total number of basic operations (for example, addition, subtraction, multiplication, division, and comparison) of a procedure is bounded by a polynomial of the problem data. We use boldface font, such as in $\boldsymbol{x}$, to represent the optimization variables with respect to which we seek to minimize a cost subject to constraints. We adopt the following format to represent an optimization problem over a subset $\mathcal{X}$ of a real coordinate space, in which we seek to minimize a cost $f:\mathcal{X}\rightarrow\mathbb{R}$ subject to an additional constraint set $\mathcal{C}$:

[TABLE]

We use $\mathcal{V}^{*}_{\eqref{prob: notation problem}}$ to denote the optimal value of (P). As a convention, a matrix $X$ is rank-one decomposable [1] at $x_{1}$ if there exist other $r-1$ vectors $x_{2},\dots,x_{r}$ such that $X=x_{1}x_{1}^{T}+x_{2}x_{2}^{T}+\dots+x_{r}x_{r}^{T}$, where $r:=\operatorname{rank}(X)$.

2 Preliminary results and concepts

We start with introducing assumptions and reviewing the key results in [1]. For the reader’s convenience, we follow the notation in [1] and rewrite (QP0) in a homogeneous quadratic form:

[TABLE]

where

[TABLE]

In the rest of the paper, we use $x$ to represent an $(n+1)$-dimensional vector concatenating a scalar $t$ and an $n$-dimensional vector $z$ as follows

[TABLE]

The semidefinite relaxation of (QP) is the following:

[TABLE]

where I_{00}=\bigl{[}\begin{smallmatrix}1&0_{1\times n}\\ 0_{n\times 1}&0_{n\times n}\end{smallmatrix}\bigr{]}.

The dual problem of (SP) is the following:

[TABLE]

Note that (SD) is also the dual of (QP).

Assumption 2.1**.**

*Problem (SP) satisfies Slater’s condition, that is, there exists a symmetric positive definite $(n+1)\times(n+1)$-dimensional real matrix $X$ such that *

[TABLE]

Remark 2.2**.**

2.1* holds when Slater’s condition holds for (QP0) [1] (and, hence, for (QP)), that is, there exists an $n$-dimensional vector $z$ such that *

[TABLE]

Assumption 2.3**.**

*Slater’s condition holds for (SD), that is, there exist a scalar $y_{0}$ and positive scalars $y_{1}$ and $y_{2}$ such that *

[TABLE]

Remark 2.4**.**

For problem data $M(q_{0}),M(q_{1}),$ and $M(q_{2})$, one can numerically check whether 2.1 and 2.3 are met by solving the feasibility problem of (SP) and (SD), respectively, using an SDP solver.

Remark 2.5**.**

*The inequality (2.5) holds, by Schur complement, if and only if there exist a scalar $y_{0}$ and positive scalars $y_{1}$ and $y_{2}$ such that *

[TABLE]

where $\xi(y_{1},y_{2}):=b_{0}+y_{1}b_{1}+y_{2}b_{2}$.

Remark 2.6**.**

*By [20, Proposition 2.1], 2.3 holds if the objective function of (QP0) is strictly convex (that is, $Q_{0}$ is positive definite), or at least one of the constraints of (QP0) is elliptical (that is, $Q_{i}\succ 0$ and $b_{i}^{T}Q_{i}^{-1}b_{i}-c_{i}>0$ for $i$ being $1$ or $2$, or both). The latter condition, which implies (2.5) by Remark 2.5, is required by Ai and Zhang in [1] which ensures that Slater’s condition holds for (SD). *

Remark 2.7**.**

2.1* and 2.3 together imply that both (SP) and (SD) have attainable optimal solutions that yield an identical optimal value. This, in turn, implies that (SP) is a tight relaxation of (QP) (that is, the optimal value of (SP) is identical to that of (QP)) if and only if strong duality holds for (QP). This observation will be relevant later on when we establish our optimality gap test.*

We denote optimal solutions of (QP), (SP), and (SD), respectively, by $x^{*},\hat{X},$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$, and their optimal values, respectively, by $\mathcal{V}^{*}_{\eqref{prob: original problem}}$, $\mathcal{V}^{*}_{\eqref{prob: semidefinite relaxation}}$, and $\mathcal{V}^{*}_{\eqref{prob: dual problem}}$. Note that a primal-dual feasible pair, $X$ and $(Z,y_{0},y_{1},y_{2})$, is optimal if and only if it satisfies the complementary conditions:

[TABLE]

Property I, which is defined below (Definition 2.8), is the key to the necessary and sufficient conditions for an optimality gap (or, equivalently, for a duality gap between (QP) and (SD)) stated in Theorem 2.9 for when $Q_{1}$ is positive definite.

Definition 2.8** ([1, Definition 4.1]).**

For $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$, a given pair of optimal solutions for (SP) and (SD), respectively, we say that this pair has Property I if all the conditions (I.1)–(I.4) specified below hold:

• (I.1)

$\hat{y}_{1}\hat{y}_{2}\neq 0$;

• (I.2)

$\operatorname{rank}(\hat{Z})=n-1$;

• (I.3)

$\operatorname{rank}(\hat{X})=2$;

• (I.4)

There is a rank-one decomposition of $\hat{X}$, of the form $\hat{X}=\hat{x}_{1}\hat{x}_{1}^{T}+\hat{x}_{2}\hat{x}_{2}^{T}$, for which the following hold:

[TABLE]

Theorem 2.9** ([1, Theorem 4.2]).**

Consider (QP) where Slater’s condition is satisfied and $Q_{1}$ is positive definite. Denote with $\mathfrak{P}$ the set of all pairs $\hat{X}^{\prime}$ and $(\hat{Z}^{\prime},\hat{y}_{0}^{\prime},\hat{y}_{1}^{\prime},\hat{y}_{2}^{\prime})$ of optimal solutions for the semidefinite relaxation (SP) and the dual problem (SD), respectively. For any given pair $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$ arbitrarily selected from $\mathfrak{P}$, $\mathcal{V}^{*}_{\eqref{prob: semidefinite relaxation}}<\mathcal{V}^{*}_{\eqref{prob: original problem}}$ holds if and only if the pair satisfies Property I.

3 Main results

We start with modifying Property I by adding an extra constraint and naming the resulting conditions Property I+ as follows:

Definition 3.1**.**

For $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$, a given pair of optimal solutions for (SP) and (SD), respectively, we say that this pair has Property I+ if all the conditions (I.1)–(I.3) and (I+.4) specified below hold:

• (I.1)

$\hat{y}_{1}\hat{y}_{2}\neq 0$;

• (I.2)

$\operatorname{rank}(\hat{Z})=n-1$;

• (I.3)

$\operatorname{rank}(\hat{X})=2$;

• (I+.4)

There is a rank-one decomposition of $\hat{X}$, of the form $\hat{X}=\hat{x}_{1}\hat{x}_{1}^{T}+\hat{x}_{2}\hat{x}_{2}^{T}$, for which not only (2.8) and (2.9) are satisfied, but the following inequality also holds:

[TABLE]

Notice that since the conditions (I.1)–(I.3) are the same in the definitions of Properties I and I+, the fact that (I+.4) is more stringent than (I.4) is what distinguishes the properties. As we shall see in the following theorem, Property I+ is the key to the necessary and sufficient condition for the optimality gap (or, equivalently, for the duality gap between (QP) and (SD)), when the positive definiteness of $Q_{1}$ is not assumed.

Theorem 3.2**.**

Consider (QP) and let 2.1 and 2.3 hold. Denote with $\mathfrak{P}$ the set of all pairs $\hat{X}^{\prime}$ and $(\hat{Z}^{\prime},\hat{y}_{0}^{\prime},\hat{y}_{1}^{\prime},\hat{y}_{2}^{\prime})$ of optimal solutions for the semidefinite relaxation (SP) and the dual problem (SD), respectively. For any given pair $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$ arbitrarily selected from $\mathfrak{P}$, ${\mathcal{V}^{*}_{\eqref{prob: semidefinite relaxation}}<\mathcal{V}^{*}_{\eqref{prob: original problem}}}$ holds if and only if the pair satisfies Property I+.

Proof.

See section 7. ∎

Remark 3.3**.**

Notice that Theorem 3.2, and its proof given in section 7, hold irrespective of the choice of the pair $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$ so long as it is in $\mathfrak{P}$. Hence, one immediate conclusion from Theorem 3.2 is that either all pairs in $\mathfrak{P}$ satisfy Property I+ or all pairs in $\mathfrak{P}$ violate Property I+. This is particularly relevant for applications because in order to use the theorem to determine whether there is an optimality gap, it suffices to check if one (arbitrarily chosen) pair in $\mathfrak{P}$ satisfies Property I+. Moreover, it will follow from our discussion in section 4, where we describe an algorithm (Algorithm 4.1) to test Property I+, that if (I+.1)–(I+.4) are checked in the order and manner specified in the algorithm then the validity of (I+.4) can be determined by checking only one rank-one decomposition of $\hat{X}$ arbitrarily chosen among those satisfying (2.8). Remark 4.4 gives a justification of this fact by leveraging our proof of Theorem 3.2 in section 7.

Notice that the test in Theorem 3.2 is tractable because it involves the solution of an SDP, and its dual, followed by a rank-one decomposition, both of which run in polynomial time [20]. Furthermore, the necessity part of the proof (section 7) of Theorem 3.2 is constructive and useful in its own right because it embeds the description of a method to construct a solution for (QP) from that of (SP) when Property I+ is violated.

Besides the numerical results that are going to be presented in the next section, Theorem 3.2 was used in [6, Theorem 2] to propose conditions under which there is no optimality gap for a control-oriented QC2QP problem.

The following remark is relevant due to the intriguing fact that, although $M(q_{1})$ and $M(q_{2})$ play identical “roles” in (QP) and (SP), they enter Definitions 2.8 and 3.1 in different ways.

Remark 3.4**.**

One can readily verify by inspection that interchanging $M(q_{1})$ and $M(q_{2})$ in (QP) and (SP) would not modify the underlying problem and, in particular, would not alter whether $\mathcal{V}^{*}_{\eqref{prob: semidefinite relaxation}}<\mathcal{V}^{*}_{\eqref{prob: original problem}}$ holds. Similarly, Theorem 3.2 would remain valid had we chosen to interchange $M(q_{1})$ and $M(q_{2})$ in Definitions 2.8 and 3.1.

We now proceed to remark on the differences between the proofs given in section 7 and [1] for Theorems 3.2 and 2.9, respectively.

Remark 3.5**.**

*We start by observing that, in section 7, we prove the sufficient condition of Theorem 3.2 by exploring the fact that $\hat{X}$ is the unique solution of (SP) when Property I+ holds, whereas the proof of the sufficient condition of Theorem 2.9 is ascertained in [1] using contradiction. The proofs of the necessary condition in [1] and section 7 construct a rank-one solution to (SP) for the possible cases in which Property I and Property I+ fail, respectively. Although the four cases enumerated in the former are also present in the latter, there are differences in the analysis. Specifically, in section 7 we need to invoke 2.3 in cases 1, 2, and 4 to show the existence of a nonzero solution, while the argument in [1] uses the positive definiteness111Recall that in our article $Q_{0}$, $Q_{1}$, and $Q_{2}$ are only required to satisfy 2.1 and 2.3. It may be the case that none is positive definite. of $Q_{1}$. In addition, we need to consider the fifth case in section 7 to account for when Property I+ fails and Property I holds. *

4 Testing Property I+ numerically

Theorem 3.2 enables a simply verifiable optimality gap test for a semidefinite relaxation of a QC2QP. This test only requires to solve one SDP with its dual and conducting a rank-one decomposition, both of which run in polynomial time. However, in general, SDP solvers (for example, SDPT3 [18] and SeDuMi [17]) give approximate, rather than exact, solutions within certain tolerance. Hence, it is useful to establish an optimality gap test utilizing Property I+ in an approximation sense. The following procedures refer to the purified $(\epsilon_{1},\epsilon_{2})$-approximation in [1]:

Let $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$ denote a pair of numerical solutions of (SP) and its dual (SD), respectively, solved by an SDP solver whose tolerance is $\epsilon_{1}>0$. Let $\epsilon_{2}>0$ be the tolerance for purification. First, conduct an eigendecomposition of $\hat{X}$ and $\hat{Z}$, that is,

[TABLE]

where $P_{1}$ and $P_{2}$ are $(n+1)\times(n+1)$-dimensional orthogonal matrices, and $\Lambda_{1}$ and $\Lambda_{2}$ are $(n+1)\times(n+1)$-dimensional diagonal matrices. Let $\Lambda_{i}:=\text{diag}(\lambda_{i1},\dots,\lambda_{i(n+1)}),$ and let

[TABLE]

Define the purified solutions by

[TABLE]

The above step essentially purifies the rank of $\hat{X}$ and $\hat{Z}$ so that

[TABLE]

Remark 4.1**.**

We use numerical $\epsilon$-rank, as defined in [11], to determine the numerical rank of a matrix with tolerance $\epsilon$. Namely, the numerical rank of a matrix $A$ in $\mathbb{R}^{m\times n}$ with tolerance $\epsilon$, denoted by $\operatorname{rank}(A,\epsilon)$, is $r$, if the singular values, $\sigma_{1}\geq\sigma_{2}\geq\dots\geq\sigma_{\text{min}(m,n)}\geq 0$, of $A$ satisfy $\sigma_{1}\geq\sigma_{2}\geq\dots\geq\sigma_{r}>\epsilon>\sigma_{r+1}\geq\dots\geq\sigma_{\text{min}(m,n)}$. For more details, see [8] and [11].

Henceforth, we use $X^{*}$ and $(Z^{*},y_{0}^{*},y_{1}^{*},y_{2}^{*})$ to denote a pair of purified $(\epsilon_{1},\epsilon_{2})$-approximate optimal solutions of (SP) and (SD), respectively.

Definition 4.2**.**

For $X^{*}$ and $(Z^{*},y_{0}^{*},y_{1}^{*},y_{2}^{*})$, a given pair of purified $(\epsilon_{1},\epsilon_{2})$-approximate optimal solutions of (SP) and (SD), respectively, we say that this pair has Property I${}^{+}(\epsilon_{2})$ if:

1. (I${}^{+}(\epsilon_{2})$.1)

$y_{1}^{*}>\epsilon_{2}$* and $y_{2}^{*}>\epsilon_{2}$;* 2. (I${}^{+}(\epsilon_{2})$.2)

$\operatorname{rank}(Z^{*},\epsilon_{2})=n-1$; 3. (I${}^{+}(\epsilon_{2})$.3)

$\operatorname{rank}(X^{*},\epsilon_{2})=2$; 4. (I${}^{+}(\epsilon_{2})$.4)

There is a rank-one decomposition of $X^{*}$, of the form $X^{*}=x_{1}^{*}(x_{1}^{*})^{T}+x_{2}^{*}(x_{2}^{*})^{T}$, for which the following hold:

[TABLE]

Now, we introduce a polynomial-time algorithm to test the optimality gap for the relaxation (SP), with $\epsilon_{1}$-precision SDP solutions $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$ as well as Property I${}^{+}(\epsilon_{2})$.

Remark 4.3**.**

Algorithm 4.1* can always find the gap when $\epsilon_{2}$ is chosen properly. The algorithm relies on Property I${}^{+}(\epsilon_{2})$ instead of Property I+ because the latter is exact, which cannot handle the round-off errors in numerical computations. Hence, $\epsilon_{2}$ must be carefully chosen so that*

The terms that are zero-valued in Property I+* will be in the $\epsilon_{2}$-neighborhood of zero subject to round-off errors. Likewise, the terms that are nonzero-valued will be out of the $\epsilon_{2}$-neighborhood of zero.* 2. 2.

Only the significant eigenvalues (those greater or equal to $\epsilon_{2}$) of $\hat{X}$ and $\hat{Z}$ are counted towards the numerical rank of each matrix.

By Remark 2.7, the test described by Algorithm 4.1 can also be applied to check whether a duality gap exists for (QP). Note that it could be difficult to check the existence of the duality gap using other methods, for example, solving for the global minimum of the primal problem. The primal problem is nonconvex, though smooth. Hence, to find the global minimum, a good initial point is necessary for the convergence of a gradient-based algorithm. However, to the best of the authors’ knowledge, there is no efficient method to pick an initial point for convergence to the global minimum.

Remark 4.4**.**

We proceed to explain why in Steps 8 and 10 of Algorithm 4.1 we test whether only one rank-one decomposition $(x_{1}^{*},x_{2}^{*})$ (arbitrarily selected among those complying with (4.9)) violates (4.10) or (4.11). A clarification is warranted because, even when it holds, the test in and of itself does not preclude the existence of another rank-one decomposition satisfying (4.9)–(4.11). In our justification, we will repurpose portions of the proof of Theorem 3.2 in section 7. Namely, if $(x_{1}^{*},x_{2}^{*})$ violates (4.10) in Step 8 then we can use $(x_{1}^{*},x_{2}^{*})$ as the rank-one pair selected in Case 3 of the necessity proof in section 7 and follow the procedure therein to establish constructively the absence of an optimality gap. Conversely, if $(x_{1}^{*},x_{2}^{*})$ satisfies (4.10), which means that $(x_{1}^{*},x_{2}^{*})$ “approximately” belongs to the set $\mathbb{O}^{+}(\hat{X})$ specified in Remark 7.2 of section 7, and if $(x_{1}^{*},x_{2}^{*})$ violates (4.11) in Step 10 then we can use $(x_{1}^{*},x_{2}^{*})$ as the rank-one pair selected in Case 5 of the proof and establish the absence of the optimality gap. Otherwise, if $(x_{1}^{*},x_{2}^{*})$ satisfies (4.10) and (4.11) in Step 12 then the pair $X^{*}$ and $(Z^{*},y^{*}_{0},y^{*}_{1},y^{*}_{2})$ satisfies Property I${}^{+}(\epsilon_{2})$, in which case we can follow the sufficiency proof in section 7 to show that there is an optimality gap.

We have implemented Algorithm 4.1 in a MATLAB script which is available online [5]222 The description and data of a numerical experiment that tests the proportion of the randomly generated feasible nonconvex QC2QP instances of which there is no optimality gap are also available online [5].. We invite the interested readers to use this script with their own QC2QP problem data.

We provide numerical examples which contain two nonconvex QC2QPs, in order to illustrate the test described in Algorithm 4.1. The optimality gap does not exist in the first example but does in the second one. Note that Theorem 2.9 cannot be applied to the optimality gap test because the constraints in both examples are nonconvex, which violates the assumption in [1] that requires at least one of the constraints to be strictly convex.

To solve SDPs, we use CVX [9][10] with solver SDPT3 [18] and the default tolerance $\epsilon_{1}=1.49\times 10^{-8}$. We set the purification tolerance $\epsilon_{2}$ to $1\times 10^{-5}$.

4.1 Example: (there is no optimality gap)

Consider the following data in (QP0):

[TABLE]

Both the objective function and constraints are hyperbolic and nonconvex. 2.1 and 2.3 are verified to hold in this example. Solving (SP) and (SD), we obtain the purified $(\epsilon_{1},\epsilon_{2})$-approximate optimal solutions:

[TABLE]

where $\operatorname{rank}(X^{*},\epsilon_{2})=1$, $\operatorname{rank}(Z^{*},\epsilon_{2})=2$, and a rank-one decomposition yields ${X^{*}=x^{*}(x^{*})^{T}}$ such that

[TABLE]

It can be verified that

[TABLE]

Denote the normalized solutions by $\hat{z}$, where $\hat{z}=z^{*}/t^{*}$ follows the partition of $x^{*}$ in the form of (2.2). Thus,

[TABLE]

is marked in Figure 1.

Since Property I${}^{+}(\epsilon_{2})$ is violated, we claim that there is no optimality gap. This can be verified since the primal optimal solution

[TABLE]

coincides with $\hat{z}$. The globally optimal value is $-54.8271061$, which is identical to that of (SP).

4.2 Example: (there is an optimality gap)

Consider the following data in (QP0):

[TABLE]

Both the objective function and constraints are hyperbolic and nonconvex. 2.1 and 2.3 are verified to hold in this example. Solving (SP) and (SD), we obtain the purified $(\epsilon_{1},\epsilon_{2})$-approximate optimal solutions:

[TABLE]

where $\operatorname{rank}(X^{*},\epsilon_{2})=2$, $\operatorname{rank}(Z^{*},\epsilon_{2})=1$, and a rank-one decomposition yields $X^{*}=x_{1}^{*}(x_{1}^{*})^{T}+x_{2}^{*}(x_{2}^{*})^{T}$ such that

[TABLE]

It can be verified that

[TABLE]

Denote the normalized solutions by $\hat{z}_{1}$ and $\hat{z}_{2}$, where $\hat{z}_{1}=z_{1}^{*}/t_{1}^{*}$ and $\hat{z}_{2}=z_{2}^{*}/t_{2}^{*}$ follow the partition of $x_{1}^{*}$ and $x_{2}^{*}$, respectively, in the form of (2.2). Thus,

[TABLE]

are marked in Figure 2.

Since Property I${}^{+}(\epsilon_{2})$ is met, we claim that there is an optimality gap. This can be verified since the primal optimal solution

[TABLE]

which is marked in Figure 2, yields the globally optimal value $-1.5335857$, whereas the optimal value of (SP) is $-3.1269177$.

5 On why Theorem 3.2 is an extension of Theorem 2.9

The following facts support the claim that Theorem 3.2 is an extension of Theorem 2.9:

• •

The assumptions required for the validity of Theorem 3.2 are weaker than those needed in Theorem 2.9 (see Remark 2.6 for more details).

• •

Property I+, which determines the test in the necessary and sufficient condition of Theorem 3.2, involves an extra condition when compared with Property I used in Theorem 2.9. Furthermore, as the following proposition establishes, Property I+ and Property I are equivalent when $Q_{1}\succ 0$, as assumed in Theorem 2.9.

In short, if the assumptions of Theorem 2.9 are satisfied then those of Theorem 3.2 are also valid and the necessary and sufficient test is equivalent. However, the opposite does not hold, as the parameters defining the examples in section 4.1 and section 4.2 illustrate.

Proposition 5.1**.**

Consider (QP) where Slater’s condition holds and $Q_{1}\succ 0$. Let $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$ denote a pair of optimal solutions for (SP) and (SD), respectively. If $\operatorname{rank}(\hat{X})=2$ and $\hat{y}_{1}\neq 0$, then there exists a rank-one decomposition of $\hat{X}$, of the form $\hat{X}=\hat{x}_{1}\hat{x}_{1}^{T}+\hat{x}_{2}\hat{x}_{2}^{T}$, which satisfies M(q_{1})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{x}_{1}\hat{x}_{1}^{T}=M(q_{1})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{x}_{2}\hat{x}_{2}^{T}=0 and M(q_{1})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{x}_{1}\hat{x}_{2}^{T}\neq 0.

Proof.

By complimentary slackness (2.7b), $\hat{y}_{1}\neq 0$ implies M(q_{1})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{X}=0. By [20, Lemma 2.2], we can always obtain a rank-one decomposition $\hat{X}=\hat{x}_{1}\hat{x}_{1}^{T}+\hat{x}_{2}\hat{x}_{2}^{T}$ that satisfies

[TABLE]

We proceed to show that the pair $(\hat{x}_{1},\hat{x}_{2})$ obtained from a rank-one decomposition of $\hat{X}$ that satisfies (5.1) also yields M(q_{1})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{x}_{1}\hat{x}_{2}^{T}\neq 0. We can conduct a change of coordinates $\tilde{x}=Rx$ to make

[TABLE]

because

[TABLE]

where

[TABLE]

and \bigl{[}\begin{smallmatrix}{t}\\ {z}\end{smallmatrix}\bigr{]} and \bigl{[}\begin{smallmatrix}\tilde{t}\\ \tilde{z}\end{smallmatrix}\bigr{]} are partitions of $x$ and $\tilde{x}$, respectively, in the form of (2.2).

By contradiction, assume M(q_{1})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{x}_{1}\hat{x}_{2}^{T}=0. Hence, adopting the partition of $\hat{x}_{1}$ and $\hat{x}_{2}$ in the form of (2.2) and using the $M(q_{1})$ in (5.2), we have

[TABLE]

By Cauchy–Schwarz inequality, the last equality implies $\hat{z}_{1}$ and $\hat{z}_{2}$ are linearly dependent, and so are $\hat{x}_{1}$ and $\hat{x}_{2}$, which contradicts the fact that $\operatorname{rank}(\hat{X})=2$ and $\hat{X}=\hat{x}_{1}\hat{x}_{1}^{T}+\hat{x}_{2}\hat{x}_{2}^{T}$. ∎

6 Conclusions and future directions

In this article, we establish in Theorem 3.2 that a general QC2QP has an optimality gap if and only if it satisfies a condition we denote as Property I+, on account of the earlier test based on the so-called Property I [1]. According to our result, there is no optimality gap in the desirable case in which Property I+ is violated, and a solution of the QC2QP can be obtained from a rank-one decomposition of a solution of a semidefinite relaxation. Our result generalizes the state of the art stated in Theorem 2.9 (reproduced from [1, Theorem 4.2]), which uses Property I to establish a similar test subject to the additional requirement that at least one of the constraints is strictly convex. In contrast to Theorem 2.9, Theorem 3.2 remains valid when none of the inequalities of the QC2QP is strictly convex, but they are equivalent otherwise.

Future directions include tightening and eliminating the optimality gap of a QC2QP that is tested affirmative of Property I+. Here, one could possibly leverage existing methods that tighten or eliminate the gap for special cases of QC2QP such as, for example, those based on lifting [19], adding SOC constraints [21], or Lagrange multipliers [15]. Another interesting possibility is to establish an $\epsilon$-version of Theorem 3.2 which would bound the optimality gap by a function of $\epsilon$, when Property I+ is violated by $\epsilon$.

7 Appendix: Proof of Theorem 3.2

In order to prove Theorem 3.2, we shall use the following lemma whose proof follows from that of the sufficient condition of [13, Theorem 2.4]. As is evident from the analysis in [13, p. 19], this result does not rely on the assumption [13, (2.1)] required in other portions of [13, Chapter 2].

Lemma 7.1**.**

Let $X=VV^{T}$ be a solution of (SP), where $V$ is a matrix in $\mathbb{R}^{n\times r}$ and $r=\operatorname{rank}(X)$. Define a linear mapping $\mathcal{A}_{V}:\mathcal{S}^{r}\rightarrow\mathbb{R}^{3}$ as

[TABLE]

and consider the following conditions:

• (Lemma 7.1-C1)* All solutions $\tilde{X}$ of (SP) have rank $\tilde{r}\leq r$.*
• (Lemma 7.1-C2)* $\mathcal{N}(\mathcal{A}_{V})=\{0_{r\times r}\}$*

If conditions (Lemma 7.1-C1) and (Lemma 7.1-C2) hold then $X$ is the unique solution of (SP).

Proof of Theorem 3.2.

We will start by proving sufficiency, and later prove necessity.

(Proof of the sufficiency portion of Theorem 3.2):

In proving sufficiency, we start by assuming that $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$ is a pair of optimal solutions for the semidefinite relaxation (SP) and dual problem (SD), respectively, and that they satisfy Property I+.

Main idea: The main tenet of our proof of sufficiency is to establish the fact that $\hat{X}$ is a unique solution of (SP). Notice that since $\hat{X}$ satisfies Property I+, and hence has rank $2$, the aforementioned fact implies that (SP) has no rank-one optimal solution, in which case an optimality gap exists. Indeed, if there was no optimality gap then one could construct a rank-one optimal solution for (SP) from that of (QP).

Proving uniqueness of $\hat{X}$: We prove uniqueness by establishing that the conditions of Lemma 7.1 hold for $r=2$.

Establishing (Lemma 7.1-C1): Let $\tilde{X}$ denote any optimal solution of (SP). Using Sylvester’s inequality and the complementary condition (2.7a) we conclude that the following holds:

[TABLE]

that is, the maximum rank of an optimal solution of (SP) is 2.

Establishing (Lemma 7.1-C2): We use the fact that (I+.4) is assumed to hold to select a rank-one decomposition based on $(\hat{x}_{1},\hat{x}_{2})$ satisfying (2.8)–(3.1). Using this choice, we form $V$ such that $\hat{X}=VV^{T}$ and we also construct a matrix $\Delta$ as follows:

[TABLE]

where $\hat{z}_{1}$ and $\hat{z}_{2}$ are $n$-dimensional real vectors and $\hat{t}_{1},\hat{t}_{2},\Delta_{1},\Delta_{2}$, and $\Delta_{3}$ are real numbers. Next, we show that $\mathcal{N}(\mathcal{A}_{V})=\{0_{2\times 2}\}$ holds. Equivalently, we need to show that the only solution $\Delta$ in $\mathcal{S}^{2}$ for the following system of equations is $\Delta=0_{2\times 2}$:

[TABLE]

From the fact that (I.1) is true, we invoke $\hat{y}_{1}\hat{y}_{2}\neq 0$ in connection with (2.7c) to conclude that the following holds:

[TABLE]

We proceed to define $\alpha$ as follows:

[TABLE]

and

[TABLE]

where $(a)$ above follows from (2.8). We proceed by using $\Gamma$ to express (7.4) as a system of linear equalities for $\Delta_{1},\Delta_{2}$ and $\Delta_{3}$:

[TABLE]

Notice that I_{00}\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{X}=1 implies $\hat{t}_{1}^{2}+\hat{t}_{2}^{2}=1$, which causes the matrix $\Gamma$ to be full rank because

[TABLE]

Here, we also used the fact that $(\hat{x}_{1},\hat{x}_{2})$ satisfying (2.9) and (3.1) implies that $\alpha\neq 0$ and M(q_{1})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{x}_{1}\hat{x}_{2}^{T}\neq 0, respectively. So $\Delta_{1},\Delta_{2}$, and $\Delta_{3}$ are all zero and the only solution of (7.4) is $\Delta=0_{2\times 2}$.

This concludes our proof of the sufficient condition.

(Proof of the necessity portion of Theorem 3.2):

In proving necessity, we start by assuming that $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$ is a pair of optimal solutions for (SP) and (SD), respectively, and that the pair violates Property I+.

We proceed to prove in a constructive way that there is no optimality gap. Specifically, we will examine five exhaustive and mutually exclusive cases for which Property I+ is violated. We will outline in each of these cases, denoted by Case $1$ through Case $5$, how to construct a rank-one solution for (SP) (and hence also (QP)), thus showing that there is no optimality gap. We shall identify each case using the labels (I.1)–(I.4) and (I+.4) of the conditions specifying Property I and Property I+ in Definitions 2.8 and 3.1, respectively. The flowchart in Fig. 3 summarizes how we identify the cases in terms of the condition labels.

(Case 1: (I.1) fails.)

In this case, $\hat{y}_{1}\hat{y}_{2}=0$. We proceed by assuming, without loss of generality, that $\hat{y}_{2}=0$, as the analysis for $\hat{y}_{1}=0$ would be analogous. Since $\hat{y}_{2}=0$, the complementary slackness equality (2.7c) indicates that $\hat{X}$ is only required to satisfy the following feasibility constraint of (SP):

[TABLE]

If $\hat{y}_{1}>0$ holds then the complementary slackness condition (2.7b) indicates that $\hat{X}$ satisfies the corresponding feasibility constraint in (SP) with equality, and the case in which $\hat{y}_{1}=0$ causes (2.7b) to be satisfied for any feasible solution:

[TABLE]

Subsequently, we follow the procedure in the proof of [20, Lemma 2.2] to compute a rank-one decomposition $\hat{X}=\hat{x}_{1}\hat{x}_{1}^{T}+\hat{x}_{2}\hat{x}_{2}^{T}+\dots+\hat{x}_{r}\hat{x}_{r}^{T}$, with $r:=\operatorname{rank}(\hat{X})$, such that the following holds:

[TABLE]

Now, notice that (7.9) implies that there exists $i$ in $\{1,2,\dots,r\}$ for which M(q_{2})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{x}_{i}\hat{x}_{i}^{T}\leq 0 holds. Hence, we can order the indices of $\{\hat{x}_{1},\ldots,\hat{x}_{r}\}$ so that $\hat{x}_{1}$ satisfies

[TABLE]

We proceed to argue that $\hat{x}_{1}\hat{x}_{1}^{T}/\hat{t}_{1}^{2}$ is an optimal solution of (SP), or, equivalently, that $\hat{x}_{1}/\hat{t}_{1}$ is an optimal solution to (QP), where $\hat{t}_{1}$ comes from the partition of $\hat{x}_{1}$ in (2.2). In order to do so, we notice that $\hat{x}_{1}$ is in the range space of $\hat{X}$, and, hence, must be in the null space of $\hat{Z}$, which implies $\hat{x}_{1}\hat{x}_{1}^{T}\hat{Z}=0_{(n+1)\times(n+1)}$. This fact, together with (7.11) and (7.12), establishes that $\hat{x}_{1}\hat{x}_{1}^{T}/\hat{t}_{1}^{2}$ satisfies (2.7). Hence, we can invoke the complementary slackness principle to argue that $\hat{x}_{1}\hat{x}_{1}^{T}/\hat{t}_{1}^{2}$ is an optimal solution of (SP). Finally, it remains to show that $\hat{t}_{1}\neq 0$. We do so by showing that ${\hat{t}_{1}=0}$ would contradict 2.3, which is Slater’s condition for (SD). A key observation is that if $\hat{t}_{1}=0$ then $\hat{X}+\alpha\hat{x}_{1}\hat{x}_{1}^{T}$ is a solution to (SP) satisfying (2.7) for every $\alpha\geq 0$. Hence, we can use \hat{Z}\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{x}_{1}\hat{x}_{1}^{T}=0 to invoke the complementary slackness principle to argue that $\mathbb{X}:=\{\hat{X}+\alpha\hat{x}_{1}\hat{x}_{1}^{T}\ |\ \alpha\geq 0\}$ is a set of optimal solutions for (SP) when $\hat{t}_{1}=0$. Given that $\hat{x}_{1}\neq 0$, we conclude that $\mathbb{X}$ is unbounded.

However, according to [16, Theorem 2.5], the unboundedness of $\mathbb{X}$ would contradict 2.3.

**(Case 2: (I.1) holds and (I.3) fails.) **

In this case, $\hat{y}_{1}\hat{y}_{2}\neq 0$ and $\operatorname{rank}(\hat{X})\neq 2$. Notice that $r:=\operatorname{rank}(\hat{X})$ is positive because, from (SP), we know that I_{00}\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{X}=1 holds. If $r=1$ then $\hat{X}=\hat{x}\hat{x}^{T}$ is already a rank-one optimal solution to (SP), or equivalently, $\hat{x}$ is an optimal solution to (QP). Hence, subsequently, we proceed to analyze the case in which $r\geq 3$.

We start by invoking the complementary condition (2.7) to conclude that $\hat{y}_{1}\hat{y}_{2}\neq 0$ implies:

[TABLE]

Hence, given that $r\geq 3$, by [1, Theorem 3.4], one can find a rank-one decomposition ${\hat{X}=\hat{x}_{1}\hat{x}_{1}^{T}+\hat{x}_{2}\hat{x}_{2}^{T}+\dots+\hat{x}_{r}\hat{x}_{r}^{T}}$ such that the following holds:

[TABLE]

We proceed to argue that $\hat{x}_{1}\hat{x}_{1}^{T}/\hat{t}_{1}^{2}$ is an optimal solution of (SP), or, equivalently, that $\hat{x}_{1}/\hat{t}_{1}$ is an optimal solution to (QP), where $\hat{t}_{1}$ comes from the partition of $\hat{x}_{1}$ in (2.2). In order to do so, we notice that $\hat{x}_{1}$ is in the range space of $\hat{X}$, and, hence, must be in the null space of $\hat{Z}$, which implies $\hat{x}_{1}\hat{x}_{1}^{T}\hat{Z}=0_{(n+1)\times(n+1)}$. This fact, together with (7.14), establishes that $\hat{x}_{1}\hat{x}_{1}^{T}/\hat{t}_{1}^{2}$ satisfies (2.7). Hence, we can invoke the complementary slackness principle to argue that $\hat{x}_{1}\hat{x}_{1}^{T}/\hat{t}_{1}^{2}$ is an optimal solution of (SP).

Finally, we proceed to prove by contradiction that $\hat{t}_{1}\neq 0$. Our strategy is to show that assuming $\hat{t}_{1}=0$ would contradict Slater’s condition (2.6a). Hence, assuming that $\hat{t}_{1}=0$ and recalling that $\hat{x}_{1}^{T}\hat{Z}\hat{x}_{1}=\operatorname{Tr}(\hat{x}_{1}\hat{x}_{1}^{T}\hat{Z})=0$, we can use the first constraint in (SD) to write:

[TABLE]

where $\hat{z}_{1}$ comes from the partition of $\hat{x}_{1}$ in (2.2). On the other hand, when we substitute $\hat{t}_{1}=0$ in (7.14), we get:

[TABLE]

Combining (7.15) and (7.16), we conclude that:

[TABLE]

However, (7.15)–(7.17) would imply that the following would hold for all $\bar{y}_{1},\bar{y}_{2}>0$:

[TABLE]

which would contradict 2.3, unless $\hat{z}_{1}=0_{n\times 1}$ or, equivalently, $\hat{x}_{1}=0_{(n+1)\times 1}$ because we are assuming $\hat{t}_{1}=0$. However, the definition of rank-one decomposability [1] guarantees that $\hat{x}_{1}\neq 0_{(n+1)\times 1}$. Hence, we reached a contradiction, from which we conclude that $\hat{t}_{1}\neq 0$.

Remark 7.2**.**

Before we proceed our analysis of the subsequent cases, we introduce the following sets:

[TABLE]

From [20, Lemma 2.2], we conclude that when $\hat{X}$ is a solution of (SP) satisfying $\operatorname{rank}(\hat{X})=2$ and M(q_{1})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{X}=0, it is always possible to obtain a rank-one decomposition $\hat{X}=\hat{x}_{1}\hat{x}_{1}^{T}+\hat{x}_{2}\hat{x}_{2}^{T}$ that complies with (2.8). This fact and (2.7b) imply that $\mathbb{O}(\hat{X})$ is nonempty when $\hat{y}_{1}\neq 0$ and $\operatorname{rank}(\hat{X})$ is $2$. Hence, we can infer that if (I.1) and (I.3) hold then $\mathbb{O}(\hat{X})$ is nonempty and the cases in which (I.4) fails or holds have the following further implications in the terms of $\mathbb{O}^{+}(\hat{X})$:

• •

*If (I.1) and (I.3) hold, and (I.4) fails, then $\mathbb{O}(\hat{X})$ is nonempty, but $\mathbb{O}^{+}(\hat{X})$ is empty, or, equivalently, (2.9) is violated for every pair $(\hat{x}_{1},\hat{x}_{2})$ in $\mathbb{O}(\hat{X})$. *

• •

*If (I.1), (I.3), and (I.4) hold then $\mathbb{O}^{+}(\hat{X})$ is nonempty. *

(Case 3: (I.1) and (I.3) hold, and (I.4) fails.)

In this case, Remark 7.2 guarantees that $\mathbb{O}(\hat{X})$ is nonempty but (2.9) is violated for every pair in it, which, together with (2.7c) and the fact that $\hat{y}_{2}\neq 0$, imply that:

[TABLE]

We start by selecting $(\hat{x}_{1},\hat{x}_{2})$ in $\mathbb{O}(\hat{X})$ and adopting the partition of $\hat{x}_{1}$ and $\hat{x}_{2}$ in (7.3). Consequently, I_{00}\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{X}=1 implies that $\hat{t}_{1}^{2}+\hat{t}_{2}^{2}=1$, that is, at least one of $\hat{t}_{1}$ and $\hat{t}_{2}$ is nonzero. Without loss of generality, assume $\hat{t}_{1}\neq 0$. Since $\hat{x}_{1}$ is in the range space of $\hat{X}$, (2.7a) implies that it must be in the null space of $\hat{Z}$, leading to $\hat{x}_{1}\hat{x}_{1}^{T}\hat{Z}=0_{(n+1)\times(n+1)}$. Now, recall that because it is an element of $\mathbb{O}(\hat{X})$, $(\hat{x}_{1},\hat{x}_{2})$ satisfies M(q_{1})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{x}_{1}\hat{x}_{1}^{T}=0. Finally, using the facts that $\hat{x}_{1}\hat{x}_{1}^{T}\hat{Z}=0_{(n+1)\times(n+1)}$, M(q_{1})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{x}_{1}\hat{x}_{1}^{T}=0 and (7.21) we can invoke (2.7) to conclude that $\hat{x}_{1}\hat{x}_{1}^{T}/\hat{t}_{1}^{2}$ is a rank-one optimal solution of (SP). Therefore, $\hat{x}_{1}/\hat{t}_{1}$ is an optimal solution of (QP).

(Case 4: (I.1), (I.3), and (I.4) hold, and (I.2) fails)

In this case, $\hat{y}_{1}\hat{y}_{2}\neq 0$, $\operatorname{rank}(\hat{Z})\neq n-1$, $\operatorname{rank}(\hat{X})=2$, and, by Remark 7.2, $\mathbb{O}^{+}(\hat{X})$ is nonempty.

We start by noticing that the following holds:

[TABLE]

which implies that the rank condition below is statisfied

[TABLE]

indicating that the following inequality holds:

[TABLE]

Now $\hat{X}+\hat{Z}$ is singular because $\operatorname{rank}(\hat{X}+\hat{Z})\leq\operatorname{rank}(\hat{X})+\operatorname{rank}(\hat{Z})$. Also, both $\hat{X}$ and $\hat{Z}$ are positive semidefinite. So there must be a nontrivial $(n+1)$-dimensional real vector $y$ in the intersection of the null space of $\hat{X}$ and the null space of $\hat{Z}$. Select a pair $(\hat{x}_{1},\hat{x}_{2})$ in $\mathbb{O}^{+}(\hat{X})$ and let

[TABLE]

Consequently, $\operatorname{rank}(X)=3$ and $X\hat{Z}=0_{(n+1)\times(n+1)}$ because $\hat{X}\hat{Z}=0_{(n+1)\times(n+1)}$ and $y$ is in the null space of $\hat{Z}$. By [1, Lemma 3.3], we know there exists an $(n+1)$-dimensional real vector $x$ such that $X$ is rank-one decomposable at $x$ and that

[TABLE]

Since $x$ is in the range space of $X$, it must be in the null space of $\hat{Z}$ which implies $xx^{T}\hat{Z}=0_{(n+1)\times(n+1)}$. Therefore, by the complementary condition (2.7a), $xx^{T}\hat{Z}=0_{(n+1)\times(n+1)}$ and (7.28) imply that $xx^{T}/t^{2}$ is an optimal solution of (SP), where $t$ comes from the partition of $x$ in (2.2). Hence, $x/t$ is an optimal solution of (QP). Note that $t\neq 0$ follows the same argument of $\hat{t}_{1}\neq 0$ in Case 2.

(Case 5: (I.1)–(I.4) hold and (I+.4) fails.)

This is the case in which $\hat{X}$ and $(\hat{Z},\hat{y}_{0},\hat{y}_{1},\hat{y}_{2})$ violate Property I+, while satisfying Property I. Notice that there is no case comparable to this in [1].

We start by observing that since (I.1)–(I.4) hold by assumption, it follows that $\hat{y}_{1}\hat{y}_{2}\neq 0$, $\operatorname{rank}(\hat{Z})=n-1$, $\operatorname{rank}(\hat{X})=2$, and Remark 7.2 guarantees that $\mathbb{O}^{+}(\hat{X})$ is nonempty. Furthermore, under the stated assumption that (I+.4) fails, (3.1) is violated for all pairs in $\mathbb{O}^{+}(\hat{X})$. Hence, the following holds:

[TABLE]

We proceed by selecting $(\hat{x}_{1},\hat{x}_{2})$ in $\mathbb{O}^{+}(\hat{X})$. Our strategy is to use such a choice to construct another rank-one decomposition $\hat{X}=\check{x}_{1}\check{x}_{1}^{T}+\check{x}_{2}\check{x}_{2}^{T}$ satisfying the following equalities:

[TABLE]

Because (7.30) will hold by construction, $(\check{x}_{1},\check{x}_{2})$ will also satisfy (7.21). Consequently, once the construction of $(\check{x}_{1},\check{x}_{2})$ is complete, we can employ the method used in Case $3$ to compute an optimal solution of (QP).

Hence, it remains to construct a pair $(\check{x}_{1},\check{x}_{2})$ that satisfies (7.30). We start by noticing that (7.29) together with (2.8), which is satisfied for our choice $(\hat{x}_{1},\hat{x}_{2})$ in $\mathbb{O}^{+}(\hat{X})$, imply that the following holds:

[TABLE]

The final step is to construct $\check{x}_{1}$ and $\check{x}_{2}$ as linear combinations of $\hat{x}_{1}$ and $\hat{x}_{2}$ so that the following equality holds:

[TABLE]

Since from (2.7c) and $\hat{y}_{2}\neq 0$ we know that M(q_{2})\mathchoice{\mathbin{\vbox{\hbox{\scalebox{0.5}{\displaystyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\textstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptstyle\bullet}}}}}{\mathbin{\vbox{\hbox{\scalebox{0.5}{\scriptscriptstyle\bullet}}}}}\hat{X}=0, we proceed to determine such $\check{x}_{1}$ and $\check{x}_{2}$ by following the procedure in the proof of [20, Lemma 2.2]. To do so, we substitute $G$, $u_{i}$ and $\bar{u}_{i}$ present in [20, p. 249] with $M(q_{2})$, $\hat{x}_{i}$ and $\check{x}_{i}$, respectively, for $i$ in $\{1,2\}$. The rank-one decomposition $\hat{X}=\check{x}_{1}\check{x}_{1}^{T}+\check{x}_{2}\check{x}_{2}^{T}$ obtained in this way will satisfy (7.30).

This concludes our proof of the necessary condition. ∎

Remark 7.3**.**

The proof of [21, Theorem 2.6] also shows the uniqueness of the solution of (SP) in the CDT subproblem when Property I holds. The proof uses a property of the boundary points of an SOC whereas we use a result on the uniqueness of the solution of a semidefinite program.

Acknowledgments

The authors would like to thank Dr. MirSaleh Bahavarnia for reading this paper carefully and providing helpful suggestions.