This paper was converted on www.awesomepapers.org from LaTeX by an anonymous user.
Want to know more? Visit the Converter page.

Constructing QCQP Instances Equivalent to Their SDP Relaxations

Masakazu Kojima Department of Data Science for Business Innovation (kojima@is.titech.ac.jp).    Naohiko Arima (nao_\_arima@me.com).    Sunyoung Kim Department of Mathematics, Ewha W. University, 52 Ewhayeodae-gil, Sudaemoon-gu, Seoul 03760, Korea (skim@ewha.ac.kr). The research was supported by NRF 2021-R1A2C1003810.   
Abstract

General quadratically constrained quadratic programs (QCQPs) are challenging to solve as they are known to be NP-hard. A popular approach to approximating QCQP solutions is to use semidefinite programming (SDP) relaxations. It is well-known that the optimal value η\eta of the SDP relaxation problem bounds the optimal value ζ\zeta of the QCQP from below, i.e., ηζ\eta\leq\zeta. The two problems are considered equivalent if η=ζ\eta=\zeta. In the recent paper by Arima, Kim and Kojima [arXiv:2409.07213], a class of QCQPs that are equivalent to their SDP relaxations are proposed with no condition imposed on the quadratic objective function, which can be chosen arbitrarily. In this work, we explore the construction of various QCQP instances within this class to complement the results in [arXiv:2409.07213]. Specifically, we first construct QCQP instances with two variables and then extend them to higher dimensions. We also discuss how to compute an optimal QCQP solution from the SDP relaxation.

Key words. QCQPs satisfying exact conditions, exact semidefinite programming relaxations, computing optimal solutions of QCQPs.

AMS Classification. 90C20, 90C22, 90C25, 90C26.

1 Introduction

We study a quadratically constrained quadratic program (QCQP) that minimizes a quadratic objective function in multiple real variables over a feasible region represented by quadratic inequalities in the variables. In general, QCQPs are NP-hard [10] and become increasingly difficult to solve as the number of variables grows. It is well-known that the optimal value ζ\zeta of a QCQP is bounded by the optimal value η\eta of its SDP relaxation from below [11, 12]. This property has frequently been utilized for (approximately) solving the QCQP since the SDP can be solved numerically. We say that the QCQP and the SDP relaxation are equivalent if η=ζ\eta=\zeta. In [3], conditions on the feasible region were presented to ensure the equivalence of a QCQP and its SDP relaxation. We first illustrate their conditions using the following figure.

    The unshaded area, including its boundary, represents the feasible region where two variables u1u_{1} and u2u_{2} can vary without limitation, while the green ellipsoidal regions indicate the restricted zones labeled 1,,71,\ldots,7. The feasible region is enclosed by the ellipsoid 8 within which 7 restricted zones are positioned. We regard the area outside of the ellipsoid 8 as a restricted zone. Given a quadratic function q(u1,u2)q(u_{1},u_{2}) in 2 variables u1,u2u_{1},u_{2}, say q(u1,u2)=u12+2u1u2+3u222u12u2q(u_{1},u_{2})=u_{1}^{2}+2u_{1}u_{2}+3u_{2}^{2}-2u_{1}-2u_{2}, we consider the optimization problem of minimizing q(u1,u2)q(u_{1},u_{2}) over the feasible region. The essential properties of this optimization problem are:

[Uncaptioned image]

(a) The objective function is a quadratic function in multiple variables.

(b) Each restricted zone is represented by a single quadratic inequality in multiple variables.

(c) Two distinct restricted zones could intersect with their boundaries but their interior never overlap (a more rigid condition is given as Condition (B)’ below).

Properties (a) and (b) imply that this problem is a QCQP, while (c) represents a condition imposed on the QCQPs that we consider throughout the paper. Since (c) imposes restrictions only on the constraints of QCQPs, the quadratic objective function can be chosen arbitrarily. It was shown in [3] that every QCQP satisfying (c) is equivalent to its SDP relaxation. The study in [3], however, focused mainly on the theoretical relationships conditions, including (c) that ensure the equivalence of QCQPs and their SDP relaxations, but did not fully address a variety of QCQP instances satisfying (c), nor the construction of such instances. This paper aims to complement the work in [3] by showing the construction of various QCQP instances satisfying (c). In addition to the instance mentioned above, Figures 1 through 13 illustrate two-dimensional feasible regions satisfying condition (c). We also generate higher-dimensional QCQP instances by extending the two-dimensional instances.

For the equivalence of a QCQP and its SDP relaxation, we need to impose some conditions on the quadratic functions involved in the QCQP. Broadly speaking, there are three types of conditions that guarantee the equivalence between QCQPs and their SDP relaxations. The first type is the convexity of quadratic objective function and quadratic inequality constraints. QCQPs satisfying this condition are called as convex quadratic programs. The second type of conditions concerns the sign pattern of the coefficient matrices in both of quadratic objective and constraint functions. A class of QCQPs satisfying this type of conditions was proposed in [17] and has since been extensively studied in the literature [4, 7, 13].

The focus of this paper is on the third type which imposes specific conditions only on quadratic inequality constraints, without any conditions on the objective function. We note that condition (c) mentioned above falls under this type. If the constraints of a QCQP satisfy this type of conditions, then we can arbitrarily choose any quadratic objective function so that the QCQP is equivalent to its SDP relaxation. QCQPs satisfying this type of conditions were studied in [1, 2, 3, 9]. Specifically, we focus on Condition (B)’ below proposed in [3]. We note that Condition (B)’ is translated into condition (c) above.

For the subsequent discussion, we describe the standard form QCQP as the following:

ζ\displaystyle\zeta =\displaystyle= inf{𝒙T𝑸𝒙:𝒙n,𝒙T𝑩𝒙0(𝑩),𝒙T𝑯𝒙=1}\displaystyle\inf\left\{\mbox{\boldmath$x$}^{T}\mbox{\boldmath$Q$}\mbox{\boldmath$x$}:\mbox{\boldmath$x$}\in\mathbb{R}^{n},\ \mbox{\boldmath$x$}^{T}\mbox{\boldmath$B$}\mbox{\boldmath$x$}\geq 0\ (\mbox{\boldmath$B$}\in\mbox{$\cal B$}),\ \mbox{\boldmath$x$}^{T}\mbox{\boldmath$H$}\mbox{\boldmath$x$}=1\right\} (1)
=\displaystyle= inf{𝑸𝒙𝒙T:𝒙n,𝑩𝒙𝒙T0(𝑩),𝑯𝒙𝒙T=1}\displaystyle\inf\left\{\mbox{\boldmath$Q$}\bullet\mbox{\boldmath$x$}\mbox{\boldmath$x$}^{T}:\mbox{\boldmath$x$}\in\mathbb{R}^{n},\ \mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}\mbox{\boldmath$x$}^{T}\geq 0\ (\mbox{\boldmath$B$}\in\mbox{$\cal B$}),\ \mbox{\boldmath$H$}\bullet\mbox{\boldmath$x$}\mbox{\boldmath$x$}^{T}=1\right\}
=\displaystyle= inf{𝑸𝑿:𝑿𝚪n,𝑩𝑿0(𝑩),𝑯𝑿=1}.\displaystyle\inf\left\{\mbox{\boldmath$Q$}\bullet\mbox{\boldmath$X$}:\mbox{\boldmath$X$}\in\mbox{\boldmath$\Gamma$}^{n},\ \mbox{\boldmath$B$}\bullet\mbox{\boldmath$X$}\geq 0\ (\mbox{\boldmath$B$}\in\mbox{$\cal B$}),\ \mbox{\boldmath$H$}\bullet\mbox{\boldmath$X$}=1\right\}.

Here

n:the n-dimensional Euclidean space of column vectors 𝒙=(x1,,xn),\displaystyle\mathbb{R}^{n}:\mbox{the $n$-dimensional Euclidean space of column vectors $\mbox{\boldmath$x$}=(x_{1},\ldots,x_{n})$},
𝕊n:the linear space of n×n symmetric matrices,\displaystyle\mathbb{S}^{n}:\mbox{the linear space of $n\times n$ symmetric matrices},
𝕊+n𝕊n:the convex cone of n×n positive semidefinite matrices,\displaystyle\mathbb{S}^{n}_{+}\subseteq\mathbb{S}^{n}:\mbox{the convex cone of $n\times n$ positive semidefinite matrices},
𝚪n={𝒙𝒙T𝕊n:𝒙n},𝑸𝕊n,𝑯𝕊n,\displaystyle\mbox{\boldmath$\Gamma$}^{n}=\left\{\mbox{\boldmath$x$}\mbox{\boldmath$x$}^{T}\in\mathbb{S}^{n}:\mbox{\boldmath$x$}\in\mathbb{R}^{n}\right\},\ \mbox{\boldmath$Q$}\in\mathbb{S}^{n},\ \mbox{\boldmath$H$}\in\mathbb{S}^{n},
:a nonempty subset of 𝕊n,\displaystyle\mbox{$\cal B$}:\mbox{a nonempty subset of $\mathbb{S}^{n}$},
𝒙T:the transposed row vector of 𝒙n,\displaystyle\mbox{\boldmath$x$}^{T}:\mbox{the transposed row vector of $\mbox{\boldmath$x$}\in\mathbb{R}^{n}$},
𝑨𝑿=i=1nj=1nAijXij:the inner product of 𝑨,𝑿𝕊n.\displaystyle\mbox{\boldmath$A$}\bullet\mbox{\boldmath$X$}=\sum_{i=1}^{n}\sum_{j=1}^{n}A_{ij}X_{ij}:\ \mbox{the inner product of $\mbox{\boldmath$A$},\ \mbox{\boldmath$X$}\in\mathbb{S}^{n}$}.

The set 𝚪n\mbox{\boldmath$\Gamma$}^{n} forms a cone in 𝕊+n\mathbb{S}^{n}_{+}; that is, for every 𝑿𝚪n\mbox{\boldmath$X$}\in\mbox{\boldmath$\Gamma$}^{n} and λ0\lambda\geq 0, it holds that λ𝑿𝚪n\lambda\mbox{\boldmath$X$}\in\mbox{\boldmath$\Gamma$}^{n}. It is not convex unless n=1n=1. We also know that 𝕊+n=co𝚪n\mathbb{S}^{n}_{+}=\mbox{co}\mbox{\boldmath$\Gamma$}^{n} (the convex hull of 𝚪n\mbox{\boldmath$\Gamma$}^{n}). We may assume that \cal B is finite in this paper, though Theorem 1.1 presented below remain valid even when the cardinality of \cal B is infinite.

For every 𝒙n\mbox{\boldmath$x$}\in\mathbb{R}^{n}, 𝒙𝒙T𝕊n\mbox{\boldmath$x$}\mbox{\boldmath$x$}^{T}\in\mathbb{S}^{n} is an n×nn\times n rank-11 positive semidefinite matrix, and 𝚪n\mbox{\boldmath$\Gamma$}^{n} can be written as 𝚪n={𝑿𝕊n:rank𝑿=1}\mbox{\boldmath$\Gamma$}^{n}=\{\mbox{\boldmath$X$}\in\mathbb{S}^{n}:\mbox{rank}\mbox{\boldmath$X$}=1\}. Hence we can rewrite (1) as

ζ\displaystyle\zeta =\displaystyle= inf{𝑸𝑿:𝑿𝕊+n,rank𝑿=1,𝑩𝑿0(𝑩),𝑯𝑿=1}.\displaystyle\inf\left\{\mbox{\boldmath$Q$}\bullet\mbox{\boldmath$X$}:\mbox{\boldmath$X$}\in\mathbb{S}^{n}_{+},\ \mbox{rank}\mbox{\boldmath$X$}=1,\ \mbox{\boldmath$B$}\bullet\mbox{\boldmath$X$}\geq 0\ (\mbox{\boldmath$B$}\in\mbox{$\cal B$}),\ \mbox{\boldmath$H$}\bullet\mbox{\boldmath$X$}=1\right\}.

If we remove rank𝑿=1\mbox{\boldmath$X$}=1 in the QCQP above (or relax 𝚪n\mbox{\boldmath$\Gamma$}^{n} by 𝕊+n𝚪n\mathbb{S}^{n}_{+}\supseteq\mbox{\boldmath$\Gamma$}^{n} in QCQP (1)), we obtain an SDP relaxation of QCQP (1):

η\displaystyle\eta =\displaystyle= inf{𝑸𝑿:𝑿𝕊+n,𝑩𝑿0(𝑩),𝑯𝑿=1}.\displaystyle\inf\left\{\mbox{\boldmath$Q$}\bullet\mbox{\boldmath$X$}:\mbox{\boldmath$X$}\in\mathbb{S}^{n}_{+},\ \mbox{\boldmath$B$}\bullet\mbox{\boldmath$X$}\geq 0\ (\mbox{\boldmath$B$}\in\mbox{$\cal B$}),\ \mbox{\boldmath$H$}\bullet\mbox{\boldmath$X$}=1\right\}. (2)

In general, ηζ\eta\leq\zeta holds.

As a special case of QCQP (1), we ocus on the case where 𝑯𝕊n\mbox{\boldmath$H$}\in\mathbb{S}^{n} is the diagonal matrix with the diagonal entries (0,,0,1)n(0,\ldots,0,1)\in\mathbb{R}^{n}, i.e., 𝑯=diag(0,,0,1)𝕊n\mbox{\boldmath$H$}=\mbox{diag}(0,\ldots,0,1)\in\mathbb{S}^{n}. Then the identity 𝑯𝒙𝒙T=1\mbox{\boldmath$H$}\bullet\mbox{\boldmath$x$}\mbox{\boldmath$x$}^{T}=1 implies either xn=1x_{n}=1 or xn=1x_{n}=-1. Since 𝑩𝒙𝒙T=𝑩(𝒙)(𝒙)T\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}\mbox{\boldmath$x$}^{T}=\mbox{\boldmath$B$}\bullet(-\mbox{\boldmath$x$})(-\mbox{\boldmath$x$})^{T} (𝑩)(\mbox{\boldmath$B$}\in\mbox{$\cal B$}) and 𝑸𝒙𝒙T=𝑸(𝒙)(𝒙)T\mbox{\boldmath$Q$}\bullet\mbox{\boldmath$x$}\mbox{\boldmath$x$}^{T}=\mbox{\boldmath$Q$}\bullet(-\mbox{\boldmath$x$})(-\mbox{\boldmath$x$})^{T} for every 𝒙n\mbox{\boldmath$x$}\in\mathbb{R}^{n}, we can fix xn=1x_{n}=1 in this case. Therefore, we can rewrite QCQP (1) with 𝑯=diag(0,,0,1)𝕊n\mbox{\boldmath$H$}=\mbox{diag}(0,\ldots,0,1)\in\mathbb{S}^{n} as

ζ\displaystyle\zeta =\displaystyle= inf{(𝒖1)T𝑸(𝒖1):𝒖n1,(𝒖1)T𝑩(𝒖1)0(𝑩)}.\displaystyle\inf\left\{\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\mbox{\boldmath$Q$}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}:\mbox{\boldmath$u$}\in\mathbb{R}^{n-1},\ \begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\mbox{\boldmath$B$}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}\geq 0\ (\mbox{\boldmath$B$}\in\mbox{$\cal B$})\right\}. (3)

To simplify the subsequent discussion, we introduce the following notation:

𝕁+(𝑩)or 𝕁0(𝑩)\displaystyle\mathbb{J}_{+}(\mbox{\boldmath$B$})\ \mbox{or }\mathbb{J}_{0}(\mbox{\boldmath$B$}) \displaystyle\equiv {𝑿𝕊+n:𝑩𝑿or =0},\displaystyle\left\{\mbox{\boldmath$X$}\in\mathbb{S}^{n}_{+}:\mbox{\boldmath$B$}\bullet\mbox{\boldmath$X$}\geq\ \mbox{or }=0\right\},
𝕁+()\displaystyle\mathbb{J}_{+}(\mbox{$\cal B$}) \displaystyle\equiv 𝑩𝕁+(𝑩)={𝑿𝕊+n:𝑩𝑿0(𝑩)},\displaystyle\bigcap_{\mbox{\boldmath$B$}\in\mbox{$\cal B$}}\mathbb{J}_{+}(\mbox{\boldmath$B$})=\left\{\mbox{\boldmath$X$}\in\mathbb{S}^{n}_{+}:\mbox{\boldmath$B$}\bullet\mbox{\boldmath$X$}\geq 0\ (\mbox{\boldmath$B$}\in\mbox{$\cal B$})\right\},
𝑩,𝑩 or 𝑩<\displaystyle\mbox{\boldmath$B$}_{\geq},\ \mbox{\boldmath$B$}_{\leq}\ \mbox{ or }\mbox{\boldmath$B$}_{<} \displaystyle\equiv {𝒖n1:(𝒖1)T𝑩(𝒖1),or <0},\displaystyle\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{n-1}:\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\mbox{\boldmath$B$}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}\geq,\ \leq\ \mbox{or }<0\right\},
\displaystyle\mbox{$\cal B$}_{\geq} \displaystyle\equiv 𝑩𝑩={𝒖n1:(𝒖1)T𝑩(𝒖1)0(𝑩)}\displaystyle\bigcap_{\mbox{\boldmath$B$}\in\mbox{$\cal B$}}\mbox{\boldmath$B$}_{\geq}=\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{n-1}:\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\mbox{\boldmath$B$}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}\geq 0\ (\mbox{\boldmath$B$}\in\mbox{$\cal B$})\right\}
(the feasible region of QCQP (3))

for every 𝑩𝕊n\mbox{\boldmath$B$}\in\mathbb{S}^{n} and 𝕊n\mbox{$\cal B$}\subseteq\mathbb{S}^{n}. Using the above notation, we can rewrite QCQP (1), its SDP relaxation (2) and QCQP (3) as

ζ\displaystyle\zeta =\displaystyle= inf{𝑸𝑿:𝑿𝚪n𝕁+(),𝑯𝑿=1},\displaystyle\inf\left\{\mbox{\boldmath$Q$}\bullet\mbox{\boldmath$X$}:\mbox{\boldmath$X$}\in\mbox{\boldmath$\Gamma$}^{n}\cap\mathbb{J}_{+}(\mbox{$\cal B$}),\ \mbox{\boldmath$H$}\bullet\mbox{\boldmath$X$}=1\right\}, (4)
η\displaystyle\eta =\displaystyle= inf{𝑸𝑿:𝑿𝕁+(),𝑯𝑿=1},\displaystyle\inf\left\{\mbox{\boldmath$Q$}\bullet\mbox{\boldmath$X$}:\mbox{\boldmath$X$}\in\mathbb{J}_{+}(\mbox{$\cal B$}),\ \mbox{\boldmath$H$}\bullet\mbox{\boldmath$X$}=1\right\}, (5)
ζ\displaystyle\zeta =\displaystyle= inf{(𝒖1)T𝑸(𝒖1):𝒖},\displaystyle\inf\left\{\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\mbox{\boldmath$Q$}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}:\mbox{\boldmath$u$}\in\mbox{$\cal B$}_{\geq}\right\}, (6)

respectively.

We now consider the following conditions on 𝕊n\mbox{$\cal B$}\subseteq\mathbb{S}^{n}.

(B)

𝕁0(𝑩)𝕁+()\mathbb{J}_{0}(\mbox{\boldmath$B$})\subseteq\mathbb{J}_{+}(\mbox{$\cal B$}) (𝑩)(\mbox{\boldmath$B$}\in\mbox{$\cal B$}).

(B)’

𝑨<𝑩=\mbox{\boldmath$A$}_{<}\cap\mbox{\boldmath$B$}_{\leq}=\emptyset for every distinct pair of 𝑨\mbox{\boldmath$A$}\in\mbox{$\cal B$} and 𝑩\mbox{\boldmath$B$}\in\mbox{$\cal B$}.

(C)’

𝑩<\mbox{\boldmath$B$}_{<}\not=\emptyset for every 𝑩\mbox{\boldmath$B$}\in\mbox{$\cal B$}.

(D)

There exist αB>0\alpha_{B}>0 (𝑩)(\mbox{\boldmath$B$}\in\mbox{$\cal B$}) such that αA𝑨+αB𝑩𝕊+n\alpha_{A}\mbox{\boldmath$A$}+\alpha_{B}\mbox{\boldmath$B$}\in\mathbb{S}^{n}_{+} for every distinct pair of 𝑨\mbox{\boldmath$A$}\in\mbox{$\cal B$} and 𝑩\mbox{\boldmath$B$}\in\mbox{$\cal B$}.

Conditions (B), (B)’ and (C)’ were introduced in [3, Section 1], and (D) is a special case of a sufficient condition discussed in [3, Lemma 3.4] for (B).

Theorem 1.1.

 

(i) Let 𝑸,𝑯𝕊n\mbox{\boldmath$Q$},\ \mbox{\boldmath$H$}\in\mathbb{S}^{n}. Assume that

𝕁+()^(𝚪n){ closed convex cone 𝕁𝕊+n:𝕁=co(𝕁𝚪n) },\displaystyle\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n})\equiv\left\{\mbox{ closed convex cone $\mathbb{J}\subseteq\mathbb{S}^{n}_{+}:\mathbb{J}=\mbox{co}(\mathbb{J}\cap\mbox{\boldmath$\Gamma$}^{n})$ }\right\},

where co(𝕁𝚪n)\mbox{co}(\mathbb{J}\cap\mbox{\boldmath$\Gamma$}^{n}) denotes the convex hull of 𝕁𝚪n\mathbb{J}\cap\mbox{\boldmath$\Gamma$}^{n}. Then

<η<η=ζ\displaystyle-\infty<\eta\ \Leftrightarrow-\infty<\eta=\zeta

in QCQP (4) and its SDP relaxation (5). (𝕁^(𝚪n)\mathbb{J}\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}) is called ROG (Rank-One Generated) in the literature [1, 6]).

(ii) Condition (B) \Rightarrow 𝕁+()^(𝚪n)\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}).

(iii) Condition (B)’ and (C)’ \Rightarrow 𝕁+()^(𝚪n)\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}).

(iv) Conditiion (D) \Rightarrow Conditions (B) and (B)’.

Proof.

See [9, Theorems 3.1] for (i), [3, Theorems 1.2] for (ii) and [3, Theorem 1.3] for (iii). (iv) can be proved easily. To prove (D) \Rightarrow (B), let 𝑩\mbox{\boldmath$B$}\in\mbox{$\cal B$} and 𝑿𝕁0(𝑩)\mbox{\boldmath$X$}\in\mathbb{J}_{0}(\mbox{\boldmath$B$}). Then we see that

0(αA𝑨+αB𝑩)𝑿=αA𝑨𝑿for every 𝑨,\displaystyle 0\leq(\alpha_{A}\mbox{\boldmath$A$}+\alpha_{B}\mbox{\boldmath$B$})\bullet\mbox{\boldmath$X$}=\alpha_{A}\mbox{\boldmath$A$}\bullet\mbox{\boldmath$X$}\ \mbox{for every }\mbox{\boldmath$A$}\in\mbox{$\cal B$},

which implies 𝑿𝕁+()\mbox{\boldmath$X$}\in\mathbb{J}_{+}(\mbox{$\cal B$}). To prove (D) \Rightarrow (B)’, assume on the contrary that 𝒖𝑨<𝑩\mbox{\boldmath$u$}\in\mbox{\boldmath$A$}_{<}\cap\mbox{\boldmath$B$}_{\leq}\not=\emptyset for some distinct 𝑨,𝑩\mbox{\boldmath$A$},\mbox{\boldmath$B$}\in\mbox{$\cal B$}. Then

0>(𝒖1)TαA𝑨(𝒖1)+(𝒖1)TαB𝑩(𝒖1)=(𝒖1)T(αA𝑨+αB𝑩)(𝒖1)0,\displaystyle 0>\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\alpha_{A}\mbox{\boldmath$A$}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}+\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\alpha_{B}\mbox{\boldmath$B$}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}=\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}(\alpha_{A}\mbox{\boldmath$A$}+\alpha_{B}\mbox{\boldmath$B$})\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}\geq 0,

which is a contradiction.

Remark 1.2.

Assertion (iii) above can be strengthened to

(iii)’ Condition (B)’ \Rightarrow 𝕁+()^(𝚪n)\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}).

In fact, we can prove that Condition (C)’ is equivalent to 𝕊+n=\mbox{$\cal B$}\cap\mathbb{S}^{n}_{+}=\emptyset. If 𝑩𝕊+n\mbox{\boldmath$B$}\in\mathbb{S}^{n}_{+} then 𝕁+(𝑩)=𝕊+n\mathbb{J}_{+}(\mbox{\boldmath$B$})=\mathbb{S}^{n}_{+} and 𝑩=n1\mbox{\boldmath$B$}_{\geq}=\mathbb{R}^{n-1}. Hence 𝕁+(\𝕊+n)=𝕁+()\mathbb{J}_{+}(\mbox{$\cal B$}\backslash\mathbb{S}^{n}_{+})=\mathbb{J}_{+}(\mbox{$\cal B$}) and (\𝕊+n)=(\mbox{$\cal B$}\backslash\mathbb{S}^{n}_{+})_{\geq}=\mbox{$\cal B$}_{\geq}. Therefore, we will not explicitly refer to Condition (C)’ in the following discussion, assuming that it is satisfied.

It should be noted that 𝕁+()^(𝚪n)\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}) does not involve 𝑸𝕊n\mbox{\boldmath$Q$}\in\mathbb{S}^{n} and 𝑯𝕊n\mbox{\boldmath$H$}\in\mathbb{S}^{n}. By Theorem 1.1, if \cal B satisfies on Condition (B)’ imposed on QCQP (6), which is a special case of QCQP (4) with 𝑯=diag(0,,0,1)𝕊+n\mbox{\boldmath$H$}=\mbox{diag}(0,\ldots,0,1)\in\mathbb{S}^{n}_{+}, then <η<η=ζ-\infty<\eta\ \Leftrightarrow-\infty<\eta=\zeta holds in QCQP (4) and its SDP relaxation (5) for every 𝑸,𝑯𝕊n\mbox{\boldmath$Q$},\ \mbox{\boldmath$H$}\in\mathbb{S}^{n}. In particular, if we take a positive definite matrix for 𝑯𝕊n\mbox{\boldmath$H$}\in\mathbb{S}^{n}, the feasible regions of both QCQP (4) and its SDP relaxation (5) are bounded; hence <η-\infty<\eta. In this case, we have η=ζ\eta=\zeta. When the feasible region of SDP relaxation (5) is unbounded, however, =η<ζ-\infty=\eta<\zeta may happen even if 𝕁+()^(𝚪n)\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}) holds. We will show such a case in Instance 2.7.

Main contribution of the paper

Our first contribution is to present various QCQP instances satisfying Condition (B)’ by introducing a systematic method to construct such QCQPs. This work is based on the authors’ previous work [2, 3, 9]. In [3], Condition (B)’ was proposed as a sufficient condition for the equivalence between QCQP and its SDP relaxation. A detailed theoretical analysis was also conducted to examine its relationship with several previously proposed sufficient conditions in [2, 9]. While the paper [3] provided some examples of QCQPs satisfying Condition (B)’, it remains unclear what kinds of problems the entire class of such QCQPs include. This work seeks to bridge this gap by presenting various QCQP instances within this class. Specifically, for two-dimensional QCQPs, we provide instances that fully utilize the geometric properties of Condition (B)’ to make them more intuitive and easier to understand. Based on these, it is demonstrated that various instances of high-dimensional QCQPs can be constructed freely. These findings deepen the understanding of Conditions (B)’ proposed in [3] and are expected to facilitate the broader application of QCQPs that satisfy Conditions (B)’.

The second contribution of this work is the introduction of a numerical method to compute the optimal solution of a QCQP satisfying Condition (B) from the optimal solution of its SDP relaxation. The method is based on [16, Lemma 2.2] and its constructive proof, and is illustrated with numerical results to demonstrate its effectiveness.

Organization of the paper

In Section 2, we deal with the case n=3n=3. In this case, (6) is a QCQP in the 22-dimensional variable vector 𝒖u. After demonstrating how to construct basic quadratic constraints in 𝒖u, we combine them to generate 77 instances of \cal B satisfying Conditions (B)’ and (D). In Section 3, we discuss how to combine those instances for constructing higher-dimensional \cal B satisfying Condition (D). In Section 4, we show how to compute an optimal solution 𝑿~𝚪n\widetilde{\mbox{\boldmath$X$}}\in\mbox{\boldmath$\Gamma$}^{n} of QCQP (4) from an optimal solution 𝑿¯𝕊+n\overline{\mbox{\boldmath$X$}}\in\mathbb{S}^{n}_{+} of its SDP relaxation (5) under Condition (B) and some additional assumption.

2 QCQP (6) with 𝒖2\mbox{\boldmath$u$}\in\mathbb{R}^{2}

Throughout this section, we assume that n=3n=3 and 𝒖2\mbox{\boldmath$u$}\in\mathbb{R}^{2} in QCQP (6). Sections 2.1 and 2.2 describe quadratic constraints for Section 2.3 where 77 instances of \cal B satisfying Conditions (B)’ and (D) are presented. We introduce 44 types of basic quadratic constraints in Section 2.1, and 33 types of linear transformations for them in Section 2.2.

2.1 Basic quadratic constraints

A disk constraint:

𝑬d(r)(10001000r2),𝑬d(r)or 𝑬d(r)={𝒖2:u12+u22r2or 0},\displaystyle\mbox{\boldmath$E$}^{\rm d}(r)\equiv\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&-r^{2}\end{pmatrix},\ \mbox{\boldmath$E$}^{\rm d}(r)_{\geq}\ \mbox{or }\mbox{\boldmath$E$}^{\rm d}(r)_{\leq}=\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:u_{1}^{2}+u_{2}^{2}-r^{2}\geq\ \mbox{or }\leq 0\right\},

where r>0r>0 denotes a parameter. See Figure 1.

Refer to caption
Refer to caption
Figure 1: The disk constraint 𝑬d(r)\mbox{\boldmath$E$}^{\rm d}(r)_{\geq}: r=1r=1 (left), r=2r=2 (right).

A hyperbola constraint:

𝑬h(r)(10001000r2),𝑬h(r)or 𝑬h(r)={𝒖2:u12+u22+r2or 0},\displaystyle\mbox{\boldmath$E$}^{\rm h}(r)\equiv\begin{pmatrix}-1&0&0\\ 0&1&0\\ 0&0&r^{2}\end{pmatrix},\ \mbox{\boldmath$E$}^{\rm h}(r)_{\geq}\ \mbox{or }\mbox{\boldmath$E$}^{\rm h}(r)_{\leq}=\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:-u_{1}^{2}+u_{2}^{2}+r^{2}\geq\ \mbox{or }\leq 0\right\},

where r0r\geq 0 denotes a parameter. See Figure 2.

Refer to caption
Refer to caption
Figure 2: The hyperbola constraint 𝑬h(r)\mbox{\boldmath$E$}^{\rm h}(r)_{\geq}: r=1r=1 (left), r=0r=0 (right).

A parabola constraint:

𝑬p(r)(001/20101/20r),𝑬p(r)or 𝑬p(r)={𝒖2:u1+u22+ror 0},\displaystyle\mbox{\boldmath$E$}^{\rm p}(r)\equiv\begin{pmatrix}0&0&-1/2\\ 0&1&0\\ -1/2&0&r\end{pmatrix},\ \mbox{\boldmath$E$}^{\rm p}(r)_{\geq}\ \mbox{or }\mbox{\boldmath$E$}^{\rm p}(r)_{\leq}=\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:-u_{1}+u_{2}^{2}+r\geq\ \mbox{or }\leq 0\right\},

where r0r\geq 0 denotes a parameter. See Figure 3.

Refer to caption
Refer to caption
Figure 3: The parabola constraint 𝑬p(r)\mbox{\boldmath$E$}^{\rm p}(r)_{\geq}: r=1r=1 (left), r=2r=2 (right).

A linear constraint:

𝑬(r)(001/20001/20r),𝑬(r)or 𝑬(r)={𝒖2:u1ror 0},\displaystyle\mbox{\boldmath$E$}^{\rm\ell}(r)\equiv\begin{pmatrix}0&0&1/2\\ 0&0&0\\ 1/2&0&-r\end{pmatrix},\ \mbox{\boldmath$E$}^{\rm\ell}_{\geq}(r)\ \mbox{or }\mbox{\boldmath$E$}^{\rm\ell}(r)_{\leq}=\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:u_{1}-r\geq\ \mbox{or }\leq 0\right\},

where rr\in\mathbb{R} denotes a parameter. See Figure 4.

Refer to caption
Refer to caption
Figure 4: The linear constraint 𝑬(r)\mbox{\boldmath$E$}^{\rm\ell}(r)_{\geq}: r=0r=0 (left), r=1r=1 (right).

2.2 Scaling, rotation and parallel transformation

If we apply a linear transformation defined by (𝒖1)(𝒖1)=𝑴(𝒗1){\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}\rightarrow{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}=\mbox{\boldmath$M$}{\scriptsize\begin{pmatrix}\mbox{\boldmath$v$}\\ 1\end{pmatrix}}, where (𝒖1)3{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}\in\mathbb{R}^{3} and 𝑴M denotes a 3×33\times 3 nonsingular matrix, to the basic quadratic constraints described in Section 2.1, we obtain general quadratic constraints in 𝒗v. Under this transformation, the quadratic function (𝒖1)T𝑩(𝒖1){\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$B$}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}} in 𝒖2\mbox{\boldmath$u$}\in\mathbb{R}^{2} is transformed into the quadratic function (𝒗1)T𝑴T𝑩𝑴(𝒗1){\scriptsize\begin{pmatrix}\mbox{\boldmath$v$}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$M$}^{T}\mbox{\boldmath$B$}\mbox{\boldmath$M$}{\scriptsize\begin{pmatrix}\mbox{\boldmath$v$}\\ 1\end{pmatrix}}. As such a linear transformation, we consider a scaling, a rotation and a parallel transformation, which are used in Section 2.3 for 33 instances of \cal B satisfying Conditions (B)’ and (D).

We describe matrices for scaling, rotation and parallel transformation.

A 3×33\times 3 scaling matrix: 𝑺(𝒔)=(1/s10001/s20001)\mbox{\boldmath$S$}(\mbox{\boldmath$s$})=\begin{pmatrix}1/s_{1}&0&0\\ 0&1/s_{2}&0\\ 0&0&1\end{pmatrix}, where 0<𝒔=(s1,s2)2\mbox{\bf 0}<\mbox{\boldmath$s$}=(s_{1},s_{2})\in\mathbb{R}^{2} denotes a parameter vector. The vector 𝒖2\mbox{\boldmath$u$}\in\mathbb{R}^{2} is transformed to 𝒗=(s1u1,s2u2)2\mbox{\boldmath$v$}=(s_{1}u_{1},s_{2}u_{2})\in\mathbb{R}^{2}.

A 3×33\times 3 rotation matrix: 𝑹(θ)=(cosθsinθ0sinθcosθ0001)\mbox{\boldmath$R$}(\theta)=\begin{pmatrix}\mbox{cos}\theta&\mbox{sin}\theta&0\\ -\mbox{sin}\theta&\mbox{cos}\theta&0\\ 0&0&1\end{pmatrix}, where θ[π,π]\theta\in[-\pi,\pi] or [0,2π][0,2\pi] denotes a parameter. The vector 𝒖2\mbox{\boldmath$u$}\in\mathbb{R}^{2} is transformed to 𝒗=(cosθu1sinθu2,sinθu1+cosθu2)2\mbox{\boldmath$v$}=(\mbox{cos}\theta u_{1}-\mbox{sin}\theta u_{2},\mbox{sin}\theta u_{1}+\mbox{cos}\theta u_{2})\in\mathbb{R}^{2}.

A 3×33\times 3 matrix for parallel transformation: 𝑷(𝒑)=(10p101p2001)\mbox{\boldmath$P$}(\mbox{\boldmath$p$})=\begin{pmatrix}1&0&-p_{1}\\ 0&1&-p_{2}\\ 0&0&1\end{pmatrix}, where 𝒑=(p1,p2)2\mbox{\boldmath$p$}=(p_{1},p_{2})\in\mathbb{R}^{2} denotes a parameter vector. The vector 𝒖2\mbox{\boldmath$u$}\in\mathbb{R}^{2} is transformed to 𝒗=(u1+p1,u2+p2)2\mbox{\boldmath$v$}=(u_{1}+p_{1},u_{2}+p_{2})\in\mathbb{R}^{2}.

Figure 5 illustrates the application of scaling, rotation and parallel transformation to the parabola constraint 𝑬p(r)\mbox{\boldmath$E$}^{\rm p}(r)_{\geq} presented in Section 2.1. Figure 6 illustrates the application of scaling, rotation and parallel transformation to the linear constraint 𝑬(r)\mbox{\boldmath$E$}^{\rm\ell}(r)_{\geq} given in Section 2.1. We note that if the order of scaling, rotation and parallel transformation is changed, then the resulting constraint may differ.

Refer to caption
Refer to caption
Refer to caption
Refer to caption
Figure 5: The application of scaling, rotation and linear transformation to the parabola constraint 𝑬p(r)\mbox{\boldmath$E$}^{\rm p}(r)_{\geq} described in Section 2.1. Here 𝑩0=𝑬p(2),𝑩1=𝑺((1,0.2))T𝑩0𝑺((1,0.2)),𝑩2=𝑹(π/4)T𝑩1𝑹(π/4),𝑩3=𝑷((1,2))T𝑩2𝑷((1,2))\mbox{\boldmath$B$}^{0}=\mbox{$\mbox{\boldmath$E$}^{\rm p}(2)$},\ \mbox{\boldmath$B$}^{1}=\mbox{\boldmath$S$}((1,0.2))^{T}\mbox{\boldmath$B$}^{0}\mbox{\boldmath$S$}((1,0.2)),\ \mbox{\boldmath$B$}^{2}=\mbox{\boldmath$R$}(\pi/4)^{T}\mbox{\boldmath$B$}^{1}\mbox{\boldmath$R$}(\pi/4),\ \mbox{\boldmath$B$}^{3}=\mbox{\boldmath$P$}((-1,-2))^{T}\mbox{\boldmath$B$}^{2}\mbox{\boldmath$P$}((-1,-2)).
Refer to caption
Refer to caption
Figure 6: The application of scaling, rotation and parallel transformation to the linear constraint 𝑬(r)\mbox{\boldmath$E$}^{\rm\ell}(r)_{\geq} given in Section 2.1. Here 𝑩1=𝑹(π/4)T𝑬(0)𝑹(π/4),𝑩2=𝑷((1,1))T𝑩1𝑷((1,1)),𝑩3=𝑷((1,1))T𝑹(5π/4)T𝑬(0)𝑹(5π/4)𝑷((1,1)).\mbox{\boldmath$B$}^{1}=\mbox{\boldmath$R$}(\pi/4)^{T}\mbox{$\mbox{\boldmath$E$}^{\rm\ell}(0)$}\mbox{\boldmath$R$}(\pi/4),\ \mbox{\boldmath$B$}^{2}=\mbox{\boldmath$P$}((-1,-1))^{T}\mbox{\boldmath$B$}^{1}\mbox{\boldmath$P$}((-1,-1)),\ \mbox{\boldmath$B$}^{3}=\mbox{\boldmath$P$}((1,1))^{T}\mbox{\boldmath$R$}(5\pi/4)^{T}\mbox{$\mbox{\boldmath$E$}^{\rm\ell}(0)$}\mbox{\boldmath$R$}(5\pi/4)\mbox{\boldmath$P$}((1,1)).

2.3 Instances of 𝕊3\mbox{$\cal B$}\subseteq\mathbb{S}^{3} satisfying Conditions (B)’ and (D)

Each 𝑩𝕊3\mbox{\boldmath$B$}\in\mbox{$\cal B$}\subseteq\mathbb{S}^{3} in Instances 2.1, 2.2 and 2.3 is obtained by applying scaling, rotation and/or parallel transformation to 𝑬d(r)\mbox{\boldmath$E$}^{\rm d}(r), 𝑬h(r)\mbox{\boldmath$E$}^{\rm h}(r) or 𝑬p(r)\mbox{\boldmath$E$}^{\rm p}(r) for some r>0r>0. Each 𝑩𝕊3\mbox{\boldmath$B$}\in\mbox{$\cal B$}\subseteq\mathbb{S}^{3} in Instances 2.4 and 2.5 is described explicitly without relying on scaling, rotation or parallel transformation. Instance 2.6 leads to an extension of 𝕊3\mbox{$\cal B$}\subseteq\mathbb{S}^{3} satisfying Condition (D) to a general 𝕊n\mbox{$\cal B$}\subseteq\mathbb{S}^{n} with n3n\geq 3 in Section 3. Throughout all instances of \cal B, 𝑩\mbox{\boldmath$B$}_{\leq} is illustrated by the shaded regions, where the unshaded regions correspond to the feasible regions \mbox{$\cal B$}_{\geq} of QCQP (6). Instance 2.7 shows a case where =η<ζ-\infty=\eta<\zeta occurs even if \cal B satisfies Condition (B)’.

Instance 2.1.

We define 𝑩k𝕊3(0k7)\mbox{\boldmath$B$}^{k}\in\mathbb{S}^{3}\ (0\leq k\leq 7) (0<r1/2)(0<r\leq 1/2) and 𝕊3\mbox{$\cal B$}\subseteq\mathbb{S}^{3} as follows:

𝑩k=𝑹(kπ/3)T𝑷((1,0))T𝑬d(r)𝑷((1,0))𝑹(kπ/3)(0k5),𝑩6=𝑬d(r),𝑩7=𝑬d(3/2),={𝑩k:0k7}.}\displaystyle\left.\begin{array}[]{l}\mbox{\boldmath$B$}^{k}=\mbox{\boldmath$R$}(k\pi/3)^{T}\mbox{\boldmath$P$}((1,0))^{T}\mbox{$\mbox{\boldmath$E$}^{\rm d}(r)$}\mbox{\boldmath$P$}((1,0))\mbox{\boldmath$R$}(k\pi/3)\ (0\leq k\leq 5),\\[3.0pt] \mbox{\boldmath$B$}^{6}=\mbox{$\mbox{\boldmath$E$}^{\rm d}(r)$},\ \mbox{\boldmath$B$}^{7}=-\mbox{\boldmath$E$}^{\rm d}(3/2),\ \mbox{$\cal B$}=\{\mbox{\boldmath$B$}^{k}:0\leq k\leq 7\}.\end{array}\right\} (9)

Figure 7 illustrates 𝑩k\mbox{\boldmath$B$}^{k}_{\leq} (0k7)(0\leq k\leq 7), where the unshaded region corresponds to \mbox{$\cal B$}_{\geq}. We took r=1/2r=1/2 in the left figure, and r=1/3r=1/3 in the right figure. We see from Figure 7 that \cal B satisfies Condition (B)’. Let αBk=1\alpha_{B^{k}}=1 (0k6)(0\leq k\leq 6) and αB7=1/3\alpha_{B^{7}}=1/3. We can verify that αBj𝑩j+αBk𝑩k𝕊+3\alpha_{B^{j}}\mbox{\boldmath$B$}^{j}+\alpha_{B^{k}}\mbox{\boldmath$B$}^{k}\in\mathbb{S}^{3}_{+} (0j<k7)(0\leq j<k\leq 7). Therefore, \cal B also satisfies Condition (D).

Refer to caption
Refer to caption
Figure 7: Illustration of Instance 2.1. The value of rr is 1/21/2 for the left figure and 1/31/3 for the right figure.
Instance 2.2.

Let 2m2\leq m\in\mathbb{Z} where \mathbb{Z} denotes the set of integers. We also let r>0r>0 and 𝒑2\mbox{\boldmath$p$}\in\mathbb{R}^{2}. Define

𝑩k=𝑷(𝒑)T𝑹(kπ/m)T𝑺((1,tan(π/(2m))))T𝑬h(r)\displaystyle\mbox{\boldmath$B$}^{k}=\mbox{\boldmath$P$}(\mbox{\boldmath$p$})^{T}\mbox{\boldmath$R$}(k\pi/m)^{T}\mbox{\boldmath$S$}((1,\mbox{tan}(\pi/(2m))))^{T}{\mbox{\boldmath$E$}^{\rm h}(r)}
𝑺((1,tan(π/(2m))))𝑹(kπ/m)𝑷(𝒑)(0km1),\displaystyle\hskip 82.51299pt\mbox{\boldmath$S$}((1,\mbox{tan}(\pi/(2m))))\mbox{\boldmath$R$}(k\pi/m)\mbox{\boldmath$P$}(\mbox{\boldmath$p$})\ (0\leq k\leq m-1),
𝑩m=𝑷(𝒑)T𝑬d(r)𝑷(𝒑),={𝑩k(0km)}.\displaystyle\mbox{\boldmath$B$}^{m}=\mbox{\boldmath$P$}(\mbox{\boldmath$p$})^{T}\mbox{$\mbox{\boldmath$E$}^{\rm d}(r)$}\mbox{\boldmath$P$}(\mbox{\boldmath$p$}),\ \mbox{$\cal B$}=\left\{\mbox{\boldmath$B$}^{k}\ (0\leq k\leq m)\right\}.

See Figure 8. In the left figure, we took m=2m=2, r=1r=1 and 𝒑=(1,1)\mbox{\boldmath$p$}=(1,1), where the hyperbola constraint

𝑩\displaystyle\mbox{\boldmath$B$}_{\leq} =\displaystyle= (𝑺((1,tan(π/(2×2))))T𝑬h(1)𝑺((1,tan(π/(2×2)))))\displaystyle(\mbox{\boldmath$S$}((1,\mbox{tan}(\pi/(2\times 2))))^{T}{\mbox{\boldmath$E$}^{\rm h}(1)}\mbox{\boldmath$S$}((1,\mbox{tan}(\pi/(2\times 2)))))_{\leq}
=\displaystyle= {𝒖2:u12+u22+120}=𝑬h(1)\displaystyle\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:-u_{1}^{2}+u_{2}^{2}+1^{2}\leq 0\right\}=\mbox{\boldmath$E$}^{\rm h}(1)_{\leq}

is shifted to the upper-right direction with the origin (0,0)(1,1)(0,0)\rightarrow(1,1) to create 𝑩0\mbox{\boldmath$B$}^{0}_{\leq}. After rotating 𝑩\mbox{\boldmath$B$}_{\leq} by π/2\pi/2, and then we move the resulting hyperbola toward the (1,1)(1,1)-direction to obtain 𝑩1\mbox{\boldmath$B$}^{1}_{\leq}. We can make a similar observation on the right figure. From Figure 8, we see that \cal B satisfies Condition (B)’. We can also verify that \cal B also satisfies Condition (D) with αBk=1\alpha_{B^{k}}=1 (0km)(0\leq k\leq m).

Refer to caption
Refer to caption
Figure 8: Illustration of Instance 2.2. The values for the parameters are m=2,r=1m=2,\ r=1 and 𝒑=(1,1)\mbox{\boldmath$p$}=(1,1) for the left figure, and m=5,r=2m=5,\ r=2 and 𝒑=(1,0)\mbox{\boldmath$p$}=(-1,0) for the right figure.
Instance 2.3.

For every 3m3\leq m\in\mathbb{Z} and r>0r>0, define

𝑪k=𝑹(2kπ/m)T𝑺((1,2(tan(π/m))r))T𝑬p(r) 𝑺((1,2(tan(π/m))r))𝑹(2kπ/m)(0km1),𝑪m=𝑬d(r),𝒞={𝑪k(0km)}.}\displaystyle\left.\begin{array}[]{l}\mbox{\boldmath$C$}^{k}=\mbox{\boldmath$R$}(2k\pi/m)^{T}\mbox{\boldmath$S$}((1,2(\mbox{tan}(\pi/m))\sqrt{r}))^{T}{\mbox{\boldmath$E$}^{\rm p}(r)}\\ \mbox{ \ }\hskip 122.34685pt\mbox{\boldmath$S$}((1,2(\mbox{tan}(\pi/m))\sqrt{r}))\mbox{\boldmath$R$}(2k\pi/m)\ (0\leq k\leq m-1),\\[3.0pt] \mbox{\boldmath$C$}^{m}=\mbox{$\mbox{\boldmath$E$}^{\rm d}(r)$},\ \mbox{$\cal C$}=\left\{\mbox{\boldmath$C$}^{k}\ (0\leq k\leq m)\right\}.\end{array}\right\} (13)

See Figure 9. In the left figure, 𝑪0\mbox{\boldmath$C$}^{0}_{\leq} is expressed as a scaled parabola constraint such that

𝑪0\displaystyle\mbox{\boldmath$C$}^{0}_{\leq} =\displaystyle= (𝑺((1,2(tan(π/3))1))T𝑬p(1)𝑺((1,2(tan(π/3))1)))\displaystyle(\mbox{\boldmath$S$}((1,2(\mbox{tan}(\pi/3))\sqrt{1}))^{T}\mbox{$\mbox{\boldmath$E$}^{\rm p}(1)$}\mbox{\boldmath$S$}((1,2(\mbox{tan}(\pi/3))\sqrt{1})))_{\leq}
=\displaystyle= {𝒖2:u1+((2tan(π/3))u2)2+10}.\displaystyle\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:-u_{1}+((2\mbox{tan}(\pi/3))u_{2})^{2}+1\leq 0\right\}.

𝑪1\mbox{\boldmath$C$}^{1}_{\leq} and 𝑪2\mbox{\boldmath$C$}^{2}_{\leq} are obtained by rotating 𝑪0\mbox{\boldmath$C$}^{0}_{\leq} by π/3\pi/3 and 2π/32\pi/3, respectively. We can make a similar observation on the right figure. We see from Figure 9 that 𝒞\cal C satisfies Condition (B)’. Let αBk=1\alpha_{B^{k}}=1 (0km1)(0\leq k\leq m-1) and αBm=1/(2r)\alpha_{B^{m}}=1/(2r). Then we can verify that αCj𝑪j+αCk𝑪k𝕊+n\alpha_{C^{j}}\mbox{\boldmath$C$}^{j}+\alpha_{C^{k}}\mbox{\boldmath$C$}^{k}\in\mathbb{S}^{n}_{+} (0j<km)(0\leq j<k\leq m) holds. Therefore, 𝒞\cal C satisfies Condition (D).

Refer to caption
Refer to caption
Figure 9: Illustration of Instance 2.3. We took (m,r)=(3,1)(m,r)=(3,1) for the left figure, and (m,r)=(7,2)(m,r)=(7,2) for the right figure

.

Instance 2.4.

In this and next instances, we construct \cal B without relying on scaling, rotation and parallel transformation. For every aa\in\mathbb{Z} and r0r\geq 0, define

𝑩(a,r)=(a21/4a0a1000r2).\displaystyle\mbox{\boldmath$B$}(a,r)=\begin{pmatrix}a^{2}-1/4&-a&0\\ -a&1&0\\ 0&0&r^{2}\end{pmatrix}.

We then see that

𝑩(a,r)or 𝑩(a,r)\displaystyle\mbox{\boldmath$B$}(a,r)_{\geq}\ \mbox{or }\mbox{\boldmath$B$}(a,r)_{\leq} =\displaystyle= {𝒖2:(u2au1)2u12/4+r2or 0}.\displaystyle\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:(u_{2}-au_{1})^{2}-u_{1}^{2}/4+r^{2}\geq\ \mbox{or }\leq 0\right\}.

It is easy to observe that, for every distinct a1,a2a_{1},\ a_{2}\in\mathbb{Z} and every r1,r2+r_{1},\ r_{2}\in\mathbb{R}_{+},

𝑩(a1,r1)+𝑩(a2,r2)\displaystyle\mbox{\boldmath$B$}(a_{1},r_{1})+\mbox{\boldmath$B$}(a_{2},r_{2}) =\displaystyle= (a12+a221/2a1a20a1a22000r12+r22)𝕊+3\displaystyle\begin{pmatrix}a_{1}^{2}+a_{2}^{2}-1/2&-a_{1}-a_{2}&0\\ -a_{1}-a_{2}&2&0\\ 0&0&r_{1}^{2}+r_{2}^{2}\end{pmatrix}\in\mathbb{S}^{3}_{+}

holds. Hence, if we take a finite subset Gh{G^{\rm h}} of ×+\mathbb{Z}\times\mathbb{R}_{+} satisfying

a1a2for every pair of distinct (a1,r1)Gh and (a2,r2)Gh,\displaystyle a_{1}\not=a_{2}\ \mbox{for every pair of distinct $(a_{1},r_{1})\in{G^{\rm h}}$ and $(a_{2},r_{2})\in{G^{\rm h}}$},

then (Gh)={𝑩(a,r):(a,r)Gh}\mbox{$\cal B$}({G^{\rm h}})=\left\{\mbox{\boldmath$B$}(a,r):(a,r)\in{G^{\rm h}}\right\} satisfies Condition (D). By Theorem 1.1 (iv), (Gh)\mbox{$\cal B$}({G^{\rm h}}) also satisfies Condition (B)’, as can be verified by Figure 10.

Refer to caption
Refer to caption
Figure 10: Illustration of Instance 2.4: We took Gh={(2,1),(1,1),(0,1),(1,1),(2,1)}{G^{\rm h}}=\{(2,1),(1,1),(0,1),(-1,1),(-2,1)\} for the left figure, and Gh={(2,0),(0,1),(1,2),(2,0.2)}{G^{\rm h}}=\{(2,0),(0,1),(-1,2),(-2,0.2)\} for the right figure.
Instance 2.5.

For every aa\in\mathbb{Z} and r0r\geq 0, define

𝑩(a,r)=(a2a1/2a101/20r).\displaystyle\mbox{\boldmath$B$}(a,r)=\begin{pmatrix}a^{2}&-a&-1/2\\ -a&1&0\\ -1/2&0&r\end{pmatrix}. (14)

Then it follows that

𝑩(a,r)or 𝑩(a,r)\displaystyle\mbox{\boldmath$B$}(a,r)_{\geq}\ \mbox{or }\mbox{\boldmath$B$}(a,r)_{\leq} =\displaystyle= {𝒖2:(au1u2)2u1+r0or 0}.\displaystyle\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:(au_{1}-u_{2})^{2}-u_{1}+r\geq 0\ \mbox{or }\leq 0\right\}.

We can easily verify that, for every distinct a1,a2a_{1},\ a_{2}\in\mathbb{Z} and every r1,r2[1,)r_{1},\ r_{2}\in[1,\infty),

𝑩(a1,r1)+𝑩(a2,r2)\displaystyle\mbox{\boldmath$B$}(a_{1},r_{1})+\mbox{\boldmath$B$}(a_{2},r_{2}) =\displaystyle= (a12+a22a1a21a1a22010r1+r2)𝕊+3\displaystyle\begin{pmatrix}a_{1}^{2}+a_{2}^{2}&-a_{1}-a_{2}&-1\\ -a_{1}-a_{2}&2&0\\ -1&0&r_{1}+r_{2}\end{pmatrix}\in\mathbb{S}^{3}_{+}

holds. Therefore, for every finite subset Gp{G^{\rm p}} of ×[1,)\mathbb{Z}\times[1,\infty) satisfying

a1a2for every pair of distinct (a1,r1)Gp and (a2,r2)Gp,\displaystyle a_{1}\not=a_{2}\ \mbox{for every pair of distinct $(a_{1},r_{1})\in{G^{\rm p}}$ and $(a_{2},r_{2})\in{G^{\rm p}}$},

(Gp)={𝑩(a,r):(a,r)Gp}\mbox{$\cal B$}({G^{\rm p}})=\left\{\mbox{\boldmath$B$}(a,r):(a,r)\in{G^{\rm p}}\right\} satisfies Condition (D). By Theorem 1.1 (iv), (Gp)\mbox{$\cal B$}({G^{\rm p}}) satisfies Condition (B)’. See Figure 11.

Refer to caption
Refer to caption
Figure 11: Illustration of Instance 2.5. We took Gp={(2,1),(1,1),(0,1),(1,1),(2,1)}{G^{\rm p}}=\{(2,1),(1,1),(0,1),(-1,1),(-2,1)\} for the left figure, and Gp={(2,1),(0,2),(1,3),(2,1)}{G^{\rm p}}=\{(2,1),(0,2),(-1,3),(-2,1)\} for the right figure.
Instance 2.6.

In this instance, we utilize \cal B in (9) (Instance 2.1) and 𝒞\cal C in (13) (Instance 2.3). Let r=1/2r=1/2 as in the left figure of Figure 7, and (m,r)=(7,2)(m,r^{\prime})=(7,2) as in the right figure of Figure 9. Then each of \cal B and 𝒞\cal C consists of 88 matrices in 𝕊3\mathbb{S}^{3}. As we have stated there, they both satisfy Condition (D);

αBj𝑩j+αBk𝑩k𝕊+3,αBj>0,αBk>0(0j<k7),\displaystyle\alpha_{B^{j}}\mbox{\boldmath$B$}^{j}+\alpha_{B^{k}}\mbox{\boldmath$B$}^{k}\in\mathbb{S}^{3}_{+},\ \alpha_{B^{j}}>0,\ \alpha_{B^{k}}>0\ (0\leq j<k\leq 7),
αCj𝑪j+αCk𝑪k𝕊+3,αCj>0,αCk>0(0j<k7).\displaystyle\alpha_{C^{j}}\mbox{\boldmath$C$}^{j}+\alpha_{C^{k}}\mbox{\boldmath$C$}^{k}\in\mathbb{S}^{3}_{+},\ \alpha_{C^{j}}>0,\ \alpha_{C^{k}}>0\ (0\leq j<k\leq 7).

Letting λ>0\lambda>0, μ>0\mu>0 and λ+μ=1\lambda+\mu=1, we consider the convex combination of αBk𝑩k\alpha_{B^{k}}\mbox{\boldmath$B$}^{k} and αCk𝑪k\alpha_{C^{k}}\mbox{\boldmath$C$}^{k} such that

𝑨k=λαBk𝑩k+μαCk𝑪k(0k7).\displaystyle\mbox{\boldmath$A$}^{k}=\lambda\alpha_{B^{k}}\mbox{\boldmath$B$}^{k}+\mu\alpha_{C^{k}}\mbox{\boldmath$C$}^{k}\ (0\leq k\leq 7).

Then

𝑨j+𝑨k\displaystyle\mbox{\boldmath$A$}^{j}+\mbox{\boldmath$A$}^{k} =\displaystyle= λαBj𝑩j+μαCj𝑪j+λαBk𝑩k+μαCk𝑪k\displaystyle\lambda\alpha_{B^{j}}\mbox{\boldmath$B$}^{j}+\mu\alpha_{C^{j}}\mbox{\boldmath$C$}^{j}+\lambda\alpha_{B^{k}}\mbox{\boldmath$B$}^{k}+\mu\alpha_{C^{k}}\mbox{\boldmath$C$}^{k}
=\displaystyle= λ(αBj𝑩j+αBk𝑩k)+μ(αCj𝑪j+αCk𝑪k)𝕊+3(0j<k7)\displaystyle\lambda(\alpha_{B^{j}}\mbox{\boldmath$B$}^{j}+\alpha_{B^{k}}\mbox{\boldmath$B$}^{k})+\mu(\alpha_{C^{j}}\mbox{\boldmath$C$}^{j}+\alpha_{C^{k}}\mbox{\boldmath$C$}^{k})\in\mathbb{S}^{3}_{+}\ (0\leq j<k\leq 7)

holds. Therefore 𝒜={𝑨k:0k7}\mbox{$\cal A$}=\{\mbox{\boldmath$A$}^{k}:0\leq k\leq 7\} satisfies Condition (D). See Figure 12. If we renumber 𝑪k\mbox{\boldmath$C$}^{k} contained in 𝒞\cal C before taking the convex combination, a different 𝒜\cal A is obtained.

Refer to caption
Refer to caption
Figure 12: Illustration of Instance 2.6: We took (λ,μ)=(0.09,0.91)(\lambda,\mu)=(0.09,0.91) for the left figure, and (λ,μ)=(0.05,0.95)(\lambda,\mu)=(0.05,0.95) for the right figure.
Instance 2.7.

Define

𝑩2=(001/2001/21/21/22)𝕊3,𝑩3=(001/2001/21/21/22),\displaystyle\mbox{\boldmath$B$}^{2}=\begin{pmatrix}0&0&1/2\\ 0&0&1/2\\ 1/2&1/2&2\end{pmatrix}\in\mathbb{S}^{3},\ \mbox{\boldmath$B$}^{3}=\begin{pmatrix}0&0&-1/2\\ 0&0&-1/2\\ -1/2&-1/2&2\end{pmatrix},

and let ={𝑩2,𝑩3}\mbox{$\cal B$}=\{\mbox{\boldmath$B$}^{2},\ \mbox{\boldmath$B$}^{3}\}. Then it follows that

𝑩2={𝒖2:u1+u2+20},𝑩3={𝒖2:u1u2+20},𝑩2𝑩3=,\displaystyle\mbox{\boldmath$B$}^{2}_{\leq}=\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:u_{1}+u_{2}+2\leq 0\},\ \mbox{\boldmath$B$}^{3}_{\leq}=\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:-u_{1}-u_{2}+2\leq 0\},\ \mbox{\boldmath$B$}^{2}_{\leq}\cap\mbox{\boldmath$B$}^{3}_{\leq}=\emptyset,
={𝒖2:2u1+u22}.\displaystyle\mbox{$\cal B$}_{\geq}=\{\mbox{\boldmath$u$}\in\mathbb{R}^{2}:-2\leq u_{1}+u_{2}\leq 2\}. (15)

See the right figure of Figure 6. By Theorem 1.1, the relation <η<η=ζ-\infty<\eta\Leftrightarrow-\infty<\eta=\zeta holds between QCQP (6) and its SDP relaxation (5) with any 𝑸𝕊3\mbox{\boldmath$Q$}\in\mathbb{S}^{3}. If the objective function (𝒖1)T𝑸(𝒖1){\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\mbox{\boldmath$Q$}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}} is convex in 𝒖2\mbox{\boldmath$u$}\in\mathbb{R}^{2}, then (6) becomes a convex QCQP, so that η=ζ\eta=\zeta holds. We now consider a nonconvex case. For example, consider the case where

𝑸=(110110000),(𝒖1)T𝑸(𝒖1)=(u1+u2)2.\displaystyle\mbox{\boldmath$Q$}=\begin{pmatrix}-1&-1&0\\ -1&-1&0\\ 0&0&0\end{pmatrix},\ \begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\mbox{\boldmath$Q$}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}=-(u_{1}+u_{2})^{2}.

Obviously, QCQP (6) attains the minimum ζ=4\zeta=-4 at every boundary points 𝒖u of \mbox{$\cal B$}_{\geq} satisfying |u1+u2|=2|u_{1}+u_{2}|=2. On the one hand, 𝑿(t)=(t010t1111)\mbox{\boldmath$X$}(t)={\scriptsize\begin{pmatrix}t&0&1\\ 0&t&1\\ 1&1&1\end{pmatrix}} is a feasible solution of SDP (5) for every t2t\geq 2, and 𝑸𝑿(t)\mbox{\boldmath$Q$}\bullet\mbox{\boldmath$X$}(t)\rightarrow-\infty as tt\rightarrow\infty. Therefore, we obtain η=<ζ=4\eta=-\infty<\zeta=4 in this case.
     We note that \mbox{$\cal B$}_{\geq} is represented by two linear inequalities. We now represent the same \mbox{$\cal B$}_{\geq} by a single quadratic inequality such that

𝑩4=(110110004),𝑩4={𝒖:04(u1+u2)2}={𝒖:2u1+u22}=.\displaystyle\mbox{\boldmath$B$}^{4}=\begin{pmatrix}-1&-1&0\\ -1&-1&0\\ 0&0&4\end{pmatrix},\ \mbox{\boldmath$B$}^{4}_{\geq}=\{\mbox{\boldmath$u$}:0\leq 4-(u_{1}+u_{2})^{2}\}=\{\mbox{\boldmath$u$}:-2\leq u_{1}+u_{2}\leq 2\}=\mbox{$\cal B$}_{\geq}.

In this case, η=ζ=4\eta=\zeta=-4 holds. In general, SDP relaxation of a QCQP depends on the representation of the QCQP. In particular, quadratic inequality representation often yields an SDP relaxation that provides a better approximate optimal value (see [5, Theorem 2.1]).

3 An extension to higher-dimensional instances

Several cases of 𝕊n\mbox{$\cal B$}\subseteq\mathbb{S}^{n} satisfying 𝕁+()^(𝚪n)\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}) were provided in [2, Section 4.1] and [3, Section 6]. In this section, we show how to construct various \cal B’s 𝕊n\subseteq\mathbb{S}^{n} that satisfy Condition (D) by combining multiple \cal B’s 𝕊3\subseteq\mathbb{S}^{3}. Recall that we have shown in Instance 2.6 that a combination of two \cal B’s 𝕊3\subseteq\mathbb{S}^{3}, which have the same cardinality and satisfy Condition (D), creates a new 𝕊3\mbox{$\cal B$}\subseteq\mathbb{S}^{3} satisfying Condition (D). We generalize this technique to recursively construct a higher-dimensional \cal B satisfying Condition (D).

Let 1p1\leq p\in\mathbb{Z}. We assume that both p={𝑩p1,,𝑩pm}𝕊np\mbox{$\cal B$}^{p}=\{\mbox{\boldmath$B$}^{p1},\ldots,\mbox{\boldmath$B$}^{pm}\}\subseteq\mathbb{S}^{n_{p}} and 𝒞p={𝑪p1,,𝑪pm}𝕊p\mbox{$\cal C$}^{p}=\{\mbox{\boldmath$C$}^{p1},\ldots,\mbox{\boldmath$C$}^{pm}\}\subseteq\mathbb{S}^{\ell_{p}} satisfy Condition (D). We note that (α𝑩)=𝑩(\alpha\mbox{\boldmath$B$})_{\geq}=\mbox{\boldmath$B$}_{\geq} for every 𝑩𝕊n\mbox{\boldmath$B$}\in\mathbb{S}^{n} and α>0\alpha>0. For simplicity of notation, we assume that αBpk=αCpk=1\alpha_{B^{pk}}=\alpha_{C^{pk}}=1 (1km)(1\leq k\leq m). Thus, our assumption indicates that

𝑩pi+𝑩pj𝕊+np(1i<jm),𝑪pi+𝑪pj𝕊+p(1i<jm).\displaystyle\mbox{\boldmath$B$}^{pi}+\mbox{\boldmath$B$}^{pj}\in\mathbb{S}^{n_{p}}_{+}\ (1\leq i<j\leq m),\ \mbox{\boldmath$C$}^{pi}+\mbox{\boldmath$C$}^{pj}\in\mathbb{S}^{\ell_{p}}_{+}\ (1\leq i<j\leq m).

Let 𝑳p\mbox{\boldmath$L$}^{p} be an (np+p)×np+1(n_{p}+\ell_{p})\times n_{p+1} matrix. Then, we define 𝑩(p+1)i\mbox{\boldmath$B$}^{(p+1)i} (1im)(1\leq i\leq m) by

𝑩(p+1)i=(𝑳p)T(𝑩pi𝑶𝑶T𝑪pi)𝑳p𝕊np+1,\displaystyle\mbox{\boldmath$B$}^{(p+1)i}=(\mbox{\boldmath$L$}^{p})^{T}\begin{pmatrix}\mbox{\boldmath$B$}^{pi}&\mbox{\boldmath$O$}\\ \mbox{\boldmath$O$}^{T}&\mbox{\boldmath$C$}^{pi}\end{pmatrix}\mbox{\boldmath$L$}^{p}\in\mathbb{S}^{n_{p+1}}, (16)

where 𝑶O denotes the np×pn_{p}\times\ell_{p} matrix of 0’s. Let p+1={𝑩(p+1)i:1im}\mbox{$\cal B$}^{p+1}=\{\mbox{\boldmath$B$}^{(p+1)i}:1\leq i\leq m\}. Then,

𝑩(p+1)i+𝑩(p+1)j\displaystyle\mbox{\boldmath$B$}^{(p+1)i}+\mbox{\boldmath$B$}^{(p+1)j} =\displaystyle= (𝑳p)T(𝑩pi𝑶𝑶T𝑪pi)𝑳p+(𝑳p)T(𝑩pj𝑶𝑶T𝑪pj)𝑳p\displaystyle(\mbox{\boldmath$L$}^{p})^{T}\begin{pmatrix}\mbox{\boldmath$B$}^{pi}&\mbox{\boldmath$O$}\\ \mbox{\boldmath$O$}^{T}&\mbox{\boldmath$C$}^{pi}\end{pmatrix}\mbox{\boldmath$L$}^{p}+(\mbox{\boldmath$L$}^{p})^{T}\begin{pmatrix}\mbox{\boldmath$B$}^{pj}&\mbox{\boldmath$O$}\\ \mbox{\boldmath$O$}^{T}&\mbox{\boldmath$C$}^{pj}\end{pmatrix}\mbox{\boldmath$L$}^{p}
=\displaystyle= (𝑳p)T(𝑩pi+𝑩pj𝑶𝑶T𝑪pi+𝑪pj)𝑳p𝕊+np+1(1i<jm)\displaystyle(\mbox{\boldmath$L$}^{p})^{T}\begin{pmatrix}\mbox{\boldmath$B$}^{pi}+\mbox{\boldmath$B$}^{pj}&\mbox{\boldmath$O$}\\ \mbox{\boldmath$O$}^{T}&\mbox{\boldmath$C$}^{pi}+\mbox{\boldmath$C$}^{pj}\end{pmatrix}\mbox{\boldmath$L$}^{p}\in\mathbb{S}^{n_{p+1}}_{+}\ (1\leq i<j\leq m)

holds. Therefore, p+1\mbox{$\cal B$}^{p+1} satisfies Condition (D). Furthermore, by replacing pp by p+1p+1, we can continue this procedure recursively. This recursive procedure is highly flexible as it allows arbitrary selection of p\mbox{$\cal B$}^{p} and 𝒞p\mbox{$\cal C$}^{p} satisfying Condition (D), as well as any (np+p)×np+1(n_{p}+\ell_{p})\times n_{p+1} matrix 𝑳p\mbox{\boldmath$L$}^{p}. Thus we can create various \cal B’s in 𝕊n\mathbb{S}^{n} by this recursive procedure.

In addition to \cal B’s 𝕊3\subseteq\mathbb{S}^{3} described in Section 2, which can be employed for the initial 1\mbox{$\cal B$}^{1} and 𝒞1\mbox{$\cal C$}^{1} and also for 𝒞p\mbox{$\cal C$}^{p} (p2)(p\geq 2), we mention some \cal B’s satisfying Condition (D).

(a)

Every 𝕊+n\mbox{$\cal B$}\subseteq\mathbb{S}^{n}_{+} satisfies Condition (D). Specifically, every +\mbox{$\cal B$}\subseteq\mathbb{R}_{+} satisfies Condition (D).

(b)

For every 𝒂n1\mbox{\boldmath$a$}\in\mathbb{R}^{n-1} and ρ>0\rho>0 , let

𝑩(𝒂,ρ)=(𝑰𝒂𝒂T𝒂T𝒂ρ2).\displaystyle\mbox{\boldmath$B$}(\mbox{\boldmath$a$},\rho)=\begin{pmatrix}\mbox{\boldmath$I$}&-\mbox{\boldmath$a$}\\ -\mbox{\boldmath$a$}^{T}&\mbox{\boldmath$a$}^{T}\mbox{\boldmath$a$}-\rho^{2}\end{pmatrix}.

Then 𝑩(𝒂,ρ)\mbox{\boldmath$B$}(\mbox{\boldmath$a$},\rho)_{\leq} forms a ball with the center 𝒂a and radius ρ\rho. For every Gn1×(0,1/2]G\subseteq\mathbb{Z}^{n-1}\times(0,1/2] satisfying

𝒂1𝒂2for every pair of distinct (𝒂1,ρ1)G and (𝒂2,ρ2)G,\displaystyle\mbox{\boldmath$a$}_{1}\not=\mbox{\boldmath$a$}_{2}\ \mbox{for every pair of distinct $(\mbox{\boldmath$a$}_{1},\rho_{1})\in G$ and $(\mbox{\boldmath$a$}_{2},\rho_{2})\in G$}, (17)

define (G)={𝑩(𝒂,ρ):(𝒂,ρ)G}.\mbox{$\cal B$}(G)=\left\{\mbox{\boldmath$B$}(\mbox{\boldmath$a$},\rho):(\mbox{\boldmath$a$},\rho)\in G\right\}. Then, we see that

(𝑰𝒂1𝒂1T𝒂1T𝒂1ρ12)+(𝑰𝒂2𝒂2T𝒂2T𝒂2ρ22)\displaystyle\begin{pmatrix}\mbox{\boldmath$I$}&-\mbox{\boldmath$a$}_{1}\\ -\mbox{\boldmath$a$}_{1}^{T}&\mbox{\boldmath$a$}_{1}^{T}\mbox{\boldmath$a$}_{1}-\rho_{1}^{2}\end{pmatrix}+\begin{pmatrix}\mbox{\boldmath$I$}&-\mbox{\boldmath$a$}_{2}\\ -\mbox{\boldmath$a$}_{2}^{T}&\mbox{\boldmath$a$}_{2}^{T}\mbox{\boldmath$a$}_{2}-\rho_{2}^{2}\end{pmatrix}
=\displaystyle= (2𝑰(𝒂1+𝒂2)(𝒂1+𝒂2)T𝒂1T𝒂1+𝒂2T𝒂2ρ12ρ22)𝕊+n\displaystyle\begin{pmatrix}2\mbox{\boldmath$I$}&-(\mbox{\boldmath$a$}_{1}+\mbox{\boldmath$a$}_{2})\\ -(\mbox{\boldmath$a$}_{1}+\mbox{\boldmath$a$}_{2})^{T}&\mbox{\boldmath$a$}_{1}^{T}\mbox{\boldmath$a$}_{1}+\mbox{\boldmath$a$}_{2}^{T}\mbox{\boldmath$a$}_{2}-\rho_{1}^{2}-\rho_{2}^{2}\end{pmatrix}\in\mathbb{S}^{n}_{+}

for every distinct (𝒂1,ρ1),(𝒂2,ρ2)G(\mbox{\boldmath$a$}_{1},\rho_{1}),\ (\mbox{\boldmath$a$}_{2},\rho_{2})\in G. In fact, every leading principal minors of the matrix above is nonnegative. (In particular, its determinant can be computed with its Schur complement). Therefore, (G)\mbox{$\cal B$}(G) satisfies Condition (D).

(c)

Suppose that both p\mbox{$\cal B$}^{p} and 𝒞p\mbox{$\cal C$}^{p} satisfy Condition (D), but the number #𝒞p\#\mbox{$\cal C$}^{p} of matrices in 𝒞p\mbox{$\cal C$}^{p} is less than #p\#\mbox{$\cal B$}^{p}. To apply (16), we must either discard #p#𝒞p\#\mbox{$\cal B$}^{p}-\#\mbox{$\cal C$}^{p} matrices from p\mbox{$\cal B$}^{p} or introduce #p#𝒞p\#\mbox{$\cal B$}^{p}-\#\mbox{$\cal C$}^{p} ‘dummy’ matrices into 𝒞p\mbox{$\cal C$}^{p} to adjust their cardinalities. If we take a sufficiently large λ0\lambda\geq 0, then λ𝑰\lambda\mbox{\boldmath$I$} serves as such a dummy matrix that 𝒞p{λ𝑰}\mbox{$\cal C$}^{p}\cup\{\lambda\mbox{\boldmath$I$}\} satisfies Condition (D). More precisely, let λmin(𝑩)\lambda_{\rm min}(\mbox{\boldmath$B$}) denote the minimum eigenvalue of 𝑩\mbox{\boldmath$B$}\in\mbox{$\cal B$}, and λmin{λmin(𝑩)(𝑩), 0}0-\lambda\leq\min\{\lambda_{\rm min}(\mbox{\boldmath$B$})\ (\mbox{\boldmath$B$}\in\mbox{$\cal B$}),\ 0\}\leq 0. Then {λ𝑰}\mbox{$\cal B$}\cup\{\lambda\mbox{\boldmath$I$}\} satisfies Condition (D).

We now show how to use (b) and (c) for the recursive formula (16) to construct p+1\mbox{$\cal B$}^{p+1}. Let σ1\sigma_{1}\in\mathbb{R} and 𝒞={σ1}\mbox{$\cal C$}=\{\sigma_{1}\}. Then 𝒞\cal C satisfies Condition (D) since it consists of a single 1×11\times 1 matrix in 𝕊1\mathbb{S}^{1}. For a finite subset GG of n1×(0,1/2]\mathbb{Z}^{n-1}\times(0,1/2] satisfying (17), let p=(G)\mbox{$\cal B$}^{p}=\mbox{$\cal B$}(G), which satisfies Condition (D). Suppose that p\mbox{$\cal B$}^{p} consists of m2m\geq 2 matrices 𝑩p1,,𝑩pm\mbox{\boldmath$B$}^{p1},\ldots,\mbox{\boldmath$B$}^{pm}. By choosing m1m-1 nonnegative numbers σimax{σ1,0}\sigma_{i}\geq\max\{-\sigma_{1},0\} (i=2,,m)(i=2,\ldots,m), we expand 𝒞={σ1}\mbox{$\cal C$}=\{\sigma_{1}\} to 𝒞p={𝑪1=σ1,𝑪2=σ2,,𝑪m=σm}\mbox{$\cal C$}^{p}=\{\mbox{\boldmath$C$}^{1}=\sigma_{1},\mbox{\boldmath$C$}_{2}=\sigma_{2},\ldots,\mbox{\boldmath$C$}^{m}=\sigma_{m}\}, which satisfies Condition (D) by (c) and contains the same number of matrices as p\mbox{$\cal B$}^{p}. Thus we can apply the recursive formula (16) to construct p+1={𝑩(p+1)1,,𝑩(p+1)m}\mbox{$\cal B$}^{p+1}=\{\mbox{\boldmath$B$}^{(p+1)1},\ldots,\mbox{\boldmath$B$}^{(p+1)m}\}. If we take σi=ri20\sigma_{i}=r_{i}^{2}\geq 0 (i=1,,m)(i=1,\ldots,m), n=2n=2, ρ=1/2\rho=1/2, G=GhG={G^{h}} and 𝑳p=(010100001)\mbox{\boldmath$L$}^{p}={\scriptsize\begin{pmatrix}0&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix}}, then p+1\mbox{$\cal B$}^{p+1} corresponds to (Gh)\mbox{$\cal B$}({G^{h}}) in Instance 2.4.

We provide two examples of 𝑳p\mbox{\boldmath$L$}^{p}. For simplicity, we assume that np=p=3n_{p}=\ell_{p}=3, but the discussion below can be generalized to any np,p3n_{p},\ \ell_{p}\geq 3 in a straightforward manner. Let λp>0,μp>0,λp+μp=1\lambda_{p}>0,\ \mu_{p}>0,\ \lambda_{p}+\mu_{p}=1. For the first example, let

𝑳p\displaystyle\mbox{\boldmath$L$}^{p} =\displaystyle= (λp000λp000λpμp000μp000μp).\displaystyle\begin{pmatrix}\sqrt{\lambda_{p}}&0&0\\ 0&\sqrt{\lambda_{p}}&0\\ 0&0&\sqrt{\lambda_{p}}\\ \sqrt{\mu_{p}}&0&0\\ 0&\sqrt{\mu_{p}}&0\\ 0&0&\sqrt{\mu_{p}}\end{pmatrix}.

Then, through the linear transformation

(𝒖z)2+13+3(𝒙1𝒙2)=𝑳p(𝒖z),\displaystyle\begin{pmatrix}\mbox{\boldmath$u$}\\ z\end{pmatrix}\in\mathbb{R}^{2+1}\rightarrow\mathbb{R}^{3+3}\ni\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix}=\mbox{\boldmath$L$}^{p}\begin{pmatrix}\mbox{\boldmath$u$}\\ z\end{pmatrix},

the quadratic function (𝒙1𝒙2)T𝑩(p+1)i(𝒙1𝒙2){\scriptsize\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix}}^{T}\mbox{\boldmath$B$}^{(p+1)i}{\scriptsize\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix}} in (𝒙1𝒙2){\scriptsize\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix}} is transformed to the quadratic function

(𝒙1𝒙2)T(𝑳p)T𝑩(p+1)i𝑳p(𝒙1𝒙2)\displaystyle\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix}^{T}(\mbox{\boldmath$L$}^{p})^{T}\mbox{\boldmath$B$}^{(p+1)i}\mbox{\boldmath$L$}^{p}\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix} =\displaystyle= (𝒖z)T(λp𝑩pi+μp𝑪pi)(𝒖z)\displaystyle\begin{pmatrix}\mbox{\boldmath$u$}\\ z\end{pmatrix}^{T}(\lambda_{p}\mbox{\boldmath$B$}^{pi}+\mu_{p}\mbox{\boldmath$C$}^{pi})\begin{pmatrix}\mbox{\boldmath$u$}\\ z\end{pmatrix}
=\displaystyle= λp(𝒖z)T𝑩pi(𝒖z)+μp(𝒖z)T𝑪pi(𝒖z)\displaystyle\lambda_{p}\begin{pmatrix}\mbox{\boldmath$u$}\\ z\end{pmatrix}^{T}\mbox{\boldmath$B$}^{pi}\begin{pmatrix}\mbox{\boldmath$u$}\\ z\end{pmatrix}+\mu_{p}\begin{pmatrix}\mbox{\boldmath$u$}\\ z\end{pmatrix}^{T}\mbox{\boldmath$C$}^{pi}\begin{pmatrix}\mbox{\boldmath$u$}\\ z\end{pmatrix}

in (𝒖z)2+1{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ z\end{pmatrix}}\in\mathbb{R}^{2+1}. Hence, if we fix zz to be 11, we obtain a convex combination of the two quadratic functions (𝒖1)T𝑩pi(𝒖1){\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$B$}^{pi}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}} and (𝒖1)T𝑪pi(𝒖1){\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$C$}^{pi}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}. Thus this case corresponds to Instance 2.6.


Now, we consider the case

𝑳p\displaystyle\mbox{\boldmath$L$}^{p} =\displaystyle= (λp00000λp0000000λp00μp00000μp00000μp).\displaystyle\begin{pmatrix}\sqrt{\lambda_{p}}&0&0&0&0\\ 0&\sqrt{\lambda_{p}}&0&0&0\\ 0&0&0&0&\sqrt{\lambda_{p}}\\ 0&0&\sqrt{\mu_{p}}&0&0\\ 0&0&0&\sqrt{\mu_{p}}&0\\ 0&0&0&0&\sqrt{\mu_{p}}\end{pmatrix}.

In this case, through the linear transformation

(𝒖1𝒖2z)2+2+13+3(𝒙1𝒙2)=𝑳p(𝒖1𝒖2z),\displaystyle\begin{pmatrix}\mbox{\boldmath$u$}^{1}\\ \mbox{\boldmath$u$}^{2}\\ z\end{pmatrix}\in\mathbb{R}^{2+2+1}\rightarrow\mathbb{R}^{3+3}\ni\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix}=\mbox{\boldmath$L$}^{p}\begin{pmatrix}\mbox{\boldmath$u$}^{1}\\ \mbox{\boldmath$u$}^{2}\\ z\end{pmatrix},

the quadratic function (𝒙1𝒙2)T𝑩(p+1)i(𝒙1𝒙2){\scriptsize\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix}}^{T}\mbox{\boldmath$B$}^{(p+1)i}{\scriptsize\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix}} in (𝒙1𝒙2){\scriptsize\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix}} is transformed to the quadratic function

(𝒙1𝒙2)T(𝑳p)T𝑩(p+1)i𝑳p(𝒙1𝒙2)\displaystyle\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix}^{T}(\mbox{\boldmath$L$}^{p})^{T}\mbox{\boldmath$B$}^{(p+1)i}\mbox{\boldmath$L$}^{p}\begin{pmatrix}\mbox{\boldmath$x$}^{1}\\ \mbox{\boldmath$x$}^{2}\end{pmatrix} =\displaystyle= (λp𝒖1λpzμp𝒖2μpz)T(𝑩pi𝑶𝑶T𝑪pi)(λp𝒖1λpzμp𝒖2μpz)\displaystyle\begin{pmatrix}\sqrt{\lambda_{p}}\mbox{\boldmath$u$}^{1}\\ \sqrt{\lambda_{p}}z\\ \sqrt{\mu_{p}}\mbox{\boldmath$u$}^{2}\\ \sqrt{\mu_{p}}z\end{pmatrix}^{T}\begin{pmatrix}\mbox{\boldmath$B$}^{pi}&\mbox{\boldmath$O$}\\ \mbox{\boldmath$O$}^{T}&\mbox{\boldmath$C$}^{pi}\end{pmatrix}\begin{pmatrix}\sqrt{\lambda_{p}}\mbox{\boldmath$u$}^{1}\\ \sqrt{\lambda_{p}}z\\ \sqrt{\mu_{p}}\mbox{\boldmath$u$}^{2}\\ \sqrt{\mu_{p}}z\end{pmatrix}
=\displaystyle= λp(𝒖1z)T𝑩pi(𝒖1z)+μp(𝒖2z)T𝑪pi(𝒖2z)\displaystyle\lambda_{p}\begin{pmatrix}\mbox{\boldmath$u$}^{1}\\ z\end{pmatrix}^{T}\mbox{\boldmath$B$}^{pi}\begin{pmatrix}\mbox{\boldmath$u$}^{1}\\ z\end{pmatrix}+\mu_{p}\begin{pmatrix}\mbox{\boldmath$u$}^{2}\\ z\end{pmatrix}^{T}\mbox{\boldmath$C$}^{pi}\begin{pmatrix}\mbox{\boldmath$u$}^{2}\\ z\end{pmatrix}

in (𝒖1𝒖2z)2+2+1{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}^{1}\\ \mbox{\boldmath$u$}^{2}\\ z\end{pmatrix}}\in\mathbb{R}^{2+2+1}. Therefore, if we fix zz to be 11, we obtain a convex combination of the quadratic function (𝒖11)T𝑩pi(𝒖11){\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}^{1}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$B$}^{pi}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}^{1}\\ 1\end{pmatrix}} in 𝒖12\mbox{\boldmath$u$}^{1}\in\mathbb{R}^{2} and the quadratic function (𝒖21)T𝑪pi(𝒖21){\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}^{2}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$C$}^{pi}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}^{2}\\ 1\end{pmatrix}} in 𝒖22\mbox{\boldmath$u$}^{2}\in\mathbb{R}^{2}.

Remark 3.1.

To add a linear equality 𝑨𝒖=𝒃\mbox{\boldmath$A$}\mbox{\boldmath$u$}=\mbox{\boldmath$b$} to QCQP (6), we introduce 𝑩B == (𝑨,𝒃)T-(\mbox{\boldmath$A$},-\mbox{\boldmath$b$})^{T} (𝑨,𝒃)(\mbox{\boldmath$A$},-\mbox{\boldmath$b$}) \in 𝕊n\mathbb{S}^{n}, where 𝑨A denotes an (n1)×(n-1)\times\ell matrix, and 𝒃\mbox{\boldmath$b$}\in\mathbb{R}^{\ell}. We see that

𝑩={𝒖n1:(𝒖1)T𝑩(𝒖1)0}={𝒖n1:𝑨𝒖𝒃20}.\displaystyle\mbox{\boldmath$B$}_{\geq}=\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{n-1}:\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\mbox{\boldmath$B$}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}\geq 0\right\}=\left\{\mbox{\boldmath$u$}\in\mathbb{R}^{n-1}:-\parallel\mbox{\boldmath$A$}\mbox{\boldmath$u$}-\mbox{\boldmath$b$}\parallel^{2}\geq 0\right\}.

Hence 𝑨𝒖=𝒃\mbox{\boldmath$A$}\mbox{\boldmath$u$}=\mbox{\boldmath$b$} if and only if 𝒖𝑩\mbox{\boldmath$u$}\in\mbox{\boldmath$B$}_{\geq}. It is known that if \cal B satisfies Condition (B), then so does ={𝑩}\mbox{$\cal B$}^{\prime}=\mbox{$\cal B$}\cup\{\mbox{\boldmath$B$}\}. See [2, Section 4.4] for more details.

4 Computing a QCQP optimal solution from an optimal solution of its SDP relaxation

Throughout this section, we assume

(a)

={𝑩1,,𝑩m}𝕊n\mbox{$\cal B$}=\{\mbox{\boldmath$B$}^{1},\ldots,\mbox{\boldmath$B$}^{m}\}\subseteq\mathbb{S}^{n} satisfies Condition (B),

(b)

The SDP relaxation (5) of QCQP (4) has an optimal solution 𝑿¯𝕊n\overline{\mbox{\boldmath$X$}}\in\mathbb{S}^{n},

(c)

𝑿=𝑿¯𝕊n\mbox{\boldmath$X$}=\overline{\mbox{\boldmath$X$}}\in\mathbb{S}^{n} satisfies the KKT (Karush-Kuhn-Tucker) stationary condition: there exists a (t¯,𝒚¯,𝒀¯)×m×𝕊n(\bar{t},\bar{\mbox{\boldmath$y$}},\overline{\mbox{\boldmath$Y$}})\in\mathbb{R}\times\mathbb{R}^{m}\times\mathbb{S}^{n} such that

𝑿𝕊+n,𝑩k𝑿0(1km),𝑯𝑿=1(primal feasibility),𝒚¯0,𝑸𝑯t¯k=1my¯k𝑩k=𝒀¯𝕊+n(dual feasibility),y¯k(𝑩k𝑿)=0(1km),𝒀¯𝑿=0(complementarity).}\displaystyle\left.\begin{array}[]{l}\mbox{\boldmath$X$}\in\mathbb{S}^{n}_{+},\ \mbox{\boldmath$B$}^{k}\bullet\mbox{\boldmath$X$}\geq 0\ (1\leq k\leq m),\ \mbox{\boldmath$H$}\bullet\mbox{\boldmath$X$}=1\ \mbox{(primal feasibility)},\\[3.0pt] \bar{\mbox{\boldmath$y$}}\geq\mbox{\bf 0},\ \mbox{\boldmath$Q$}-\mbox{\boldmath$H$}\bar{t}-\sum_{k=1}^{m}\bar{y}_{k}\mbox{\boldmath$B$}^{k}=\overline{\mbox{\boldmath$Y$}}\in\mathbb{S}^{n}_{+}\ \mbox{(dual feasibility)},\\[3.0pt] \bar{y}_{k}(\mbox{\boldmath$B$}^{k}\bullet\mbox{\boldmath$X$})=0\ (1\leq k\leq m),\ \overline{\mbox{\boldmath$Y$}}\bullet\mbox{\boldmath$X$}=0\ \mbox{(complementarity)}.\end{array}\right\} (21)

In (c), (t¯,𝒚¯,𝒀¯)×m×𝕊n(\bar{t},\bar{\mbox{\boldmath$y$}},\overline{\mbox{\boldmath$Y$}})\in\mathbb{R}\times\mathbb{R}^{m}\times\mathbb{S}^{n} corresponds to an optimal solution of the dual of SDP (5). We note that (c) \Rightarrow (b). If 𝑯𝕊+n\mbox{\boldmath$H$}\in\mathbb{S}^{n}_{+}, in particular, if 𝑯=diag(0,,0,1)𝕊+n\mbox{\boldmath$H$}=\mbox{diag}(0,\ldots,0,1)\in\mathbb{S}^{n}_{+} as in QCQP (6), then (b) \Rightarrow (c) [8, Theorem 2.1]. By Theorem 1.1, the optimal values of SDP (5) and QCQP (4) coincide, i.e., η=ζ\eta=\zeta.

We will describe a numerical method for computing an optimal solution 𝑿~𝚪n\widetilde{\mbox{\boldmath$X$}}\in\mbox{\boldmath$\Gamma$}^{n} of QCQP (4) from 𝑿¯𝕊+n\overline{\mbox{\boldmath$X$}}\in\mathbb{S}^{n}_{+}. Recall that all 22-dimensional QCQP instances in Section 2.3 satisfy Condition (D) and that a recursive procedure is provided for constructing higher-dimensional QCQP instances satisfying Condition (D) in Section 3. Since Condition (D) implies Condition (B) by Theorem 1.1 (iv), we can apply the method to those instances.

Let K0={k:𝑩k𝑿¯=0}K_{0}=\{k:\mbox{\boldmath$B$}^{k}\bullet\overline{\mbox{\boldmath$X$}}=0\}. Then we have either

(i) kK0k\in K_{0}\not=\emptyset for some k{1,,m}k\in\{1,\ldots,m\}.

(ii) K0=K_{0}=\emptyset.

We first deal with case (i). In this case, the method is based on the following lemma and its constructive proof.

Lemma 4.1.

([16, Lemma 2.2], see also [15, Proposition 3]) Let 𝐁𝕊n\mbox{\boldmath$B$}\in\mathbb{S}^{n} and 𝐗¯𝕊+n\overline{\mbox{\boldmath$X$}}\in\mathbb{S}^{n}_{+} with rank𝐗¯=r\overline{\mbox{\boldmath$X$}}=r. Suppose that 𝐁𝐗¯0\mbox{\boldmath$B$}\bullet\overline{\mbox{\boldmath$X$}}\geq 0. Then, there exists a rank-1 decomposition of 𝐗¯\overline{\mbox{\boldmath$X$}} such that 𝐗¯=i=1r𝐱i𝐱iT\overline{\mbox{\boldmath$X$}}=\sum_{i=1}^{r}\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T} and 𝐁𝐱i𝐱iT0\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}\geq 0 (1ir)(1\leq i\leq r). If, in particular, 𝐁𝐗¯=0\mbox{\boldmath$B$}\bullet\overline{\mbox{\boldmath$X$}}=0, then 𝐁𝐱i𝐱iT=0\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}=0 (1ir)(1\leq i\leq r).

By Lemma 4.1, there exists a rank-1 decomposition of 𝑿¯\overline{\mbox{\boldmath$X$}} such that 𝑿¯=i=1r𝒙i𝒙iT\overline{\mbox{\boldmath$X$}}=\sum_{i=1}^{r}\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T} and 𝑩kxixiT=0\mbox{\boldmath$B$}^{k}\bullet x_{i}x_{i}^{T}=0 (i.e., 𝒙i𝒙iT𝕁0(𝑩k)\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}\in\mathbb{J}_{0}(\mbox{\boldmath$B$}^{k})) (1ir)(1\leq i\leq r). By assumption (a), 𝒙i𝒙iT𝕁+()(1ir)\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}\in\mathbb{J}_{+}(\mbox{$\cal B$})\ (1\leq i\leq r). Since 1=𝑯𝑿¯=i=1r𝑯𝒙i𝒙iT1=\mbox{\boldmath$H$}\bullet\overline{\mbox{\boldmath$X$}}=\sum_{i=1}^{r}\mbox{\boldmath$H$}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}, there exist a τ1/r\tau\geq 1/r and a j{1,,r}j\in\{1,\ldots,r\} such that 𝑯𝒙j𝒙jT=τ\mbox{\boldmath$H$}\bullet\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}=\tau. Let 𝑿~=𝒙j𝒙jT/τ\widetilde{\mbox{\boldmath$X$}}=\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}/\tau. Since 𝒙j𝒙jT𝕁+()\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}\in\mathbb{J}_{+}(\mbox{$\cal B$}), 𝑿¯𝕁+()\overline{\mbox{\boldmath$X$}}\in\mathbb{J}_{+}(\mbox{$\cal B$}). We also see that 𝑯𝑿~=𝑯(𝒙jT𝒙j/τ)=1\mbox{\boldmath$H$}\bullet\widetilde{\mbox{\boldmath$X$}}=\mbox{\boldmath$H$}\bullet(\mbox{\boldmath$x$}_{j}^{T}\mbox{\boldmath$x$}_{j}/\tau)=1. Hence 𝑿~𝚪n\widetilde{\mbox{\boldmath$X$}}\in\mbox{\boldmath$\Gamma$}^{n} is a rank-11 feasible solution of SDP (5). Furthermore, we see from 𝒀¯𝕊+n\overline{\mbox{\boldmath$Y$}}\in\mathbb{S}^{n}_{+} and 𝒙i𝒙iT𝕊+n\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}\in\mathbb{S}^{n}_{+} (1ir)(1\leq i\leq r) that

0𝒀¯𝑿~=𝒀¯𝒙j𝒙jTτ𝒀¯i=1m𝒙i𝒙iTτ=𝒀¯𝑿¯τ=0.\displaystyle 0\leq\overline{\mbox{\boldmath$Y$}}\bullet\widetilde{\mbox{\boldmath$X$}}=\frac{\overline{\mbox{\boldmath$Y$}}\bullet\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}}{\tau}\leq\frac{\overline{\mbox{\boldmath$Y$}}\bullet\sum_{i=1}^{m}\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}}{\tau}=\frac{\overline{\mbox{\boldmath$Y$}}\bullet\overline{\mbox{\boldmath$X$}}}{\tau}=0.

Hence, 𝑿~𝚪n\widetilde{\mbox{\boldmath$X$}}\in\mbox{\boldmath$\Gamma$}^{n} is a rank-1 optimal solution of SDP (5) and it is an optimal solution of QCQP (4) with the same objective value 𝑸𝑿¯\mbox{\boldmath$Q$}\bullet\overline{\mbox{\boldmath$X$}}.

We now describe how to compute the rank-11 decomposition of 𝑿¯\overline{\mbox{\boldmath$X$}} such that 𝑿¯=i=1r𝒙i𝒙iT\overline{\mbox{\boldmath$X$}}=\sum_{i=1}^{r}\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T} and 𝑩𝒙i𝒙iT=0\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}=0 (1ir)(1\leq i\leq r) based on the constructive proof of Lemma 4.1 in [15, 16]. Let 𝑿¯=i=1r𝒙i𝒙iT\overline{\mbox{\boldmath$X$}}=\sum_{i=1}^{r}\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T} be an arbitrary rank-11 decomposition. If 𝑩𝒙i𝒙iT=0\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}=0 (1ir)(1\leq i\leq r), then we are done. Otherwise, there exist ii and jj such that 𝑩𝒙i𝒙iT<0<𝑩𝒙j𝒙jT\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}<0<\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}, say i=1i=1 and j=2j=2. We consider the following quadratic equation in ss:

0\displaystyle 0 =\displaystyle= 𝑩(s𝒙1+𝒙2)(s𝒙1+𝒙2)T=(𝑩𝒙1𝒙1T)s2+2(𝑩𝒙1𝒙2T)s+𝑩𝒙2𝒙2T.\displaystyle\mbox{\boldmath$B$}\bullet(s\mbox{\boldmath$x$}_{1}+\mbox{\boldmath$x$}_{2})(s\mbox{\boldmath$x$}_{1}+\mbox{\boldmath$x$}_{2})^{T}=(\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{1}\mbox{\boldmath$x$}_{1}^{T})s^{2}+2(\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{1}\mbox{\boldmath$x$}_{2}^{T})s+\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{2}\mbox{\boldmath$x$}_{2}^{T}.

Since 𝑩𝒙1𝒙1T×𝑩𝒙2𝒙2T<0\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{1}\mbox{\boldmath$x$}_{1}^{T}\times\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{2}\mbox{\boldmath$x$}_{2}^{T}<0, this equation has two distinct roots with opposite signs. Let s¯\bar{s} be one of the roots. Let

𝒙¯1=s¯s¯2+1𝒙1+1s¯2+1𝒙2and 𝒙¯2=1s¯2+1𝒙1+s¯s¯2+1𝒙2.\displaystyle\bar{\mbox{\boldmath$x$}}_{1}=\frac{\bar{s}}{\sqrt{\bar{s}^{2}+1}}\mbox{\boldmath$x$}_{1}+\frac{1}{\sqrt{\bar{s}^{2}+1}}\mbox{\boldmath$x$}_{2}\ \mbox{and }\bar{\mbox{\boldmath$x$}}_{2}=-\frac{1}{\sqrt{\bar{s}^{2}+1}}\mbox{\boldmath$x$}_{1}+\frac{\bar{s}}{\sqrt{\bar{s}^{2}+1}}\mbox{\boldmath$x$}_{2}.

Then, we have

𝑩𝒙¯1𝒙¯1T=0,𝒙¯1𝒙¯1T+𝒙¯2𝒙¯2T=𝒙1𝒙1T+𝒙2𝒙2T.\displaystyle\mbox{\boldmath$B$}\bullet\bar{\mbox{\boldmath$x$}}_{1}\bar{\mbox{\boldmath$x$}}_{1}^{T}=0,\ \bar{\mbox{\boldmath$x$}}_{1}\bar{\mbox{\boldmath$x$}}_{1}^{T}+\bar{\mbox{\boldmath$x$}}_{2}\bar{\mbox{\boldmath$x$}}_{2}^{T}=\mbox{\boldmath$x$}_{1}\mbox{\boldmath$x$}_{1}^{T}+\mbox{\boldmath$x$}_{2}\mbox{\boldmath$x$}_{2}^{T}.

Now replace 𝒙1\mbox{\boldmath$x$}_{1} with 𝒙¯1\bar{\mbox{\boldmath$x$}}_{1} and 𝒙2\mbox{\boldmath$x$}_{2} with 𝒙¯2\bar{\mbox{\boldmath$x$}}_{2}. Then we still have the rank-11 decomposition 𝑿¯=i=1r𝒙i𝒙iT\overline{\mbox{\boldmath$X$}}=\sum_{i=1}^{r}\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}. If 𝑩𝒙i𝒙iT0\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}\not=0 for some i2i\geq 2, we continue this procedure recursively till all 𝑩𝒙i𝒙iT=0\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}=0 (0ir)(0\leq i\leq r) are attained. To compute an optimal solution 𝑿~𝚪n\widetilde{\mbox{\boldmath$X$}}\in\mbox{\boldmath$\Gamma$}^{n} of QCQP (4), we can terminate the procedure once we find 𝒙jn\mbox{\boldmath$x$}_{j}\in\mathbb{R}^{n} such that 𝑩𝒙j𝒙jT=0\mbox{\boldmath$B$}\bullet\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}=0 and τ𝑯𝒙j𝒙jT1/r\tau\equiv\mbox{\boldmath$H$}\bullet\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}\geq 1/r. In this case, 𝑿~=𝒙j𝒙jT/τ\widetilde{\mbox{\boldmath$X$}}=\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}/\tau is an optimal solution of QCQP (4).

We now consider case (ii). In the KKT condition (21), 𝑿=𝑿¯\mbox{\boldmath$X$}=\overline{\mbox{\boldmath$X$}} satisfies 𝑩k𝑿¯>0\mbox{\boldmath$B$}^{k}\bullet\overline{\mbox{\boldmath$X$}}>0 and y¯k=0\bar{y}_{k}=0 (1km)(1\leq k\leq m). This implies that 𝑿¯\overline{\mbox{\boldmath$X$}} is an optimal solution of a simple SDP with the single equality constraint 𝑯𝑿=1\mbox{\boldmath$H$}\bullet\mbox{\boldmath$X$}=1:

η~=inf{𝑸𝑿:𝑿𝕊+n,𝑯𝑿=1}.\displaystyle\tilde{\eta}=\inf\left\{\mbox{\boldmath$Q$}\bullet\mbox{\boldmath$X$}:\mbox{\boldmath$X$}\in\mathbb{S}^{n}_{+},\ \mbox{\boldmath$H$}\bullet\mbox{\boldmath$X$}=1\right\}. (22)

The KKT stationary condition for SDP (22) is written as

𝑿𝕊+n,𝑯𝑿=1(primal feasibility)𝑸𝑯t¯=𝒀¯𝕊+n(dual feasibility)𝒀¯𝑿=0(complementarity)},\displaystyle\left.\begin{array}[]{l}\mbox{\boldmath$X$}\in\mathbb{S}^{n}_{+},\ \mbox{\boldmath$H$}\bullet\mbox{\boldmath$X$}=1\ \mbox{(primal feasibility)}\\[3.0pt] \mbox{\boldmath$Q$}-\mbox{\boldmath$H$}\bar{t}=\overline{\mbox{\boldmath$Y$}}\in\mathbb{S}^{n}_{+}\ \mbox{(dual feasibility)}\\[3.0pt] \overline{\mbox{\boldmath$Y$}}\bullet\mbox{\boldmath$X$}=0\ \mbox{(complementarity)}\end{array}\right\}, (26)

for some (t~,𝒀~)×𝕊n(\tilde{t},\widetilde{\mbox{\boldmath$Y$}})\in\mathbb{R}\times\mathbb{S}^{n}, which serves as a sufficient condition for 𝑿𝕊n\mbox{\boldmath$X$}\in\mathbb{S}^{n} to be an optimal solution of SDP (22). Now we compute a rank-11 optimal solution of SDP (22) from 𝑿¯\overline{\mbox{\boldmath$X$}}. Let r=rank𝑿¯r=\mbox{rank}\overline{\mbox{\boldmath$X$}}. If r=1r=1, then we have done. So assume r2r\geq 2. Let 𝑿¯=i=1r𝒙i𝒙iT\overline{\mbox{\boldmath$X$}}=\sum_{i=1}^{r}\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T} be a rank-11 decomposition of 𝑿¯\overline{\mbox{\boldmath$X$}}. It follows from (21) with 𝑿=𝑿¯\mbox{\boldmath$X$}=\overline{\mbox{\boldmath$X$}} that

0=𝒀¯𝑿¯=𝒀¯(i=1r𝒙i𝒙iT)=i=1r(𝒀¯𝒙i𝒙iT).\displaystyle 0=\overline{\mbox{\boldmath$Y$}}\bullet\overline{\mbox{\boldmath$X$}}=\overline{\mbox{\boldmath$Y$}}\bullet(\sum_{i=1}^{r}\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T})=\sum_{i=1}^{r}(\overline{\mbox{\boldmath$Y$}}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}).

Since 𝒀¯𝕊+n\overline{\mbox{\boldmath$Y$}}\in\mathbb{S}^{n}_{+} and 𝒙i𝒙iT𝕊+n\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}\in\mathbb{S}^{n}_{+}, 𝒀¯𝒙i𝒙iT0\overline{\mbox{\boldmath$Y$}}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}\geq 0 (1ir)(1\leq i\leq r). Hence the identity above implies 𝒀¯𝒙i𝒙iT=0\overline{\mbox{\boldmath$Y$}}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}=0 (1ir)(1\leq i\leq r). We also see from 1=𝑯𝑿¯=i=1r(𝑯𝒙i𝒙iT)1=\mbox{\boldmath$H$}\bullet\overline{\mbox{\boldmath$X$}}=\sum_{i=1}^{r}(\mbox{\boldmath$H$}\bullet\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T}) that there exists a jj such that τ𝑯𝒙j𝒙jT1/r\tau\equiv\mbox{\boldmath$H$}\bullet\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}\geq 1/r. Define 𝑿~=𝒙j𝒙jT/τ\widetilde{\mbox{\boldmath$X$}}=\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}/\tau. Then 𝑿=𝑿~\mbox{\boldmath$X$}=\widetilde{\mbox{\boldmath$X$}} satisfies (26). Thus 𝑿~\widetilde{\mbox{\boldmath$X$}} is a rank-11 optimal solution of SDP (22). If 𝑩k𝑿~0\mbox{\boldmath$B$}^{k}\bullet\widetilde{\mbox{\boldmath$X$}}\geq 0 (1km)(1\leq k\leq m) then 𝑿~\widetilde{\mbox{\boldmath$X$}} is a rank-11 optimal solution of SDP (5), hence an optimal solution of QCQP (4). Otherwise, we can take an optimal solution 𝑿^\widehat{\mbox{\boldmath$X$}} of SDP (5) such that {k:𝑩k𝑿^=0}\{k:\mbox{\boldmath$B$}^{k}\bullet\widehat{\mbox{\boldmath$X$}}=0\}\not=\emptyset as a convex combination of 𝑿¯\overline{\mbox{\boldmath$X$}} and 𝑿~\widetilde{\mbox{\boldmath$X$}}, which leads to case (i).

Refer to caption
Figure 13: The unshaded region represents the feasible region \mbox{$\cal B$}_{\geq} of QCQP (27).

To illustrate how the method works, we consider the following instance.

Instance 4.2.

We simultaneously describe 6 QCQP problems with quadratic objective functions qk:2q^{k}:\mathbb{R}^{2}\rightarrow\mathbb{R} (1k6)(1\leq k\leq 6) over a common quadratic inequality feasible region:

η\displaystyle\eta =\displaystyle= inf{qk(𝒖):𝒖2,22u1u22<=4,(u11)2+u221}\displaystyle\inf\left\{q^{k}(\mbox{\boldmath$u$}):\mbox{\boldmath$u$}\in\mathbb{R}^{2},\ -2\leq 2u_{1}-u_{2}^{2}<=4,\ (u_{1}-1)^{2}+u_{2}^{2}\geq 1\right\} (27)
=\displaystyle= inf{(𝒖1)T𝑸k(𝒖1):𝒖}\displaystyle\inf\left\{\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}^{T}\mbox{\boldmath$Q$}^{k}\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}:\mbox{\boldmath$u$}\in\mbox{$\cal B$}_{\geq}\right\}
=\displaystyle= inf{𝑸k𝑿:𝑿𝚪3𝕁+(),diag(0,0,1)𝑿=1},\displaystyle\inf\left\{\mbox{\boldmath$Q$}^{k}\bullet\mbox{\boldmath$X$}:\mbox{\boldmath$X$}\in\mbox{\boldmath$\Gamma$}^{3}\cap\mathbb{J}_{+}(\mbox{$\cal B$}),\ \mbox{diag}(0,0,1)\bullet\mbox{\boldmath$X$}=1\right\}, (28)

where

𝑩1=(001010102),𝑩2=(0010+10104),𝑩3=(1010+10100),={𝑩1,𝑩2,𝑩3},\displaystyle\mbox{\boldmath$B$}^{1}={\scriptsize\begin{pmatrix}0&0&1\\ 0&-1&0\\ 1&0&2\end{pmatrix}},\ \mbox{\boldmath$B$}^{2}={\scriptsize\begin{pmatrix}0&0&-1\\ 0&+1&0\\ -1&0&4\end{pmatrix}},\ \mbox{\boldmath$B$}^{3}={\scriptsize\begin{pmatrix}1&0&-1\\ 0&+1&0\\ -1&0&0\end{pmatrix}},\ \mbox{$\cal B$}=\{\mbox{\boldmath$B$}^{1},\mbox{\boldmath$B$}^{2},\mbox{\boldmath$B$}^{3}\},
q1(𝒖)=(u12)2+(u21)2=(𝒖1)T𝑸1(𝒖1),𝑸1=(102011215),\displaystyle q^{1}(\mbox{\boldmath$u$})=(u_{1}-2)^{2}+(u_{2}-1)^{2}={\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$Q$}^{1}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}},\ \mbox{\boldmath$Q$}^{1}={\scriptsize\begin{pmatrix}1&0&-2\\ 0&1&-1\\ -2&-1&5\end{pmatrix}},
q2(𝒖)=(u1+3)2+u22=(𝒖1)T𝑸2(𝒖1),𝑸2=(103010309),\displaystyle q^{2}(\mbox{\boldmath$u$})=(u_{1}+3)^{2}+u_{2}^{2}={\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$Q$}^{2}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}},\ \mbox{\boldmath$Q$}^{2}={\scriptsize\begin{pmatrix}1&0&3\\ 0&1&0\\ 3&0&9\end{pmatrix}},
q3(𝒖)=2u1=(𝒖1)T𝑸3(𝒖1),𝑸3=(001000100),\displaystyle q^{3}(\mbox{\boldmath$u$})=2u_{1}={\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$Q$}^{3}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}},\ \mbox{\boldmath$Q$}^{3}={\scriptsize\begin{pmatrix}0&0&1\\ 0&0&0\\ 1&0&0\end{pmatrix}},
q4(𝒖)=0=(𝒖1)T𝑸4(𝒖1),𝑸4=𝑶,\displaystyle q^{4}(\mbox{\boldmath$u$})=0={\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$Q$}^{4}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}},\ \mbox{\boldmath$Q$}^{4}=\mbox{\boldmath$O$},
q5(𝒖)=(u1+4u24)2=(𝒖1)T𝑸5(𝒖1),𝑸5=(1444161641616)\displaystyle q^{5}(\mbox{\boldmath$u$})=(u_{1}+4u_{2}-4)^{2}={\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$Q$}^{5}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}},\ \mbox{\boldmath$Q$}^{5}={\scriptsize\begin{pmatrix}1&4&-4\\ 4&16&-16\\ -4&-16&16\end{pmatrix}}
q6(𝒖)=(u13)2=(𝒖1)T𝑸6(𝒖1),𝑸6=(103000309).\displaystyle q^{6}(\mbox{\boldmath$u$})=(u_{1}-3)^{2}={\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}}^{T}\mbox{\boldmath$Q$}^{6}{\scriptsize\begin{pmatrix}\mbox{\boldmath$u$}\\ 1\end{pmatrix}},\ \mbox{\boldmath$Q$}^{6}={\scriptsize\begin{pmatrix}1&0&-3\\ 0&0&0\\ -3&0&9\end{pmatrix}}.

See Figure 13 for the feasible region \mbox{$\cal B$}_{\geq} of QCQP (27).

Table 1 presents a summary of the numerical results obtained for solving QCQP (27). The optimal solutions of QCQP (27) can be easily obtained from Figure 13 for k=1,,6k=1,\ldots,6, as shown in Table 1. The optimal solutions 𝑿¯\overline{\mbox{\boldmath$X$}} of the SDP relaxation were computed by SeDuMi [14]. For k=1k=1 or k=2k=2, QCQP (27) has a unique optimal solution, and a rank-11 solution 𝑿¯\overline{\mbox{\boldmath$X$}} was obtained by just solving the SDP relaxation. For k=3k=3, QCQP (27) has a unique optimal solution (1,0)(-1,0), but the computed optimal solution 𝑿¯\overline{\mbox{\boldmath$X$}} of the SDP relaxation has rank-22 and case (i) occurred. For k=4k=4, 55 or 66, the SDP relaxation as well as QCQP (27) have multiple optimal solutions, and rank𝑿¯2\overline{\mbox{\boldmath$X$}}\geq 2. For k=4k=4 or k=5k=5, the rank-11 decomposition 𝑿¯=i=13𝒙i𝒙iT\overline{\mbox{\boldmath$X$}}=\sum_{i=1}^{3}\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T} in case (ii) led to an optimal solution 𝑿~=𝒙j𝒙jT/τ\widetilde{\mbox{\boldmath$X$}}=\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}/\tau of QCQP (28) for some jj and τ=diag(0,0,1)𝒙j𝒙jT1/3\tau=\mbox{diag}(0,0,1)\bullet\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}\geq 1/3. For k=6k=6, case (ii) occurred at the optimal solution 𝑿¯\overline{\mbox{\boldmath$X$}} of the SDP relaxation, but its rank-11 decomposition 𝑿¯=i=13𝒙i𝒙iT\overline{\mbox{\boldmath$X$}}=\sum_{i=1}^{3}\mbox{\boldmath$x$}_{i}\mbox{\boldmath$x$}_{i}^{T} yielded no feasible rank-11 solution 𝑿~\widetilde{\mbox{\boldmath$X$}} and the recursive procedure of case (i) was carried out from a convex combination of 𝒙j𝒙jT/(diag(0,0,1)𝒙j𝒙jT)\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}/(\mbox{diag}(0,0,1)\bullet\mbox{\boldmath$x$}_{j}\mbox{\boldmath$x$}_{j}^{T}) for some jj and 𝑿¯\overline{\mbox{\boldmath$X$}}.

QCQP SDP, 𝑿¯\overline{\mbox{\boldmath$X$}}: Computed Optimal Solution
kk Opt.Sol. Opt.Val. Opt.Val Rank 𝑩1𝑿¯\mbox{\boldmath$B$}^{1}\bullet\overline{\mbox{\boldmath$X$}} 𝑩2𝑿¯\mbox{\boldmath$B$}^{2}\bullet\overline{\mbox{\boldmath$X$}} 𝑩3𝑿¯\mbox{\boldmath$B$}^{3}\bullet\overline{\mbox{\boldmath$X$}} Case
1 𝒖=(2,1)\mbox{\boldmath$u$}=(2,1) 0 0.000.00 1 5.00 1.00 1.00 (ii)
2 𝒖=(1,0)\mbox{\boldmath$u$}=(-1,0) 4 4.004.00 1 0.000.00 6.00 3.00 (i)
3 𝒖=(1,0)\mbox{\boldmath$u$}=(-1,0) -2 2.00-2.00 2 0.000.00 6.00 5.26 (i)
4 𝒖S4\forall\mbox{\boldmath$u$}\in S^{4} 0 0.000.00 3 2.00 3.99 2.28 (ii)
5 𝒖S5\forall\mbox{\boldmath$u$}\in S^{5} 0 0.000.00 2 1.68 4.324.32 1.42 (ii)
6 𝒖S6\forall\mbox{\boldmath$u$}\in S^{6} 0 0.000.00 2 3.45 2.55 7.55 (ii) \rightarrow (i)
Table 1: S4=S^{4}=\mbox{$\cal B$}_{\geq}. S5={𝒖:u1+4u2=4}S^{5}=\left\{\mbox{\boldmath$u$}\in\mbox{$\cal B$}_{\geq}:u_{1}+4u_{2}=4\right\}. S6={𝒖:u1=3}S^{6}=\left\{\mbox{\boldmath$u$}\in\mbox{$\cal B$}_{\geq}:u_{1}=3\right\}.

5 Concluding remarks

If 𝕊n\mbox{$\cal B$}\subseteq\mathbb{S}^{n} satisfies 𝕁+()^(𝚪n)\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}), then QCQP (4) is equivalent to its SDP relaxation (5) whose optimal value and solution can be easily computed. In Section 3, we have shown a recursive procedure for constructing various \cal B’s 𝕊n\subseteq\mathbb{S}^{n} satisfying Condition (D). The class of QCQPs with such \cal B’s appears to be quite broad, at least in theory. It should be noted that Condition (D) is merely sufficient for 𝕁+()^(𝚪n)\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}). For example, we could relax Condition (D) to

Condition (D)’:

If 𝑨,𝑩\mbox{\boldmath$A$},\ \mbox{\boldmath$B$}\in\mbox{$\cal B$} and 𝑨𝑩\mbox{\boldmath$A$}\not=\mbox{\boldmath$B$} then α𝑨+β𝑩𝕊+n\alpha\mbox{\boldmath$A$}+\beta\mbox{\boldmath$B$}\in\mathbb{S}^{n}_{+} for some nonzero (α,β)2(\alpha,\beta)\in\mathbb{R}^{2} ([1, Proposition 1]).

There is still a gap, however, between Condition (D)’ and the condition 𝕁+()^(𝚪n)\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}) ([1, Theorem 1] and [3, Example 6.1]). See [3, Section 3] where various sufficient conditions for 𝕁+()^(𝚪n)\mathbb{J}_{+}(\mbox{$\cal B$})\in\mbox{$\widehat{\mbox{$\cal F$}}$}(\mbox{\boldmath$\Gamma$}^{n}) and their relationships are discussed, under moderate assumptions including the Slater constraint qualification that 𝕁+()𝕊+n\mathbb{J}_{+}(\mbox{$\cal B$})\subseteq\mathbb{S}^{n}_{+} contains a positive definite matrix.

References

  • [1] C. J. Argue, F. Kilinç-Karzan, and A.L. Wang. Necessary and sufficient conditions for rank-one-generated cones. Math. Oper. Res., 48(1):100–126, 2023.
  • [2] N. Arima, S. Kim, and M. Kojima. Further development in convex conic reformulation of geometric nonconvex conic optimization problems. SIAM J. Optim., 34(4):3194–3211, August 2024.
  • [3] N. Arima, S. Kim, and M. Kojima. Exact SDP relaxations for a class of quadratic programs with finite and infinite quadratic constraints. Technical Report arXiv:2409.07213, September 2024.
  • [4] G. Azuma, Fukuda M., S. Kim, and M. Yamashita. Exact SDP relaxations for quadratic programs with bipartite graph structures. J. of Global Optim., 86:671–691, 2023.
  • [5] T. Fujie and M. Kojima. Semidefinite programming relaxation for nonconvex quadratic programs. J. of Global Optim., 10:367–368, 1997.
  • [6] R. Hildebrand. Spectrahedral cones generated by rank 1 matrices. J. Global Optim, 64:349–397, 2016.
  • [7] S. Kim and M. Kojima. Exact solutions of some nonconvex quadratic optimization problems via SDP and SOCP relaxations. Comput. Optim. Appl., 26(2):143–154, 2003.
  • [8] S. Kim and M. Kojima. Strong duality of a conic optimization problem with a single hyperplane and two cone constraints strong duality of a conic optimization problem with a single hyperplane and two cone constraints. Optimization, 74(1):33–53, 2025.
  • [9] S. Kim, M. Kojima, and K. C. Toh. A geometrical analysis of a class of nonconvex conic programs for convex conic reformulations of quadratic and polynomial optimization problems. SIAM J. Optim., 30:1251–1273, 2020.
  • [10] K. G. Murty and S. N. Kabadi. Some NP-complete problems in quadratic and non-linear programming. Math. Program., 39:117–129, 1987.
  • [11] N. Z. Shor. Quadratic optimization problems. Soviet Journal of Computer and Systems Sciences, 25:1–11, 1987.
  • [12] N. Z. Shor. Dual quadratic estimates in polynomial and boolean programming. Ann. Oper. Res., 25:163–168, 1990.
  • [13] S. Sojoudi and J. Lavaei. Exactness of semidefinite relaxations for nonlinear optimization problems with underlying graph structure. SIAM J. Optim., 24(4):1746–1778, 2014.
  • [14] J. F. Sturm. SeDuMi 1.02, a MATLAB toolbox for optimization over symmetric cones. Optim. Methods and Softw., 11&12:625–653, 1999.
  • [15] J. F. Sturm and S. Zhang. On cones of nonnegative quadratic functions. Math. Oper. Res., 28(2):246–267, 2003.
  • [16] Y. Ye and S. Zhang. New results on quadratic minimization. SIAM J. Optim., 14:245–267, 2003.
  • [17] S. Zhang. Quadratic optimization and semidefinite relaxation. Math. Program., 87:453–465, 2000.