On Equivalence of Semidefinite Relaxations for Quadratic Matrix Programming

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1 On Equivalence of Semidefinite Relaxations for Quadratic Matrix Programming Yichuan Ding Dongdong Ge Henry Wolkowicz April 28, 2010 Contents 1 Introduction Preliminaries Outline Quadratic Matrix Programming: Case I Lagrangian Relaxation Equivalence of Vector and Matrix Lifting for QMP Quadratic Matrix Programming: Case II Equivalence of Vector and Matrix Lifting for QMP Proof of (Main) Theorem Unbalanced Orthogonal Procrustes Problem An Extension to QMP with Conic Constraints Graph Partition Problem Conclusion 19 References 23 Index 23 Abstract In this paper, we analyze two popular semidefinite programming (SDP) relaxations for quadratically constrained quadratic programs (QCQP) with matrix variables. These are based on vector-lifting and on matrix lifting and are of different size and expense. We prove, under mild assumptions, that these two relaxations provide equivalent bounds. Thus, our results provide a theoretical guideline for how to choose an inexpensive SDP relaxation and still obtain a strong bound. Our results also shed important insights on how to simplify large-scale SDP constraints by exploiting the particular sparsity pattern. Technical report CORR ; URL: orion.math.uwaterloo.ca/ hwolkowi/reports/abstracts.html Department of Management Science & Engineering, Stanford University. y7ding@stanford.edu Antai College of Economics and Management, Shanghai Jiao Tong University. ddge@sjtu.edu.cn Research supported by Natural Sciences Engineering Research Council Canada. hwolkowicz@uwaterloo.ca 1

2 Keywords: Semidefinite Programming Relaxations, Quadratic Matrix Programming, Quadratic Constrained Quadratic Programming, Hard Combinatorial Problems. 1 Introduction In this paper, we aim to provide theoretical insights on how to compare different semidefinite programming (SDP) relaxations. In particular, we study a vector lifting relaxation and compare it to a significantly smaller matrix lifting relaxation and show that the resulting two bounds are equal. Many hard combinatorial problems can be formulated as a quadratically constrained quadratic program (QCQP) with matrix variables. If the resulting formulated problem is nonconvex, then SDP relaxations provide an efficient and successful approach for computing approximate solutions and strong bounds. Finding strong and inexpensive bounds is essential for branch and bound algorithms for solving large hard combinatorial problems. However, there can be many different SDP relaxations for the same problem, and it is usually not obvious which relaxation is overall optimal with regard to both computational efficiency and bound quality, e.g., 12]. For examples of using SDP relaxations for QCQP arising from hard problems, see e.g., quadratic assignment (QAP) 11, 12, 23, 26, 35], graph partitioning (GPP) 34], sensor network localization (SNL) 9, 10, 21], and the more general Euclidean distance matrix completions 1]. 1.1 Preliminaries The concept of quadratic matrix programming (QMP)was introduced by Beck in 6], where it refers to a special instance of QCQP with matrix variables. Because we include the study of more general problems, we denote the model discussed in 6] as the first case of quadratic matrix programming, denoted (QMP 1 ), (QMP 1 ) µ P1 := min trace(xt Q 0 X) + 2trace(C0 TX) + β 0 s.t. trace(x T Q j X) + 2trace(Cj TX) + β j 0, j = 1,2,...,m X M nr, where M nr denotes the set of n by r matrices, Q j S n,j = 0,1,...,m, S n is the space of n n symmetric matrices, and C j M nr. Throughout this paper, we use the trace inner-product (dot product) C X := trace C T X. The applicability of QMP 1 is limited when compared to the more general class QCQP. However, many applications use QCQP models in the form of QMP 1, e.g., robust optimization 7] and sensor network localization 1]. In addition, many combinatorial problems are formulated with orthogonality constraints in one of the two forms XX T = I, X T X = I. (1) When X is square, the pair of constraints in (1) are equivalent to each other, in theory. However, relaxations that include both forms of the constraints rather than just one, can be expected to obtain stronger bounds. For example, Anstreicher and Wolkowicz 4] proved that strong duality holds for a certain relaxation of QAP when both forms of the orthogonality constraints in (1) are included; however, there can be a duality gap if only one of the forms is used. Motivated by this result, we extend our scope of problems so that the objective and constraint functions can include 2

3 both forms of quadratic terms X T Q j X and XP j X T. We now define the second case of quadratic matrix programming (QMP 2 ) problems as (QMP 2 ) min trace(x T Q 0 X) + trace(xp 0 X T ) + 2trace(C T 0 X) + β 0 s.t. trace(x T Q j X) + trace(xp j X T ) + 2trace(C T j X) + β j 0,j = 1,...,m X M nr, (2) where Q j, P j are symmetric matrices of appropriate sizes. Both QMP 1 and QMP 2 can be vectorized into the QCQP form using trace(x T QX) = vec(x) T (I r Q)vec(X), trace(xpx T ) = vec(x) T (P I n )vec(x), (3) where denotes the Kronecker product, and vec(x) vectorizes X by stacking columns of X on top of each other. The difference in the Kronecker products (I r Q),(P I n ) shows that there is a difference in the corresponding Lagrange multipliers and illustrates why the bounds from Lagrangian relaxation will be different for these two sets of constraints. The SDP relaxation for the vectorized QCQP is called the vector-lifting semidefinite relaxation (V SDR). Under a constraint qualification assumption, V SDR for QCQP is equivalent to the dual of classical Lagrangian relaxation, see e.g., 24, 33, 5]. From (3), we get trace(x T QX) = trace(i r Q)Y, if Y = vec(x)vec(x) T trace(xpx T ) = trace(p I n )Y, if Y = vec(x)vec(x) T. (4) V SDR is derived using (4) with the relaxation Y vec(x)vec(x) T. A Schur complement argument, e.g., 25, 22], implies the equivalence of this relaxation to the large matrix variable constraint 1 vec(x T ] ) 0. A similar result holds for trace(xpx vec(x) Y T ) = vec(x) T (P I n )vec(x). Alternatively, from (3), we get the smaller system trace(x T QX) = trace QY, if Y = XX T trace(xpx T ) = trace PY, if Y = X T X. (5) The matrix-lifting semidefinite relaxation M SDR is derived using (5) with the relaxation Y XX T. A Schur complement argument ] now implies the equivalence of this relaxation to the smaller I X T matrix variable constraint 0. Again, a similar result holds for trace(xpx X Y T ). Intuitively, one expects that V SDR should provide stronger bounds than M SDR. Beck 6] proved that V SDR is actually equivalent to MSDR for QMP 1 if both SDP relaxations attain optimality and have a zero duality gap, e.g., when a constraint qualification, such as the Slater condition, holds for the dual program. In this paper we strengthen the above result by dropping the constraint qualification assumption. Then, we present our main contribution, i.e., we show the equivalence between MSDR and V SDR for the more general problem QMP 2 under a constraint qualification. This result is of more interest because QMP 2 does not possess the same nice structure (chordal pattern) as QMP 1. Moreover, QMP 2 encompasses a much richer class of problems and therefore has more significant applications. 3

4 1.2 Outline In Section 2 we present the equivalence of the corresponding V SDR and M SDR formulations for QMP 1 and prove Beck s result without the constraint qualification assumption, see Theorem 2.1. Section 3 proves the main result that V SDR and MSDR generate equivalent lower bounds for QMP 2, under a constraint qualification assumption, see Theorem 3.1. Numerical tests are included. Section 4 provides concluding remarks. 2 Quadratic Matrix Programming: Case I We first discuss the two relaxations for QMP 1. We denote the matrices in the relaxations obtained from vector and matrix lifting by M(qj V ( )) := β j vec(c j ) T ], vec(c j ) I r Q j ] βj M(qj M( )) := r I r Cj T. We let y = C j ( ) x0 R nr+1, Y = vec(x) and denote the quadratic and homogenized quadratic functions Q j ( ) X0 M (r+n)r, X q j (X) := trace(x T Q j X) + 2trace(C T j X) + β j q V j (X,x 0) := trace(x T Q j X) + 2trace(C T j Xx 0) + β j x 2 0 = y T M(q V j ( ))y q M j (X,X 0) := trace(x T Q j X) + 2trace(X T 0 CT j X) + trace β j r XT 0 I rx 0 = trace Y T M(q M j ( ))Y 2.1 Lagrangian Relaxation As mentioned above, under a constraint qualification, V SDR for QCQP is equivalent with the dual of classical Lagrangian relaxation. We include this result for completeness and to illustrate the role of a constraint qualification in the relaxation. We follow the approach in 24, Pg. 403] and use the strong duality of the trust region subproblem 31] to obtain the Lagrangian relaxation (or 4

5 dual) for QMP 1 as an SDP. µ L := max min q 0(X) + m λ 0 X i=j λ jq j (X) = max min q V λ 0 X,x 2 0 =1 0 (X,x 0) + m j=1 λ iqj V (X,x 0) ( = max min y T M(q V λ 0,t y 0 ( )) + ) m j=1 λ jm(qj V ( )) y + t(1 x 2 0 ( ) = max min trace M(q V λ 0,t y 0 ( )) + ) m j=1 λ jm(qj V ( )) yy T + t(1 x 2 0 ) max ] t t 0 = (DV SDR 1 ) s.t. m 0 0 j=1 λ jm(qj V ( )) M(qV 0 ( )) λ R m +,t R. (6) As illustrated in (6), Lagrangian relaxation is the dual program (denoted by DV SDR 1 ) of the vector-lifting relaxation V SDR 1 given below. Hence, under a constraint qualification, the Lagrangian relaxation is equivalent with the vector-lifting semidefinite relaxation. The usual constraint qualification is the Slater condition, i.e., λ R m +, s.t. M(q V 0 ( )) + m λ j M(qj V ( )) 0. (7) j=1 2.2 Equivalence of Vector and Matrix Lifting for QMP 1 Recall that the dot product refers to the trace inner-product, C X = trace C T X. The vector-lifting relaxation is (V SDR 1 ) Thus the constraint matrix is blocked as Z V = The matrix-lifting relaxation is µ V 1 := min M(qV 0 ( )) Z V s.t. M(qj V ( )) Z V 0, j = 1,2,...,m (Z V ) 1,1 = 1 Z V 0. 1 vec(x) T vec(x) Y V µ M1 := min M(qM 0 ( )) Z M s.t. M(qj M (MSDR 1 ) ( )) Z M 0, j = 1,2,...,m (Z M ) 1:r,1:r = I r Z M 0. Ir X Thus the constraint matrix is blocked as Z M = T ]. X Y M V SDR 1 is obtained by relaxing the quadratic equality constraint Y V = vec(x)vec(x) T to Y V vec(x)vec(x) T 1 vec(x), and then formulating this as Z V = T ] 0. MSDR vec(x) Y 1 is V obtained by relaxing the quadratic equality constraint Y M = XX T to Y M XX T and then Ir X reformulating this to the linear conic constraint Z M = T ] 0. V SDR X 1 involves O((nr) 2 ) 5 Y M ].

6 variables and O(m) constraints which is often at the complexity of O(nr), whereas the smaller problem MSDR 1 has only O((n + r) 2 ) variables. The equivalence of relaxations using vector and matrix-liftings is proved in 6, Theorem 4.3] by assuming a constraint qualification for the dual programs. We now present our first main result and prove the above mentioned equivalence without any constraint qualification assumptions. The proof is of interest in itself in that we use the chordal property and matrix completions to connect the two relaxations. Theorem 2.1. As numbers in the extended real line,+ ], the optimal values of the two relaxations obtained using vector and matrix-liftings are equal, i.e., µ V 1 = µ M1. Proof. The proof follows by showing that both V SDR 1 and MSDR 1 generate the same optimal values as the following program. µ V 1 := min Q 0 r j=1 Y jj + 2C 0 X + β 0 (V SDR 1 ) s.t. Q j r j=1 Y jj + 2C j X + β j 0, j = 1,2,...,m ] 1 x T Z jj = j 0, j = 1,2,...,r, x j where x j, j = 1,2,...,r, are the columns of matrix X, and Y jj, j = 1,2,,r, represent the corresponding quadratic parts x j x T j. We first show that the optimal values of V SDR 1 and V SDR 1 are equal, i.e., that Y jj µ V 1 = µ V 1. (8) The equivalence of the two optimal values can be established by showing for each program, for each feasible solution, one can always construct a corresponding feasible solution to the other program with the same objective value. ] 1 x First, suppose V SDR T 1 has a feasible solution Z jj = j, j = 1,2,...,r. Construct the partial symmetric matrix x j Y jj 1 x T 1 x T 2... x T r x 1 Y 11??? Z V = x 2? Y 22??,..??..? x r??? Y rr where the entries denoted by? are unknown/unspecified. By observation, the unspecified entries of Z V are not involved in the constraints or in the objective function of V SDR 1. In other words, adding values to the unspecified positions will not change the constraint function values and the objective value. Therefore, any positive semidefinite completion of the partial matrix Z V is feasible for V SDR 1 and has ] the same objective value. The feasibility of Z jj (j = 1,2,...,r) for 1 x V SDR T 1 implies j 0 for each j = 1,2,...,r. So all the specified principal submatrices x j Y jj of Z V are positive semidefinite, and hence Z V is a partial positive semidefinite matrix (See references 2, 15, 16, 18, 32] for the specific definitions of partial positive semidefinite, chordal graph, semidefinite completion.) It is not difficult to verify the chordal graph property for the sparsity 6

7 pattern of Z V. Therefore Z V has a positive semidefinite completion by the classical completion result 15, Theorem 7]. Thus we have constructed a feasible solution to V SDR 1 with the same objective value as the feasible solution from V SDR 1. i.e., this shows that µ V 1 µ V 1. Conversely, suppose V SDR 1 has a feasible solution 1 x T 1 x T 2... x T r x 1 Y 11 Y Y 1r Z V = x 2 Y 21 Y Y 2r x r Y r1 Y r2... Y rr ] 1 x T Now we construct Z jj := j, j = 1,2,...,r. Because each Z jj is a principal submatrix x j Y jj of the positive semidefinite matrix Z V, we have Z jj 0. The feasibility of Z V for V SDR 1 also implies M(q V i ( )) Z V 0, i = 1,2,...,m. (9) It is easy to check that r Q i Y jj + 2C i X + β i = M(qi V ( )) Z V 0, i = 1,2,...,m, (10) j=1 where X = ] x 1 x 2 x r. Therefore, Zjj, j = 1,2,...,r, is feasible for V SDR 1, and also generates the same objective value for V SDR 1 as Z V for V SDR 1 by (10), i.e., this shows that µ V 1 µ V 1. This completes the proof of (8). Next we prove that the optimal values of MSDR 1 and V SDR 1 are equal, i.e., that µ M1 = µ V 1. (11) The proof is similar to the one for (8). First suppose V SDR 1 has a feasible solution Z jj = ] 1 x T j, j = 1,2,,m. Let X = ] x x j Y 1 x 2 x r, and YM = r j=1 Y jj. Now we construct jj Ir X Z M := T ]. Then by Z X Y jj 0, j = 1,2,...,r, we have Y M = r j=1 Y jj r j=1 x jx T j = XXT, M which implies ZM 0 by the Schur Complement 22, 25]. And, because r M(qj M ( )) Z M = Q j Y ii + 2C j X + β j, j = 1,...,m, (12) i=1 we get Z M is feasible for MSDR and it generates the same objective value as the one by Z jj, j = 1,2,,m, for V SDR 1, i.e., µ M1 µ V 1. Ir X Conversely, suppose Z M = T ] 0 is feasible for MSDR X Y 1, and X = ] x 1 x 2 x r. M Let Y ii = x i (x i ) T for i = 1,2,...,r 1, and let Y rr = x r x T r + (Y M XX T ). As a result, Y ii x i x T i for i = 1,2,...,r, and ] r 1 x T i=1 Y ii = Y M. So, by constructing Z jj = j,j = 1,2,,r, it is easy to show that Z jj is feasible for V SDR 1 and generates an objective value equal to the objective value of MSDR with Z M, i.e., µ M1 µ V 1. This completes the proof of (11). Combining this with (8) completes the proof of the Theorem. 7 x j Y jj

8 Though MSDR 1 is significantly less expensive, Theorem 2.1 implies that the quality of the MSDR 1 bound is no weaker than that from V SDR 1. This tells us to always choose MSDR 1 if the problem can be formulated as in QMP 1. Example 2.1 (Sensor Network Localization Problem). The Sensor Network Localization, SNL, problem is one of the most studied problems in Graph Realization, e.g., 20, 21, 30]. In this problem one is given a graph with m known points (anchors) a k R d, k = 1,2,,m, and n unknown points (sensors) x j R d, j = 1,2,,n, where d is the embedding dimension. A Euclidean distance ˆd kj between a k and x j or distance ˆd ij between x i and x j is also given for some pairs of two points. The goal is to seek estimates of the positions for all unknown points. One possible formulation of the problem is as follows. min 0 s.t. trace(x T (E ii + E jj 2E ij )X) = d ij, (i,j) N x trace(x T E ii X) 2trace( a T j 0 ] X) + a T j a j = d ij, (i,j) N a X M nr, (13) where N x,n a refers to sets of known distances. This formulation is a QMP 1, so we can develop both its V SDR 1 and MSDR 1 relaxations. min 0 s.t. I (E ii + E jj 2E ij ) Y = d ij, (i,j) N x I E ii Y 2a T j x i + a T j a j = d ij, (i,j) N 1 ] a x T 0 x Y (14) min 0 s.t. (E ii + E jj 2E ij ) Y = d ij, (i,j) N x E ii Y 2 a T j 0 ] X + a T j a j = d ij, (i,j) N ] a I X T 0 X Y Theorem 2.1 implies that the MSDR 1 relaxation always provides the same lower bound as the V SDR 1 one, while the number of variables for MSDR 1 (O((n+d) 2 )) is significantly smaller than the number for V SDR 1 (O(n 2 d 2 )). The quality of the bounds combined with a lower computational complexity explains why MSDR 1 is a favourite relaxation for researchers. 3 Quadratic Matrix Programming: Case II In this section, we move to the main topic of our paper, i.e., the equivalence of the vector and matrix relaxations for the more general QMP 2. (15) 8

9 3.1 Equivalence of Vector and Matrix Lifting for QMP 2 We first propose the Vector-Lifting Semidefinite Relaxation V SDR 2 for QMP 2. From applying both equations in (4), we get the following. µ V 2 := min β 0 vec(c 0 ) T ] Z vec(c 0 ) I r Q 0 + P 0 I V n β s.t. j vec(c j ) T ] Z (V SDR 2 ) vec(c j ) I r Q j + P j I V 0, j = 1,2,...,m n (Z V ) 1,1 = 1 Z V S rn+1 + ( 1 vec(x) Z V = T ]) vec(x) Y V Matrix Y V is nr nr and can be partitioned into exactly r 2 block matrices Y ij V, i,j = 1,2,...,r, where each block is n n. From applying both equations in (5), we get the smaller Matrix- Lifting Semidefinite Relaxation MSDR 2 for QMP 2. (We add the additional constraint trace Y 1 = trace Y 2 since trace XX T = trace X T X.) (MSDR 2 ) µ M2 := min Q 0 Y 1 + P 0 Y 2 + 2C 0 X + β 0 s.t. Q j Y 1 + P j Y 2 + 2C j X + β j 0, j = 1,2,...,m ( Y 1 XX T S+ n Ir X Z 1 := T ] ) 0 X Y 1 ( ] ) Y 2 X T X S+ r In X Z 2 := X T 0 Y 2 trace Y 1 = trace Y 2. V SDR 2 has O((nr) 2 ) variables, whereas MSDR 2 has only O((n + r) 2 ) variables. The computational advantage of using the smaller problem MSDR 2 motivates the comparison of the corresponding bounds. The main result is interesting and surprising, i.e., that V SDR 2 and MSDR 2 actually generate the same bound under a constraint qualification assumption. In general, the bound from MSDR 2 is at least as strong as the bound from V SDR 2. Define the block-diag and block-offdiag transformations, respectively, B 0 Diag(Q) : S n S rn+1, O 0 Diag(P) : S r S rn+1, ] ] B Diag(Q) :=, O Diag(P) :=. 0 I r Q 0 P I n (See 35] for the r = n case.) It is clear that Q,P 0 implies that both B 0 Diag(Q) 0,O 0 Diag(P) 0. The adjoints b 0 diag,o 0 diag are, respectively, Y 1 Y 2 = B 0 Diag (Z V ) = b 0 diag(z V ) := r j=1 Y jj V, = O 0 Diag (Z V ) = o 0 diag(z V ) := (trace Y ij V ) i,j=1,2,...,r. Lemma 3.1. Let X M nr be given. Suppose that one of the following two conditions hold. 1. Let Y v be given and Z V defined as in V SDR 2. Let the pair Z 1,Z 2 in MSDR 2 be constructed using Y 1 = b 0 diag(z V ),Y 2 = o 0 diag(z V ). (17) (16) 9

10 2. Let Y 1,Y 2 be given with trace Y 1 = trace Y 1, and let Z 1,Z 2 be defined as in MSDR 2. Let Y V,Z V for V SDR 2 be constructed from Y 1,Y 2 as follows. 1 V 1 n (Y 1 2) 12 I n... n (Y 2) 1r I n 1 n (Y 1 2) 12 I n V 2... n (Y 2) 2r I n Y V = n (Y 2) (r 1)r I n, (18) V r with r V i = Y 1, trace V i = (Y 2 ) ii,i = 1,...,r. (19) i=1 Then, Z v satisfies the linear inequality constraints in V SDR 2 if, and only if, Z 1,Z 2 satisfies the linear inequality constraints in MSDR 2. Moreover, the values of the objective functions with the corresponding variables are equal. Proof. 1. Note that β vec(c) T ] Z vec(c) I r Q + P I V = β + 2C X + ( B 0 Diag(Q) + O 0 Diag(P) ) Z V n = β + 2C X + Q b 0 diag(z V ) + P o 0 diag(z V ) = β + 2C X + Q Y 1 + P Y 2, by (17). (20) 2. Conversely, we note that trace Y 1 = trace Y 2 is a constraint in MSDR 2. And, Z V as constructed using (18) satisfies (17). In addition, the n + r assignment type constraints in (19) on the rn variables in the diagonals of the V i,i = 1,...,r, can always be solved. We can now apply the argument in (20) again. Lemma 3.1 guarantees the equivalence of the feasible sets of the two relaxations with respect to the linear inequality constraints and the objective function. However, this ignores the semidefinite constraints. The following result partially addresses this deficiency. Corollary 3.1. If the feasible set of V SDR 2 is nonempty, then the feasible set of MSDR 2 is also nonempty and µ M2 µ V 2. (21) Proof. Suppose Z V = 1 vec(x) T vec(x) Y V and can be partitioned into exactly r 2 block matrices Y ij ] is feasible for V SDR 2. Recall that Matrix Y V is nr nr V, i,j = 1,2,...,r. As above, we set Y 1,Y 2 Ir X following (17), and we set Z 1 = T ] ] In X,Z X Y 2 = 1 X T. Y 2 Denote the j-th column of X by X :j, j = 1,2,...,r. Now Z V 0 implies Y jj V X :jx:j T 0. Therefore, r j=1 Y jj V r j=1 X :jx:j T = Y 1 XX T 0, i.e. Z 1 0. Similarly, denote the k-th row of X by X k:, k = 1,2,...,n. Let (Y ij V ) kk denote the k-th diagonal entry of Y ij V, and define the r r matrix Y k := ((Y ij V ) kk) i,j=1,2,,r. Then Z V 0 implies Y k Xk: TX k: 0. Therefore, n k=1 Y k n k=1 XT k: X k: = Y 2 X T X 0, i.e. Z 2 0. The proof now follows from Lemma

11 Corollary 3.1 holds because MSDR 2 only restricts the sum of some principal submatrices of Z V (i.e., b 0 diag(z V ), o 0 diag(z V )) to be positive semidefinite; while V SDR 2 restricts the whole matrix Z V positive semidefinite. So the semidefinite constraints in MSDR 2 are not as strong as in V SDR 2. Moreover, the entries of Y V involved in b 0 diag( ),o 0 diag( ) form a partial semidefinite matrix which is not chordal and does not necessarily have a semidefinite completion. Therefore, the semidefinite completion technique we used to prove the equivalence between V SDR 1 and MSDR 1 is not applicable here. Instead, we will prove the equivalence of their dual programs. It is well known that the primal equals the dual when the generalized Slater condition holds 17, 28], and in this case we will then conclude that V SDR 2 and MSDR 2 generate the same bound. Definition 3.1. For λ R m, let: and let C λ,q λ,p λ be defined similarly. β λ := β 0 + m λ j β j ; After substituting α β λ α, we see that the dual of V SDR 2 is equivalent to j=1 (DV SDR 2 ) max β λ α α vec(c s.t. λ ) T ] 0 vec(c λ ) I r Q λ + P λ I n α R,λ R m +. And, the dual of MSDR 2 is (DMSDR 2 ) max β λ trace S 1 trace S 2 S1 R s.t. 1 T ] 0 R 1 Q λ ti n ] S2 R 2 R2 T 0 P λ + ti r R 1 + R 2 = C λ λ R m +,S 1 S r,s 2 S n,r 1,R 2 M nr,t R. The Slater condition for DV SDR 2 is equivalent to the following: λ R m +, s.t. I r Q λ + P λ I n 0. (22) The corresponding constraint qualification condition for DMSDR 2 is t R, λ R m +, s.t. Q λ ti r 0,P λ + ti n 0. (23) These two conditions are equivalent due to the following lemma, which will also be used in our subsequent analysis. Lemma 3.2. Let Q S n, P S r. Then I r Q + P I n 0, (resp. 0) if, and only if, t R, s.t. Q ti n 0, P + ti r 0, (resp. 0.) 11

12 Proof. Suppose the symmetric matrices Q, P have spectral decomposition Q = V Λ Q V T, P = UΛ P U T, where V,U are orthogonal, and the eigenvalues are in the diagonal matrices Λ Q = Diag (( λ 1 (Q) λ 2 (Q)... λ n (Q) )), Λ P = Diag (( λ 1 (P) λ 2 (P)... λ r (P) )). Therefore, we get the equivalences: I r Q + P I n 0 if, and only if, (V U)(I r Λ P + P I n )(V U) T 0 if, and only if, I r Λ Q + Λ P I n 0 if, and only if, minλ i (Q) + min j λ j (P) > 0 i if, and only if, min λ i (Q) t > 0,min λ j (P) + t > 0, for some t R. i j The equivalences hold if the strict inequalities, 0 and > are replaced by the inequalities 0 and, respectively. Now we state the main theorem of this paper on the equivalence of the two SDP relaxations for QMP 2. Theorem 3.1. Suppose that DV SDR 2 is strictly feasible. As numbers in the extended real line (,+ ], the optimal values of the two relaxations V SDR 2, MSDR 2, obtained using vector and matrix-liftings, are equal, i.e., µ V 2 = µ M Proof of (Main) Theorem 3.1 Since DV SDR 2 is strictly feasible, Lemma 3.2 implies that both dual programs satisfy constraint qualifications. Therefore, both programs satisfy strong duality, see e.g., 28]. Therefore, both have zero duality gaps, i.e., the optimal values of DV SDR 2, DMSDR 2, are µ V 2,µ M2, respectively. Now assume that λ is feasible for DV SDR 2. (24) Lemma 3.2 implies that λ is also feasible for DMSDR 2, i.e., there exists t R such that Q := Q λ ti n 0, P := P λ + ti r 0. (25) (To simplify notation, we use Q, P to denote these dual slack matrices.) The spectral decomposition of Q, P can be expressed as ] ] Q = V Λ Q V T ΛQ + 0 = V 1 V 2 ] V V 2 ] T, P = UΛ P U T ΛP + 0 = U 1 U 2 ] U U 2 ] T, where the columns of the submatrices U 1, V 1 form an orthonormal basis that spans the range spaces R(P) and R(Q), respectively; while the columns of U 2, V 2 span the orthogonal complements R(P) and R(Q), respectively. Λ Q + is a diagonal matrix where diagonal entries are nonzero eigenvalues of matrix Q, and Λ P + is defined similarly. Let {σ i }, {θ i } denote the eigenvalues of P, Q, respectively. We similarly simplify the notation C := C λ, c := vec(c), β := β λ. (26) Let A denote the Moore-Penrose pseudoinverse of matrix A. The following lemma allows us to express µ V 2 as a function of Q, P, c and β. 12

13 Lemma 3.3. Let λ,p,q,c,β be defined as above in (24), (25), (26). Let α := c T ((I r Q + P I n ) )c. (27) Then, α,λ is a feasible pair for DV SDR 2. And, for any pair α,λ feasible to DV SDR 2, we have α + β α + β. Proof. A general quadratic ] function f(x) = x T Qx+2 c T x+ β is nonnegative for any x R n if, and β c T only if, the matrix 0, e.g., 8, Pg. 163]. Therefore, c Q if, and only if, For a fixed α, this is further equivalent to ] ] α c T α c T = 0 (28) c I r Q λ + P λ I n c I r Q + P I n x T (I r Q + P I n )x + 2c T x + α 0, x R nr. α min Therefore, we can choose α as in (27). x x T (I r Q + P I n )x + 2c T x = c T ((I r Q + P I n ) )c. To further explore the structure of (27), we note that c can be decomposed as c = (U 1 V 1 )r 11 + (U 1 V 2 )r 12 + (U 2 V 1 )r 21 + (U 2 V 2 )r 22. (29) The validity of such an expression follows from the fact that the columns of U 1 U 2 ] V 1 V 2 ] form an orthonormal basis of] R nr. Furthermore, the dual feasibility of DV SDR 2 includes the constraint α c T 0, which implies c R(I c I r Q + P I r Q + P I n ). This range space is spanned by n the columns in the matrices U 1 V 1, U 2 V 1 and U 1 V 2, which implies that c has no component in R(U 2 V 2 ), i.e., r 22 = 0 in (29). The following lemma provides a key observation for the connections between the two dual programs. It deduces that if c is in R(U 1 V 1 ), then the α component of the objective value of DV SDR 2 in Lemma 3.3 has a specific representation. Lemma 3.4. If c R(U 1 V 1 ), then the α component of the objective value of DV SDR 2 in Lemma 3.3 satisfies α = c T ((I r Q + P I n ) )c max vec(r 1 ) T (I r Q) vec(r 1 ) vec(r 2 ) T (P I n ) vec(r 2 ) (30) = s.t. R 1 + R 2 = C R 1,R 2 M nr 13

14 Proof. We can eliminate R 2 and express the maximization problem on the right hand side of the equality as max R1 φ(r 1 ), where φ(r 1 ) := vec(r 1 ) T ((I r Q) + (P I n ) )vec(r 1 ) +2c T (P I n ) vec(r 1 ) c T (P I n ) c. (31) Since P, Q are both positive semidefinite, we know (I r Q) + (P I n ) 0. Hence φ is concave. It is not difficult to verify that (P I n ) c R((I r Q) + (P I n ) ). Therefore, the maximum of the quadratic concave function φ(r 1 ) is finite and attained at R 1, vec(r 1 ) = ((I r Q) + (P I n ) ) (P I n ) c = (P Q + PP I n ) c; (32) and this corresponds to a value φ(r 1 ) = ct ((P Q + P P I n ) (P I n ) (P I n ) )c = c T (U V )ˆΛ(U V ) T c, where ˆΛ := Λ P I n (Λ 2 P Λ Q + Λ P I n ). Matrix ˆΛ is diagonal. Its diagonal entries can be calculated as { 1 σ ˆΛ(i,j) = i +θ j if σ i > 0,θ j > 0 0 if σ i = 0 or θ j = 0. We now compare φ(r 1 ) with ct (I r Q + P I n ) c. Let Λ := (I r Q + P I n ) = (U V )(I r Λ Q + Λ P I n ) (U V ) T = (U V )((I σ+ Λ Q + Λ P I θ+ ) + I σ0 Λ Q + Λ P I θ 0 )(U V ) T, (33) where matrix I σ+ (resp. I σ0 ) is r r, diagonal, and zero, except for the i-th diagonal entries that are equal to one if σ i > 0 (resp. σ i = 0); and, matrix I θ+ (resp. I θ0 ) is defined in the same way. Hence we know that matrix Λ is also diagonal. Its diagonal entries can be calculated as Λ(i,j) = 1 σ i +θ j if σ i > 0,θ j > 0 1 σ i if σ i > 0,θ j = 0 1 θ j if σ i = 0,θ j > 0 0 if σ i = 0,θ j = 0. By assumption, c = (U 1 V 1 )r 11, for some r 11 of appropriate size. Note that (U 1 V 1 )r 11 is orthogonal to the columns in U 2 V 1 and U 1 V 2. Thus only the part (I σ+ Λ Q + Λ P I θ+ ) in the diagonal matrix is involved in computations, i.e., c T (I r Q + P I n ) c = r T 11 (U 1 V 1 ) T (U V ) Λ(U V ) T (U 1 V 1 )r 11 = r T 11 (U 1 V 1 ) T (U V )(I σ+ Λ Q + Λ P I θ+ ) (U V ) T (U 1 V 1 )r 11 = r T 11 (U 1 V 1 ) T (U V )ˆΛ(U V ) T (U 1 V 1 )r 11 = φ(r 1 ). (34) 14

15 Now, for the given feasible α,λ of Lemma 3.3, we will construct a feasible solution for DMSDR 2 that generates the same objective value. Using Lemma 3.2, we choose t R satisfying Q = Q λ ti n 0, P = P λ + ti r 0. We can now find a lower bound for the optimal value of DMSDR 2. Proposition 3.1. Let λ,t,p,q,c,c,β be as above. Let R 1 denote the maximizer of φ(r 1) in the proof of Lemma 3.4. Construct R 1 R 2 as follows: vec(r 1 ) = vec(r 1 ) + (U 2 V 1 )r 21 vec(r 2 ) = vec(r 2 ) + (U 1 V 2 )r 12. (35) Then we obtain a lower bound for the optimal value of DMSDR 2. µ M2 vec(r 1 ) T (I r Q) vec(r 1 ) vec(r 2 ) T (P I n ) vec(r 2 ) + β. (36) Proof. Consider the subproblem that maximizes the objective with λ, t, R 1 and R 2 defined as above. max β trace S 1 trace S 2 S1 R s.t. 1 T ] 0 R 1 Q ] (37) S2 R 2 R2 T 0 P S 1 S r S 2 S n, Because λ, t, R 1, and R 2 are all feasible for DMSDR 2, this subproblem will generate a lower bound for µ M2. We now invoke ] a result from 6], i.e., that there exists a symmetric matrix S such S C T that trace S γ and 0 if, and only if, f(x) = trace(x C Q T QX + 2C T X) + γ 0 for any X M nr. This is equivalent to γ min X M trace(xt QX + 2C T X) = trace(c T Q C). nr Therefore, the subproblem (37) can be reformulated as max β trace S 1 trace S 2 s.t. trace S 1 trace(r T 1 Q R 1 ) trace S 2 trace(r 2 Q R T 2 ) S 1 S r,s 2 S n. (38) Hence, the optimal value of (38) has an explicit expression and provides a lower bound for DMSDR 2 µ M2 vec(r 1) T (I r Q) vec(r 1 ) vec(r 2 ) T (P I n ) vec(r 2 ) + β. With all the above preparations, we now complete the proof of (main) Theorem

16 Proof. (of Theorem 3.1) Now we compare µ V 2 from the expression (27) with µ M2 based on the lower bound expression in (36). By writing c in the form of (29), we get µ v2 = r T 11 (U 1 V 1 ) T + r T 12 (U 1 V 2 ) T + r T 21 (U 2 V 1 ) T] (I r Q + P I n ) (U 1 V 1 )r 11 + (U 1 V 2 )r 12 + (U 2 V 1 )r 21 ] +β. Consider the crossterm such as r T 11 (U 1 V 1 ) T (I r Q + P I n ) (U 1 V 2 )r 12. Since (U 1 V 1 )r 11 is orthogonal to (U 1 V 2 )r 12, and (I r Q + P I n ) is diagonalizable by U 1 U 2 ] V 1 V 2 ], this term is actually zero. Similarly, we can verify that the other crossterms equal zero. As a result, only the following sum of three quadratic terms remain, which we label using C1,C2,C3, respectively. (39) µ v2 = r11 T (U 1 V 1 ) T (I r Q + P I n ) (U 1 V 1 )r 11 r21 T (U 2 V 1 ) T (I r Q + P I n ) (U 2 V 1 )r 21 r12 T (U 1 V 2 ) T (I r Q + P I n ) (U 1 V 2 )r 12 + β =: C1 + C2 + C3 + β. (40) We can also formulate the lower bound for µ M2 based on (36): µ M2 vec(r 1 ) T (I r Q) vec(r 1 ) vec(r 2 ) T (P I n ) vec(r 2 ) + β = (vec(r 1 ) + (U 2 V 1 )r 21 ) T (I r Q) vec(r 1 + (U 2 V 1 )r 21 ) (vec(r 2 ) + (U 1 V 2 )r 12 ) T (P I n ) (vec(r 2 ) + (U 1 V 2 )r 12 ) + β. (41) Since vec(r1 ) and vec(r 2 ) are both in R(U 1 V 1 ), and this is orthogonal to both (U 1 V 2 )r 12 and (U 2 V 1 )r 21, and both matrices (I r Q) and (P I n ) are diagonalizable by U 1 U 2 ] V 1 V 2 ], we conclude that the crossterms, such as vec(r1 )T (I r Q) (U 2 V 1 )r 21, all equal zero. Therefore, the lower bound for µ M2 can be reformulated as µ M2 (vec(r1 )(I r Q) vec(r1 ) + vec(r 2 )(P I n) vec(r2 )) r21 T (U 2 V 1 ) T (I r Q) (U 2 V 1 )r 21 r12 T (U 1 V 2 ) T (P I n ) (U 1 V 2 )r 12 + β. =: T1 + T2 + T3 + β. (42) As above, denote the first three quadratic terms by T1,T2,T3, respectively. We will show that term C1, C2 and C3 equal T1, T2 and T3, respectively. The equality between C1 and T1 follows from Lemma 3.4. For the other terms, consider C2 first. Write (I r Q+P I n ) as the diagonal matrix Λ by (33). Note that (U 2 V 1 )r 21 is orthogonal with the columns in U 1 V 1 and U 1 V 2. Thus only the part I σ0 Λ Q in diagonal matrix Λ is involved in computing term C2, i.e., r21 T (U 2 V 1 ) T (I r Q + P I n ) (U 2 V 1 )r 21 = r21 T (U 2 V 1 ) T (U V )(I σ0 Λ Q )(U V (43) )T (U 2 V 1 )r 21. Similarly, since (U 2 V 1 )r 21 is orthogonal with eigenvectors in U 1 V 2, we have r T 21 (U 2 V 1 ) T (I r Q) (U 2 V 1 )r 21 = r T 21 (U 2 V 1 ) T (U V )(I σ+ Λ Q + I σ 0 Λ Q )(U V )T (U 2 V 1 )r 21 = r T 21 (U 2 V 1 ) T (U V )(I σ0 Λ Q )(U V )T (U 2 V 1 )r 21. (44) 16

17 By (43) and (44), we conclude that that the term C2 equals T2. We may use the same argument to prove the equality of C3 and T3. Therefore, we conclude that c T ((I r Q + P I n ) )c + β = vec(r 1 ) T ((I r Q) )vec(r 1 ) vec(r 2 ) T ((P I n ) )vec(r 2 ) + β. (45) Then by (27) and (36), we have established µ V 2 µ M2. The other direction has been proved in Corollary 3.1. This completes the proof of the theorem Unbalanced Orthogonal Procrustes Problem Example 3.1. In the Unbalanced Orthogonal Procrustes Problem 14] one seeks to solve the following minimizing problem. min AX B 2 F s.t. X T X = I r, (46) X M nr, where A M nn, B M nr, and n r. The balanced case n = r can be solved efficiently 29] and this special case also admits a QMP 1 relaxation, 6]. Note that the unbalanced case is a typical QMP 2. Its V SDR 2 can be written as: min trace((i r A T A)Y ) trace(2b T AX) s.t. trace((e ] ij I n )Y ) = δ i,j, i,j = 1,2,,r, (47) 1 x x T 0. Y It is easy to check that the SDP in (47) is feasible and its dual is strictly feasible, which implies the equivalence between MSDR 2 and V SDR 2. Thus we can obtain a nontrivial lower bound from its MSDR 2 relaxation: min trace(a T AY 2B T AX) Ir X s.t. T ] 0, X Y ] In X 0, X T I r trace Y = r. We also run preliminary computational experiments to compare the efficiency of the two SDP relaxations, see Table 3.1. All matrices in 5 instances are randomly generated and the problems are solved by SeDuMi 1.1 on a notebook computer(cpu: Intel Core 2 Duo, 2.53GHZ/Memory: 3GB). Table 3.1 exhibits the computational advantage of MSDR 2 over V SDR 2. (48) Table 1: The solution time(cpu seconds) of two SDP relaxations on the orthogonal Procrustes problem. Problem Scale (n,r) (15,5) (20,10) (30,10) (30,15) (40,20) V SDR 2 (CPU time) MSDR 2 (CPU time)

18 3.2 An Extension to QMP with Conic Constraints Some QMP problems include conic constraints such as X T X ( )S, where S is a given positive semidefinite matrix. We can prove that the corresponding MSDR 2 and V SDR 2 are still equivalent for such problems. Consider the following general form of QMP 2 with conic constraints: (QMP 3 ) min trace(x T Q 0 X) + trace(xp 0 X T ) + 2trace(C T 0 X) + trace(ht 0 Z) s.t. trace(x T Q j X) + trace(xp j X T ) + 2trace(C T j X) + trace(ht j Z) + β j 0, j = 1,2,...,m X M nr, Z K where K can be the direct sum of convex cones (e.g., second-order cones, semidefinite cones). Note that the constraint X T X ( )S can be formulated as trace(x T XE ij ) + ( )trace(ze ij ) = S ij Z 0. (49) The formulations of V SDR 2 and MSDR 2 for QMP 3 are the same as for QMP 2 except for the additional term H j Z and the conic constraint Z K. Correspondingly, the dual programs DV SDR 2 and DMSDR 2 for QMP 3 will both have an additional constraint H 0 m λ j H j K. (50) j=1 If a dual solution λ is feasible for DV SDR 2, then it satisfies the constraint (50) in both DV SDR 2 and DMSDR 2. Therefore, we can follow the proof of Theorem 3.1, and construct a feasible solution for DMSDR 2 with λ which generates the same objective value as µ V 2. This yields the following. Corollary 3.2. Assume V SDR 2 for QMP 3 is strictly feasible and its dual DV SDR 2 is feasible. Then DV SDR 2 and DMSDR 2 both attain their optimum at the same λ and generate the same optimal value µ V 2 = µ M Graph Partition Problem Example 3.2 (Graph Partition Problem, (GPP)). GPP is an important combinatorial optimization problem with broad applications in network design and floor planning 3, 27]. Given a graph with n vertices, the problem is to find an r-partition S 1,S 2,...,S r of the vertex set, such that S i = m i with m := (m i ) i=1,...,r given cardinalities of subsets, and the total number of edges across different subsets is minimized. Define matrix X M nr to be the assignment of vertices, i.e., X ij = 1 if vertex i is assigned to subset j; X ij = 0 otherwise. With L the Laplacian matrix, the GPP can be formulated as an optimization problem: µ GPP = min 1 2 trace(xt LX) s.t. X T X = Diag(m) diag(xx T ) = e n X 0 (51) 18

19 This formulation involves quadratic matrix constraints of both types trace(x T E ii X) = 1, i = 1,...,n and trace(xe ij X T = m i δ(i,j), i,j = 1,...,r. Thus it can be formulated as a QMP 2 but not a QMP 1. Anstreicher and Wolkowicz 5] proposed a semidefinite program relaxation with O(n 4 ) variables and proved that its optimal value equals the so-called Donath-Hoffman lower bound 13]. This SDP formulation can be written in a more compact way as Povh 27] suggested: µ DH = min 1 2 trace((i r L)V ) s.t. r i=1 1 m i V ii + W = I n, trace(v ij ) = m i δ i,j, i,j = 1,...,r trace((i E ii )V ) = 1, i = 1,...,n V S rn +, W S+ n, (52) where V has been partitioned into r 2 square blocks, with each block size of n by n, and V ij is the (i,j)-th block of V. Note that formulation (52) reduces the number of variables to O(n 2 r 2 ). An interesting application is the graph equipartition problem in which m i (= m 1 ) s are all the same. Povh s SDP formulation is actually a V SDR 2 for QMP 3 : min 1 2 trace(xt LX) s.t. trace( 1 m 1 X T E ij X + E ij W) = δ i,j, i,j = 1,...,n trace(xe ij X T ) = m 1 δ i,j, i,j = 1,...,r trace(x T E ii X) = 1, i = 1,...,n W S n +. It is easy to check that (52) is feasible and its dual is strictly feasible. Hence by Corollary 3.2, the equivalence between MSDR 2 and V SDR 2 for QMP 3 implies that the Donath-Hoffman bound can be computed by solving a small MSDR 2 : µ DH = min 1 2 L Y 1, s.t. Y 1 m 1 I n, Y 2 = m 1 I r, diag(y 1 ) = e n, Ir X T ] 0, X Y 1 ] In X X T 0 Y 2 Because X and Y 2 do not appear in the objective, formulation (54) can be reduced to a very simple form: µ DH = min 1 2 L Y 1 s.t. diag(y 1 ) = e n (55) 0 Y 1 m 1 I n. This MSDR formulation has only O(n 2 ) variables, which is a significant reduction from O(n 2 r 2 ). This result coincides with Karish and Rendl s result 19] for the graph equipartition. Their proof derives from the particular problem structure, while our result is based on the general equivalence of V SDR 2 and MSDR 2. 4 Conclusion This paper proves the equivalence of two SDP bounds for the hard QCQP in QMP 2. Thus, it is clear that a user should use the smaller/inexpensive MSDR bound from matrix-lifting, rather (53) (54) 19

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23 Index A, Moore-Penrose pseudoinverse, 12 C λ, 11 P λ, 11 Q λ, 11 Y ij V, blocks of Y V, 9 B 0 Diag, Block-diag, 9 M nr, set of n by r matrices, 2 O 0 Diag, Block-offdiag, 9 R( ), range space, 13 S n, the space of n n symmetric matrices, 2 β λ, 11 b 0 diag, block-diag adjoint, 9 vec(x), 3 o 0 diag, block-offdiag adjoint, 9 (X), homogenized quadratic matrix function, q M j q V j 4 (X), homogenized quadratic vector function, 4 q j (X), quadratic function, 4 M SDR, matrix-lifting semidefinite relaxation, 3 MSDR 2, vector-lifting relaxation, 9 QMP, quadratic matrix programming, 2 QMP 1, first case of quadratic matrix programming, 2 QMP 2, second case of quadratic matrix programming, 3, 8 V SDR, vector-lifting semidefinite relaxation, 3 V SDR 1, vector-lifting relaxation, 5 V SDR 2, vector-lifting relaxation, 9 MSDR 1, matrix-lifting relaxation, 5 block-diag adjoint, b 0 diag, 9 Block-diag, B 0 Diag, 9 block-offdiag adjoint, o 0 diag, 9 Block-offdiag, O 0 Diag, 9 blocks of Y V, Y ij V, 9 chordal graph, 7 homogenized quadratic matrix function, q M j (X), 4 homogenized quadratic vector function, q V j (X), 4 Kronecker product, 3 lifted matrices for QMP 1, 4 matrix-lifting relaxation, MSDR 1, 5 matrix-lifting semidefinite relaxation, M SDR, 3 Moore-Penrose pseudoinverse, A, 12 partial symmetric matrix, 6 quadratic function, q j (X), 4 quadratic matrix programming, QMP, 2 quadratically constrained quadratic program, QCQP, 2 range space, R( ), 13 Schur complement, 3 second case of quadratic matrix programming, QMP 2, 3, 8 sensor network localization problem, SNL, 8 set of n by r matrices, M nr, 2 SNL, sensor network localization problem, 8 the space of n n symmetric matrices, S n, 2 trace dot product, 2, 5 trace inner-product, 2 vector-lifting relaxation, MSDR 2, 9 vector-lifting relaxation, V SDR 1, 5 vector-lifting relaxation, V SDR 2, 9 vector-lifting semidefinite relaxation, V SDR, 3 first case of quadratic matrix programming, QMP 1, 2 23