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New Results on Hermitian Matrix Rank-One Decomposition
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1 New Results on Hermitian Matrix Rank-One Decomposition Wenbao Ai Yongwei Huang Shuzhong Zhang June 22, 2009 Abstract In this paper, we present several new rank-one decomposition theorems for Hermitian positive semidefinite matrices, which generalize our previous results in [18, 2]. The new matrix rank-one decomposition theorems appear to have wide applications in theory as well as in practice. On the theoretical side, for example, we show how to further extend some of the classical results including a lemma due to Yuan [27], the classical results on the convexity of the joint numerical ranges [23, 4], and the so-called Finsler s lemma [9, 4]. On the practical side, we show that the new results can be applied to solve two typical problems in signal processing and communication: one for radar code optimization and the other for robust beamforming. The new matrix decomposition theorems are proven by construction in this paper, and we demonstrate that the constructive procedures can be implemented efficiently, stably, and accurately. The URL of our Matlab programs is given in this paper. We strongly believe that the new decomposition procedures, as a means to solve non-convex quadratic optimization with a few quadratic constraints, are useful for many other potential engineering applications. Keywords: positive semidefinite Hermitian matrix, matrix rank-one decomposition, joint numerical range, quadratic optimization. Mathematics Subject Classification: 90C33, 90C51, 90C05. School of Science, Beijing University of Posts and Telecommunications, Beijing , People s Republic of China. Project supported by Chinese NSFC. Department of Systems Engineering and Engineering Management, The Chinese University of Hong Kong, Shatin, Hong Kong. Department of Systems Engineering and Engineering Management, The Chinese University of Hong Kong, Shatin, Hong Kong. Research supported by Hong Kong RGC Earmarked Grants CUHK and CUHK
2 1 Introduction In a series of recently papers ([25, 18, 2]) we have developed the technique of decomposing a positive semidefinite matrix into the sum of rank-one matrices with some desirable properties. This matrix rank-one decomposition approach proves to be useful in a variety of ways. Most noticeably, it can be used to derive the rank-one optimal solutions for an Semidefinite Program when the number of constraints is small; see [26]. In that way, the method has found wide applications in signal processing and communication; see [12, 13]. The first such type matrix rank-one decomposition method was introduced by Sturm and Zhang in [25] as a means to characterize the matrix cone whose quadratic form is co-positive over a given domain. This naturally connects to the S-lemma of Yakubovich, since the S-lemma is concerned with the positivity of a quadratic form over the domain defined by the level set of another quadratic form. As a matter of fact, these results can be viewed as a duality pair. The matrix rank-one decomposition procedure proposed in [25] is easy to implement, and can be considered as a constructive proof for the S-lemma. For a survey on the S-lemma, we refer to Polik and Terlaky [24]. Applications of such matrix decomposition technique can be quite versatile. For instance, Sturm and Zhang in [25] showed how one can solve the quadratic optimization problem with the constraint set being the intersection of an ellipsoid and a half-plane, via semidefinite programing (SDP) relaxation followed by a rank-one decomposition procedure. Following the approaches adopted in Sturm and Zhang [25], Huang and Zhang [18] generalized the constructive technique to the domain of Hermitian PSD matrices, proving the complex version of Yakubovich s S-lemma, and also deriving an upper bound on the lowest rank among all the optimal solutions for a standard complex SDP problem. Pang and Zhang [23] and Huang and Zhang [18] gave alternative proofs for some convexity properties of the joint numerical ranges ([10, 4]) using the matrix rank-one decomposition techniques. Up to that point, the matrix rank-one decomposition was meant to be a complete decomposition. Ai and Zhang [2] obtained a partial rank-one decomposition result for the real symmetric positive semidefinite matrices, and used this result to fully characterize the condition under which the strong duality holds for the (nonconvex) quadratic optimization problem over the intersection of two ellipsoids: the problem is known as the CDT subproblem in the literature needed by the trust region method for nonlinear programming. In the current paper, we present some new rank-one decomposition results for the Hermitian positive semidefinite matrices, generalizing the matrix decomposition theorems in [18, 2]. We illustrate the potentials of the new decomposition results from both practical and theoretical aspects. On the practical side, we shall demonstrate two applications of the new decomposition theorems in signal processing, viz. the optimal radar code selection problem and the robust beamforming problem. On the theoretical side, we present a generalization of Yuan s lemma [27], and show the connections to the famous convexity results with regard to the joint numerical ranges (cf. [23, 4]) and a generalized 2
3 Finsler s lemma (cf. [9, 4]). Finally, we discuss the numerical performance of the matrix rank-one decomposition algorithms. The URL of our Matlab programs can be found at Section 6. We believe that the algorithms are useful in many other applications. This paper is organized as follows. In Section 2, we present three new Hermitian p.s.d. matrix rank-one decomposition theorems and some initial analysis of the results. Section 3 is devoted to the proofs of these three new theorems. To showcase the potential applications of the new results, we illustrate in Section 4 applications of the decomposition theorems arising from signal processing. To put our new results in perspective, we present in Subsection 5.1 an extension of Yuan s lemma [27], and in Subsection 5.2 the equivalence between one of the new decomposition theorems and some other well-established results in the literature. All of our proofs are constructive, and they can be efficiently implemented in the form of algorithms. In order to evaluate these algorithms, in Section 6 we present the performance of the constructive procedures; we pay special attention to the numerical stability and accuracy of those solution procedures. Notation. Throughout, we denote ā to be the conjugate of a complex number a, C n to be the space of n-dimension complex vectors, i to be the imaginary unit, i.e., i 2 = 1. For a given vector z C n, z H denotes the conjugate transpose of z. The space of n n real symmetric and complex Hermitian matrices are denoted by S n and H n, respectively. For a matrix Z H n, we denote Re Z and Im Z as the real and imaginary part of Z respectively. Matrix Z being Hermitian implies that Re Z is symmetric and Im Z is skew-symmetric. We denote by S+ n (S++) n and H+ n (H++) n the cones of real symmetric positive semidefinite (positive definite) and complex Hermitian positive semidefinite (positive definite) matrices respectively. The notation Z ( 0) means that Z is positive semidefinite (positive definite). For two complex matrices Y and Z, their inner product Y Z = Re (tr Y H Z) = tr [ (Re Y ) T (Re Z) + (Im Y ) T (Im Z) ], where tr denotes the trace of a matrix and T denotes the transpose of a matrix, and H denotes the conjugate transpose of a matrix. For a square matrix M, the notation rank(m), Range(M), Null(M) stand for, respectively, the rank, the range space, and the null space of M. 2 The Main Results Let us start by presenting the main results of this paper, in the form of three theorems. The first theorem is about decomposing a Hermitian positive semidefinite matrix into the sum of rank-one matrices, with a specific condition, as we see below. Theorem 2.1. Let A 1, A 2, A 3 H n, and X H+ n be a nonzero Hermitian positive semidefinite matrix. If r = rank (X) 3, then one can find in polynomial-time a rank-one decomposition of X, 3
4 X = r i=1 x ix H i, such that A 1 x i x H i = A 1 X/r, i = 1,..., r; A 2 x i x H i = A 2 X/r, i = 1,..., r; A 3 x i x H i = A 3 X/r, i = 1,..., r 2. The real-case counterpart of the above theorem is in Ai and Zhang [2]. In the real case, the result actually appears to be weaker. The next theorem mainly deals with the situation where Theorem 2.1 does not apply; in particular, when rank (X) = 2. In Theorem 2.2, we no longer seek for a complete rank-one decomposition of X. Rather, we look for a rank-one matrix solution to a system of linear matrix equations, within a slightly expanded range space (of X). Theorem 2.2. Suppose n 3. Let A 1, A 2, A 3 H n, and X H n + be a nonzero Hermitian positive semidefinite matrix of rank r. y Range (X) such that If r 3, then one can find in polynomial-time a nonzero vector A 1 yy H = A 1 X, A 2 yy H = A 2 X, A 3 yy H = A 3 X, with X 1 r yyh 0 and rank (X 1 r yyh ) r 1. If r = 2, then for any z / Range (X) there exists y C n in the linear subspace spanned by z and Range (X), such that A 1 yy H = A 1 X, A 2 yy H = A 2 X, A 3 yy H = A 3 X, with X + zz H 1 r yyh 0 and rank (X + zz H 1 r yyh ) 2. Although Theorem 2.2 is best possible in general, with one more regularity condition we will be able to extend the result to satisfy four matrix equations. Theorem 2.3. Suppose n 3. Let A 1, A 2, A 3, A 4 H n, and X H+ n be a nonzero Hermitian positive semidefinite matrix of rank r. Furthermore, suppose that (A 1 Y, A 2 Y, A 3 Y, A 4 Y ) (0, 0, 0, 0), for any nonzero matrix Y H+. n If r 3, then one can find in polynomial-time a nonzero vector y Range (X) such that A 1 yy H = A 1 X, A 2 yy H = A 2 X, A 3 yy H = A 3 X, A 4 yy H = A 4 X. 4
5 If r = 2, then for any z / Range (X) there exists y C n in the linear subspace spanned by z and Range (X), such that A 1 yy H = A 1 X, A 2 yy H = A 2 X, A 3 yy H = A 3 X, A 4 yy H = A 4 X. Before discussing the proofs, let us start by presenting the following three examples, demonstrating that the conditions in these theorems are all necessary. Examples 1 and 2 show that the assumption n 3 is necessary for Theorems 2.1 and 2.3, and Example 3 shows that the assumption (A 1 Y, A 2 Y, A 3 Y, A 4 Y ) (0, 0, 0, 0) for all nonzero Y 0 is necessary for Theorem 2.3. Example 1. Consider A 1 = [ ], A 2 = [ ], A 3 = [ 0 i i 0 ], X = I 2, where I 2 represents the 2 2 identity matrix. We have A 1 X = A 2 X = A 3 X = 0. Notice that the system A 1 yy H = A 2 yy H = A 3 yy H = 0 is equivalent to y 1 2 y 2 2 = 0, Re (ȳ 1 y 2 ) = 0, Im (ȳ 1 y 2 ) = 0, which imply that y 1 = y 2 = 0. Thus there can be no rank-one matrix X to the system of equations: A 1 X = A 2 X = A 3 X = 0, X 0. This shows that n 3 is necessary for Theorems 2.1 and 2.2. Example 2. Consider [ A 1 = ], A 2 = [ ], A 3 = [ 1 i i 1 ], A 4 = [ ]. It is easily verified that all the assumptions in Theorem 2.3 are satisfied, except for n 3. Notice that A 1 I 2 = A 2 I 2 = A 3 I 2 = A 4 I 2 = 2. Assume that there is a two-dimensional complex vector y = [y 1, y 2 ] T such that A 1 yy H = A 2 yy H = A 3 yy H = A 4 yy H = 2. These equalities amount to y 1 2 = 1, y y Re (ȳ 1 y 2 ) = 2, y y Im (ȳ 1 y 2 ) = 2, y y 2 2 = 2, which lead to y 1 = y 2 = 1, ȳ 1 y 2 = 0, a clear contradiction. Hence there can be no y C 2 \ {0} satisfying A i yy H = A i I 2, i = 1, 2, 3, 4. 5
6 Example 3. Consider A 1 = , A 2 = , A 3 = 1 i 0 i , A 4 = / /2, and X = /2. Notice that A i I 3 = 0, i = 1, 2, 3, 4, and A i X = 1, i = 1, 2, 3, and A 4 X = 5/4. Suppose that there is a three-dimension complex vector y = [y 1, y 2, y 3 ] T satisfying A i yy H = 1, i = 1, 2, 3, and A 4 yy H = 5/4. Then, these equalities would lead to y 1 2 y 2 2 = 1, y 1 2 y Re (ȳ 1 y 2 ) = 1, y 1 2 y Im (ȳ 1 y 2 ) = 1, y y y 3 2 = 5/4, which is clearly impossible. 3 Proofs of the Main Results In this section we shall present proofs for the three theorems that we have just discussed above. To accomplish this task, we need the following two technical results, Lemma 3.1 and Proposition 3.2, as the main tools. Lemma 3.1. For any positive numbers c 1 > 0, c 0 > 0, any complex numbers a i, b i, c i, i = 1, 2, 3, and any real numbers a 4, b 4, c 4, the following system of equations Re (a 1 xy) + Re (a 2 xz) + Re (a 3 ȳz) + a 4 z 2 = 0, Re (b 1 xy) + Re (b 2 xz) + Re (b 3 ȳz) + b 4 z 2 = 0, c 1 x 2 c 0 y 2 + Re (c 1 xy) + Re (c 2 xz) + Re (c 3 ȳz) + c 4 z 2 = 0, always admits a non-zero complex-valued solution. Our proof for the above lemma is constructive, but it is rather tedious. In order not to distract the flow of our discussion, we delegate the detailed proof to Appendix A. The following proposition is due to Huang and Zhang [18] (Theorem 2.1), where the proof is based on a simple (polynomial-time) construction. 6
7 Proposition 3.2. Suppose that A 1, A 2 H n and X H+. n Then there exists a rank-one decomposition of X, X = x 1 x H x rx H r, such that A 1 x i x H i = A 1 X/r, A 2 x i x H i = A 2 X/r, i = 1,..., r, where r = rank (X). 3.1 Proof of Theorem 2.1 To simply the notation, in this and subsequent proofs we shall denote δ i = A i X, for all index i. It follows from Proposition 3.2 that there is a rank-one decomposition X = r i=1 x ix H i such that A 1 x i x H i = δ 1 /r, A 2 x i x H i = δ 2 /r, i = 1,..., r. If A 3 x i x H i = δ 3 /r for i = 1,..., r 2, then the theorem would follow. Otherwise, let us suppose that A 3 x 1 x H 1 δ 3/r > 0 and A 3 x 2 x H 2 δ 3/r < 0. Let y = α 1x 1 + α 2 x 2 + α 3 x 3 α1 2 + α α 3, 2 where α i, i = 1, 2, 3, are three complex-valued parameters to be specified. Then the equation (with respect to α 1, α 2, α 3 ) A 1 yy H = δ 1 /r, A 2 yy H = δ 2 /r, A 3 yy H = δ 3 /r, is equivalent to 0 = Re (2x H 1 A 1x 2 ᾱ 1 α 2 ) + Re (2x H 2 A 1x 3 ᾱ 2 α 3 ) + Re (2x H 3 A 1x 1 ᾱ 3 α 1 ), 0 = Re (2x H 1 A 2x 2 ᾱ 1 α 2 ) + Re (2x H 2 A 2x 3 ᾱ 2 α 3 ) + Re (2x H 3 A 2x 1 ᾱ 3 α 1 ), 0 = (x H 1 A 3x 1 δ 3 /r) α (x H 2 A 3x 2 δ 3 /r) α Re (2x H 1 A 3x 2 ᾱ 1 α 2 ) +Re (2x H 2 A 3x 3 ᾱ 2 α 3 ) + Re (2x H 3 A 3x 1 ᾱ 3 α 1 ) + (x H 3 A 3x 3 δ 3 /r) α 3 2. By Lemma 3.1 we know that the above system of equations has a nonzero solution (α1 0, α0 2, α0 3 ). Let us normalize the solution to be α := (α1, 0 α2, 0 α3) 0 T / α α α One easily verifies that I 3 αα H is a positive semidefinite matrix of rank 2, and x 1 x H 1 + x 2 x H 2 + x 3 x H 3 yy H = [x 1, x 2, x 3 ](I 3 αα H )[x 1, x 2, x 3 ] H. Therefore X yy H remains positive semidefinite and rank (X yy H ) = r 1. (3.1) 7
8 Let us update x 1 and X by x 1 := y and X := X yy H, and then repeat the above procedure until rank (X) = 2. Then we have A i x 1 x H 1 = = A i x r 2 x H r 2 = δ i/r, i = 1, 2, 3. Finally, applying Proposition 3.2 to the resulting X leads to A i x r 1 x H r 1 = A i x r x H r = δ i /r, i = 1, 2. The proof is thus complete. We remark that Theorem 2.1 is a generalization of Theorem 3.4 of [2], Theorem 2.1 of [18] and Corollary 4 of [25]. 3.2 Proof of Theorem 2.2 If rank (X) 3, then applying Theorem 2.1, we have that there is a rank-one decomposition X = r i=1 x ix H i such that A 1 x i x H i = δ 1 /r, A 2 x i x H i = δ 2 /r, i = 1,..., r; A 3 x i x H i = δ 3 /r, i = 1,..., r 2, where r = rank (X). Observing that x H i z = 0, for all z Null(X), hence x i Range(X), i = 1,..., r. Take y = rx 1. We have A 1 yy H = δ 1, A 2 yy H = δ 2, A 3 yy H = δ 3. Also, clearly in this case X 1 r yyh 0 and rank (X 1 r yyh ) r 1, and the proof is complete. If rank (X) = 2, then by Proposition 3.2 there exists a rank-one decomposition of X, X = x 1 x H 1 + x 2x H 2, such that A 1 x 1 x H 1 = A 1 x 2 x H 2 = δ 1 /2; A 2 x 1 x H 1 = A 2 x 2 x H 2 = δ 2 /2. If A 3 x 1 x H 1 = δ 3/2, then choosing y = 2x 1 would complete the proof. Otherwise, in the remainder let us consider the case that (A 3 x 1 x H 1 δ 3/2)(A 3 x 2 x H 2 δ 3/2) < 0. Since n 3, there must be a vector x 3 which is linearly independent of x 1 and x 2. We claim that there exists a nonzero (α 1, α 2, α 3 ), such that y = α 1x 1 + α 2 x 2 + α 3 x 3 α1 2 + α α 3 ; A 2 1 yy H = δ 1 /2, A 2 yy H = δ 2 /2, A 3 yy H = δ 3 /2. Indeed, the above can be equivalently expanded to the following system of equations: Re (2x H 1 A 1x 2 ᾱ 1 α 2 ) + Re (2x H 2 A 1x 3 ᾱ 2 α 3 ) + Re (2x H 3 A 1x 1 ᾱ 3 α 1 ) + (x H 3 A 1x 3 δ 1 /2) α 3 2 = 0, Re (2x H 1 A 2x 2 ᾱ 1 α 2 ) + Re (2x H 2 A 2x 3 ᾱ 2 α 3 ) + Re (2x H 3 A 2x 1 ᾱ 3 α 1 ) + (x H 3 A 2x 3 δ 2 /2) α 3 2 = 0, (x H 1 A 3x 1 δ 3 /2) α (x H 2 A 3x 2 δ 3 /2) α (x H 3 A 3x 3 δ 3 /2) α 3 2 +Re (2x H 1 A 3x 2 ᾱ 1 α 2 ) + Re (2x H 2 A 3x 3 ᾱ 2 α 3 ) + Re (2x H 3 A 3x 1 ᾱ 3 α 1 ) = 0. (3.2) Following the constructive proof of Lemma 3.1, we can find such (α 1, α 2, α 3 ). Thus, letting z := x 3 and y := 2y produces the required solution. 8
9 Remark. The mere existence of y in Theorem 2.2 follows from a result of Bohnenblust [9]; however, Bohnenblust s proof is nonconstructive. In the case that (A 1 X, A 2 X, A 3 X) (0, 0, 0), one may also prove Theorem 2.2 constructively by using some purification techniques (see e.g. Theorem 5.1 of [18], or Appendix A of [5], or [21]). However, if A i X = 0 for all i = 1, 2, 3, then these purification techniques do not work. Even if (A 1 X, A 2 X, A 3 X) (0, 0, 0), such y obtained by the purification technique may not satisfy the property that rank ( X 1 r yyh) r 1. Here is such an example. Suppose that A 1 = , A 2 = , A 3 = 1 i 0 i The system of matrix equations A 1 X = δ 1 = 2, A 2 X = δ 2 = 0, A 3 X = δ 3 = 2, has a matrix solution X = I 3, the 3 3 identity matrix. procedure (e.g. [18]), we obtain from I 3 a rank-one solution x 1 x H 1 = (3.3) Following a standard rank reduction However, one verifies that rank ( X 1 ) 3 x 1x H 1 = rank = 3. If we apply the procedure as in the proof of Theorem 2.2 to the above example, then the output of our Matlab program (to be introduced later) is i y = i, and the rank of X 1 3 yyh is indeed i 3.3 Proof of Theorem 2.3 Since (δ 1, δ 2, δ 3, δ 4 ) 0, without loss of generality, let us assume δ 4 0. Then we have ( A i δ ) i A 4 X = 0, i = 1, 2, 3. δ 4 9
10 It follows from Theorem 2.2 that there exists a nonzero n-dimensional complex vector y C n such that ( A i δ ) i A 4 yy H = 0, i = 1, 2, 3. δ 4 Denote t = A 4 yy H /δ 4. Then we have A i yy H = tδ i, i = 1, 2, 3, 4. It is clear that t 0, since (A 1 Y, A 2 Y, A 3 Y, A 4 Y ) (0, 0, 0, 0) for all nonzero Y 0. Furthermore, t > 0, because if t < 0 then by setting Y = X 1 t yyh 0 we will have A i Y = 0 for all i = 1, 2, 3, 4, which violates the condition of the theorem. Letting y := 1 t y, we get A i yy H = δ i, i = 1, 2, 3, 4. 4 Applications in Signal Processing and Communication In this section, we will illustrate applications of the matrix decomposition results presented in Section 2 to solve some specific problems in signal processing and communication. In particular, we shall study one application from the radar code optimization and another from robust beamforming. These two case-studies will be presented in the next two subsections, to show how non-convex (complex) quadratic optimization models arise in engineering applications, and how our new results help to solve such non-convex quadratic programs. 4.1 Application in Radar Space-Time Adaptive Processing (STAP) Various algorithms for radar signal design have been proposed in the recent years. One approach in the radar signal design, known as the radar coding, relies on the modulation of a pulse train parameters (amplitude, phase, and frequency), so as to synthesize waveforms to satisfy some specified properties. A substantial amount of work has been done and recorded in the literature on this topic; the interested reader is referred to our recent paper and the references therein [14]. In this subsection, we introduce the code optimization problem for radar STAP in the presence of colored Gaussian disturbances. Our main goal, however, is to demonstrate how the decomposition theorems developed in Section 2 of the paper help solve the resulting optimization problems. The problem is to design an optimal radar code, which is used to modulate transmitted signals belonging to the class of coded pulse trains, in such a way that the detection performance of radar is maximized under a control on the achievable values of the temporal and spatial Doppler estimation 10
11 accuracy, as well as on the degree of similarity with a pre-fixed radar code. The resulting optimization problem is formulated as follows: max c C n c H R 0 c s.t. c H c = 1, (QP) c H R t c δ t, c H R s c δ s, c ĉ 2 ɛ. The objective function is the output Signal-to-Interference-plus-Noise Ratio (SINR), with respect to which the detection probability (P d ) of radar is increasing, and R 0 0. The first and fourth constraints are normalized code constraint and similarity constraint with a pre-fixed code ĉ (ĉ H ĉ = 1), respectively, and the second and third constraints rule the region of achievable temporal and spatial Doppler estimation accuracies, respectively. For the detailed derivation, one is referred to [14]. It is easy to verify that (QP) is tantamount to the following homogenous quadratic programming: max c C n c H R 0 c s.t. c H c = 1, (HQP) c H R t c δ t, c H R s c δ s, c H C 0 c δ ɛ, where C 0 = ĉĉ H and δ ɛ = (1 ɛ/2) 2, in the sense that they have the same optimal value and c = ce i arg( ch ĉ) is an optimal solution of (QP) if c is optimal to (HQP). For this reason, from now on we shall focus on (HQP). Note that the SDP relaxation for (HQP) and its dual are: max R 0 C s.t. I C = 1, R t C δ t, min y 1 y 2 δ t y 3 δ s y 4 δ ɛ (SDR) and (DSDR) s.t. y 1 I y 2 R t y 3 R s y 4 C 0 R 0, R s C δ s, y 1 R, y 2 0, y 3 0, y 4 0, C 0 C δ ɛ, C 0, respectively. Suppose that both the primal and dual SDPs are solvable. (To this end, it suffices to assume that the initial code ĉ is strictly feasible to (QP), which ensures (SDR) to be strictly feasible, while (DSDR) is always strictly feasible). Suppose that the primal-dual feasible pair (C, y 1, y 2, y 3, y 4 ) is 11
12 optimal to (SDR) and (DSDR) respectively. Consequently, it satisfies the complementary conditions: (y1i y2r t y3r s y4c 0 R 0 ) C = 0, (4.1) (R t C δ t )y2 = 0, (4.2) (R s C δ s )y3 = 0, (4.3) (C 0 C δ ɛ )y4 = 0. (4.4) If rank (C ) 3, then apply Theorem 2.3 to (C, I, R t, R s, C 0 ) we find a vector c Range (C ) such that I c c H = I C, R t c c H = R t C, R s c c H = R s C, C 0 c c H = C 0 C. Since the primal-dual pair (c c H, y1, y 2, y 3, y 4 ) is feasible and satisfies the complementary conditions, we conclude that c c H is optimal to (SDR) and c is optimal to (HQP). If rank (C ) = 2 and one of the inequality constraints of (SDR) is non-binding (non-active) at optimality, then a rank-one solution of (SDR) can be obtained by resorting to Proposition 3.2. We remarked in [14, Section IV-C] that when we pick the parameters R 0, R t, and R s from the standard Knowledge Aided Sensor Signal Processing and Expert Reasoning (KASSPER) datacube, and set ĉ to be a generalized Barker sequence, and randomly choose δ t, δ s and δ ɛ from proper intervals respectively, our simulation shows that only 10 out of 10,000 instances of (QP) can be classified as in the hard case, meaning that the output solution of the corresponding (SDR) is exactly of rank 2 and all inequality constraints are active at the optimality. For these instances classified as the hard case above, we then apply Theorem 2.3 to get a high quality feasible solution for (HQP); specifically, we may choose any z vector as required in Theorem 2.3 to produce a feasible solution based on that vector z. The z vector can be picked randomly and we can choose the best one among the random samples. Our simulation results show that this heuristic for the hard case instances works extremely well in practice. To summarize, in this approach the main steps to solve the radar code optimization problem involve: (1) using an SDP solver (e.g. SeDuMi) to solve (SDR); (2) implementing the appropriate decomposition theorem to further process the solution of (SDR) as we discussed above; (3) assembling an optimal solution (or high quality feasible solution) for (HQP). In the last section of the paper, we will test the numerical performance of the decomposition procedures stipulated in Theorems Application in Triply Constrained Robust Capon Beamformer Design Beamforming is an important task in array signal processing. The so-called Capon beamformer selects the beamforming weight vector adaptively so as to minimize the array output power subject to the 12
13 linear constraint that the signal of interest does not suffer from any distortion ([19, 20]). As long as the array steering vector (i.e., the response of the array to an arriving plane wave, termed also the array manifold in the literature) corresponding to the signal of interest is known, the Capon beamformer has a better performance than the standard data-independent beamformer; see [20]. However, like other adaptive beamformers, the Capon beamformer is sensitive to the steering vector to the signal of interest, and the performance of the Capon beamformer will be substantially degraded if the mismatch between the presumed steering vector and the true steering vector occurs. In practical situations, this type of mismatch can easily occur due, for instance, to look directions and signal pointing errors, among other possible causes. To cope with the performance degradation caused by the mismatch and improve the robustness of the adaptive beamformers, engineers have proposed various remedies in the past three decades (cf. [19] and the references therein). In this subsection, we shall focus on the robust Capon beamforming problem as introduced in [20], and we shall demonstrate how to resolve this class of problems using the decomposition theorems presented in this paper. Consider an array of n sensors, and denote by R the theoretical covariance matrix of the array output vector. Assume that R has the following form: K R = σ 2 aa H + σk 2 b kb H k k=1 + Q, (4.5) where (σ 2, σ1 2,..., σ2 K ) are the powers of the (K + 1) uncorrelated signals impinging on the array, a is the steering vector to the signal of interest, (b 1,..., b K ) are the steering vectors to the K signals treated as interference, and Q 0 is the noise covariance. In practical applications, R is replaced by the sample covariance of recently received samples of the array output, e.g., ˆR = 1 N N y n yn H. n=1 The robust Capon beamforming problem that we deal with is to determine the power of the signal of interest (i.e., σ 2 in (4.5)), where the only knowledge about its steering vector a is that it belongs to an uncertainty set. More specifically, we assume the uncertainty set is given by A = {a C n a 2 = n, (a a 1 ) H D 1 (a a 1 ) ɛ 1, (a a 2 ) H D 2 (a a 2 ) ɛ 2 }. (4.6) In the above set, the norm constraint is imposed to avoid ambiguities (see [20]); in the second constraint (i.e., the similarity constraint), D 1 0 and ɛ 1 > 0 define the similarity scale, and a 1 is a presumed nominal steering vector; by imposing the third constraint, the steering vector is forced to be away from the prefixed steering vector a 2 corresponding to a known co-existing source emitting contaminating (or interfering) signals, and the parameters D 2 0 and ɛ 2 0. In particular, if D 1 = I and ɛ 2 = 0, then the uncertainty region is exactly the same as the one considered in [20]. 13
14 Using the reformulation of the robust Capon beamforming problem in [20], to which the uncertainty set (4.6) is appended, we obtain a robust estimate of σ 2, without calculating the beamforming weight vector: (RCB) { max σ 2, a σ 2 s.t. R σ 2 aa H 0, a A. It is easily checked that the optimal value is attained ˆσ 2 = 1/(a H R 1 a ), where a is an optimal solution of the following equivalent problem min a H R 1 a s.t. a 2 = n, (RCB) (a a 1 ) H D 1 (a a 1 ) ɛ 1, (a a 2 ) H D 2 (a a 2 ) ɛ 2. Therefore, all we need to do now is to solve (RCB), which can be further homogenized as: where min x H R 0 x s.t. x H R 1 x = n, (HRCB) x H R 2 x ɛ 1, x H R 3 x ɛ 2, x H R 4 x = 1, [ ] [ ] [ ] a R 1 0 n 1 I 0 n 1 x =, R 0 =, R 1 =, t 0 1 n n 0 [ ] [ ] [ ] D 1 D 1 a 1 D 2 D 2 a 2 0 n n 0 n 1 R 2 = a H 1 D 1 a H 1 D, R 3 = 1a 1 a H 2 D 2 a H 2 D, R 4 =. 2a n 1 In other words, (RCB) and (HRCB) have the same optimal value, and a /t is an optimal solution if x = [a T, t T ] T is an optimal solution of (HRCB). The SDP relaxation of (HRCB) and its dual are: min R 0 X s.t. R 1 X = n, R 2 X ɛ 1, max ny 1 + ɛ 1 y 2 + ɛ 2 y 3 + y 4 (SDR2) and (DSDR2) s.t. R 0 y 1 R 1 y 2 R 2 y 3 R 3 y 4 R 4 0, R 3 X ɛ 2, y 1 R, y 2 0, y 3 0, y 4 R. R 4 X = 1, X 0, Suppose that both (SDR2) and (DSDR2) are solvable, and let (X, y1, y 2, y 3, y 4 ) be an optimal primal-dual pair. Similarly as in the previous application case, if the rank of X is at least three, then an optimal solution of (HRCB) can be found using Theorem 2.3; if the rank of X is two, and one of 14
15 the two inequality constraints of (SDR2) is non-binding at the point X, then an optimal solution of (HRCB) can be found by Proposition 3.2; if the rank of X is two, and the two inequality constraints of (SDR2) are binding at the point X, then a good feasible solution of (HRCB) can be found by resorting to Theorem 2.3, where one may randomly choose the dimension extending direction z. Here we shall briefly summarize our findings in this section. As far as optimization formulations are concerned, our new matrix decomposition results can be applied, in combination of SDP relaxation, to solve any non-convex, complex-valued, homogeneous, quadratically constrained quadratic programs, where the number of constraints is at most four and the number of variables is at least three. Our approach solves a vast majority of the instances to optimality, leaving only one possible unsettled case, whose frequency of occurrence is exceedingly small. Even when the unsettled case does occur, our decomposition theorem (Theorem 2.3) can still be applied to yield a good feasible solution (though not necessarily optimal anymore in this case). Such non-convex quadratic optimization models appear to be useful for a variety of practical applications, and according to our numerical experiences the approach proposed in this paper works extremely well, for instance, in solving problems arising from signal processing and communication. 5 Theoretical Implications of the New Results 5.1 Extension of Yuan s lemma and the S-lemma Yuan [27] proved the following result (Lemma 2.3 in Yuan [27]) which turns out to be useful in several applications. Theorem 5.1. Let A 1 and A 2 be in S n. Then the following are equivalent: (i) max{x T A 1 x, x T A 2 x} 0 for all x R n (resp. > 0 for all x 0). (ii) There exist µ 1 0, µ 2 0, µ 1 + µ 2 = 1 such that µ 1 A 1 + µ 2 A 2 0 (resp. 0). Our new results lead to the following extension for Yuan s theorem. Theorem 5.2. Suppose that n 3, A i H n, i = 1, 2, 3, 4, and there are λ i R, i = 1, 2, 3, 4, such that λ 1 A 1 + λ 2 A 2 + λ 3 A 3 + λ 4 A 4 0. If max{z H A 1 z, z H A 2 z, z H A 3 z, z H A 4 z} 0, z C n (resp. > 0, z 0), then there are µ i 0, i = 1, 2, 3, 4, such that µ 1 + µ 2 + µ 3 + µ 4 = 1 and µ 1 A 1 + µ 2 A 2 + µ 3 A 3 + µ 4 A 4 0 (resp. 0). 15
16 Proof. If the conditions of the theorem hold, then max{a 1 Z, A 2 Z, A 3 Z, A 4 Z} 0 for all Z 0. To see this, suppose by contradiction that there is Z 0 such that A i Z < 0, for i = 1, 2, 3, 4. It follows from Theorem 2.3 that there is nonzero y C n, with A i yy H < 0, i = 1, 2, 3, 4, which contradicts to the condition that max{z H A 1 z, z H A 2 z, z H A 3 z, z H A 4 z} 0, z C n. Therefore, the optimal value of the following SDP problem min t s.t. t A i Z 0, i = 1, 2, 3, 4 I Z = 1, Z 0 is nonnegative. At the same time, the dual of the above problem is max µ 5 s.t. µ 1 A 1 + µ 2 A 2 + µ 3 A 3 + µ 4 A 4 µ 5 I, 1 µ 1 µ 2 µ 3 µ 4 = 0, µ 1 0, µ 2 0, µ 3 0, µ 4 0. It is easily seen that both the primal and dual problems have a strictly feasible solution. Hence the strong duality holds, with both primal and dual problems having attainable optimal solutions, to be denoted by (t, Z ) and (µ 1, µ 2, µ 3, µ 4, µ 5 ) respectively. Note that µ i 0, i = 1, 2, 3, 4, µ 1 + µ 2 + µ 3 + +µ 4 = 1, and µ 1 A 1 + µ 2 A 2 + µ 3 A 3 + µ 4 A 4 µ 5 I = t I 0, completing the proof. We remark that in the same vein, Yuan s result (Theorem 5.1) can be proven analogously (but then in the real domain), using Proposition 3 of Sturm and Zhang [25]. In fact, there are several different ways to prove Theorem 5.1, including of course the original one presented in Yuan [27]. Remark here that Hiriart-Urruty and Torki [16] used a separation argument to prove Theorem 5.1, based on a convexity result of the joint numerical range. The situation is similar for the proofs of the S-lemma (see Pólik and Terlaky [24]). Particularly, in Section 2.4 of [24], the authors used Theorem 5.1 to prove the S-lemma. 5.2 Joint numerical ranges To put the results in perspective, we present in this subsection some other connections between the convexity of the joint numerical ranges, generalized Finsler s lemma, and our new rank-one decomposition theorem. Theorem 5.3. The following statements are all equivalent and correct: 1) Theorem 2.2, where A i X = 0, i = 1, 2, 3. 16
17 2) Suppose n 3, and A 1, A 2, A 3 H n, then the set W = { (z H A 1 z, z H A 2 z, z H A 3 z) z H z = 1, z C n} is convex. 3) Suppose n 3, and A 1, A 2, A 3 H n satisfy ρ(z) = (z H A 1 z) 2 + (z H A 2 z) 2 + (z H A 3 z) 2 > 0, for all z with z = 1. Then the set S := {( z H A 1 z/ρ(z), z H A 2 z/ρ(z), z H A 3 z/ρ(z) ) z H z = 1 } is a closed region contained in a half-sphere of the 3-dimensional unit ball, with the property that x S = x S. 4) Suppose n 3, and A 1, A 2, A 3 H n satisfy (z H A 1 z, z H A 2 z, z H A 3 z) (0, 0, 0), z C n \ {0}. Then there exist λ 1, λ 2, λ 3 R such that λ 1 A 1 + λ 2 A 2 + λ 3 A 3 0. Proof. 1) = 2). Let W = {(A 1 Z, A 2 Z, A 3 Z) Z I = 1, Z 0}. Clearly W is a convex set, and W W. We shall show W W. Let (v 1, v 2, v 3 ) = (A 1 Z, A 2 Z, A 3 Z) W, and A i := A i v i I, i = 1, 2, 3. Then A i Z = 0, i = 1, 2, 3. It follows by Theorem 2.2 that there is a non-zero vector y C n such that A i yyh = 0, i = 1, 2, 3; hence ( ) y y H A i = v i, i = 1, 2, 3, y H y y T y implying that (v 1, v 2, v 3 ) belongs to W as well, and consequently W = W. 2) = 3). We begin by showing that S does not contain opposite points. Suppose that this is not the case. Let (a, b, c) S, while (a, b, c) S; that is, there are x, y C n with x = y = 1 and ρ(x) (a, b, c) = (x H A 1 x, x H A 2 x, x H A 3 x), ρ(y) (a, b, c) = (y H A 1 y, y H A 2 y, y H A 3 y). Since W is convex, we have and so ρ(y) ρ(x) + ρ(y) (xh A 1 x, x H A 2 x, x H ρ(x) A 3 x) + ρ(x) + ρ(y) (yh A 1 y, y H A 2 y, y H A 3 y) W, (0, 0, 0) = ρ(y) ρ(x) ρ(x)(a, b, c) + ρ(y)( a, b, c) W. ρ(x) + ρ(y) ρ(x) + ρ(y) Hence there is z with z = 1 such that (z H A 1 z, z H A 2 z, z H A 3 z) = (0, 0, 0), i.e., ρ(z) = 0, which is a contradiction. 17
18 Observe that S is the image of W on the sphere of the unit ball, projected from the origin. Since W is a compact convex set which does not contain the origin, its image S must be contained strictly in one half of the unit ball. 3) = 4). Since S is a closed spherical region contained strictly in one half of the 3-dimensional unit ball, and does not contain two opposite points with respect to the origin, its convex hull S must not contain the origin either. By the separation theorem, there is a plane going through 0, with S lying properly on one half of the plane. That is, if we let (a 0, b 0, c 0 ) the the normal direction of the separating plane, then the angle between any vector in S and the direction (a 0, b 0, c 0 ) should be strictly less than θ < π/2. In other words, cos θ = a 0 z H A 1 z/ρ(z) + b 0 z H A 2 z/ρ(z) + c 0 z H A 3 z/ρ(z) > 0, z : z = 1. This implies that a 0 A 1 + b 0 A 2 + c 0 A ) = 1). A key step in the proof of Theorem 2.2 is that the system of equations (3.2) has a nontrivial solution. Recall that (3.2) can be explicitly written as 0 x H 1 A 1x 2 x H 3 A 1x 1 (ᾱ 1, ᾱ 2, ᾱ 3 ) 0 x H 2 A 1x 3 α 1 α 2 = 0 (ᾱ 1, ᾱ 2, ᾱ 3 ) x H 3 A 1x 3 δ x H 1 A 2x 2 x H 3 A 2x 1 0 x H 2 A 2x 3 α 3 α 1 α 2 = 0 (ᾱ 1, ᾱ 2, ᾱ 3 ) x H 3 A 2x 3 δ 2 2 α 3 x H 1 A 3x 1 δ 3 2 x H 1 A 3x 2 x H 3 A 3x 1 x H 2 A 3x 2 δ 3 2 x H 2 A 3x 3 x H 3 A 3x 3 δ 3 2 α 1 α 2 α 3 = 0, where for clarity we only displayed the upper part of the matrices the lower part is determined by (Hermitian) symmetry. The first two diagonal elements of these three matrices are (0, 0), (0, 0), and (x H 1 A 3x 1 δ 3 2, x H 2 A 3x 2 δ 3 2 ). The last vector has one positive element and one negative element, and so it is impossible to combine these vectors into a positive vector, let alone the three entire matrices. By 4), we know that there is a non-zero solution of (3.2). The rest of the proof is identical to that in the proof of Theorem 2.2. The essential difference between this argument, and the proof presented in Section 2 is that the latter is constructive; i.e., following that proof we can actually produce a rankone solution of a system of linear matrix equations, which makes the result more useful in practical engineering applications. Another immediate consequence of Theorem 2.3 is the following result, which generalizes the famous theorems of Hausdorff [15] and Brickman [10]. 18
19 Theorem 5.4. Suppose n 3. Let A 1, A 2, A 3, A 4 H n satisfy (A 1 Y, A 2 Y, A 3 Y, A 4 Y ) (0, 0, 0, 0), for any nonzero matrix Y H+. n Then, {(z H A 1 z, z H A 2 z, z H A 3 z, z H A 4 z) z C n } is a convex cone. Proof. Let us denote K 1 := {(z H A 1 z, z H A 2 z, z H A 3 z, z H A 4 z) z C n }, and K 2 := {(A 1 Z, A 2 Z, A 3 Z, A 4 Z) Z H n +}. Evidently, K 1 K 2. Next we shall show the other containing relation, thus establishing the equality. Take any 0 Z H+. n If rank (Z) = 1, then (A 1 Z, A 2 Z, A 3 Z, A 4 Z) K 1 ; if rank (Z) 3, then Theorem 2.3 asserts that there is z Range(Z) such that (A 1 Z, A 2 Z, A 3 Z, A 4 Z) = (z H A 1 z, z H A 2 z, z H A 3 z, z H A 4 z) K 1 ; if rank (Z) = 2, then, since n 3 we have C n \Range(Z), and so by taking any x C n \Range(Z), Theorem 2.3 guarantees the existence of z which is in the span of x and Range(Z), such that (A 1 Z, A 2 Z, A 3 Z, A 4 Z) = (z H A 1 z, z H A 2 z, z H A 3 z, z H A 4 z) K 1. This shows that K 2 K 1, and consequently K 1 = K 2 as a result. In case A 4 = I n, then the condition of the theorem (regarding the A i matrices) holds. It thus follows that {(z H A 1 z, z H A 2 z, z H A 3 z) z H z = 1} is a convex set whenever n 3, which is an enhancement of Hausdorff s theorem [15], and also a generalization of Brickman s theorem [10] (see also Part 2 of Theorem 5.3). The real number counterpart of Theorem 5.4 includes a result of Polyak [22] as a special case; in Polyak s paper, a seemingly different and stronger condition is used: the A i matrices can be linearly combined into a positive definite one. Interestingly, the statements as in Theorem 5.4 can be slightly strengthened along this line as well. 19
20 Theorem 5.5. Suppose n 3, and A 1, A 2, A 3, A 4 H n. If (A 1 Y, A 2 Y, A 3 Y, A 4 Y ) (0, 0, 0, 0), for any nonzero matrix Y H n +, then there exist α i R, i = 1, 2, 3, 4 such that 4 i=1 α ia i 0, and is a pointed, closed, convex cone. {(z H A 1 z, z H A 2 z, z H A 3 z, z H A 4 z) z C n } Proof. Let us denote K := {(z H A 1 z, z H A 2 z, z H A 3 z, z H A 4 z) z C n } which is a convex cone by Theorem 5.4. Moreover, K is also pointed, because, if there is 0 d K, with d K, then by denoting d = (z1 H A 1 z 1, z1 H A 2 z 1, z1 H A 3 z 1, z1 H A 4 z 1 ) and d = (z2 H A 1 z 2, z2 H A 2 z 2, z2 H A 3 z 2, z2 H A 4 z 2 ) we would have ( A1 (z 1 z1 H + z 2 z2 H ), A 2 (z 1 z1 H + z 2 z2 H ), A 3 (z 1 z1 H + z 2 z2 H ), A 4 (z 1 z1 H + z 2 z2 H ) ) = (0, 0, 0, 0) which contradicts the condition of the theorem. We proceed to show that K is a closed set. For this purpose, let us take any sequence v k K with lim k v k = ˆv. Clearly, if ˆv = 0 then ˆv K. Let us consider the nontrivial case where ˆv 0. Suppose that v k = (zk H A 1z k, zk H A 2z k, zk H A 3z k, zk H A 4z k ) ( ( ) H ( = z k 2 zk z k A 1 z k z k, zk z k ) H A 2 z k z k, Without losing generality we may assume lim k z k / z k = ẑ. Therefore, ( ) H ( ) ) zk z H k A 3 z k z k, zk z k A 4. z k z k ˆv = lim k vk = lim k z k 2 (ẑ H A 1 ẑ, ẑ H A 2 ẑ, ẑ H A 3 ẑ, ẑ H A 4 ẑ). By the condition of the theorem, we know that (ẑ H A 1 ẑ, ẑ H A 2 ẑ, ẑ H A 3 ẑ, ẑ H A 4 ẑ) 0, and therefore lim k z k 2 exists and is finite; hence ˆv K. This shows that K is closed. To summarize, we have shown that K is a pointed, closed, and convex cone. Using the separation theorem, we conclude that there is α i R, i = 1, 2, 3, 4, such that 4 i=1 α iz H A i z > 0 for any 0 z C n. Stated differently, 4 i=1 α ia i 0. 20
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AOE has an expert-level understanding of every element of ACCME accreditation criteria, and they remained flexible throughout the process to deliver what we needed. We earned Accreditation with Commendation, and AOE delivered everything to us on time or before it was due. What was promised was exactly what was delivered.Carly Harrington
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|Message view
|« Date » · « Thread »
|« Date » · « Thread »
|"Pinaki Poddar" <ppod...@bea.com>
|RE: Testing an OpenJPA module
|Tue, 10 Apr 2007 21:02:42 GMT
> such an exhaustive list of possible problems one might run into. Murphy's law: 'If something can go wrong, it will". Extension to Murphy's law: "Murphy was an optimist". :) Pinaki Poddar BEA Systems 415.402.7317 -----Original Message----- From: firstname.lastname@example.com [mailto:firstname.lastname@example.com] On Behalf Of Jacek Laskowski Sent: Tuesday, April 10, 2007 4:00 PM To: firstname.lastname@example.com Subject: Re: Testing an OpenJPA module On 4/10/07, Pinaki Poddar <email@example.com> wrote: > The error message could have been more specific in the following way: > a) no META-INF/persistence.xml has not been found in classpath > b) META-INF/persistence.xml has been found but there is no 'ode-store' > unit defined in it. > c) META-INF/persistence.xml has been found but provider can not be > loaded/invoked Thanks! I've never thought about such an exhaustive list of possible problems one might run into. Nice to keep it handy. > When I first encountered this error, my interpreation was (b) from the > way the message was worded. Perhaps it's because I'm a mere developer so the only answer I'd have given would've been c) yet I knew others were possible too (just I don't see them so often). Jacek -- Jacek Laskowski http://www.JacekLaskowski.pl Notice: This email message, together with any attachments, may contain information of BEA Systems, Inc., its subsidiaries and affiliated entities, that may be confidential, proprietary, copyrighted and/or legally privileged, and is intended solely for the use of the individual or entity named in this message. If you are not the intended recipient, and have received this message in error, please immediately return this by email and then delete it.
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Le Clandestin (edition 2012)
by John Grisham (Author), Patrick Berthon (Traduction)
Work details
The Broker by John Grisham
Legal Stories (23)
To Read (59)
No current Talk conversations about this book.
@ a wheeler - dealer — six years in jail — gets a pardon — running for his life — gets better
In his final hours in the Oval Office, the outgoing President grants a controversial last-minute pardon to Joel Backman, a notorious Washington power broker who has spent the last six years hidden away in a federal prison. What no one knows is that the President issues the pardon only after receiving enormous pressure from the CIA. It seems Backman, in his power broker heyday, may have obtained secrets that compromise the world's most sophisticated satellite surveillance system.
Backman is quietly smuggled out of the country in a military cargo plane, given a new name, a new identity, and a new home in Italy. Eventually, after he has settled into his new life, the CIA will leak his whereabouts to the Israelis, the Russians, the Chinese, and the Saudis. Then the CIA will do what it does best: sit back and watch. The question is not whether Backman will survive, there is no chance of that. The question the CIA needs answered is, who will kill him?
My only other Grisham was [b:The Firm|5358|The Firm|John Grisham|http://photo.goodreads.com/books/1165518895s/5358.jpg|38270], so I don't much to compare to.
This book was bland, boring and filled with repeated, ad nauseum, references to drinking coffee and talking in Italian.
I'm never reading another [a:John Grisham|721|John Grisham|http://photo.goodreads.com/authors/1198174022p2/721.jpg] book. I'm 50/50, so I don't want to chance it getting worse...
Good story, let down by the rushed ending which deflated the tension much too quickly.
I was pretty much captivated from the get-go with this novel.
The story whilst a little far-fetched both in technology and initial events is actually excellent. The building tension as you know something is about to happen but not sure when it's going to or where it's coming from. Then suddenly there's a diversion which changes where you thought things were going.... the lead up to the end is probably one of the better hastily travel dialogues of read.
There's also some great information about the scenery of Bolonga, in northern Italy there too which as a good little side bonus.
My only compliant is there's not a part 2, although I can certainly see the value in ending the plot where it finishes with an air of limitless possibility.
I'll definitely be reading some more Grisham in the near future.
I think maybe John Grisham bought some Rosetta Stone software to try to learn Italian, got frustrated with it, and then spent the rest of the day imagining a situation in which someone would actually be motivated to learn Italian quickly. Presto: you have The Broker, in which a felon is pardoned and whisked off to Italy in a sortof ersatz witness-protection operation, where his life depends on passing for a native Italian. There are pages and pages and pages of description of his Italian lessons and his motivation to learn (he keeps making the tutor start lessons earlier and earlier), and we're inside his head as he recites the Italian word for everything in his apartment.
Now, I like languages and occasionally mumbled some of the words out loud to myself while reading just to test my Italian accent, so I enjoyed the weird preoccupation with learning Italian. However, you have to admit it's a bizarrely academic preoccupation for a run-for-your-life thriller--especially when, in the context of the story, it's wholly unnecessary to the point of being unbelievable. Surely if the point is for this American guy to blend in while in hiding, you'd stash him in... Canada, Australia, England? You know, where the language skills he already has will be useful? And since it turns out that the US government us just waiting to see whether it's the Russians, the Chinese, or the Israelis who will find and kill our fugitive first, I don't understand why they'd bother paying a tutor a full-time salary to teach the guy flawless Italian. So I'm back to where I started: all I can figure is that Grisham really wanted to write a story about learning to speak Italian.
To make a weak plot even weaker, Beckman is utterly unsympathetic.
I had a very good time with The Broker, found Backman believable and charming and interesting, got a few laughs and felt my pulse thumping as the climax approached. But there's a rather hasty aspect to the book: too many short paragraphs, too many unnecessary exclamation points, a rushed and contrived ending.
Zippy but uneventful, the book tastes like something Robert Ludlum left sitting on his stove when he died.
Readers looking for a non-stop thrill ride won't find it in The Broker. For those ready to relax and settle into an almost sensuous pace, The Broker is benissimo.
Is contained in
Is abridged in
Reader's Digest Select Editions: The Broker • Solo • Blood Memory • Mosaic by Reader's Digest
Reader's Digest Select Editions - The Broker / Sam's Letters To Jennifer / Blood Memory / Mosaic by Reader's Digest
Reader's Digest: De deal; Regen en sterren; Je bent nooit alleen; Jacht op de jager by Reader's Digest
References to this work on external resources.
Wikipedia in English (1)
Amazon.com Amazon.com Review (ISBN 0385340540, Paperback)Before he was sent to federal prison for treason (among other things), Joel Backman was an extremely powerful man. Known as "the broker," Backman was a high roller--a lawyer making $10 million a year who could "open any door in Washington." That is, until he tried to broker a deal selling access to the world's most powerful satellite surveillance system to the highest bidder. When caught, Backman accepted prison as the one option that would keep him safe and alive, since the interested parties (the Israelis, the Saudis, the Russians, and the Chinese) were all itching to get their hands on his secrets at any cost. Little does he know that his own government has designs on accessing that information--or at least letting it die with him. Now, six years after his incarceration, the director of the CIA convinces a lame duck president to pardon Backman, and the broker becomes a free man--and an open target.
The Broker marries the best of John Grisham's many talents--his ability to immerse himself in the culture of small town life (in this case, Bologna, Italy), and his uncanny mastery of the chase. The first half of the book focuses on Backman's transformation from infamous power broker to helpless victim in his own game. Upon his release from prison, Backman is taken into "protective custody" and whisked off to Italy where he is assigned a new identity, and a tutor to help him blend in. Sure he is on the run, but some readers may feel that Backman's time spent in Bologna is a bit too leisurely--readers join him on an almost cinematic tour through the Italian town, complete with language and history lessons. Impatient readers will be happy to know that the final half of the novel is classic Grisham--a fast-paced, thrilling cat and mouse chase pitting Backman against the numerous agencies that want him dead--as the broker makes a move to take back his life. --Daphne Durham
Exclusive Video Interview with John Grisham
Grisham: The Books
A Time to Kill, 1989The Firm, 1991The Pelican Brief, 1992The Client, 1993 The Chamber, 1994 The Rainmaker, 1995
The Runaway Jury, 1996The Partner, 1997The Street Lawyer, 1998The Testament, 1999 The Brethren, 2000 A Painted House, 2001
Skipping Christmas, 2001The Summons, 2002The King of Torts, 2003Bleachers, 2003 The Last Juror, 2004The Broker, 2005
Essential Grisham
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(retrieved from Amazon Thu, 12 Mar 2015 18:02:19 -0400)
With fourteen years left on a twenty-year sentence, notorious Washington power broker, Joel Blackman, receives a surprise pardon from a lame-duck president. He is smuggled out of the country on a military cargo plane, given a new identity, and tucked away in a small town in Italy. But Blackman has serious enemies from his past. As the CIA watches him closely, the question is not whether he will be killed, but rather who will kill him first.… (more)
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JEWISH COLONIES MAN 'BURMA ROAD'; New Settlements Dot Wider Jerusalem Corridor -- One Is Named for Morgenthau
Special to THE NEW YORK TIMES. ();
October 28, 1948,
, Section , Page 7, Column , words
MORGENTHAU, Palestine, Oct. 27 -- Morgenthau is not yet on the map but its Hebrew equivalent soon will be. It is another Jewish settlement, founded today just inside the "Burma Road" corridor that the Israelis carved through Arab territory from Israel proper to Jerusalem.
October 28, 1948
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Full Form of BASIC:
Beginner All-purpose Symbolic Instruction Code
BASIC Full Form is Beginner All-purpose Symbolic Instruction Code. BASIC, refers to a host of high-level, general-purpose programming languages the design of which emphasizes on the ease of use. In the year 1964, Thomas E. Kurtz and John G. Kemeny created the original BASIC language in New Hampshire. They wanted to allow students in fields apart from mathematics and science for the purpose of using computers. Back at that time, all computers used complicated custom software for programming which only mathematicians and scientists understood.
The use of BASIC became popular on microcomputers during the 1980s period. The advantage of BASIC was that it allowed people to learn basic programming and greatly helped professionals, hobbyists, business owners, and consultants for the development of custom software. In the year 2010, BASIC continued to be a popular computer dialect and influenced new programming languages such as the Visual Basic of the Microsoft's. In the year 2006, 59 percent of developers of the .NET Framework made use of Visual Basic.NET. Typical BASIC keywords have been given in the following bullets to give an idea about the same to the readers:
- DATA: It is used for Data manipulation. It holds an enumeration of values that are assigned sequentially with the help of READ command.
- LET: It assigns a value to a variable.
- IF….THEN….ELSE: The command is used for program flow control. It is made use of in the performance of comparisons or decision-making.
- DO…LOOP {WHILE} or {UNTIL}: The command is related to program flow control and is used to reiterate a section of code eternally or until the fulfillment of a specific condition. The condition so stipulated may be determined before or after each iteration of the loop.
- GOTO: The command is used to switch to a labeled or numbered line in a program.
- ON…GOTO/GOSUB: The command is used for choosing where to switch on the basis of stipulated conditions.
- LIST: It is an Input/Output command used to display all inputted code.
- INPUT: The command is used for asking the user to input a value of the variable.
- PRINT: The command is used to display or print a message on a screen.
- REM: It is command used to hold the comment of a programmer or ReMark.
- TROFF: It is a command used to switch off the display of the number in each command line.


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The Whitehall Hotel
105 East Delaware Place, Chicago, IL
0,6 km från
947 kr+
I would have preferred a discount on the Parking for staying in the hotel.
Great location a few steps from the best shopping and restaurants in Chicago. Very friendly and professional staff.
Great location, beautiful room and good staff. Except for my wife losing her nose ring and no thin sheets were available; our stay was outstanding.
The rooms were nice, but our room had not been properly cleaned. There was an empty wine bottle in the trash, a stain on the carpet, and dirt on the bedspread. The location is great for exploring the city. A block from Michigan Ave, the beach, and Rush street. The customer service was not good. We waited 30-45 minutes for our bags to be brought up to our room and finally had to go down to get them. There were two bellmen standing in the lobby doing nothing and our luggage was just sitting there. As mentioned the room was not properly cleaned which is never a good feeling when you are staying in a hotel. I requested a late check-out and was immediately told "No due to someone else staying in our room and they needed to clean it". He then asked what room we were staying in. He again said it wasn't possible as someone was staying in that room and they would need to clean it. When we checked out I forgot a brand new shirt I bought while staying there. I rarely hang things in the closet and had done so since it was a new shirt. I realized this the next day and called and again was given horrible customer service. They told me it was not in the room and that nobody stayed in that room the night before so they couldn't help me. Obviously, the maid who cleaned that room took it, but they accepted no responsibility for that. Not to mention they couldn't give us a late check-out due to another guest needing the room. It's a shame the customer service was so bad because it is a nice hotel in a great area of the city.
Large rooms and bathrooms. Plenty of fast elevators. Great location for the north end of the Magnificent Mile. More attention to detail could be provided to the cleanliness of the shower area. Several areas of black mildew in the grout of the shower wall.
Rooms too small
Great spot. Conveniently located. Pleased overall with the property.
The French ambiance. A bit expensive.
beautiful hotel bed was too soft
Great location, great value for the price. Staff need to stop acting so high fallutin. Yes it is the Sofitel, but you still work for me.
we were traveling with our toddler. the hotel crib came with complimentary johnson & johnson toiletries and a stuffed bear. the room was clean, comfortable and spacious enough to be tolerable with a toddler in the same room. the bathroom was spacious, in excellent condition and well-appointed. we were on a high floor and had a great view of the city. it would've been great to have iPhone charging docks/stations. we forgot to bring one charger.
Very good location.
The bed is gorgeusly confortable Many problems with shower, room cleaning, front desk attendece, etc.
when I make check out from the hotel they told me i have to pay more than what booking website show me,after desscution they agree to refund 110 $ to me.
Location/spa and swimming pool None
They had absolutely no trace of our reservation on arrival, but after some spirited discussion more than made good the upheaval.
Very good welcome and staff performance An elevator alarm went off in the middle of the night and it took over one hour to fix it - not conducive to comfortable sleep.
Comfortable, elegant, convenient excellent accomodations. Food very good but somewhat overpriced. The hotel tower is very tall and our initial room on the 40th floor was very noisy in the wind and rain storm. The swaying was not noticable but the sound of creaking wall and ductwork was uncomfatably loud. The staff did move us to a lower floor to a large apartment suit which was very quiet.
Great restaurant recommendation and service!
Location! And they left a small treat for my child every afternoon.
Hotel has great location, very nice accomodations. There was broken glass in the bar and I cut my finger pretty badly on our first night at the hotel. Although hotel staff was very nice and they did call security, I thought they could have done more in way of compensation. In hindsight, I probably needed stitches and other than call to my room by the manager (who had already left by the time we got back to the room), they really didn't even follow-up.
Beautiful. Comfortable, luxurious. Staff assistance was hit or miss. Very frustrating. Some staff was very helpful. Others were not to the point of carelessness or rudeness. At this hotel I had expected service to be exquisite. It was not. For the price there's other hotels with better service who value their customers more highly.
Excellent ...one of the nicest hotels I have ever stayed at. Professional staff. When coming to Chicago it will be the only Hotel I will stay at.
We loved everything about it. From the very beginning (checkin) till the very last second before we left the hotel. Everyone was so helpful and friendly. The Concierge made arrangements for a boat trip on the Chicago River. Housekeeping kept our room in perfect shape. We had a room on the 21st floor which had the perfect view towards the chicago skyline. It was awesome. I gladly recommend to visit the Thompson Chicago and will for sure also book on other occasions other Thompson hotels. Thank you for this very wonderful and perfect vacation, Thompson. nothing!
We chose the Thompson Hotel for its location and value on Kayak. If I was traveling alone on business it would have been a better choice. They are super accomaditing allowing us to check in at 11:00 am without any problems. They also have a great online precheck in service. There is a Starbucks next door and good restaurants all around it. The precheck in service is a great idea however, it is as if you actually request something like an extra blanket or extra towels which were questions on the precheck in system they do not fulfill them. The elevator system requires you to use your room key to get up to your floor however, there is only one reader per elevator and it was virtually impossible to get to it when using the elevator. So many people were talking about riding the elevator up and down trying to get to their floors. The showers were never hot just warm regardless of the time of day. The A/C was all over the the place, hot / cold. The room refresh was seriously below average and actually just a really bad bed make up with fresh towels. They basically missed at the attempt to be a chic boutique hotel. There is no real identity to this property.
Body hair on the bathroom sink. stains on toilet. Kind of gross for the price.
Fantastic location, outstanding room. Would go back in a minute.
We go downtown every Christmas and this was the first time we stayed here. We loved it. Friendly staff, wonderful accommodation, and great location. We will certainly be back.
We had a great weekend staying at The Thompson! The location, staff and room were outstanding! We will definitely be staying here again!
Very hospitable. Staff very forthcoming. Nice space!
The Thompson Hotel Staff were amazing, very friendly and accommodating. We were upgraded to a corner king suite as it was our wedding night so that was a pleasant surprise as well. Overall wonderful experience. Would definitely go back. Nothing!
People are friendly Restaurant is very good Design is current Beds are comfortable air conditioning unit in room was loud
comfy bed, great location, nice lobby. Front desk lady was not very helpful. Toilet smelled like sewage. They called me from front desk to tell me someone was at my door from hotel personnel when I had privacy sign on...that woke me up...What kind of "privacy on" is this?
Nice Interior Design. The staff was very pleasant and helpful. This is a much better location than other hotels I have stayed at further south. We took a very nice walk north to Lincoln Park. It is a beautiful neighborhood and does not even feel like you are in the city. Too pricey for me to use other than business. I did not need or use "facilities" so that is the only reason I did not give that an "excellent" score. We were not impressed with the breakfast food. We were very pleased with our lunch at Gibson's across the street.
The beautiful view of lake Michigan. Close proximity to great restaurants and shopping. Nico Osteria restaurant within the hotel was excellent. The room was beautifully decorated, very comfortable bed, reading light ( unexpected), and huge bathroom and shower. The staff was extremely friendly and helpful. One of the late afternoons we needed to call for extra towels,since our room had not been cleaned,and we had been out all day. I do realize that this weekend was a home Bears foorball game and was very busy.
Personal service. They actually get to know your name. You sort of feel a bit like family. They care about your stay. Were willing to go the extra bit to make our time there great. Thanks for a fabulous trip!
Large Rooms, Great beds The restaurant was small with limited options but it appears they are renovating.
Very nice staff and great location It was perfect
From check in everyone was very nice. We were going for a short trip to another city in the US and were going to stay again at Talbott before heading home. We didn't want to carry one bag & a set of golf clubs so asked if we could leave it behind. Talbott was very understanding and stored it till return. All hotels do it but the attitude of the staff at Talbott made the difference. ..
Pretty much everything....very boutique...great staff....room maintenance five stars...great location No complaints
This was our second stay at the Talbott. It has a great location, the rooms are large, and the staff is welcoming and efficient. We particularly like the cozy parlor areas with fireplaces. There was nothing we did not like.
The hotel staff went above and beyond to make sure our stay was special. I will definately be returning
We loved the location and the staff Never received maid service without calling for the three days we stayed there. Disappointed I had to pay $4 for coffee in the morning. Should have a coffee bar out early in the lobby. Would be nice to have a fitness room in the hotel.
I liked almost everything about the hotel. The room was great, the day staff was great, the location is better than several in the area we stayed at in the past. Room amenities were also very good. The female employee who checked us out in the morning acted liked she was bothered by having to leave the rear office to deal with us. She was sullen and sighed and didn't even look up as I talked to her. She is NOT a good employee to have at this wonderful hotel.
Great staff, very helpful and friendly
Unexpected personal communication, services, everything top end. Technology and use of it to make the most of your stay. Very accommodating for our Shitzu. Great location. Room Service was as expected for similar hotels. Pricey, Ave quality, the usual limited items. Wouldn't take much to provide better then expected.
Staff was very nice and helpful. Never did we feel like we were bothering them with questions. The site was clean, well maintained. I would stay there again. The location is great if u are a tourist without a vehicle. We used the bus for everything. It is near the bus lines. Great hotel.
Fantastic location, and the staff is very friendly. It's gotten quite shabby.
It was so comfortable! And the staff did all they could to make it more so. We arrived at 6 AM after driving through a storm. They found a room for us right away so that we were able to test before our meeting. The concierge was available evenings and weekends, and was able to advise us about the best way to use our time. Nothing
Great location, but the whole is a little worn out.
The staff members were very friendly, helpful and professional. The building is in a great location!
Had a meeting at Ditka's. Great staff, comfortable room, quiet. Great location. Enjoyed Ditka's radio show too!
The best I would say is, great location, nice staff and nice bed. Spoiler alert. Don't read this if you got a great rate and don't want to be persuaded not to stay. It's an old building. What Tremont really needs to do is spend a lot of money and gut out old plumbing etc. New carpet and a nice bed doesn't, a nice room, make. They really tried to help us but when you come in to a room, the toilet and sink need to work. People really demand this of the hotel they stay at and on the third attempt to make the toilet work properly they switched our room. Our travel companion's a/c wasn't working either. The place doesn't need to look Ritz, but it does need to function. Sorry Tremont, I think you intend to do right by your customers. Good luck.
Nothing I've never stayed at some hotels like this!! No warm water for shower!! We reported the problem to front desk, and they said they would fix it, if not, they can change room for us. After we got back to hotel at night, they told us it's fixed. When I turned on the water, it was too cold to take a shower!!! And they even told us we cannot change room because all rooms were full! That's ridiculous. Definitely don't book this hotel.
My overall stay was great!
The 'free' wifi was only free in the lobby. Otherwise the stay was excellent and the location can't be beat!
Nothing Everything
Had some water in one area of the carpet next to the bed. Not sure how it got there


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|Oleg Kalnichevski <ol...@bluewin.ch>
|RE: Commons server infrastructure proposal
|Mon, 12 Apr 2004 17:41:35 GMT
> AFAICT even 1.5 NIO doesn't support SSL directly. > The trick is to use the SSLEngine and drop the SSLServerSocket altogether. Well, it does. http://java.sun.com/j2se/1.5.0/docs/guide/security/jsse/jsse-tiger-beta1.html#SSLENG The trouble is that it may takes years for 1.5 to become mainstream. Beat with a (biiiiiiiiig) stick, I still doubt 1.5 JDKs will be truly useble in heavy duty server applications until release 1.5.2. Otherwise, I am certainly +1 on the proposal, even though the size of that plus pretty much depends upon my workload with HttpClient Oleg > > Also you said that the Geronimo can be the lead, but I doubt that it would > provide the hookable protocol stack. There are lots of ASF projects that > already have it's own server side code like you said. > Each one of them has at least two common things, network layer and > threadpool, but also each of the component lacks the things like connection > hooks and protocol filters, and one cannot use any of them without pulling > out and rewriting the code. > > Generally what I'd like to do is to make the generic server infrastructure > that will inherently have SSL support and will be able to group the servers > together. > It will have at least couple of things like physical layer, threadpool, > workers, connection hooks, protocol actors and most importantly it will not > be bound to the tcp/ip protocol stack, so that one can build embedded > servers just defining different physical communication layer. > > Having something like would IMO ease the building of server side > applications. > > MT. > > > --------------------------------------------------------------------- > To unsubscribe, e-mail: email@example.com > For additional commands, e-mail: email@example.com > --------------------------------------------------------------------- To unsubscribe, e-mail: email@example.com For additional commands, e-mail: email@example.com
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