Trace Minimization Principles for Positive Semi-Definite Pencils

Transcription

1 Trace Minimization Principles for Positive Semi-Definite Pencils Xin Liang Ren-Cang Li Zhaojun Bai Technical Report

2 Trace Minimization Principles for Positive Semi-Definite Pencils Xin Liang Ren-Cang Li Zhaojun Bai September 5, 2012 Abstract This paper is concerned with inf trace(x H AX) subject to X H BX = J for a Hermitian matrix pencil A λb, where J is diagonal and J 2 = I (the identity matrix of apt size). The same problem was investigated earlier by Kovač-Striko and Veselić (Linear Algebra Appl., 216: , 1995) for the case in which B is assumed nonsingular. But in this paper, B is no longer assumed nonsingular, and in fact A λb is even allowed to be a singular pencil. It is proved, among others, that the infimum is finite if and only if A λb is a positive semi-definite pencil (in the sense that there is a real number λ 0 such that A λ 0 B is positive semi-definite). The infimum, when finite, can be expressed in terms of the finite eigenvalues of A λb. Sufficient and necessary conditions for the attainability of the infimum are also obtained. Key words. Hermitian matrix pencil, positive semi-definite, trace minimization, eigenvalue, eigenvector. AMS subject classifications. Primary 15A18. Secondary 15A22, 65F15. 1 Introduction Consider Hermitian matrix A C n n. Denote its eigenvalues by λ i (i = 1, 2,..., n) in the ascending order: λ 1 λ 2 λ n. (1.1) One, among numerous others, well-known result for a Hermitian matrix is the following trace minimization principle 4, p.191 min trace(x H AX) = X H X=I k k λ i, (1.2) School of Mathematical Science, Peking University, Beijing, , P. R. China. liangxinslm@pku.edu.cn. Support in part by China Scholarship Council. This author is currently a visiting student at Department of Mathematics, University of Texas at Arlington, Arlington, TX Department of Mathematics, University of Texas at Arlington, P.O. Box 19408, Arlington, TX rcli@uta.edu. Supported in part by NSF grants DMS and DMS Department of Computer Science and Department of Mathematics, University of California, Davis, CA bai@cs.ucdavis.edu. Supported in part by NSF grants OCI and DMS , and DOE grant DE-FC02-06ER

3 where I k is the k k identity matrix, and X C n k is implied by size compatibility in matrix multiplications. Moreover for any minimizer X min of (1.2), i.e., trace(x H min AX min) = k λ i, its columns span A s invariant subspace 1 associated with the first k eigenvalues λ i, i = 1, 2,..., k. Equation (1.2) can be proved by using Cauchy s interlacing property, for example, and is also a simple consequence of the more general Wielandt s theorem 12, p.199. This minimization principle (1.2) can be extended to the generalized eigenvalue problem for a matrix pencil A λb, where A, B C n n are Hermitian and B is positive definite. Abusing the notation, we still denote the eigenvalues of A λb by λ i (i = 1, 2,..., n) in the ascending order as in (1.1). The extended result reads 10 min trace(x H AX) = X H BX=I k k λ i. (1.3) Moreover for any minimizer X min of (1.3), there is a Hermitian A 0 C k k whose eigenvalues are λ i, i = 1, 2,..., k such that AX min = BX min A 0. The result (1.3), seemingly more general than (1.2), is in fact implied by (1.2) by noticing that the eigenvalue problem for A λb is equivalent to the standard eigenvalue problem for B 1/2 AB 1/2, where B 1/2 = (B 1/2 ) 1 and B 1/2 is the unique positive definite square root of B. The next question is how far we can go in extending (1.2). In 1995, Kovač-Striko and Veselić 5 obtained a few surprising results in this regard. To explain their results, we first give the following definition. Definition 1.1. A λb is a Hermitian pencil of order n if both A, B C n n are Hermitian. A λb is a positive (semi-)definite matrix pencil of order n if it is a Hermitian pencil of order n and if there exists λ 0 R such that A λ 0 B is positive (semi-)definite. Note that this definition does not demand anything on the regularity of A λb, i.e., a Hermitian pencil or a positive semi-definite matrix pencil can be either regular (meaning det(a λb) 0) or singular (meaning det(a λb) 0 for all λ C). Kovač-Striko and Veselić 5 focused on a Hermitian 2 pencil A λb with B always nonsingular but possibly indefinite. That B is invertible ensures det(a λb) 0 and thus the regularity of A λb. Denote by n + and n the numbers of positive and negative eigenvalues of B, respectively, and let k + and k be two nonnegative integers such that k + n + and k n and set Ik+ J k = C k k, k = k + + k. (1.4) I k J k will have this assignment for the rest of this paper. Since B is nonsingular, n = n + +n. The following remarkable results are obtained in 5. 1 This invariant subspace is unique if λ k < λ k+1. This is also true for the deflating subspace spanned by the columns of the minimizer for (1.3). 2 Although Kovač-Striko and Veselić 5 were concerned about real symmetric matrices, but their arguments can be easily modified to work for Hermitian matrices. 2

4 Theorem 1.1 (Kovač-Striko and Veselić 5). Let A λb be a Hermitian pencil of order n and suppose that B is nonsingular. 1. Suppose that A λb is positive semi-definite, and denote by λ ± i the eigenvalues 3 of A λb arranged in the order: λ n λ 1 λ+ 1 λ+ n +. (1.5) Let X C k k satisfying X H BX = J k, and denote by µ ± i λx H BX arranged in the order: the eigenvalues of X H AX µ k µ 1 µ+ 1 µ+ k +. (1.6) Then λ + i µ + i λ + i+n k, for 1 i k +, (1.7) λ j+n k µ i λ i, for 1 j k, (1.8) where we set λ + i = for i > n + and λ j = for j > n. 2. If A λb is positive semi-definite, then inf trace(x H AX) = X H BX=J k k + λ + i k λ i. (1.9) (a) The infimum is attainable, if there exists a matrix X min that satisfies Xmin H BX min = J k and whose first k + columns consist of the eigenvectors associated with the eigenvalues λ + j for 1 j k + and whose last k columns consist of the eigenvectors associated with the eigenvalues λ i for 1 i k. (b) If A λb is positive definite or positive semi-definite but diagonalizable 4, then the infimum is attainable. (c) When the infimum is attained by X min, there is a Hermitian A 0 C k k whose eigenvalues are λ ± i, i = 1, 2,..., k ± such that X H minbx min = J k, AX min = BX min A A λb is a positive semi-definite pencil if and only if inf trace(x H AX) >. (1.10) X H BX=J k 4. If trace(x H AX) as a function of X subject to X H BX = J k has a local minimum, then A λb is a positive semi-definite pencil and the minimum is global. 3 Positive semi-definite pencil A λb with nonsingular B always has only real eigenvalues implied by 5, Proposition 4.1 and 3, Theorem See also Lemma 3.8 later. 4 Hermitian pencil A λb of order n is diagonalizable if there exists a nonsingular n n matrix W such that both W H AW and W H BW are diagonal. 3

5 Item 1 of this theorem is 5, Theorem 2.1, item 2 is 5, Theorem 3.1 and 5, Corollary 3.4, item 3 is 5, Corollary 3.8, and item 4 is 5, Theorem 3.5. They are proved with the prerequisite that B is nonsingular. In 5, footnote 1 on p.140, Kovač-Striko and Veselić wrote it seems plausible that many results of this paper are extendable to pencils with B singular, but det(a λb) not identically zero. As yet we know of no simple way of doing it. One of the aims of this paper is to confirm this suspicion that the nonsingularity assumption is indeed not necessary. Moreover in an attempt of being even more general, we cover singular pencils, as well. We point out that the Courant-Fisher min-max principle 12, p.201 (for a single eigenvalue, instead of sums of several eigenvalues like traces) has been generalized to arbitrary Hermitian pencils, include semi-definite ones 1, 2, 6, 8, 9. Equation (1.9) for k = 1 can be considered as a special case of those. Lancaster and Ye 6, Theorem 1.2 defined a positive definite pencil by requiring that β 0 A α 0 B be positive definite for some α 0, β 0 R. This definition is less restrictive than ours. 1. If β 0 0, we let λ 0 = α 0 /β 0 to see that Lancaster s and Ye s definition of a definite pencil includes A λ 0 B being either positive or negative definite. Definition 1.1, on the other hand, requires A λ 0 B be positive definite. 2. If β 0 = 0, then α 0 0 and thus Lancaster and Ye 6 require that B be either positive or negative definite. In this case, A λb is also positive definite by Definition 1.1 because we can always pick some λ 0 R so that A λ 0 B is positive definite. Even more general but closely related is the concept of a definite pencil which is defined by the existence of a complex linear combination of A and B being positive definite 7, 11, 13, 14. But to serve our purpose in this paper, we will stick to Definition 1.1. The rest of this paper is organized as follows. Section 2 presents our first set of main results which are essentially those summarized in Theorem 1.1 but without the nonsingularity assumption on B, while another main result of ours will be given in section 4 and it is about a sufficient and necessary condition on the attainability for the infimum of the trace function in terms of the eigen-structure of A λb. All proofs related to the main results in section 2 are grouped in section 3 for readability. Conclusions are given in section 5 Notation. Throughout this paper, C n m is the set of all n m complex matrices, C n = C n 1, and C = C 1. R is set of all real numbers. I n (or simply I if its dimension is clear from the context) is the n n identity matrix, and e j is its jth column. For a matrix X, N (X) = {x : Xx = 0} denotes X s null space and R(X) denotes X s column space, the subspace spanned by its columns. X H is the conjugate transpose of a vector or matrix. A 0 (A 0) means that A is Hermitian positive (semi-)definite, and A 0 (A 0) if A 0 ( A 0). Re (α) is the real part of α C. For matrices or scalars 4

6 X i, both diag(x 1,..., X k ) and X 1 X k denote the same matrix X 1... X k. 2 Main results Throughout the rest of this paper, A λb is always a Hermitian pencil of order n. It may even be singular, i.e., possibly det(a λb) 0 for all λ C. In particular, B is possibly indefinite and singular. The integer triplet (n +, n 0, n ) is the inertia of B, meaning B has n + positive, n 0 0, and n negative eigenvalues, respectively. Necessarily We say µ is a finite eigenvalue of A λb if r := rank(b) = n + + n. (2.1) rank(a µb) < max rank(a λb), (2.2) λ C and x C n is a corresponding eigenvector if 0 x N (A) N (B) satisfies Ax = µbx, (2.3) or equivalently, 0 x N (A µb)\(n (A) N (B)). To state our main results, for the moment we will take it for granted that a positive semi-definite pencil A λb has only r = rank(b) real finite eigenvalues, but we will prove this claim later in Lemma 3.8. Denote these finite eigenvalues by the same notations λ ± i as in section 1 for the case of a nonsingular B and arrange them in the order as (1.5): λ n λ 1 λ+ 1 λ+ n +. (1.5) throughout the rest of this paper. What we have to keep in mind that now n + + n may possibly be less than n. Also in Lemma 3.8, we will see that if λ 0 R such that A λ 0 B 0 as in Definition 1.1, then for all i, j λ i λ 0 λ + j. (2.4) Theorem 2.1. In Theorem 1.1, the condition that B is nonsingular can be removed. We emphasize again that Theorem 2.1 covers not only the case when A λb is a regular pencil and B is singular but also A λb is a singular pencil. Remark 2.1. In both Theorems 1.1 and 2.1, the infimum is taken subject to X H BX = J k. It is not difficult to see this restriction can be relaxed to X H BX is unitarily similar to J k, or equivalently X H BX is unitary and has the eigenvalue 1 with multiplicity k + and 1 with multiplicity k. A necessary condition for a Hermitian pencil A λb to be definite is that it must be regular. The next theorem extends two other results: Corollary 3.7 and Theorem 3.10 of 5 to a regular pencil. 5

7 Theorem 2.2. Let A λb be a Hermitian matrix pencil of order n, and suppose it is regular, i.e., det(a λb) 0. Suppose also that n + 1 and n A necessary and sufficient condition for A λb to be positive definite is that both infimums t + 0 = inf x H Ax, t x H 0 = inf x H Ax (2.5) Bx=1 x H Bx= 1 are attainable and t t 0 > 0. In this case ( t 0, t+ 0 ) is the positive definiteness interval of A λb, i.e., A µb 0 for any µ ( t 0, t+ 0 ). 2. Suppose 1 k + n + and 1 k n and that the positive definiteness intervals of pencils X H AX λj k, taken for all X satisfying X H BX = J k, have a nonvoid intersection I. Then A λb is positive definite, and I is the definiteness interval of A λb. Another main result of ours to be given in section 4 is a sufficient and necessary condition for the attainability of the infimum in the terms of the eigen-structure of the pencil A λb. Remark 2.2. Lancaster and Ye 6, Theorem 1.2 gave a different characterization of a positive definite pencil with nonsingular B. To state their result, we will characterize each finite real eigenvalue µ of regular Hermitian pencil A λb as of the positive type or the negative type according to whether x H Bx > 0 or x H Bx < 0, where x is a corresponding eigenvector. For a multiple eigenvalue µ with the same algebraic and geometric multiplicity, we can choose a basis of the associated eigenspace and pair each copy of µ with one basis vector and define the type of each copy accordingly. Theorem 1.2 of 6 says that A λb is positive definite if and only if it is diagonalizable, has all eigenvalues real, and the smallest finite eigenvalue of the positive type is bigger than the largest finite eigenvalue of the negative type. This result, too, can be extended to include the case when B is singular, using our proving techniques here. 3 Proofs All notations in section 2 will be adopted in whole. We will also use integer triplet (i + (H), i 0 (H), i (H)) for the inertia of a Hermitian matrix H, where i + (H), i 0 (H), and i (H) are the number of positive, zero, and negative eigenvalues of H, respectively. In particular, i + (B) = n +, i 0 (B) = n r, i (B) = n. The eventual proofs of Theorems 2.1 and 2.2 relay on a series of lemmas below. Lemma 3.1. There is a unitary U C n n such that U H BU = r n r r B 1 n r 0, U H AU = r n r r A 11 A 12 n r A 21 A 22. (3.1) where A H ij = A ji, and B H 1 = B 1 C r r is nonsingular. 6

8 Lemma 3.1 can be proved by noticing that there is a unitary U C n n to transform B as in the first equation in (3.1). The second equation there is simply due to partition U H AU accordingly for the convenience of our later use. Now if A 21 = A H 12 in (3.1) can be somehow annihilated, the situation is then very much reduced to the case studied by Kovač-Striko and Veselić 5, namely a nonsingular B. Finding a way to annihilate A 21 = A H 12 is the key to our whole proofs in this section. Lemma 3.2. Let A λb be a Hermitian matrix pencil of order n, and let P B be the orthogonal projection onto R(B). If R(I P B AP B ) R(I P B AI P B ), (3.2) then there exists a nonsingular Y C n n such that Y H AY = r n r r A 1 n r A 2, Y H BY = r n r r B 1 n r 0, (3.3) where B H 1 = B 1 is invertible, and A H i = A i. Moreover A λb has r finite eigenvalues which are the same as the eigenvalues of A 1 λb 1. Proof. We have (3.1) by Lemma 3.1. The condition (3.2) is equivalent to R(A 21 ) R(A 22 ). Thus A 22 Z = A 21 = A H 12 has solutions one of which is Z = A 22 A 21, where A 22 is the Moore-Penrose inverse of A 22. Define Ir 0 I r 0 C = = Z I n r A 22 A. (3.4) 21 I n r It can be verified that C H U H A11 A AUC = 12 A 22 A 21 0, C H U H BUC = 0 A 22 Take A 1 = A 11 A 12 A 22 A 21, A 2 = A 22, and Y = UC to get (3.3). B1 0. (3.5) 0 0 Although the condition (3.2) seems a bit of mysterious, it is always true for positive semi-definite matrix pencils as confirmed by the next lemma. Lemma 3.3. If A λb is a positive semi-definite matrix pencil of order n, then the condition (3.2) is satisfied and thus the equations in (3.3) hold for some nonsingular Y C n n, and moreover, A 2 0 and A 1 λb 1 is a positive semi-definite matrix pencil of order n r. Proof. There exists λ 0 R such that  := A λ 0B 0. We have (3.1) by Lemma 3.1, and then U H ÂU = U H (A λ 0 B)U = r n r r A 11 λ 0 B 1 A n r A 21 A 22 Thus R(A 21 ) R(A 22 ) which is (3.2), as expected. Finally, A 2 0 and that A 1 λb 1 is positive semi-definite are due to Y H (A λ 0 B)Y 0. 7

9 The decompositions in (3.3), if exist, are certainly not unique. The next lemma says the reduced pencils A 1 λb 1 and A 2 λ 0 are unique, up to nonsingular congruence transformation. Lemma 3.4. Let A λb be a Hermitian matrix pencil of order n, and suppose it admits decompositions in (3.3), where r = rank(b). Suppose it also admits r n r Ỹ H AỸ = r à 1 n r à 2, Ỹ H BỸ = r r n r B1 n r 0, (3.6) where Ỹ Cn n is nonsingular. C (n r) (n r) such that Then there exist nonsingular M 1 C r r and M 2 à 1 λ B 1 = M H 1 (A 1 λb 1 )M 1, à 2 = M H 2 A 2 M 2. Proof. Partition Y = Y 1, Y 2 and Ỹ = Ỹ1, Ỹ2 with Y 1, Ỹ1 C n r. Since BY 2 = BỸ2 = 0, we have R(Ỹ2) = N (B) = R(Y 2 ) and thus Ỹ2 = Y 2 M 2 for some nonsingular M 2 C (n r) (n r). Set M = Y 1 Ỹ 1 and partition M to get Ỹ 1 = Y M, M = r r M 1 Z Hence Ỹ = Ỹ1, Ỹ2 M1 0 = Y 1, Y 2 which implies M Z M 1 must be nonsingular. We have 2 by (3.3) and (3.6) 0 = Ỹ 1 H AỸ2 = M H Y H AY 2 M 2 = M H 0 M A 2 M H 0 = 0, 2 A 2 à 1 = Ỹ 1 H AỸ1 = M H Y H AY M = M H A1 0 M = M H A1 0 M = M1 H A 0 A M 1, B 1 = Ỹ 1 H BỸ1 = M H Y H BY M = M H B1 0 M = M1 H B M 1, à 2 = Ỹ H 2 AỸ2 = M H 2 Y H 2 AY 2 M 2 = M H 2 A 2 M 2, as expected. Lemma 3.5. Let M C l l be Hermitian and nonsingular, and let 0 y C l. Then there exists x C l such that both x H Mx 0 and x H y 0. In the case when M is indefinite, the chosen x can be made either x H Mx > 0 or x H Mx < 0 as needed. Proof. If M is positive or negative definite, taking x = y will do. Suppose M is indefinite. There is a nonsingular matrix Z C l l such that Z H MZ = diag(i l+, I l ), where l ± 1. Partition Z H y1 y =, where y 1 C l +. We may take x by y 2 either Z 1 x = y1 0 8 n r. or Z 1 0 x =, (3.7) y 2

10 depending on if y i = 0 or not. Because at least one of y i is nonzero, one of the choices in (3.7) will make both x H Mx 0 and x H y 0. It can also be done to ensure x H Mx > 0 regardless. In fact, if y 1 0, the first choice in (3.7) will do. But if y 1 = 0, then y 2 0. Take Z 1 (y H x = 2 y 2 + 1) 1/2 e 1. Then x H Mx = 1 and x H y = y H 2 y 2. Similarly we can ensure x H Mx < 0 if needed. Lemma 3.6. Let A λb be a Hermitian matrix pencil of order n. If then the condition (3.2) holds. y 2 inf trace(x H AX) >, X H BX=J k Proof. We have (3.1) by Lemma 3.1. Now for any X C n k, write X = U H X = r n r X1 X2. (3.8) We have X H BX = X H U H BU X = X H 1 B 1 X1, (3.9) trace(x H AX) = trace( X H 1 A 11 X1 ) + 2Re ( trace( X H 1 A 12 X2 ) ) + trace( X H 2 A 22 X2 ). (3.10) The condition (3.2) is equivalent to R(A 21 ) R(A 22 ) which we will prove. Assume to the contrary that R(A 21 ) R(A 22 ), or equivalently N (A 12 ) = N (A H 21) = R(A 21 ) R(A 22 ) = N (A H 22) = N (A 22 ), i.e., there exists 0 x 2 C n r such that A 22 x 2 = 0 but y := A 12 x 2 0. By Lemma 3.5, there is x 1 C r such that x H 1 B 1x 1 0 and x H 1 y 0. For our purpose, we will make x H 1 B 1x 1 > 0 if k + > 0 and x H 1 B 1x 1 < 0 otherwise. Scale x 1 so that x H 1 B 1x 1 = 1. B 1 induces an indefinite-inner product in C r and since x H 1 B 1x 1 = 1, we can extend x 1 to an orthonormal basis with respect to this B 1 -indefinite-inner product 3, p.10: x 1, x 2,..., x r, i.e., x H i B 1x j = 0 for i j and x H i B 1x i = ±1. Suppose for the moment x H 1 B 1x 1 = 1. Pick k x i out of all: x j1, x j2,..., x jk with j 1 = 1 (i.e., x 1 is included in), such that among x H i j B 1 x ij for 1 j k there are k + of them +1s and k of them 1s. Now consider those X in (3.8) with X 1 = x j1, x j2,..., x jk Π, X2 = ξ y, 0,..., 0, where ξ C, and Π is the r r permutation matrix such that X H 1 B 1 X 1 = J k and x 1 is in the first column of X 1. Then by (3.10), trace(x H AX) = trace( X H 1 A 11 X1 ) + 2Re ( ξx H 1 y ) which can be made arbitrarily small towards, contradicting that trace(x H AX) as a function of X restricted to X H BX = J k is bounded from below. Therefore R(A 21 ) R(A 22 ). The case for x H 1 B 1x 1 = 1 is similar. The proof is completed. 9

11 The standard involutary permutation matrix (SIP) of size n is the n n identity matrix with its columns rearranged from the last to the first: (3.11) The next lemma presents the well-known canonical form of a Hermitian pencil A λb with a nonsingular B under nonsingular congruence transformations. Lemma 3.7 (3, Theorem ). Let A λb be a Hermitian matrix pencil of order n, and suppose that B is nonsingular. Then there exists a nonsingular W C n n such that W H 0 Kp+1 0 Kq AW = s 1 K 1 s p K p Kp+1 H 0 Kq H, (3.12a) 0 W H 0 Sp+1 0 Sq BW = s 1 S 1 s p S p Sp+1 H 0 Sq H, (3.12b) 0 where α i α i 1 K i = , (3.13) α i 1 α i 1 α i R for 1 i p; α i C is nonreal for p + 1 i q, and s i = ±1 for 1 i p; S i is a SIP whose size is the same as that of K i for all i. The representations in (3.12) are uniquely determined by the pencil A λb, up to a simultaneous permutation of the corresponding diagonal block pairs. Lemma 3.8. Let A λb be a positive semi-definite matrix pencil of order n, and suppose that λ 0 R such that A λ 0 B There exists a nonsingular W C n n such that W H AW = n 1 Λ 1 n 1 r n 1 n r r n 1 Λ 0 n r Λ, W H BW = n 1 Ω 1 n 1 r n 1 n r r n 1 Ω 0 n r 0, (3.14) where (a) Λ 1 = diag(s 1 α 1,..., s l α l ), Ω 1 = diag(s 1,..., s l ), s i = ±1, and Λ 1 λ 0 Ω 1 0; 10

12 (b) Λ 0 = diag(λ 0,1,..., Λ 0,m+m0 ) and Ω 0 = diag(ω 0,1,..., Ω 0,m+m0 ) with Λ 0,i = t i λ 0, Ω 0,i = t i = ±1, for 1 i m, 0 λ0 0 1 Λ 0,i =, Ω λ 0 1 0,i =, for m + 1 i m + m (c) Λ = diag(α r+1,..., α n ) 0 with α i {1, 0} for r + 1 i n. The representations in (3.14) are uniquely determined by A λb, up to a simultaneous permutation of the corresponding 1 1 and 2 2 diagonal block pairs (s i α i, s i ) for 1 i l, (Λ 0,i, Ω 0,i ) for 1 i m + m 0, and (α i, 0) for r + 1 i n. 2. A λb has only real finite eigenvalues and the number of the finite eigenvalues is r = rank(b) = n + + n. Denote these finite eigenvalues by λ ± i and arrange them in the order as in (1.5). Write m = m + + m, where m + is the number of those 1 1 diagonal blocks in Λ 0 with s i = 1 and m is that of those with s i = 1. The respective sources of these finite eigenvalues are source 1. the 1 1 block pairs (Λ 0,j, Ω 0,j ) with t j = 1 produce λ i = λ 0 for 1 i m ; source 2. the 1 1 block pairs (Λ 0,j, Ω 0,j ) with t j = +1 produce λ + i = λ 0 for 1 i m + ; source 3. the 2 2 block pairs (Λ 0,m+i, Ω 0,m+i ) for 1 i m 0 produce λ m +i = λ 0 and λ + m + +i = λ 0; source 4. the diagonal matrix pair (Λ 1, Ω 1 ) produces λ ± i m 0 + m ± i n ±. (according to s j = ±1) for Each eigenvalue from sources other than source 3 has an eigenvector x that satisfies x H Bx = +1 for λ + i and x H Bx = 1 for λ j, while for source 3, each pair (λ m +i, λ+ m + +i ) of eigenvalues shares one eigenvector x that satisfies xh Bx = 0. To be more specific than (1.5), we can order these finite eigenvalues as λ n λ m 0 +m +1 < λ 0 = = λ }{{} 0 = λ 0 = = λ }{{} 0 m 0 m = = λ 0 = = λ }{{} 0 = λ 0 = = λ 0 < λ }{{} + m 0 +m + +1 λ+ n +. m + m 0 In particular λ i = λ 0 for 1 i m 0 + m and λ + i = λ 0 for 1 i m 0 + m +. (3.15) 3. {γ R A γb 0} = λ 1, λ+ 1. Moreover, if A λb is regular, then A λb is a positive definite pencil if and only if λ 1 < λ+ 1, in which case {γ R A γb 0} = (λ 1, λ+ 1 ). 4. Let µ = (λ 1 + λ+ 1 )/2. For γ > µ, let n(γ) be the number of the eigenvalues of the matrix pencil A λb in µ, γ), where µ, if an eigenvalue, is counted i + (Ω 0 ) times. 11

13 For γ < µ, let n(γ) be the number of the eigenvalues of the matrix pencil A λb in (λ, µ, where µ, if an eigenvalue, is counted i (Ω 0 ) times. Then n(γ) = i (A γb). Proof. In Lemma 3.3, A 1 λb 1 is a positive semi-definite matrix pencil with B 1 nonsingular. Such a pencil can be transformed by congruence so that Y1 HA 1Y 1 and Y1 HB 1Y 1 are in their canonical forms as given in the right-hand sides of (3.12a) and (3.12b), respectively, where Y 1 C r r is nonsingular. We now use the positive semi-definiteness to describe all possible diagonal blocks in the right-hand sides. There are a few cases to deal with: Case 1. No K i (1 i p) is 3 3 or larger. For a 3 3 K i with α i R, the right-bottom corner 2 2 submatrix of K i µs i αi µ 1 αi µ 1 0 nor for any µ R. For a k k K i with α i R and k 4, the submatrix of K i µs i, consisting of the intersections of its row 2 and k and its column 2 and k is always the 2 2 SIP which is indefinite. Case 2. No 2 2 K i (1 i p) is with s i = 1. This is because for s i = 1 0 αi α s i µs α i 1 i = i + µ 0 for any µ R. 1 0 α i + µ 1 Case 3. The α i for any 2 2 K i (1 i p), if any, is λ 0. This is because 0 αi α µ = i µ 0 if and only if µ = α α i α i µ 1 i. Case 4. K i (1 i p) with α i λ 0 is 1 1. This is a result of Case 1 and Case 3 above. Case 5. The blocks associated with nonreal α i cannot exist. This is because the submatrix consisting of the intersections of the first and last row and the first and last column of 0 Ki µ 0 Si Si H 0 Ki H 0 0 α is i µ which is never semi-definite for any µ R. ᾱ i µ 0 Together, they imply Y H 1 A 1 Y 1 = diag(λ 1, Λ 0 ), Y H 1 B 1 Y 1 = diag(ω 1, Ω 0 ), (3.16) where Λ 1, Λ 0, Ω 1, Ω 0 as described in the lemma. Since A 2 0, there exists a nonsingular Y 2 C (n r) (n r) such that Y H 2 A 2 Y 2 = diag(α r+1,..., α n ) 12

14 with α i {1, 0} for r + 1 i n. Now set W = Y diag(y 1, Y 2 ) to get (3.14). The uniqueness of the representations in (3.14), up to simultaneous permutation, is a consequence of the uniqueness claims in Lemma 3.7 and that in Lemma 3.4 up to congruence transformation. For item 2, we note that the finite eigenvalues of A λb are the union of the eigenvalues of Λ 1 λω 1 and these of Λ 0 λω 0. The rest are a simple consequence of item 1. For item 3, we note Λ 1 λ 0 Ω 1 = diag(s i (α i λ 0 )) 0. Obviously α i, i = 1,, l are some eigenvalues of A λb. If s i = 1, α i > λ 0, and thus α i = λ + j for some j > m + + m 0. Similarly, if s i = 1, α i = λ k for some k > m + m 0. Hence Also, and Λ 1 γω 1 = diag(s i (α i γ)) 0 (3.17) λ k γ λ+ j for all k > m + m 0, j > m + + m 0. Λ 0,i γω 0,i = t i (λ 0 γ) 0 for i = 1,, m (3.18) λ k γ λ+ j for all 1 k m, 1 j m +, Λ 0,i γω 0,i 0 for i = m + 1, m + m 0 γ = λ 0. (3.19) Putting all together, we have A γb 0 λ 1 γ λ+ 1. For A λb to be regular and positive semi-definite, Λ 0. Now if A λb is a positive definite pencil, then there exists γ such that the inequalities in (3.17), (3.18) and (3.19) are strict. This can only happen when m 0 = 0 and λ 1 < λ+ 1, in which case A γb 0 λ 1 < γ < λ+ 1. On the other hand, if λ 1 < λ+ 1, then m 0 = 0 and only one of m + and m can be bigger than 0, or equivalently only one of λ 1 and λ+ 1 can possibly be λ 0 but not both. So for λ 1 < γ < λ+ 1, the inequalities in (3.17) and (3.18) are strictly, and therefore A γb 0. Item 4 can be proved by separately considering four cases: 1) λ 1 < λ 0 < λ + 1 ; 2) λ 1 < λ 0 = λ + 1 ; 3) λ 1 = λ 0 < λ + 1 ; and 4) λ 1 = λ 0 = λ + 1. Detail is omitted. Lemma 3.9. Suppose B is nonsingular. A λb is a positive semi-definite matrix pencil if inf X H BX=J k trace(x H AX) >. Proof. This is part of 5, Corollary 3.8, where the proof is rather sketchy with claims that, though true, were not obvious and substantiated. What follows is a more detailed proof. If either B 0 or B 0, then there is λ 0 R such that A λ 0 B 0, and thus no proof is necessary. Suppose in what follows that B is indefinite. If the infimum is attainable, then trace(x H AX) as a function of X restricted to X H BX = J k has a (local) minimum. By item 2 of Theorem 1.1, A λb is a positive semi-definite matrix pencil. Consider the case when the infimum is not attainable. Perturb A to A ϵ := A + ϵi, where ϵ > 0, and define f ϵ (X) := trace(x H A ϵ X) = trace(x H AX) + ϵ X 2 F trace(x H AX), 13

15 where X F is X s Frobenius norm. We have for any given ϵ > 0 inf f ϵ (X) inf trace(x H AX) >. (3.20) X H BX=J k X H BX=J k We claim inf f ϵ (X) subject to X H BX = J k can be attained. In fact, let X (i) be a sequence such that (X (i) ) H BX (i) = J k, lim f ϵ (X (i) ) = inf f ϵ (X). (3.21) i X H BX=J k {X (i) } is a bounded sequence; otherwise lim f ϵ(x (i) ) i inf trace(x H AX) + lim sup ϵ X (i) 2 X H F = +, BX=J k contradicting (3.20) and (3.21). So for any given ϵ > 0, A ϵ λb is a positive semi-definite pencil, which means for every ϵ > 0, there is λ ϵ R such that A ϵ λ ϵ B 0. Pick a sequence {ϵ i > 0} that converges to 0 as i. We claim that {λ ϵi } is a bounded sequence which then must have a convergent subsequence converging to, say λ 0. Through renaming, we may assume the sequence itself is the subsequence. Then let i on A ϵi λ ϵi B 0 to conclude that A λ 0 B 0, i.e., A λb is a positive semi-definite matrix pencil. We have to show that {λ ϵi } is bounded. To this end, it suffices to show {λ ϵ : 0 < ϵ 1} is bounded. Since A ϵ λb is a positive semi-definite matrix pencil of order n, its eigenvalues are real and can be ordered as, by Lemma 3.8, i λ n (ϵ) λ 1 (ϵ) λ+ 1 (ϵ) λ+ n + (ϵ), and λ 1 (ϵ) λ ϵ λ + 1 (ϵ). Therefore for 0 < ϵ 1 as was to be shown. λ ϵ B 1 A ϵ F B 1 A F + B 1 F, Proof of Theorem 2.1. To prove item 1 (which is the item 1 of Theorem 1.1 without assuming A λb is regular, let alone B is nonsingular), we complement 5 X by X c to a nonsingular X 1 = X, X c C n n such that X H 1 BX 1 = k n k k J k 0 n k 0 B c, X H 1 AX 1 = k n k k A 11 A 12 n k A 21 A 22. (3.22) For any γ R that makes A 11 γj k nonsingular, let Ik (A Z = 11 γj k ) 1 A 12, 0 I n k 5 X c can be found as follows. Since X has full column rank, we can expand it to a nonsingular Y = X, ˆX c C n n. Partition Y H Jk B 12 BY = and let Y H I 1 = k 0 to get B 21 B 22 B 21J k I n k Y1 H Y H Jk 0 BY Y 1 =. 0 B 22 B 21 J k B 12 Notice Y Y 1 = X, ˆX c XJ k B 12. Set B c = B 22 B 21 J k B 12 and X c = ˆX c XJ k B 12 and thus X 1 = Y Y 1 to get the first equation in (3.22). The second equation is simply obtained by partitioning X H 1 AX 1 accordingly. 14

16 then Z H X H 1 (A γb)x 1 Z = diag(a 11 γj k, Â22), where Â22 = A 21 (A 11 γj k ) 1 A 12 + A 22 γb c. Thus, i (A 11 γj k ) i (A γb) = i (A 11 γj k ) + i (Â22) (3.23) i (A 11 γj k ) + n k. (3.24) Assume µ + i < λ + i for some i. Then there exists γ (µ + i, λ+ i ) such that A 11 γj k is nonsingular. The number n(γ) for A 11 λj k as defined in item 3 of Lemma 3.8 is at least i, and therefore i (A 11 γj k ) i, and n(γ) for A λb is at most i 1, and therefore i (A γb) i 1. This contradicts the inequality in (3.23). Assume µ + i > λ + i+n k for some i. Then there exists γ (λ+ i+n k, µ+ i ) such that A 11 γj k is nonsingular. The number n(γ) for A 11 λj k as defined in item 3 of Lemma 3.8 is at most i 1, and therefore i (A 11 γj k ) i 1, and n(γ) for A λb is at least i + n k, and therefore i (A γb) i + n k. This contradicts the inequality in (3.24). This proves (1.7), and (1.8) can be proved in a similar way. For item 2, the condition of Lemma 3.3 is satisfied by A λb here. So we have (3.3) in which A 2 0 and A 1 λb 1 is a positive semi-definite pencil with B 1 nonsingular. Now for any X C n k, write X = Y 1 X = r n r X1 X2, (3.25) which gives X H BX = X H Y H BY X = X H 1 B 1 X 1, having nothing to do with X 2. Since the mapping X X is one-one, we have H A1 X1 inf trace(x H 0 X1 AX) = inf trace( ) X H BX=J k X 1 HB 1 X 1 =J k X 2 0 A 2 X 2 = inf trace( X X 1 HB 1 X 1 H A 1 X1 ) + trace( X 2 H A 2 X2 ) 1 =J k = inf X H 1 B 1 X 1 =J k trace( X H 1 A 1 X1 ) + inf X 2 trace( X H 2 A 2 X2 ) = inf X H 1 B 1 X 1 =J k trace( X H 1 A 1 X1 ). (3.26) The last equality is due to A 2 0 and is attained by any X 2 satisfying R( X 2 ) N (A 2 ). Theorem 1.1 is applicable to A 1 λb 1 and the application gives, by (3.26), inf trace(x H AX) = X H BX=J k inf trace( X X 1 HB 1 X 1 H A 1 X1 ) 1 =J k = k + λ + i k λ i, as expected. Track each equal sign in the above equations to conclude the claims in items 2(a,b,c). This proved item 2. 15

17 For item 3, item 2 implies that the condition (1.10) is necessary. We have to prove that it is sufficient, too. Suppose (1.10) is true. By Lemma 3.6, the condition (3.2) of Lemma 3.2 is satisfied. So we have (3.3), (3.25), and inf trace(x H AX) = X H BX=J k which is bounded from below. Therefore inf trace( X X 1 HB 1 X 1 H A 1 X1 ) + inf trace( X 2 H A 2 X2 ) 1 =J k X 2 A 2 0, inf trace( X X 1 HB 1 X 1 H A 1 X1 ) >. (3.27) 1 =J k Since B 1 is nonsingular, Lemma 3.9 says that A 1 λb 1 is a positive semi-definite matrix pencil by the second inequality in (3.27). Therefore Y H AY λy H BY is, too; so is A λb. Now we turn to item 4. In what follows, we first use Lagrange s multiplier method, similar to 5 in proving its Theorem 3.5 there, to show that A λb is a positive semidefinite pencil. Since X H BX = J k provides k 2 independent constraints on X (in R), we can use a k k Hermitian matrix Λ which has k 2 degrees of freedom to express Lagrange s function as 6 L (X) = trace(x H AX) Λ, X H BX J k. The gradient of L at X is L (X) = 2(AX BXΛ). Therefore for any local minimal point X 0, there exists a group of Lagrange s multipliers, i.e., some Hermitian Λ 0 C k k such that AX 0 = BX 0 Λ 0, X H 0 BX 0 = J k. (3.28) Without loss of generality, we may assume that Λ 0 is diagonal. Here is why. Pre-multiply the first equation in (3.28) by X0 H to get XH 0 AX 0 = J k Λ 0. Therefore J k Λ 0 = (J k Λ 0 ) H = Λ 0 J k which implies Λ 0 is block diagonal, i.e., Λ 0 = Λ 0+ Λ 0, where Λ 0± C k ± k ± are Hermitian. Hence there exists a block diagonal unitary V = V 0+ V 0 such that V H Λ 0 V is diagonal, where V 0± C k ± k ± are unitary. So V H J k V = J k, and thus we have by (3.28) A(X 0 V ) = B(X 0 V )(V H Λ 0 V ), (V X 0 ) H B(X 0 V ) = J k. It can also be seen that X 0 V is a local minimal point, too. Assume Λ 0 is diagonal, and write Λ 0 = diag(ω + 1,..., ω+ k +, ω k,..., ω 1 ), (3.29a) ω k ω 1, ω+ 1 ω+ k +. (3.29b) Since X 0 is a local minimal point as assumed, the second derivative D 2 L (X) at X 0, taken as a quadratic form and restricted to the tangent space of S = {X C n k X H BX = J k }, 6 The standard inner product X, Y for matrices of compatible sizes is defined as X, Y = Re (trace(x H Y )), the real part of trace(x H Y ). 16

18 must be nonnegative, i.e., for any W C n k satisfying trace(w H AW ) Λ 0, W H BW 0 (3.30) X H 0 BW + W H BX 0 = 0. (3.31) Complement X 0 by X c to a nonsingular X 1 = X 0, X c C n n such that X1 H Jk 0 BX 1 =. (3.32) 0 B c Thus Xc H BX 0 = 0 and Xc H AX 0 = Xc H BX 0 Λ 0 = 0 by (3.28). Therefore X1 H X H AX 1 = 0 AX 0 0 Jk Λ 0 Xc H = 0 0 AX c 0 Xc H. (3.33) AX c Rewrite (3.31) as X H 0 X H 1 X H 1 BX 1X 1 1 W + W H X H 1 X H 1 BX 1X 1 1 X 0 = 0 and partition X1 1 W = k Ŵ 1 n k Ŵ 2, X 1 1 X 0 = Ik. 0 to get J k Ŵ 1 + Ŵ H 1 J k = 0 (3.34) which says S := J k Ŵ 1 is skew-hermitian. We have Ŵ1 = J k S which gives all possible Ŵ1 that satisfies (3.34) as S runs through all possible k k skew-hermitian matrices. From (3.30), we have for any Ŵ2 and S = S H 0 trace(w H AW ) Λ 0, W H BW = trace(w H X H 1 X H 1 AX 1 X 1 1 W ) Λ 0, W H X H 1 X H 1 BX 1 X 1 1 W = trace(ŵ H 1 (J k Λ 0 )Ŵ1) + trace(ŵ H 2 (X H c AX c )Ŵ2) Λ 0, Ŵ H 1 J k Ŵ 1 + Ŵ H 2 B c Ŵ 2 = trace(sλ 0 J k S) + trace(ŵ H 2 X H c AX c Ŵ 2 ) Λ 0, SJ k S + Ŵ H 2 B c Ŵ 2. (3.35) This is true for any Ŵ2 and S = S H. Recall (3.29). For any given i k + and j k, set Ŵ2 = 0 and S = e i e H k+1 j e k+1 je H i in (3.35) to get 0 trace(sλ 0 J k S) + trace(λ 0 SJ k S) = 2(ω + i ω j ). Therefore for any ω 0 such that ω 1 ω 0 ω + 1, X H 0 AX 0 ω 0 J k = J k Λ 0 ω 0 J k = J k (Λ 0 ω 0 I) 0. On the other hand, for any given w C n k and i k, set S = 0 and Ŵ2 = we H i in (3.35) to get 0 trace(e i w H X H c AX c we H i ) Re ( trace(λ 0 e i w H B c we H i ) ) = w H (X H c AX c ωb c )w, 17

19 where ω = e H i Λ 0e i which is one of ω ± j. Since i and w are arbitrary, XH c AX c ωb c 0 for any ω {ω ± j, 1 j k ±}. This 7 implies X H c AX c ωb c 0 for any ω k ω ω + k +. In particular, Xc H AX c ω 0 B c 0. By (3.32) and (3.33), we conclude that A ω 0 B 0 for ω1 ω 0 ω 1 +. That means A λb is a positive semi-definite pencil. It remains to show that X 0 is also a global minimizer. Since A λb is a positive semi-definite pencil, by Lemma 3.3, we have (3.3). Define the one-one mapping between X and X by (3.25). We have trace(x H AX) = trace( X H 1 A 1 X1 ) + trace( X H 2 A 2 X2 ). Notice { } X1 {X C n k : X H BX = J k } = Y C X n k : XH1 B 1 X1 = J k 2 which places no constraint on X 2. If trace(x H AX) as a function of X restricted to X H BX = J k has a local minimum, then either r = n or r < n and A 2 0. In the case r = n, B is invertible and the theorem is already proved in 5 (see Theorem 1.1). Suppose r < n and thus A 2 0. At any local minimizer X min, the corresponding X min is X min = Y 1 X min = r n r Xmin,1 Xmin,2. We have X H min,2 A 2 X min,2 = 0. Consequently X min,1 is a local minimizer of trace( X H 1 A 1 X 1 ) as a function of X 1 restricted to X H 1 B 1 X 1 = J k. Since B 1 is nonsingular, item 4 of Theorem 1.1 is applicable and leads to that X min,1 is a global minimizer for trace( X H 1 A 1 X 1 ). This in turn implies that X min is a global minimizer for trace(x H AX) as a function of X restricted to X H BX = J k. Proof of Theorem 2.2. The basic idea is to essentially reduce the current case to the case in which B is nonsingular. For item 1, we note that if A λb is positive definite, then we have (3.3) with A 2 0. for any x C n, write ˆx = Y 1 x = r ˆx 1 n r ˆx 2, (3.36) which gives x H Bx = ˆx H Y H BY ˆx = ˆx H 1 B 1ˆx 1. Since the mapping x ˆx is one-one and since A 2 0, we have inf x H Ax = inf ˆx H 1 A 1ˆx 1, x H Bx=1 ˆx H 1 B 1 ˆx 1 =1 inf x H Ax = inf ˆx H 1 A 1ˆx 1. (3.37) x H Bx= 1 ˆx H 1 B 1 ˆx 1 = 1 7 For two Hermitian matrices M and N of the same size and α < β, if M γn 0 for γ = α and γ = β, then M γn 0 for any α γ β. In fact, any α γ β can be written as γ = tα + (1 t)β for some 0 t 1 and therefore M γn = t(m αn) + (1 t)(m βn) 0. 18

20 On the other hand, if the infimums in (2.5) are attainable, then A λb is positive semidefinite by Theorem 2.1 and thus we also have (3.3) with A 2 0 and thus (3.37). But A λb is assumed regular; A 2 must not be singular and so A 2 0. Either way, the problem is reduced to the one about A 1 λb 1. Apply 5, Corollary 3.7 to conclude the proof. For item 2, pick a λ 0 I, then X H AX λ 0 J k 0 for all X satisfying X H BX = J k. Therefore inf trace(x H AX λj k ) 0 X H BX=J k inf trace(x H AX) λ 0 (k + k ) >, X H BX=J k implying that A λb is positive semi-definite by Theorem 2.1. Hence we have (3.3) with A 2 0. But A λb is assumed regular; A 2 must not be singular and so A 2 0. Again the problem is reduced to the one about A 1 λb 1. Apply 5, Theorem 3.10 to conclude the proof. 4 A sufficient and necessary condition for infimum attainability Both Theorems 1.1 and 2.1 implies that for a positive semi-definite pencil A λb the infimum is attainable if and only there is an eigenvector matrix X min C n k such that X H minbx min = J k, AX min = BX min diag(λ + k +,..., λ + 1, λ 1,..., λ k ). In this section, we shall use the indices in the canonical form of A λb as given in Lemma 3.8 to derive another sufficient and necessary condition. Throughout this section, A λb is a Hermitian positive semi-definite pencil of order n. Recall, in Lemma 3.8, the finite eigenvalues of A λb are λ n λ m 0 +m +1 < λ 0 = = λ }{{} 0 = λ 0 = = λ }{{} 0 m 0 m = = λ 0 = = λ }{{} 0 = λ 0 = = λ 0 < λ }{{} + m 0 +m + +1 λ+ n +. m + m 0 (3.15) In particular λ i = λ 0 for 1 i m 0 + m and λ + i = λ 0 for 1 i m 0 + m +. By Lemma 3.8, m 0 and m ± are uniquely determined by A λb. Lemma 4.1. Suppose A λb is regular. Let Y C n l that satisfies Y H BY = I l be an eigenvector matrix of A λb associated with λ 0 (i.e., each column of Y is an eigenvector). Then l m +. Proof. By Lemma 3.8, A λb has m + + m + m 0 linearly independent eigenvectors associated with λ 0. One set of them can be chosen according to the three sources: x 1,..., x m from source 1, x + 1,..., x+ m + from source 2, and x 1,..., x m0 from source 3 such that X H BX = I m+ ( I m ) 0 m0 m 0, 19

21 where X = x + 1,..., x+ m +, x 1,..., x m x 1,..., x m0. Any eigenvector matrix Y C n l associated with λ 0 can be expressed as Y = XZ for some Z C (m ++m +m 0 ) l. Then Y H BY = I l is equivalent to I m+ 0 0 Z H 0 I m 0 Z = I l which implies l m +. Theorem 4.1. Let A λb be a Hermitian positive semi-definite pencil of order n. Then inf trace(x H AX) = X H BX=J k k + λ + i k λ i is attainable if and only if m 0 = 0 or k ± m ± in the case of m 0 > 0. Proof. We have (3.25) and (3.26). It can be seen that the infimums in inf trace(x H AX), X H BX=J k inf trace( X X 1 HB 1 X 1 H A 1 X1 ) 1 =J k are either both attainable or neither is. Also m 0 and m ± are the same for A λb and the reduced A 1 λb 1. So without loss of generality, we assume B is nonsingular. Suppose m 0 = 0 or k ± m ± in the case of m 0 > 0. The above analysis indicates that there are k + + k eigenvectors associated with the eigenvalues λ i, λ+ j for 1 i k, 1 j k +. Put these eigenvectors side-by-side with those for λ + j first and then those for to give a matrix X that satisfies X H BX = J k and at the same time λ i trace(x H AX) = k + λ + i k λ i. Suppose now the infimum is attainable. For any X C n k, partition X = X +, X, where X ± C n k ±. X H BX = J k is equivalent to X H +BX + = I k+, X H BX = I k, and X H +BX = 0. We have k + λ + i k λ i = inf trace(x H AX) (4.1) X H BX=J k = inf trace(x H + AX + ) + trace(x AX H ) X + H BX + =I k, X H + BX = I k X H + BX =0 inf trace(x H + AX + ) + trace(x AX H ) X+ H BX +=I k+, X HBX = I k = inf trace(x+ax H + ) + inf trace(x AX H ) (4.2) X+ H BX +=I k+ X HBX = I k = k + λ i + k λ i. 20

22 Therefore for the infimum in (4.1) to be attainable, both infimums in (4.2) must be attainable. We claim that when m 0 > 0, if k + > m +, inf X H + BX + =I k+ trace(x H +AX + ) is not attainable; similarly when m 0 > 0, if k > m, inf X H BX = I k trace(x AX H ) is not attainable. The claim implies the necessity of the condition m 0 = 0 or k ± m ± in the case of m 0 > 0. We shall consider the + case only; the other one is similar. Suppose that m 0 > 0 and k + > m + and assume to the contrary that there existed an X + C k + k + such that X+BX H + = I k+ and trace(x+ax H + ) = k + λ + i. Since X + is a global minimizer, by Theorem 2.1 there existed a Hermitian Λ + C k + k + such that AX + = BX + Λ +, X H +BX + = I k+. As a result, X+AX H + = Λ +. Let Λ + = U H ΩU be its eigendecomposition, where U is unitary, Ω = diag(ω 1,..., ω k+ ), and ω 1 ω k+. Write Y = X + U = (y 1,, y k+ ). We have AY = BY Ω, Y H BY = I k+, which implies ω i is an eigenvalue of A λb and y i is a corresponding eigenvector. Since k + λ + i = trace(x H +AX + ) = trace(y H AY ) = trace(ω) = k + ω i and λ + i ω i for 1 i k + by 5, Theorem 2.1, we have λ + i = ω i for 1 i k +. Let l = min{k +, m + + m 0 } and Y 1 = Y (:,1:l), the submatrix consisting the first l columns of Y. Since m 0 > 0 and k + > m +, l > m +. Y 1 is an eigenvector matrix associated with λ 0 with more than m + columns, and Y1 HB 1Y = I l, contradicting Lemma 4.1. Thus inf X H + BX + =I k+ trace(x+ax H + ) is not attainable if m 0 > 0 and k + > m +. 5 Conclusions Given a Hermitian matrix pencil A λb of order n, we are interested in when inf X H BX=J k trace(x H AX) (5.1) is finite, attainable, and what it is when it is finite. The same questions were investigated in detail with remarkable results by Kovač-Striko and Veselić 5 for the case when B is nonsingular. They suspected that their results would be true without the nonsingularity assumption on B but with A λb being regular. Our first contribution here is to confirm that indeed the nonsingularity assumption on B is not needed, but we also have gone further to allow the singular pencil into the picture. Our second contribution is a sufficient necessary condition for the attainability of the infimum in (5.1) in terms of certain indices in the canonical representation of the pencil. 21

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