POSITIVE SEMIDEFINITE INTERVALS FOR MATRIX PENCILS

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1 POSITIVE SEMIDEFINITE INTERVALS FOR MATRIX PENCILS RICHARD J. CARON, HUIMING SONG, AND TIM TRAYNOR Abstract. Let A and E be real symmetric matrices. In this paper we are concerned with the determination of the values of t for which A + te is positive semidefinite. That is, we want to find the positive semidefinite interval for the matrix pencil t A + te. We first consider the cases when A is positive definite and when A is negative definite. In both of these cases, the positive semidefinite interval is determined from the eigenvalues of the matrix A 1 E. The results are extended to the case when A is semidefinite with range containing that of E and are combined to handle the indefinite case. 1. Introduction Let A and E be real symmetric n n matrices. The (linear) matrix pencil A + te is the function G : t A + te (t R). We consider the problem of determining the set T of those t for which G(t) is positive semidefinite. Since the set of positive semidefinite matrices is a closed convex cone in the space of n n matrices, T is a closed (possibly unbounded) interval. We call T the positive semidefinite interval. One source of interest in this problem is its connection to mathematical programming. Consider first the parametric quadratic program ([BC86], [Rit67], [Väl85]) min{c(t) x x C(t)x a i (t) x b i (t),i=1,,m}. If C(t) =A+tE, then the quadratic program is convex if and only if A+tE is positive semidefinite. The positive semidefinite interval gives the values of t for which the Karush-Kuhn-Tucker points are global minimizers. Consider also a semidefinite program ([WSV00], [VB96]) where the contraints require that the matrices n F j (x) :=A j 0 + x i A j i 0, j =1,,q i=1 Date: 21Feb05 4:16p. This research was supported by the National Sciences and Engineering Research Council. 1

2 2 RICHARD J. CARON, HUIMING SONG, AND TIM TRAYNOR be positive semidefinite. An application of a hit-and-run algorithm [BBRK + 87] for finding necessary constraints requires, at each iteration, the determination of T for each of the matrix pencils F j (ˆx + tv) =F j (ˆx)+t n i=1 v i A j i, j =1,,q where ˆx is the current iterate and v is a unit vector. If A is positive semidefinite and E is of rank one or two, then explicit expressions for the endpoints of T, and a method of computing them, are given in [CG86]. Valiaho ([Väl88]) extended these results, presenting a method of determining the inertia (the ordered triple of the numbers of positive, negative and zero eigenvalues) of A + te as a function of the parameter t. In this paper, motivated by applications to determination of constraint boundaries in semidefinite programs, we are only concerned with the determination of the endpoints of T. Section 2 gives some preliminary material. Section 3 discusses the case when A is nonsingular. Explicit expressions of the endpoints of the positive semidefinite interval are given when A is positive definite and when A is negative definite. In these cases, the endpoints are determined from the eigenvalues of A 1 E. Then we can combine these results to obtain the positive semidefinite interval when A is nonsingular and indefinite. Section 4 discusses the case when A is singular. Section 5 presents remarks on implementation and includes examples. 2. Preliminaries Throughout this paper, we shall denote the range space of a matrix M by R(M), its null space by N(M), and its rank by r(m). A real symmetric n n matrix M is positive definite (M 0), if x Mx > 0, for all x R n,x 0;M is positive semidefinite (M 0) if x Mx 0, for all x R n. It is well known that M is positive definite if and only if all eigenvalues of M are positive and M is positive semidefinite if and only if all eigenvalues of M are nonnegative. Since x Mx is continuous in x and affine in M, the set of positive definite matrices is an open convex set and the set of positive semidefinite matrices is a closed convex cone. It follows that the set T = {t R : G(t) 0} is an open interval, the positive definite interval (of G) and T = {t R : G(t) 0} =[t, t] is a closed interval, 1 the positive semidefinite interval. The only time T is the whole real line is if E = 0. Indeed, if x is an eigenvector of E with a non-zero eigenvalue, x x G(t)x is a straight line with a non-zero slope x Ex, so must be somewhere negative. Thus, except in the case E = 0, one of t and t is finite. If T is not empty, then its closure is T. Indeed, if (say) t 0 T and t is finite, then for each non-zero x in R n, x G(t 0 )x>0 and x G( t)x 0, so x G(t)x >0 for all t (t 0, t). 1 Here, and in the sequel, we will use [a, b] for {t R : a t b}. For example, if a = and b R, then [a, b] =(,b].

3 POSITIVE SEMIDEFINITE INTERVALS FOR MATRIX PENCILS 3 Given two real symmetric n n matrices A and E with A nonsingular, if det(a + te) =0, (1) then det( 1 t I A 1 E)=0; so 1 t is an eigenvalue of A 1 E. The solutions to equation (1), that is, the roots of the determining polynomial g(t) = det G(t), will determine the intervals T and T, as long as g is not identically 0. This is the case, for example, if one of A and E is non-singular. If g has no root, then T is either empty or the whole real line; if it has at least one root, then T must be bounded above or below or both. Let T 0 be the set of those roots t of the determining polynomial g for which G(t) 0. Clearly, T 0 consists of at most 2 points, the boundary points of T. Theorem 2.1. Suppose g 0. Then, the positive semidefinite interval T are determined as follows. (1) If T 0 =, then T =. (2) If T 0 consists of 2 points, then T is the closed interval they determine. (3) If T 0 consists of a single point t 0 and T is bounded, then T = T 0.In particular, this holds if t 0 is between 2 other roots of g, or any other 2 points a and b where G(a) and G(b) are not positive definite. (4) If T 0 consists of a single point t 0 and for some real a>t 0, and G(a) 0, then T =[t 0, + ). (5) If T 0 consists of a single point t 0 and for some real a<t 0, G(a) 0, then T =(,t 0 ]. If T is unbounded, then one of (4) or (5) holds (or E =0). Proof. The proof is straightforward. Consider, for example, (4). Then T is non-empty and T is its closure. Since there are no other possible endpoints, T =[t 0, + ). If E is positive definite, then A + te is positive definite for sufficiently large t. Lemma 2.2. Let E be positive definite. If max x =1 x ( A)x t> min x =1 x Ex, then A + te 0. Hence, T =[t, + ), where t is the largest root of g. Proof. Since E 0, for every x R n with x =1,x Ex > 0. Thus, for t> max x =1 x ( A)x min x =1 x Ex,

4 4 RICHARD J. CARON, HUIMING SONG, AND TIM TRAYNOR we have t min x =1 x Ex > max x =1 x ( A)x. Hence, for each x of norm 1, tx Ex > x ( A)x; that is, x (A + te)x >0; hence, A + te 0. Since T is unbounded on the right, it is non-empty of the form (t, + ) and T =[t, + ), its closure. Since T can contain no roots of g, t is the largest one. By writing A + te = A +( t)( E), we obtain the corresponding result for the case that E is negative definite. Lemma 2.3. Let E be negative definite. Then T =(, t ] and t is the smallest root of g. The eigenvalues λ i (t) ofg(t), (listed in descending order) are continuous functions of t. This follows, for example, from the Wielandt-Hoffman Theorem ([Wil65], page 104). From this it is clear that R is divided into intervals on which the inertia is constant. However, with our approach, we will have no need of this result. 3. The Positive Semidefinite Interval when A is Nonsingular In case A is nonsingular, the determining polynomial is not identically zero, so has finitely many roots, each negative inverses of eigenvalues of A 1 E. We present the results in three subsections, according to whether A is positive definite, negative definite, or indefinite A is positive definite. When A is positive definite, 0 T and T =[t, t] is its closure. Thus, t is either the smallest positive zero of g or is + ; t is either the largest negative root of g or is. Translating this into a statement about eigenvalues of A 1 E, we obtain: Theorem 3.1. Suppose that A is positive definite. Let δ n be the largest eigenvalue of A 1 E and δ 1 the smallest. The positive semidefinite interval for A + te is T =[t, t] where { 1 t = δ n, if δ n > 0, otherwise { t 1 = δ 1, if δ 1 < 0 +, otherwise. Special cases: If E is positive semidefinite, it is clear that T is not bounded above, that is, t =+. If E 0, we know therefore that T = [ 1/δ n, + ). If E is negative definite, then T =(, 1/δ 1 ], by Lemma 2.3.

5 POSITIVE SEMIDEFINITE INTERVALS FOR MATRIX PENCILS A is negative definite. From Lemmas 2.2 and 2.3 we know the form of T when E is positive definite or negative definite. If E is neither and A is negative definite, we find T is empty. Lemma 3.2. Suppose that A is negative definite. If E is singular or indefinite, then T =. Proof. If E is singular, there exists y R n, y 0, such that It follows that for every t, y Ey =0. y (A + te)y <0, so that A + te is not positive semidefinite. If E is indefinite, then there exist u, v R n, such that u Eu > 0, v Ev < 0. Then, for t 0, u (A + te)u <0; and for t>0, v (A + te)v <0, so again there is no t for which G(t) is positive definite: T =. In terms of the eigenvalues of A 1 E, we obtain the following complete description of the T in the case A is negative definite. Theorem 3.3. Suppose A is negative definite. Let δ n be the largest eigenvalue of A 1 E and δ 1 the smallest. Then the positive semidefinite interval T is given as follows. (1) T =[ 1 δ n, + ), ifδ n < 0, (2) T =(, 1 δ 1 ],ifδ 1 > 0, and (3) T =, otherwise. Proof. There are three mutually exclusive exhaustive subcases. If E 0, then by Lemma 2.2, T =[t, + ), where t is the largest root of g. Since A = G(0), is negative definite, G(t) is negative definite for all t 0; hence all the roots of g are positive, or equivalently, all the eigenvalues of A 1 E are negative. The largest root of g corresponds to the largest eigenvalue of A 1 E, so we obtain (1). Similarly, E 0 (negative definite) corresponds to all the roots of g positive that is, all the eigenvalues of A 1 E negative and, using Lemma 2.3, we obtain (2). In the remaining case, where E is neither positive definite, nor negative definite, T is empty, by Lemma 3.2

6 6 RICHARD J. CARON, HUIMING SONG, AND TIM TRAYNOR 3.3. A is nonsingular and indefinite. We now combine the results in Section 3.1 and Section 3.2 to obtain information about the positive semidefinite interval when A is nonsingular and indefinite. Let Q be an orthogonal n n matrix such that Q D1 0 AQ =, (2) 0 D 2 where D 1,D 2 are diagonal and positive definite. Let Q E1 E EQ = 3 E3, E 2 where E 1 and E 2 have the same sizes as D 1 and D 2. Then Q D1 + te (A + te)q = 1 te 3 te3. D 2 + te 2 By Theorem 3.1 and Theorem 3.3, we can determine the positive semidefinite intervals for D 1 + te 1 and D 2 + te 2. We denote them by T 1 and T 2 and obtain: Theorem 3.4. T T 1 T 2 and if E 3 =0, T = T 1 T 2. If E 3 is not a zero matrix, then we may need to use brute force, relying on Theorem 2.1. Theorem 3.4 helps, since we know the endpoints (elements of T 0 ) must be in T 1 T 2. For example, we could first determine T 1 and T 2, then search inside T 1 T 2, for roots of the determining polynomial, stopping when 2 such are found or when one such is between points where G(t) is not positive definite. 4. The Positive Semidefinite Interval when A is Singular In this section, we are concerned with the positive semidefinite interval for the matrix pencil A + te when A is singular. Choose an orthogonal matrix Q for which Q AQ = D 1 0 D 2 0, (3) 0 where D 1,D 2 are diagonal and positive definite, extending the notation of the diagonalization (2). The corresponding decomposition of Q EQ has an extra column and row of blocks. E 1 E 3 E 4 Q EQ = E3 E 2 E 5. E4 E5 E 6 Let us write these as Q A1 0 AQ = and Q Ē1 Ē EQ = 2 Ē2 (4) E 6

7 POSITIVE SEMIDEFINITE INTERVALS FOR MATRIX PENCILS 7 Then Q A1 + tē1 tē2 (A + te)q =. tē 2 te 6 D1 0 Since A 1 = is nonsingular, the positive semidefinite interval 0 D 2 for the matrix pencil G 1 : t A 1 + tē1 can be found using Theorem 3.1, Theorem 3.3, or Theorem 3.4. Let T 3 denote the positive semidefinite interval for the matrix pencil te 6. Then, (1) E 6 =0 = T 3 = R. (2) E 6 0,E 6 0 = T 3 =[0, + ). (3) E 6 0,E 6 0 = T 3 =(, 0]. (4) E 6 is indefinite = T 3 = {0}. Note that if G(t) =A + te is positive semidefinite, then both A 1 + tē1 and te 6 are positive semidefinite. However, in general, the intersection of their positive semidefinite intervals are not that of G Example 4.1. Let A = and E =. Then, the positive semidefinite interval for G is T = {0}. But those of G 1 and t te 6 are both 1 0 (, + ) The case R(E) R(A). We now restrict to the case R(E) R(A), or what is the same, by the symmetry of A, N(A) N(E). Lemma 4.2. R(E) R(A) if and only if there exists X (necessarily unique) with R(X) R(A) and AX = E. X1 0 With the notation of equation (4), this X is given by Q Q, where X 1 is A 1 1 Ē 1. Proof. First, R(E) R(A) if and only if for some X, AX = E. (5) We need to show X can be chosen so that also R(X) R(A). Using the diagonalization (4), we can write (5) as A1 0 X1 X 4 Ē1 Ē = 2 X 3 X 2 Ē2, E 6 where Q X1 X XQ = 4. X 3 X 2 A 1 1 Thus, Ē 2 = 0 and E 6 = 0 and for such X, X 1 = A 1 1 Ē 2 = 0; hence, A1 0 X1 0 = X 3 X 2 ] [Ē1 0. Ē 1, and X 4 =

8 8 RICHARD J. CARON, HUIMING SONG, AND TIM TRAYNOR Changing X 3 and X 2 arbitrarily will not destroy (5). By an argument similar to that for E, we see that R(X) R(A) if and only if also X 3 =0 and [ X 4 = ] 0. Thus, the unique X with AX = E and R(X) R(A) is X1 0 Q Q, where X 1 is A 1 1 Ē 1. It follows from Lemma 4.2 that, for the case R(E) R(A), ] [Ē1 Q 0 EQ =. Hence, Q (A + te)q = A1 + tē1 0 ; that is, Q G1 (t) 0 G(t)Q =. Since G(t) 0 if and only if G 1 (t) 0, we have: Theorem 4.3. If R(E) R(A), the positive semidefinite interval for G is that of G 1. Since A 1 is nonsingular, we can use Theorem 3.1, Theorem 3.3, or Theorem 3.4 to determine its interval. The determining polynomial becomes g 1 (t) = det(g 1 (t)). X1 0 Note, however, that the nonzero eigenvalues of X = Q Q, are the same as those of X 1 = A 1 1 Ē 1. The largest eigenvalue of X is positive iff that of X 1 is, etc. Hence, for example, Theorem 3.1 extends to the case A positive semidefinite. Corollary 4.4. Let A be positive semidefinite, R(E) R(A) and let X be the unique solution of AX = E for which R(X) R(E). If δ 1 is the smallest eigenvalue of X and δ n the largest then, T =[t, t], where { 1 t = δ n, if δ n > 0, otherwise { t 1 = δ 1, if δ 1 < 0 +, otherwise. The corresponding result for A negative semidefinite is a little more awkward to state because of the possibility of zero eigenvalues. If the rank of E is smaller than that of A, then T is empty. Otherwise, we can use the non-zero eigenvalues of X to determine the interval T Theorem 3.3.

9 POSITIVE SEMIDEFINITE INTERVALS FOR MATRIX PENCILS 9 5. Implementation and Examples Our results indicate that the calculation of T can be accomplished with the orthogonal diagonalization of A and the calculation of the eigenvalues of X = A 1 E. In practice, the diagonalization is often not needed. When A is positive semidefinite, we first determine whether or not R(E) R(A). This can be done by checking the consistency of AX = E. One approach is to first determine a decomposition of A using Choleski factorization with symmetric pivoting ([DMBS79]). If r(a) = r, there exists a nonunique permutation matrix P and a triangular matrix R (unique for a given P ) such that P AP = R R, where R11 R R = 12, and where R 11 is a nonsingular upper triangular r r matrix and R 12 is an r (n r) matrix. For simplicity, we assume that P is the identity matrix. Now, AX = E is equivalent to R RX = E. Set Y = RX. We first solve R Y = E for the matrix Y. Set Y11 Y Y = 12 E11 E and E = 12 Y 21 Y 22 E12. E 22 Then we can write R Y = E as R 11 0 Y11 Y 12 E11 E R12 = 12 0 Y 21 Y 22 E12. E 22 Since R 11 is nonsingular, Y 11 and Y 12 are uniquely determined by R 11Y 11 = E 11 and R 11Y 12 = E 12. The equation AX = E is consistent only if Y 11 and Y 12 satisfy R 12Y 11 = E 12 and R 12Y 12 = E 22. The equation Y = RX can be written as [ R11 R 12 ] X11 X 12 = X 21 X 22 Y11 Y 12. Y 21 Y 22 This is consistent if and only if Y 21 = 0 and Y 22 = 0 and then all solutions of AX = E are determined by R 11 X 11 = Y 11 R 12 X 12 and R 11 X 21 = Y 12 R 12 X 22. (6) In section 4, we showed that X 21 and X 22 can be chosen so that the solution X satisfies R(X) R(A). To do this, we can again use the Choleski factorization, obtaining X 12 = R12(R 11 + R11 1 R 12R12)R 11 2 E 11 X 22 = R 12(R 11 + R 1 11 R 12R 12)R 2 11 E 12 (7)

10 10 RICHARD J. CARON, HUIMING SONG, AND TIM TRAYNOR Example 5.1. Let A = and E = The Choleski decomposition of A yields R11 R R = 12 = We solve the equation R11 Y 11 = E 11 and R11 Y 12 = E 12 obtaining 1/2 0 1/2 Y 11 = and Y 0 1/3 12 =. 1/3 Observe that R12 Y 11 = E12 and R 12 Y 12 = E 22 ; hence, have R(E) R(A). Set Y 21 = 0 and Y 22 = 0 for consistency. Solving for the X with R(X) R(A), as in (6) and (7), we obtain 1/6 1/27 11/54 X = 1/12 2/27 17/108, 1/12 1/27 5/108 with eigenvalues 1/4, 1/9, 0 and the positive semidefinite interval is [ 4, 9], obtained from the negative inverses of those which are non-zero. Example 5.2. Consider now the matrices A = and E = , in which A is indefinite. In the notation of section 3.3, the matrix pencils D 1 + te 1 = + t and D te 2 =( 1) + t(2) have positive semidefinite intervals T 1 =[ 1, 1] and T 2 =[1/2, + ), yielding T 1 T 2 =[1/2, 1]. The eigenvalues δ i of A 1 E are δ 1 = 1, δ 2 = , δ 3 = , so the corresponding roots t i = 1/δ i of the determining polynomial g are t 1 =1,t 2 = and t 3 = Thus, T 0 = {1, } and the positive semidefinite interval for A + te is T =[( 1+ 5)/2, 1].

11 POSITIVE SEMIDEFINITE INTERVALS FOR MATRIX PENCILS 11 References [BBRK + 87] H. C. P. Berbee, C. G. E. Boender, A. H. G. Rinnooy Kan, C. L. Scheffer, R. L. Smith, and J. Telgen, Hit-and-run algorithms for the identification of nonredundant linear inequalities, Math. Programming 37 (1987), no. 2, [BC86] M. J. Best and R. J. Caron, A parameterized Hessian quadratic programming problem, Ann. Oper. Res. 5 (1986), no. 1-4, [CG86] R. J. Caron and N. I. M. Gould, Finding a positive semidefinite interval for a parametric matrix, Linear Algebra Appl. 76 (1986), [DMBS79] J. J. Dongarra, C. B. Moler, J. R. Bunch, and G. W. Stewart, LINPACK users guide, Society for Industrial and Applied Mathematics, Philadelphia, [Rit67] Klaus Ritter, A method for solving nonlinear maximum-problems depending on parameters, Naval Res. Logist. Quart. 14 (1967), [Väl85] Hannu Väliaho, A unified approach to one-parametric general quadratic programming, Math. Programming 33 (1985), no. 3, [Väl88] H. Väliaho, Determining the inertia of matrix pencil as a function of the parameter, Linear Algebra Appl. 106 (1988), [VB96] L. Vandenberghe and S. Boyd, Semidefinite programming, SIAM Review 38 (1996), [Wil65] J. H. Wilkinson, The algebraic eigenvalue problem, Clarendon Press, Oxford, MR 32 #1894 [WSV00] Henry Wolkowicz, Romesh Saigal, and Lieven Vandenberghe (eds.), Handbook of semidefinite programming, International Series in Operations Research & Management Science, 27, Kluwer Academic Publishers, Boston, MA, Department of Mathematics and Statistics, University of Windsor, Windsor, Ontario, N9B 3P4, Canada.

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