Contour integral solutions of Sylvester-type matrix equations

Transcription

1 Contour integral solutions of Sylvester-type matrix equations Harald K. Wimmer Mathematisches Institut, Universität Würzburg, Würzburg, Germany Abstract The linear matrix equations AXB CXD = E, AX X D = E, AXB X = E are studied. In the case of uniqueness the solutions are expressed in terms of contour integrals. Keywords: Generalized Sylvester equation, -Sylvester equation, -Stein equation, contour integrals, matrix pencils. 200 MSC: 5A24, 5A22, 47A60.. Introduction In this note we use contour integrals to study the matrix equations AXB CXD = E () AX X D = E (2) AXB X = E. (3) We assume A, C C m m B, D C n n E C m n, m = n in (2) (3). The generalized Sylvester equation () appears in sensitivity problems of eigenvalues [2], in numerical solutions of implicit ordinary differential equations [6], in a Lyapunov theory of descriptor systems []. Equation (2) - where X denotes the conjugate transpose of X - plays a role in palindromic eigenvalue problems [2, 9]. The main reference to (3) address: wimmer@mathematik.uni-wuerzburg.de (Harald K. Wimmer) Preprint submitted to Linear Algebra its Applications December 3, 205

2 is [3]. Besides (2) (3) equations of the form AX X T D = E X CX T D = E - where X T is the transpose of X - have been studied e.g. in [5], [3] [3]. Conditions for uniqueness of solutions of () - (3) involve matrix pencils their spectra. Let the pencil zc A be regular, that is det(λc A) 0 for some λ C. We set σ(zc A) = {λ; det(λc A) = 0} {λ; det(c λ A) = 0}. Then 0 σ(zc A) is equivalent to det A = 0. By convention / = 0 /0 =. Thus σ(zc A) if only if det C = 0. A contour in the complex plane will always mean a positively oriented simple closed curve. 2. Explicit solutions by contour integrals 2.. The generalized Sylvester equation It is known (see [4]) that equation () has a unique solution if only if the matrix pencils zc A zb D are regular, (4) σ(zc A) σ(zb D) =. (5) In the following we give an explicit description of the unique solution of (). Theorem 2.. If the conditions (4) (5) are satisfied then there exists a contour such that σ(zc A) is in the interior of σ(zb D) is outside of. The unique solution of () is given by X = (zc A) E(zB D) dz. (6) Proof. Let T W be nonsingular complex n n matrices that transform the regular pencil zb D into Weierstrass canonical form (see [7, Section XII]). Then ( ) ( ) T (zb D)W = z B D Ir 0 D0 0 = z, 0 N 0 I n r 2

3 where N is nilpotent. We have σ(zb D) = σ(z B D). Set X = XT Ẽ = EW. Then X is a solution of () if only if X is a solution of A X B C X D = Ẽ, it is not difficult to see that we can assume that B D have the form ( ) ( ) Ir 0 D0 0 B = D =, 0 N 0 I n r where σ(n) = {0}. If the pencils zc A zb D are regular then (see [0], [8]) the identity () is equivalent to (zc A) CX XB(zB D) = (zc A) E(zB D). (7) Because of (5) we have / σ(zc A) σ(zb D). Suppose / σ(zc A), that is det C 0. Then (7) can be written as ( ) (zi C A) (zir D X X 0 ) 0 0 N(zN I n r ) = (zc A) E(zB D). (8) The assumption (5) implies that there exists a contour surrounding the set σ(c A) = σ(zc A) leaving σ(zb D) in the exterior. In particular, there is no eigenvalue of D 0 in the interior of. Note that (zn I n r ) is a polynomial matrix. Hence the spectral calculus for matrices yields (zi r D 0 ) dz = 0 (zi C A) dz = I n, (zn I n r ) dz = 0. Then (6) follows immediately from (8). Now suppose / σ(zb D). Then det B 0. In that case we consider the equation D T X T C T B T X T A T = E T represent X T in accordance with (6). We obtain X = (zc A) E(zB D) dz, β where β is a contour such that σ(zb D) is in the interior of β σ(zc A) is in the exterior of β. Let β denote the simple closed curve orientated in the opposite direction of β. Then we have (6) with = β. 3

4 The Sylvester equation AX XD = E (9) is a special case of (). Suppose σ(a) σ(c) = let be a contour with σ(a) in the interior σ(d) in the outside. Then (6) implies (see e.g. []) that X = (zi A) E(zI D) dz is the solution of (9). The following result on the discrete-time Sylvester equation (= Stein equation) is also a consequence of Theorem 2.. AXB X = E (0) Corollary 2.2. Suppose / σ(a)σ(b). Let be a contour with σ(a) in the interior {/λ; λ σ(b), λ 0} in the exterior. Then the unique solution of (0) is given by X = (zi A) E(zB I) dz The (X, X )-Sylvester equation In [2] [9, p. 223] it was shown that the (X, X )-Sylvester equation has a unique solution if only if the pencil AX X D = E () A zd is regular, (2) σ(a zd ) σ(d za ) =. (3) Condition (3) means that λ σ(a zd ) implies / λ / σ(a zd ). The following result gives a contour integral solution of (). Theorem 2.3. Let (2) (3) be satisfied. Then there exists a contour such that σ(d za ) is in the interior of σ(a zd ) is outside of. The unique solution of () is given by X = (zd A) (E + ze )(D za ) A dz. (4) 4

5 Proof. Note that () is equivalent to (A zd )X X (D za ) = E + ze. (5) Hence, if A zd is regular then (5) can be written as X(zA D) (zd A) X = (zd A) (E +ze )(D za ). (6) Suppose det A 0. Then (3) implies that there exists a contour such that σ(za D) = σ ( zi D(A ) ) is in the interior of σ(a zd ) is outside of. Then X(zA D) dz = X (A ) ( zi D (A ) ) dz = X (A ) (zd A) X dz = 0. Hence (4) follows from the identity (6). Suppose det A = 0. Then we obtain (4) by a perturbation argument. We consider the equation ÃX X D = E where à = A + ɛi, ɛ > 0. If ɛ is sufficiently small then det à 0, à zd is regular, σ(ã zd ) σ(d zã ) = The (X, X )-Stein equation Zhou, Lam Duan [3] investigated the equation AXB X = E. (7) Using Kronecker products the vec-permutation matrix the following result was proved in [3, p.390, Lemma 3]. Lemma 2.4. The condition σ(zi B A) σ(i zba ) =, or equivalently, / σ(ab )σ(a B), (8) is necessary sufficient for the existence of a unique solution of (7). We apply Corollary 2.2 to derive a contour integral solution of (7). 5

6 Theorem 2.5. Assume (8). Let be a contour that has the eigenvalues of B A in the interior the set {/ λ; λ σ(b A), λ 0} in the exterior. Then X = (zi B A) (E + B EA )(zba I) dz (9) is the unique solution of (7). Proof. Let W denote the right-h side of (9). Then W satisfies the Stein equation B AW BA W = E + B EA. (20) (by Corollary 2.2). It follows from [3, Theorem 7], which deals with equation (20), that W is a solution of (7). We can show this directly as follows. Define P = AW B Q = W. Then P = is the solution of (20) yields [ (zi AB ) A(E + B EA )B(zA B I) dz (AB )P (A B) P = A(E + B EA )B, (AB )Q(A B) Q = E + AE B. Set S = P Q E. Then (AB )S(A B) S = 0. Therefore (8) implies S = 0. Hence AW B W E = 0. References [] D. J. Bender, Lyapunov-like equations reachability/observability Gramians for descriptor systems, IEEE Trans. Automat. Control 32(4) (987), [2] R. Byers D. Kressner, Structured condition numbers for invariant subspaces, SIAM J. Matrix Anal. Appl. 28(2) (2006), [3] Ch.-Y. Chiang, A note on the -Stein matrix equation, Abstr. Appl. Anal. 203, Article ID 82464, 8 pages. 6

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