Solving the Sylvester Equation AX-XB=C when $\sigma(a)\cap\sigma(b)\neq\emptyset$

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1 Electronic Journal of Linear Algebra Volume 35 Volume 35 Article Solving the Sylvester Equation AX-XBC when $\sigma(a)\cap\sigma(b)\neq\emptyset$ Nebojša Č. Dinčić Faculty of Sciences and Mathematics, University of Niš, ndincic@hotmail.com Follow this and additional works at: Part of the Algebra Commons Recommended Citation Dinčić, Nebojša Č.. (219), "Solving the Sylvester Equation AX-XBC when $\sigma(a)\cap\sigma(b)\neq\emptyset$", Electronic Journal of Linear Algebra, Volume 35, pp DOI: This Article is brought to you for free and open access by Wyoming Scholars Repository. It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository. For more information, please contact scholcom@uwyo.edu.

2 SOLVING THE SYLVESTER EQUATION AX XB C WHEN σ(a) σ(b) NEBOJŠA Č. DINČIĆ Abstract. The method for solving the Sylvester equation AX XB C in the complex matrix case, when σ(a) σ(b), by using Jordan normal form, is given. Also, the approach via the Schur decomposition is presented. Key words. Sylvester equation, Jordan normal form, Schur decomposition. AMS subject classifications. 15A An introduction. The Sylvester equation (1.1) AX XB C, where A C m m, B C n n and C C m n are given, and X C m n is an unknown matrix, is a very popular topic in linear algebra, with numerous applications in control theory, signal processing, image restoration, engineering, solving ordinary and partial differential equations, etc. It is named for famous mathematician J. J. Sylvester, who was the first to prove that this equation in matrix case has unique solution if and only if σ(a) σ(b) 17. One important special case of the Sylvester equation is the continuous-time Lyapunov matrix equation (AX + XA C). A C In 1952, Roth 15 proved that for operators A, B on finite-dimensional spaces is similar to B A if and only if the equation AX XB C has a solution X. Rosenblum 14 proved that if A B and B are operators such that σ(a) σ(b), then the equation AX XB Y has a unique solution X for every operator Y. If we define the linear operator T on the space of operators by T (X) AX XB, the conclusion of the Rosenblum theorem can be restated: T is invertible if σ(a) σ(b). Kleinecke showed that when A and B are operators on the same space and σ(a) σ(b), then the operator T is not invertible, see 14. For the exhaustive survey on these topics, please see 2 and 18 and the references therein. In recent paper of Li and Zhou, an extensive review of literature can be found, with a brief classification of existing methods for solving the Sylvester equation. Drazin 6 recently gave another equivalent condition. Theorem For any field F, any r, s N and any square matrices A M r (F ), B M s (F ) whose eigenvalues all lie in F, the following three properties of the pair A, B are equivalent: i) for any given polynomials f, g F t, there exists h F t such that f(a) h(a) and g(b) h(b); ii) A and B share no eigenvalue in common; iii) for r s matrices X over F, AX XB X. Received by the editors on January 1, 218. Accepted for publication on December 2, 218. Handling Editor: Froilán Dopico. Faculty of Sciences and Mathematics, University of Niš, PO Box 224, 18 Niš, Serbia (ndincic@hotmail.com). The author was supported by the Ministry of Science, Republic of Serbia, grant no

3 Nebojša Č. Dinčić 2 We recall some results where the solution to the Sylvester equation is given in various forms. Proposition 1.2. (2, p. 9) Let A and B be operators such that σ(b) {z : z < ρ} and σ(a) {z : z > ρ} for some ρ >. Then the solution of the equation AX XB Y is X A Y B n. n We will often use the following special form of the previous result. Proposition 1.3. In the complex matrix case, the Sylvester equation AX XB C is consistent if and only if σ(a) σ(b) and the solution is given by for invertible A, and by for invertible matrix B. X A 1 X A k CB k A k CB k B 1 Proposition Let A and B be operators whose spectra are contained in the open right half plane and the open left half plane, respectively. Then the solution of the equation AX XB Y can be expressed as X e ta Y e tb dt. Proposition Let Γ be a union of closed contours in the plane, with total winding numbers 1 around σ(a) and around σ(b). Then the solution of the equation AX XB Y can be expressed as X 1 (A ζ) 1 Y (B ζ) 1 dζ. 2πi Γ Solving the Sylvester equation in the case when σ(a) σ(b) is rather complicated and is not thoroughly studied yet. According to 3, only Ma 13, and Datta and Datta 4 treated the particular cases when B( A) has only simple eigenvalues or C and B( A T ) are unreduced Hessenberg matrices, respectively. In the paper 3, two special cases, when both A and B are symmetric or skew-symmetric matrices, were considered, by using the notion of the eigenprojection. In 11, it is shown that in the case σ(a) σ(b) (A R m m, B R n n ), the solution of the Sylvester equation is X q(a) 1 η(a, C, B), which is a polynomial in the matrices A, B and C. Also, it is shown that when σ(a) σ(b), the solution set is contained in that of the linear algebraic equation q(a)x η(a, C, B). Recall that q(s) s n + β s + + β 1 s + β is the characteristic polynomial of B and η(a, C, B) n k1 β kη(k 1, A, C, B), where η(k, A, C, B) k i Ak i CB i. In 5, the equation AX X B is considered by using the Kronecker canonical form of the matrix pencil A + λb, where X denotes either transpose or the conjugate transpose of X. Recently, Li and Zhou used the method based on spectral decompositions of the matrices A and B. Note that this method also works in the case when spectra of A and B are not disjoint.

4 3 Solving the Sylvester Equation AX XB C When σ(a) σ(b) In the paper 18, homogeneous and nonhomogeneous Sylvester-conjugate matrix equation of the forms AX + BY XF and AX + BY XF + R are considered. In this paper, we are dealing with the Sylvester equation in complex matrix case, when σ(a) σ(b). Because the operator T (X) AX XB is not invertible in this case, as Kleinecke showed, we must find the consistency condition, and general solution in the case of consistency. We will use the Jordan normal form for the matrices A and B (similar method as Gantmacher used in 7, Ch. VIII, but for homogeneous case only!), so the Sylvester equation will become the set of the simpler equations of the form J m (λ)z ZJ n (µ) W, where W C m n is given matrix, Z C m n is unknown matrix, and λ, µ are complex numbers. In the paper, we are dealing mainly with such simpler equations, discussing the consistency condition and describing the algorithm for finding the general solution. Finally, we give the result characterizing the most general case, as well as the approach via Schur decomposition. 2. Reducing the problem to the most elementary equations. Let the matrices A C m m and B C n n be given, and suppose that W σ(a) σ(b) {λ 1,..., λ s }. It is a well-known fact (see e.g., 1) that any square complex matrix can be reduced to the Jordan canonical form. Therefore, there exist regular matrices S C m m and T C n n such that A SJ A S 1 and B T J B T 1, or, more precisely, JA1 (2.2) A S J A2 S 1, B T JB1 J B2 T 1, where J A1 (respectively, J B1 ) consists of just those Jordan matrices corresponding to the eigenvalues from the σ(a) \ W (respectively, σ(b) \ W ), and J A2 and J B2 are those Jordan matrices corresponding to the eigenvalues from the set W : J A2 diag{j(λ 1 ; p 11, p,..., p 1,k1 ),..., J(λ s ; p s1, p s2,..., p s,ks )}, J B2 diag{j(λ 1 ; q 11, q,..., q 1,l1 ),..., J(λ s ; q s1, q s2,..., q s,ls )}. Here p ij, j 1, k i, i 1, s, and q ij, j 1, l i, i 1, s, are natural numbers, and k i and l i are geometric multiplicities of the eigenvalue λ i, i 1, s, of A and B, respectively; their algebraic multiplicities are given by m i p i1 + p i2 + + p i,ki and n i q i1 + q i2 + + q i,li, where i 1, s. We will use the notation J(λ; t 1,..., t k ) diag{j t1 (λ),..., J tk (λ)} J t1 (λ) J tk (λ), where J ti (λ), i 1, k, is the Jordan block J ti (λ). λ 1 λ λ 1 λ Recall that a matrix A is called non-derogatory if every eigenvalue of A has geometric multiplicity 1, if and only if corresponding to each distinct eigenvalue is exactly one Jordan block 1, p If we substitute (2.2) in (1.1), and denote S 1 XT Y and S 1 CT D, the equation becomes JA1 JB1 (2.3) Y Y D. J A2 J B2 t i t i.

5 Nebojša Č. Dinčić 4 We will partition the matrices Y Y ij 2 2 and D D ij 2 2 in accordance with the partition of J A and J B. Now (2.3) becomes: JA1 J A2 Y11 Y Y 21 Y 22 Y11 Y Y 21 Y 22 which reduces to the following four simpler Sylvester equations: JB1 J A1 Y 11 Y 11 J B1 D 11, J A1 Y Y J B2 D, J A2 Y 21 Y 21 J B1 D 21, J A2 Y 22 Y 22 J B2 D 22. J B2 D11 D D 21 D 22 By Proposition 1.3, the first three equations have the unique solutions, because σ(a 1 ) σ(b 1 ), σ(a 1 ) σ(b 2 ), σ(a 2 ) σ(b 1 ), given by the appropriate power series. The fourth equation does not have a unique solution (when it is consistent), because σ(a 2 ) σ(b 2 ) W, so we turn our attention to this case. Therefore, throughout the paper, we can, without loss of generality, consider the Sylvester equation in the form AX XB C, where A and B are square complex matrices (in general of different size) already reduced to their Jordan forms and such that σ(a) σ(b). If we partition X X ij s s and C C ij s s in accordance with already given decompositions for A and B, and put them in the equation AX XB C, we have s 2 simpler Sylvester equations of the form: (2.4) J(λ i ; p i1,..., p i,ki )X ij X ij J(λ j ; q j1,..., q j,lj ) C ij, i, j 1, s. For i j we have σ(j(λ i ; p i1,..., p i,ki )) {λ i } {λ j } σ(j(λ j ; q j1,..., q j,lj )), and by Proposition 1.3, this case also has the unique solution. Therefore, among s 2 equations in (2.4) there are s 2 s of them which are uniquely solvable, and remaining s equations (for i j) after appropriate translation reduce to s i1 (k i l i ) simple Sylvester equations of the form, J pi,u ()X (ii) u,v X (ii) u,v J qi,v () C (ii) u,v, u 1, k i, v 1, l i, i 1, s. Because of that, we will investigate this important particular case in detail Solving the equation J m ()X XJ n () C. From the previous consideration, we see that Jordan matrices play important role in the paper. Recall that multiplying some matrix C C m n by the Jordan block (or transposed Jordan block) corresponding to acts like shifting the rows (or columns) and inserting a row (or a column) of all zeros where it is necessary: J m ()C shifts up, CJ n () shifts to the right, J T m()c shifts down, while CJ T n () shifts to the left. More precisely, the following Lemma holds. Lemma 2.1. For given matrix C c ij C m n we have: { ci+1,j, i 1, m 1, i) J m ()C ij, i m. { ci,j 1, j 2, n, ii) CJ n () ij, j 1. { iii) J m () T ci 1,j, i 2, m, C ij, i 1.

6 5 Solving the Sylvester Equation AX XB C When σ(a) σ(b) iv) CJ n () T ij { ci,j+1, j 1, n 1,, j n. Moreover, for α, m 1 and β, n 1, { v) J m () α CJ n () β ci+α,j β, i 1, m α and j β + 1, n, ij, i m α + 1, m or j 1, β. { vi) (J m () T ) α CJ n () β ci α,j β, i α + 1, m and j β + 1, n, ij, i 1, α or j 1, β. Example 2.2. We consider the case A J 4 (), B J 3 (), i.e., the equation (2.5) J 4 ()X XJ 3 () C. Because of σ(a) σ(b) {} we expect no unique solution. Let us see for which C there are solutions, and then let us characterize all of them. The matrix equation (2.5) (with C c ij, X x ij C 4 3 ) gives us the following linear system: x 21 c 11, x 22 x 11 c, x 23 x c 13, x 31 c 21, x 32 x 21 c 22, x 33 x 22 c 23, x 41 c 31, x 42 x 31 c 32, x 43 x 32 c 33, c 41, x 41 c 42, x 42 c 43. We see that some conditions must be imposed to the entries of the matrix C: c 41, c 31 + c 42, c 21 + c 32 + c 43. Therefore, any matrix C for which the equation (2.5) is consistent is of the form: The solutions of (2.5) are X C c 11 c c 13 c 21 c 22 c 23 c 31 c 32 c 33 c 31 c 21 c 32. x 11 x x 13 c 11 x 11 + c x + c 13 c 21 c 11 + c 22 x 11 + c + c 23 c 31 c 21 + c 32 c 11 + c 22 + c 33 where x 11, x, x 13 are arbitrary complex numbers. We can rewrite this general solution X in the following more informative form: x 11 x x 13 x 11 x x 11 + c 11 c c 13 c 21 c 22 c c 31 c 32 c 33 c 11 c c 21 c 22, c 11 By Lemma 2.1, we can express this as: p 2 (J 3 ()) X J 4 () T ( J4 () T ) k CJ3 () k, where p 2 (J 3 ()) x 11 I 3 + x J 3 () + x 13 J 3 () 2 is arbitrary polynomial of the matrix J 3 (), with complex coefficients in general, and the degree at most 2.

7 Nebojša Č. Dinčić 6 This example motivates us to prove general result concerning this topic. Theorem 2.3. The Sylvester equation (2.6) J m ()X XJ n () C, when m n, is consistent if and only if (2.7) or, equivalently, (2.8) J m () m 1 k CJ n () k, p 1 c m k,p k, p 1, n, and its general solution (which depends on n complex parameters) is given as p (J n ()) (2.9) X + J m () T ( Jm () T ) k CJn () k, (m n) n where p is an arbitrary polynomial of the degree at most n 1. Proof. First, we show that the conditions (2.7) and (2.8) are equivalent. Let us denote S S k where S k J m () m 1 k CJ n () k, k, n 1. By Lemma 2.1.v), for α m k 1 and β k, we have { ci+m 1 k,j k, i 1, k + 1 and j k + 1, n, S k ij, i k + 2, m or j 1, k. It is easy to see that S k ij for i > j and S k ij S k+1 i+1,j+1 ; therefore, it is enough to consider only the first row (or n th column). We have { cm k,j k, j k + 1, n, S k 1j, j 1, k. Therefore, j 1 S 1j S k 1j c m k,j k, k 1, n. Note that the summation index is changed to k, j 1 because of k + 1 j n and k n 1. Hence, we proved (2.7) (2.8). ( ) : Suppose that the equation (2.6) is consistent, which means for some Ĉ there is X such that J m () X XJ n () Ĉ. We have: J m () m 1 k ĈJ n () k J m () m 1 k (J m () X XJ n ())J n () k J m () m k XJn () k J m () m 1 k XJn () k+1 J m () m X + Jm () m 1 XJn () + + J m () m n+1 XJn () J m () m 1 XJn () J m () m n+1 XJn () J m () m n XJn () n J m () m X Jm () m n XJn () n,

8 7 Solving the Sylvester Equation AX XB C When σ(a) σ(b) so the consistency condition (2.7) holds. ( ) : Now we prove, by immediate checking, that X given by (2.9) is indeed the solution under the consistency condition (2.7). Because of p (J n ()) p (J n ()) J m () J n () (m n) n (m n) n Jn () p (J n ()) p (J n ()) J n () J m n () (m n) n (m n) n J n () p (J n ()) p (J n ()) J n () m n, (m n) n (m n) n (by we denoted some submatrix which is not important now) we have: J m ()X XJ n () C J m ()J m () T ( Jm () T ) k CJn () k J m () T ( Jm () T ) k CJn () k J n () C J m ()J m () T C + ( J m ()(J m () T ) 2 J m () T ) CJ n () + + ( J m ()(J m () T ) n (J m () T ) ) CJn () ( J m ()J m () T ) ( ) I m (J m () T ) k CJ n () k + C C. Indeed, let us denote T k (J m () T ) k CJ n () k, k, n 1, and T T k. By Lemma 2.1.vi), for α β k, we have: T k ij { ci k,j k, i k + 1, m and j k + 1, n,, i 1, k or j 1, k. If we obtain that all entries in the m th row of matrix T are zero, we have the proof. We proceed: { T k mj (J m () T ) k CJ n () k cm k,j k, j k + 1, n, mj, j 1, k, and therefore, j 1 T mj T k mj c m k,j k, j 1, n. Note that again the summation index is changed to k, j 1 because of k + 1 j n and k n 1. The condition (2.7), or equivalently (2.8), ensures that T mj, j 1, n, so X given by (2.9) is, under the condition (2.7), or equivalently (2.8), indeed the solution, so the equation (2.6) is consistent. Remark 2.4. For the given matrix C C m n, m n, the set of its entries c ij such that i j m p for given p 1, n will be called p th small subdiagonal of matrix C. The condition (2.8) actually means that

9 Nebojša Č. Dinčić 8 the sum over each of n small subdiagonals must be zero, which is easy to check. However, for constructing the solutions of the more general Sylvester equations from the most simple ones, the condition (2.7) is more appropriate. We remark that the particular solution in (2.9), X p J m () T ( Jm () T ) k CJn () k, is completely determined by the matrix C, more precisely, by the independent part of C (upper (m 1) n submatrix), and it can be expressed in equivalent, but for computational purposes more practical form as:, i 1, X p ij min{i 1,j} 1 c i 1 k,j k, i 2, m. Let us step away a bit, and consider solving the equation (2.6) by the well-known Kronecker product method. We recall the vectorization operator vec given by vec : C m n C mn, vec(a) vec(a ij ) a 1 a 2 a n T, where a k denotes the k th column of the matrix A. Also, recall that the Kronecker product of A C m n and B C p q is mp nq block matrix A B a ij B m n. If we solve the equation J m ()X XJ n () C, m n, by using the well-known method with the operator vec and the Kronecker product, we have (2.1) (I n J m () J n () T I m ) vec(x) vec(c); this linear system is consistent if and only if r(i n J m () J n () T I m, C) r(i n J m () J n () T I m ). The system (2.1) in matrix form (zero submatrices are omitted) is: J m () I m J m () I m I m J m () I m J m () X 1 X 2. X n C 1 C 2. C n, i.e., J m ()X 1 C 1, J m ()X 2 C 2 + X 1,..., J m ()X n C n + X ().

10 9 Solving the Sylvester Equation AX XB C When σ(a) σ(b) We can infer the consistency condition in the following way: J m () m X n J m () m 1 (J m ()X n ) J m () m 1 (C n + X () ) n J m () m +k C k. k1 We shall not pursue this approach further, because the obtained matrix of the system is an mn mn sparse matrix, and for the consistency of the equation we need to check rank conditions and to use some generalized inverses. For Example 2.2, where m 4, n 3, the previous formula gives: 3 J 4 () k C k k1 c 21 c 31 c 41 + c 32 c 42 + c 43, which is in accordance with the conclusion from Example 2.2. We return to the main flow of the paper. Theorem 2.5. The Sylvester equation (2.11) J m ()X XJ n () C, when m n, is consistent if and only if (2.) m 1 J m () k CJ n () k, and its general solution is (2.13) X m (n m) q m 1 (J m ()) m 1 J m () k C ( J n () T ) k Jn () T, where q m 1 is an arbitrary polynomial of the degree at most m 1. Proof. If we take transpose of the original equation (2.11), and then multiply it by 1, we have J n () T X T X T J m () T C T, which is similar, but not the same as the equation (2.6), so we cannot directly apply Theorem 2.3. But if we notice that for any k N 1 1 J k () T P 1 k J k()p k, where P k P 1 k k k

11 Nebojša Č. Dinčić 1 is so-called exchange matrix, the equation becomes P n J n () T X T P m P n X T J m () T P m P n C T P m, P n J n () T P n P n X T P m P n X T P m P m J m () T P m P n C T P m, J n ()Y Y J m () D, where P n X T P m Y and P n C T P m D. Only now we can apply Theorem 2.3 to the equation J n ()Y Y J m () D. We have the consistency condition (2.) because of m 1 m 1 J n () n k 1 DJ m () k J n () k P n C T P m J m () k ( m 1 ) P n (P n J n ()P n ) k C T (P m J m ()P m ) k P m ( m 1 ) P n (J n () T ) k C T (J m () T ) k P m ( m 1 P n J m () k CJ n () n k 1 ) T P m. General solution of J n ()Y Y J m () D is, by Theorem 2.3: p m 1 (J m ()) ( Y + Jn () T ) k+1 DJm () k. (n m) m By the substitutions X (P n Y P m ) T, D P n C T P m, we have the solution (2.13): X P m p m 1 (J m () T ) P m p m 1 (J m () T ) m (n m) m 1 m (n m) P n + P m (J m () T ) k D T J n () k+1 P n P m P n m m 1 P m (J m () T ) k (P m CP n )J n () k+1 P n m 1 P m m (n m) p m 1 (J m () T )P m (P m J m () T P m ) k C(P n J n ()P n ) k+1 m 1 m (n m) p m 1 (J m ()) J m () k C(J n () T ) k+1. The condition (2.) actually means that the sum of elements over each of m small subdiagonals (c ij, i j p 1, where p 1, m) must be zero. We also remark that the particular solution in (2.13), m 1 X p J m () k C ( J n () T ) k Jn () T, is completely determined by the matrix C, more precisely, by the independent part of C (rightmost m (n 1) submatrix), and it can be expressed in equivalent, but for computational purposes, a more

12 11 Solving the Sylvester Equation AX XB C When σ(a) σ(b) practical form as: X p ij min{m i, j} Corollary 2.6. The homogeneous Sylvester equation is consistent and its general solution is c i+k,j+1+k, j 1, n 1,, j n. J m ()X XJ n () p (J n ()), m n, X (m n) n m (n m) p m 1 (J m ()), m n. where p is an arbitrary polynomial of the degree at most n 1. The rank of the solution X is given by r(x) r(p s 1 (J s ())) s min,s 1 {k : p (k) () }, s min{m, n}. Corollary 2.7. All matrices commuting with J m () are given by X p m 1 (J m ()). Any polynomial such that p m 1 () gives a nonsingular X. Also s 1 σ(x) {p m 1 ()}. From Theorem 2.3 and Theorem 2.5 in the case m n, we have the following result. Corollary 2.8. The Sylvester equation is consistent if and only if and its general solution is J n ()X XJ n () C, J n () k CJ n () k J n () k CJ n () k, n 2 X p (J n ()) J n () k C ( J n () T ) k+1 n 2 ( q (J n ()) + Jn () T ) k+1 CJn () k, where p and q are arbitrary polynomials of the degree at most n Solving the equation J m (λ)x XJ n (µ) C for λ, µ C. If we observe that J k (λ) λi k + J k (), k N, the equation (2.14) J m (λ)x XJ n (µ) C

13 Nebojša Č. Dinčić becomes (2.15) (λ µ)x + J m ()X XJ n () C. Case λ µ: Reduces to already solved equation (2.6) or (2.11). Case λ µ: Suppose, without loss of generality, that λ. By Proposition 1.3, we have the solution X J m (λ) 1 J m (λ) k CJ n (µ) k. We will try to avoid this infinite sum, as well as finding all k th powers of the inverse of a Jordan matrix. If we rewrite (2.15) as (J m () + (λ µ)i m ) X XJ n () C, i.e., J m (λ µ)x XJ n () C, J m (λ µ) is nonsingular, so we have the unique solution by Proposition 1.3. Theorem 2.9. The Sylvester equation (2.14) for λ µ has the unique solution X (J m (λ µ)) (k+1) CJ n () k m 1 J m () k C(J n (µ λ)) (k+1). Remark 2.1. Finding the inverses of the powers of nonsingular Jordan blocks appearing in previous theorem is not hard. Indeed, by using some matrix functional calculus, it is easy to prove that m 1 ( ) k + s 1 J m (α) k ( 1) s α (k+s) J m () s, k N, α. s s 3. The case when σ(a) σ(b) {}. In this section, we solve the Sylvester equation AX XB C in the case A diagj m1 (), J m2 (),..., J mp () and B diagj n1 (), J n2 (),..., J nq (), where m 1 m 2 m p > and n 1 n 2 n q > and m m p m, n n q n. Lemma 3.1. Let A J m1 () J mp (), m 1 m p >, and define We have: A k : J m1 () m1 1 k J mp () mp 1 k, k 1, m p 1. i) A 1, ii) A k+1 A AA k+1 A k, k 1, m p 2, iii) A k I ( i 1, p) m i k + 1, iv) (I ± A k ) 1 I A k, k 1, m p /2 1. The proof follows by the definition of A k. Note that from (ii) it particularly follows A A AA and A 1 A AA 1 A. Remark 3.2. Recall that J 1 () I 1, because of 1. In the sequel, N m denotes the set {1, 2,..., m}.

14 13 Solving the Sylvester Equation AX XB C When σ(a) σ(b) Definition 3.3. Let M and N be two disjoint subsets of N m N n such that M N N m N n. The matrix mask given by the set M is mapping Π M : C m n C m n, Π M (A) A M, which maps any matrix A a ij C m n to the matrix A M whose entries are given by { aij, (i, j) M, A M ij, (i, j) N. If A C m n is partitioned on submatrices as A A ij p q, and the set M N p N q is given (therefore, N is also known), then Π M (A) A M, where { Aij, (i, j) M, A M ij, (i, j) N. Note that the mapping Π M is linear and idempotent. Also, when M is known, it means we know N too, because N is the complement of M. The binary relation described by the set M will frequently be represented by a matrix also denoted by M, with entries M m ij given as m ij 1 if (i, j) M, m ij if (i, j) N. Note that M by construction has upper quasitriangular structure. The matrix mask method is the subject of another my paper which is still not finished, so we will not pursue the topic here more than it is necessary. We will be interested in the case when A diagj m1 (), J m2 (),..., J mp () and B diagj n1 (), J n2 (),..., J nq () are given and then the sets M and N depend only on the sizes of appropriate Jordan blocks: (3.16) M {(i, j) N p N q : m i n j }, N {(i, j) N p N q : m i < n j }. Lemma 3.4. Suppose that A diagj m1 (), J m2 (),, J mp () C m m and B diagj n1 (), J n2 (),, J nq () C n n, where m 1 m 2 m p > and n 1 n 2 n q > and C C m n. For the set M N p N q (equivalently, matrix M C p q ) given by (3.16), we have: i) ((A T ) s CB t ) M (A T ) s C M B t, (A s C(B T ) t ) M A s C M (B T ) t, s, t N, ii) (AC CB) M AC M C M B. Proof. Suppose C is decomposed in accordance with A and B as C C ij p q. i) The element in the position (i, j) for both left and right hand matrix is the same, namely { (Jmi () T ) s C ij J nj () t, (i, j) M,, (i, j) N, so the equality holds. The second equality is proven analogously. ii) We have desired equality because the element in the position (i, j) of the left and right hand matrix is the same: { Jmi ()C ij C ij J nj (), (i, j) M,, (i, j) N. Let us split the matrix C as C C M + C N, where C M Π M (C) and C N Π N (C); in the analogous way X X M + X N. The reason for such splitting is the fact that solving simple Sylvester equation (2.6) is different whether m n or m < n, as shown in Theorem 2.3 and Theorem 2.5.

15 Nebojša Č. Dinčić 14 Theorem 3.5. Let A J m1 () J mp (), m 1 m p >, and B J n1 () J nq (), n 1 n q >. Suppose M and N to be as in (3.16), C M Π M (C), C N Π N (C) and d min{m 1, n 1 }. The equation AX XB C is consistent if and only if (3.17) or, in more condensed form, d 1 A k C M B k, d 1 A k A k d 1 A k C N B k, CM C N The particular solution is given as X p X M + X N, where (3.18) X M (A T ) k+1 C M B k, X N n 1 1 B k B k m 1 1. A k C N (B T ) k+1, and the solutions of the homogeneous equation AX XB are given by p nj 1(J nj ()) X h Xij h p q, Xij h, (i, j) M, (mi n j) n j mi (n j m i) p mi 1(J mi ()), (i, j) N. Proof. ( ) : Let X be a solution, so A X XB C. By Lemma 3.4, we have: C M (A X XB) M A X M X M B, C N (A X XB) N A X N X N B. Now we have d 1 d 1 A k C M B k (A k A X M B k A k XM B k+1) d 1 (A k 1 XM B k A k XM B k+1) A 1 XM A d 1 XM B d, because of Lemma 3.1. In the analogous way one can prove the second equality in (3.17). ( ) : We show that X p X p M + Xp N, where Xp M and Xp N are given by (3.18), is the solution of AX XB C provided that (3.17) holds. By using Lemma 3.4, we have: n1 1 X p M n1 1 ij (A T ) k+1 C M B k ( (A T ) k+1 CB k) M The first condition in (3.17) says n j 1 ij (J mi () T ) k+1 C ij J nj () k, (i, j) M. n j 1 J mi () k C ij J nj () k, (i, j) M, ij

16 15 Solving the Sylvester Equation AX XB C When σ(a) σ(b) which by Theorem 2.3 implies that J mi ()X p M ij X p M ijj nj () C ij, (i, j) M, i.e., AX p M Xp M B C M. In the analogous way AX p N Xp N B C N can be proven (there, Theorem 2.5 is used). Therefore, we have AX p X p B AX p M Xp M B + AXp N Xp N B C M + C N C, so the equation is consistent provided that (3.17) holds. The homogeneous equation AX XB is equivalent to the set of equations of the form J mi ()X ij X ij J ni (), i 1, p, j 1, q, so by Corollary 2.6, we have the expressions for appropriate submatrices of X h. Remark 3.6. It may happen that some expressions for the consistency condition (3.17) include negative powers of J k (), but all of them will be multiplied by some zero terms, so there is no actual problem. The next theorem immediately follows from Proposition 1.3, but we restate it here because of the results from the next section. Theorem 3.7. Suppose that A(λ) J m1 (λ) J mp (λ) C m m and B(µ) J n1 (µ) J nq (µ) C n n and a max{m 1,..., m p }, b max{n 1,..., n q }. The equation A(λ)X XB(µ) C, λ µ, is consistent and its unique solution is b 1 a 1 X A(λ µ) (k+1) CB() k A() k CB(µ λ) (k+1). 4. The general case when σ(a) σ(b). Let us consider general case AX XB C when σ(a) σ(b) {λ 1,..., λ s }, C C ij s s and suppose that the matrices are in their Jordan forms: A diag{j(λ 1 ; p 11, p,..., p 1,k1 ),..., J(λ s ; p s1, p s2,..., p s,ks )}, B diag{j(λ 1 ; q 11, q,..., q 1,l1 ),..., J(λ s ; q s1, q s2,..., q s,ls )}, where p i1 p i,ki >, q j1 q j,lj >, i, j 1, s. It is not hard to see that the equations on which solvability depends are precisely of the form J(λ i ; p i1, p i2,..., p i,ki )X ii X ii J(λ i ; q i1, q i2,..., q i,li ) C ii, i 1, s, and, after translation for λ i, they reduce to the form for which Theorem 3.5 is applicable. J(; p i1, p i2,..., p i,ki )X ii X ii J(; q i1, q i2,..., q i,li ) C ii, i 1, s, The results from the previous two sections can be summed up in the following theorem. We remark the notation C ij C (ij) uv u1,ki,v1,l j C pi qj. Theorem 4.1. Suppose that A diag{j(λ 1 ; p 11, p,..., p 1,k1 ),..., J(λ s ; p s1, p s2,..., p s,ks )}, B diag{j(λ 1 ; q 11, q,..., q 1,l1 ),..., J(λ s ; q s1, q s2,..., q s,ls )}, where p i1 p i,ki >, and q j1 q j,lj >, i, j 1, s. Let us denote: d i min{p i1, q i1 }, C (ii) Π Mi (C (ii) ), C (ii) N i Π Ni (C (ii) ), where M i {(u, v) : p iu q iv }, N i {(u, v) : p iu < q iv } N ki N li, i 1, s. M i

17 Nebojša Č. Dinčić 16 The Sylvester equation AX XB C is consistent if and only if d i 1 where we used the notation Ai () k A i () k C (ii) M i C (ii) N i Bi () k B i () k, i 1, s, A i () diagj pi1 (),..., J pi,ki (), B i () diagj qi1 (),..., J qi,li (), i 1, s. In that case, the particular solution X p X p (ij) s s is given by: d i 1 X p (ii) (A i () T ) k+1 C (ii) M i B i () k d i 1 q j1 1 X p (ij) A i (λ i λ j ) (k+1) C (ij) B j () k A i () k C (ii) N i (B i () T ) k+1, i 1, s, p i1 1 A i () k C (ij) B j (λ j λ i ) (k+1) C ki lj, i j; the homogeneous solution X h diagx (ii) h X (ii) h uv, i 1, s, is given by pqiv 1(J qiv()) (piu q iv) q iv, (u, v) M i, piu (q iv p iu) p piu 1(J piu ()), (u, v) N i. Remark 4.2. The consistency condition from the previous theorem can be written in more condensed form as: d 1 A() k CM B() k A() k C N B() k, where d min{max{p i1 : i 1, s}, max{q j1 : j 1, s}}, M M 1 M s, N N 1 N s, and we used the notation A() diaga 1 (),..., A s (), B() diagb 1 (),..., B s (). Also, the diagonal blocks in X p can be given by single explicit formula X diag p d 1 ( (A() T ) k+1 C M B() k A() k C N (B() T ) k+1). The previous theorem has several important corollaries which deal with the cases when matrices A and B are non-derogatory or diagonalizable. Corollary 4.3. Suppose that matrices A and B are non-derogatory, i.e., and d i min{p i, q i }, C (ii) M i is consistent if and only if d i 1 A J p1 (λ 1 ) J ps (λ s ), B J q1 (λ 1 ) J qs (λ s ), Jpi () pi 1 k J pi () k C (ii) if p i q i, C (ii) N i C (ii) M i C (ii) N i C (ii) if p i < q i, i 1, s. The equation AX XB C Jqi () k J qi () qi 1 k, i 1, s,

18 17 Solving the Sylvester Equation AX XB C When σ(a) σ(b) and its particular solution X p X p (ij) s s is given by: X p (ii) (J pi () T ) k+1 C (ii) J qi () k, p i q i, d i 1 di 1 J pi () k C (ii) (J qi () T ) k+1, p i < q i ; q j 1 X p (ij) J pi (λ i λ j ) (k+1) C (ij) J qj () k p i 1 J pi () k C (ij) J qj (λ j λ i ) (k+1) C pi qj, i j; the homogeneous solution X h diagx (ii) h, i 1, s, is given by X (ii) h pqi 1(J qi ()) (pi q i) q i, p i q i, pi (q i p i) p pi 1(J pi ()), p i < q i. Corollary 4.4. Suppose that A and B are diagonalizable matrices, i.e., A λ 1 I k1 λ s I ks, B λ 1 I l1 λ s I ls. Then the Sylvester equation AX XB C is consistent if and only if C (ii), i 1, s, its homogeneous solution X h X (11) h X (ss) h is arbitrary block-diagonal matrix and the particular solution is X p s s with X p (ii), i 1, s, and X (ij) p X (ij) p (λ i λ j ) 1 C (ij), i j. The fact that AX XA I n for A, X C n n is well-known result, usually proven by the trace of a matrix: n tr(i n ) tr(ax XA) tr(ax) tr(xa) tr(ax) tr(ax). This result appears as the corollary of Theorem 4.1. Theorem 4.5. The matrix equation AX XA I is inconsistent. Proof. According to Theorem 4.1, if B A then d i AX XA I is consistent if and only if p i1, C (ii) M i I, i 1, s, and the equation p i1 1 p i1 1 diagj pi1 () pi1 1 k,..., J pi,ki () p i,k i 1 k diagj pi1 () k,..., J pi,ki () k diagj pi1 () pi1 1,..., J pi,ki () p i,k i 1, i 1, s, which is not true. Therefore, the equation AX XA I is inconsistent. Recall that the similar result holds on any unital Banach algebra 16, p. 351.

19 Nebojša Č. Dinčić 18 Example 4.6. Let A J 2 (1) J 1 (1) J 2 (), B J 2 (1) J 2 (1) J 3 () J 1 () (so s 2, λ 1 1, λ 2, p 11 2, p 1, p 21 2, q 11 2, q 2, q 21 3, q 22 1). We have: A 1 () J 2 () J 1 (), A 2 () J 2 (), B 1 () J 2 () J 2 (), B 2 () J 3 () J 1 (). Also d 1 min{p 11, q 11 } 2, d 2 min{p 21, q 21 } 2. We decompose matrices C and X in accordance with the forms of A and B: C (11) C () C C (21) C (22) X (11) X (), X X (21) X (22) and then further decompose C (11) and C (22) in accordance with the mask matrices M 1 p 11 q 11 and p 11 q ) and M 2 1 (because p 21 q 22 ): 1 1 (because C (11) C (11) 11 C (11) C (11) 21 C (11) 22 C (11) 11 C (11) + C (11) 21 C (11) 22 C (11) M 1 + C (11) N 1, C (22) C (22) 11 C (22) C (22) + C (22) 11 C (22) M 2 + C (22) N 2 ; similar is done for the matrix X. By Theorem 4.1, consistency conditions are (we simply write J k instead of J k () and omit zero block-matrices): 1 J 1 k 2 J k 1 J k 2 J k 1 C (11) 11 C (11) C (11) 21 C (11) 22 J k 2 J k 2 J 1 k 2 J 1 k 2, 1 J 1 k 2 J k 2 C (22) C (22) 11 J k 3 J k 1 J 2 k 3 J k 1, i.e., J 2 C (11) 11 + C (11) 11 J 2, J 2 C (11) + C (11) J 2, C (11) 21 J 2, C (11) 22 J 2, J 2 C (22), C (22) 11 J J 2 C (22) 11 J 3. If we put C c ij, X x ij C 5 8, consistency condition says that all matrices C for whom the equation is solvable are of the form C (11) C () C C (21) C (22) c 11 c c 13 c 14 c 15 c 16 c 17 c 18 c 11 c 13 c 25 c 26 c 27 c 28 c 32 c 34 c 35 c 36 c 37 c 38 c 41 c 42 c 43 c 44 c 45 c 46 c 47 c 48 c 51 c 52 c 53 c 54 c 45 c 57.

20 19 Solving the Sylvester Equation AX XB C When σ(a) σ(b) By Theorem 4.1, the particular solution X p X p (ij) is: X (11) p X (22) p 1 (A 1 () T ) k+1 C (11) B 1 () k X () p X (21) p M 1 1 J 2 () T C (11) 11 J 2 () T C (11) C (11) 21 J 2() T C (11) 22 J 2() T 1 (A 2 () T ) k+1 C (22) B 2 () k M 2 1 A 1 () k C (11) N 1 (B 1 () T ) k+1, A 2 () k C (22) N 2 (B 2 () T ) k+1 C (22) 11 J 3() T J 2 ()C (22) 11 (J 3() T ) 2 J 2 () T C (22) 2 A 1 (1) (k+1) C () B 2 () k J 2 (1) 1 C () 11 + J 2 (1) 2 C () 11 J 3() + J 2 (1) 3 C () 11 J 3() 2 J 2 (1) 1 C () C () 21 + C () 21 J 3() + C () 21 J 3() 2 C () 22 1 A 2 () k C (21) B 1 (1) (k+1) C (21) 11 J 2(1) 1 J 2 ()C (21) 11 J 2(1) 2 C (21) J 2(1) 1 J 2 ()C (22) J 2(1) 2.,, Therefore, X (11) p c 11 c c 13 c 14 c 32 c 34, X (22) p c46 c 57 c 47 c 45 c 57 c 48 X p () c 15 c 25 c 15 + c 16 2c 25 c 26 c 15 + c 16 + c 17 3c 25 2c 26 c 27 c 18 c 28 c 25 c 25 + c 26 c 25 + c 26 + c 27 c 28 c 35 c 35 + c 36 c 35 + c 36 + c 37 c 38 X p (21) c41 c 51 c c 51 c 42 c 52 c 43 c 53 c 43 c c 53 c 54 c 51 c 51 c 52 c 53 c 53 c 54 The homogeneous solution is given by X h X (ii) h or, in extended form: X (11) h where: p 1 (J 2 ()) p 1 (J 2 ()) 1 1 p (J 1 ()) 1 1 p (J 1 ()), X (22) h 2 1 p 1 (J 2 ()) p (J 1 ()) 1 1 X (11) h x 11 x x 13 x 14 x 11 x 13 x 32 x 34, X (22) h x46 x 47 x 48 x 46.

21 Nebojša Č. Dinčić 2 for arbitrary complex x 11, x, x 13, x 14, x 32, x 34, x 46, x 47 and x 48. Therefore, we found all solutions of the equation in the case when it is solvable. 5. On the dimension of the space of solutions. Let A C m m and B C n n be given matrices. For the linear operator T (X) AX XB : C m n C m n we consider the following subspaces: R(T ) {C : AX XB C for some X}, N (T ) {X : AX XB }. The subspace R(T ) is actually the set of all C such that the Sylvester equation is consistent, while N (T ) consists precisely of all solutions of the homogeneous Sylvester equation. The general solution can be seen as the set of all homogeneous solutions translated by the particular solution, so the dimension of the space of solutions is the same as the dimension of the space of all solutions to homogeneous equation. By the basic theorem for the vector space homomorphism, we have: dim R(T ) + dim N (T ) dim(c m n ) mn. This fact helps us a lot in finding the dimension of R(T ) when necessary, because it is significantly easier to deal with N (T ), as follows. If A J m (), B J n (), then by Corollary 2.6, we have: p (J n ()), m n, N (T ) (m n) n m (n m) p m 1 (J m ()), m n, and therefore, dim N (T ) min{m, n} (so dim R(T ) mn min{m, n}) and its basis is B N (T ) { In Jn (), Jn (),..., }, m n, B N (T ) { I m, Jm (),..., J m () m 1 }, m n. For the case A J m1 () J mp () and B J n1 () J nq (), by Theorem 3.5, we have N (T ) {X X ij p q : J mi ()X ij X ij J nj (), i 1, p, j 1, q}, so dim N (T ) (i,j) M n j + (i,j) N m i. In the most general case, described by Theorem 4.1, we have s dim N (T ) q kj + k1 (i,j) M k (i,j) N k p ki For the non-derogatory case, Corollary 4.3 gives s dim N (T ) min{p i, q i }, i1.

22 21 Solving the Sylvester Equation AX XB C When σ(a) σ(b) while in the diagonalizable case, given in Corollary 4.4, we have s dim N (T ) k i l i. If we test this on Example 4.6, then we have dim N (T ) ( ) + ( ) 9 which coincides with the number of independent parameters in the general solution. 6. The Schur decomposition approach. Finding the Jordan form of a given matrix having some multiple eigenvalues can be numerically unstable, so in this section, which is not closely related with previous considerations, we present some useful recursive methods for solving the Sylvester matrix equation AX XB C in the case when A and B have common eigenvalue(s). We present the consistency conditions and the algorithm for finding the solutions, based on the Schur matrix decomposition. For numerically solving the Sylvester equations, the most used methods are Bartels-Stewart 1 and Golub-Nash-van Loan 8, and their various improvements. Bartels-Stewart method uses QR-algorithm for the Schur decomposition of matrices A and B, while Golub-Nash-van Loan method uses the Hessenberg decomposition of matrices A and B. We emphasize that both methods assume disjointness of the spectra of A and B. The main idea of the Bartels-Stewart algorithm is to apply the Schur decomposition to transform Sylvester equation into a triangular linear system which can be solved efficiently by forward or backward substitutions. It is known that A C n n can be expressed as A QUQ 1, where Q is unitary matrix (i.e., Q 1 Q ), and U is upper triangular matrix, which is called a Schur form for A. Since U is similar to A, it has the same multiset of eigenvalues, and since it is triangular, those eigenvalues are the diagonal entries of U. If A is real and σ(a) R then U can be chosen to be real and orthogonal. For details on this topic, see e.g., 1, p. 79. By using the idea of the constructive proof for Schur decomposition, suppose λ σ(a) σ(b) and there exist some unitary matrices S and T such that (a and b are the multiplicities of λ in σ(a) and σ(b), respectively; U λ and V λ are appropriate eigenspaces): S λia A AS : U λ Uλ U λ Uλ, A 22 T λib B BT : V λ Vλ V λ Vλ. B 22 Note that λ / σ(a 22 ) σ(b 22 ). Now the Sylvester equation becomes: λia A S S λib B X XT T C A 22 B 22 λia A S XT S λib B XT S CT, A 22 B 22 so the equation, with S XT Y Y ij and S CT D D ij, becomes λia A Y11 Y Y11 Y λib B D11 D A 22 Y 21 Y 22 Y 21 Y 22 B 22 D 21 D 22 i1,

23 Nebojša Č. Dinčić 22 which is equivalent to the following system: (6.19) (6.2) (6.21) (6.22) A Y 21 D 11, Y (λi B 22 ) Y 11 B D A Y 22, (A 22 λi)y 21 D 21, A 22 Y 22 Y 22 B 22 D 22 + Y 21 B. From (6.19) and (6.21) we conclude the consistency condition: Also from (6.21) we have A (A 22 λi) 1 D 21 D 11. Y 21 (A 22 λi) 1 D 21. Note that A 22 and B 22 (and consequently A 22 λi and λi B 22 ) are upper triangular matrices, so some known numerical method can be applied in computing their inverses. Case 1: If σ(a) σ(b) {λ}, i.e., if λ is the only common eigenvalue (and therefore, σ(a 22 ) σ(b 22 ) ), because of (6.22), by Proposition 1.3, we have the unique Y 22 (which can be computed by using some already known numerical method for solving the Sylvester equation), while from (6.2) we have Y via Y 11 : Therefore, we found the family of solutions. Y (D A Y 22 + Y 11 B )(λi B 22 ) 1. Case 2: If there is some other common eigenvalue µ for A and B (which means µ σ(a 22 ) σ(b 22 )), the described method is applied to the equation (6.22), i.e., A 22 Y 22 Y 22 B 22 D 22 + (A 22 λi) 1 D 21 B. It remains to do backward substitution and the equation is fully solved. Because any complex matrix from C n n has at most n different eigenvalues, the algorithm terminates after at most n passes. Acknowledgments. The author thanks to his colleague Miloš Cvetković for a valuable discussion on some topics presented in the paper. The author would like to express his gratitude to the two anonymous referees and the editor, Prof. Froilán M. Dopico, for their insightful comments and suggestions which significantly improved the paper. REFERENCES 1 R.H. Bartels and G.W. Stewart. A solution of the equation AX + XB C. Comm. ACM, 15:82 826, R. Bhatia and P. Rosenthal. How and why to solve the operator equation AX XB Y. Bull. London Math. Soc., 29:1 21, Y. Chen and H. Xiao. The explicit solution of the matrix equation AX XB C. Appl. Math. Mech. (English Ed.), 16: , B.N. Datta and K. Datta. The matrix equation XA A T X and an associated algorithm for solving the inertia and stability problems. Linear Algebra Appl., 97:13 119, 1987.

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