哈密頓矩陣與辛矩陣

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(1)國立高雄大學應用數學系碩士論文. Hamiltonian matrices and symplectic matrices 哈密頓矩陣與辛矩陣. 研究生：蔡弘霖撰指導教授：郭岳承. 中華民國一百零三年六月.

(2) Hamiltonian matrices and symplectic matrices. by Hung-Lin Tsai Advisor Yueh-Cheng Kuo. Department of Applied Mathematics, National University of Kaohsiung Kaohsiung, Taiwan 811, R.O.C. June 2014.

(3) Contents 1 Introduction 1.1 Hamiltonian matrices and Symplectic matrices . . . . . . . . 1.2 Hamiltonian pencils and Symplectic pencils . . . . . . . . . 1.3 Some further observations of symplectic matrices and pencils 1.4 The structure-preserving doubling algorithm . . . . . . . . .. . . . .. 1 1 3 5 7. 2 A canonical form for Hamiltonian matrices and pencils 10 2.1 A canonical form for Hamiltonian matrices . . . . . . . . . . . 10 2.2 Hamiltonian Kronecker canonical form . . . . . . . . . . . . . 20 3 Real Hamiltonian logarithm of a real symplectic matrices. 25. References. 27.

(4) 哈密頓矩陣與辛矩陣. 指導教授：郭岳承教授國立高雄大學應用數學系. 學生：蔡弘霖國立高雄大學應用數學系摘要哈密頓矩陣與辛矩陣可應用於解代數 Riccati 方程，並在其他科學計算上有許多理論在。在這篇論文裡，我們將探討這兩類矩陣的標準型並探討部分細節。. 關鍵字：哈密頓矩陣、辛矩陣、標準型.

(5) Hamiltonian matrices and symplectic matrices Advisor: Professor Yueh-Cheng Kuo Institute of Department Applied Mathematics National University of Kaohsiung. Student: Hung-Lin Tsai Institute of Department Applied Mathematics National University of Kaohsiung. ABSTRACT Hamiltonian matrices and symplectic matrices can be applied to solve the algebraic Riccati equation, and also have a lot of theories in the other science topics. In this paper, we will discuss a canonical form and some details of the two type matrices.. Keywords: Hamiltonian matrix, symplectic matrix, Jordan canonical form.

(6) 1. Introduction. 1.1. Hamiltonian matrices and Symplectic matrices. First, we introduce the Hamiltonian matrix and the symplectic matrix and some properties of them. Let J =. . 0 In −In 0. . .. (1.1). And then we immediately have J −1 = −J = J H . Definition 1.1. A matrix H ∈ C2n×2n is said to be Hamiltonian if (J H)H = J H.. (1.2). According to the above definition, it is easy to see that a Hamiltonian matrix H takes the form H1 H2 , (1.3) H= H3 −H1H where H2H = H2 ∈ Cn×n and H3H = H3 ∈ Cn×n . A fact that if λ is an eigenvalue of H, because of H = −J HH J −1 , then ¯ is also an eigenvalue of H. −λ Definition 1.2. A matrix S ∈ C2n×2n is called symplectic if SH J S = J .. (1.4). Note that the above definitions are equivalent to (HJ )H = HJ and SJ S H = J . Since det(S)det(J )det(S H ) = det(SJ S H ) = det(J ) 6= 0, symplectic matrices are invertible. Moreover, from (S −1 )H J S −1 = J , 1.

(7) we know that S −1 is still symplectic. Actually, let S=. . S11 S12 S21 S22. . ,. (1.5). we obtain, according to the definition of symplectic matrix, H H H S11 S12 = (S11 S12 ) ,. (1.6). H S21 S22. (1.7). =. H H (S21 S22 ) ,. and H H S11 S22 − S12 S21 = In ,. (1.8). where S11 , S12 , S21 , S22 ∈ Cn×n . From (1.4), we have S −1 = −J S H J , and hence. S. −1. =. . H H S22 −S12 H H −S21 S11. . .. (1.9). Furthermore, S = J −1 S −H J ,. ¯ is also an we can conclude that if λ ∈ C is an eigenvalue of S, then 1/λ eigenvalue of S. The following gives the algebraic properties of the two matrix types. Proposition 1.1. Let H be Hamiltonian and S1 , S2 , S be symplectic. Then 1. S −1 HS is Hamiltonian. 2. S1 S2 is symplectic. Proof.. 1. For symplectic S and Hamiltonian H, (J S −1 HS)H = S H HH S −H J. = −S H HH (J S). = S H (J H)S = J S −1 HS.. Hence, S −1 HS is Hamiltonian. 2.

(8) 2. For symplectic S1 and S2 , (S1 S2 )J (S1 S2 )H = S1 S2 J S2H S1H = S1 J S H 1 = J.. Hence, S1 S2 is symplectic.. 1.2. Hamiltonian pencils and Symplectic pencils. An eigengenvalue problem Ax = λx can be considered as a generalized eigenvalue problem Ax = λBx for B = I. The following we consider the generalized eigenvalue problem for the pencil A − λB. Definition 1.3. Let M, L ∈ C2n×2n . 1. A pencil M − λL is Hamiltonian if MJ LH = −LJ MH. (1.10). 2. A pencil M − λL is symplectic if MJ MH = LJ LH. (1.11). From the above definition, if L is invertible, then the matrix L−1 M is Hamiltonian/symplectic. The following we give two types of a symplectic pencil. Definition 1.4. Given a symplectic pair M − λL. 1. The pair is called in first standard symplectic form (SSF-1) if it takes the form A 0 (1.12) M= −H In and L=. . with G = GH and H = H H . 3. In G 0 AH. . ,. (1.13).

(9) 2. The pair is called in second standard symplectic form (SSF-2) if it takes the form A 0 (1.14) M= Q −In and L=. . −P In AH 0. . ,. (1.15). with P = P H and Q = QH . From the above definition, we get a conlusion about the symplectic matrices and symplectic pencils. Theorem 1.2. Let S be a 2n×2n matrix partitioned as in (1.5). Assume that S22 is nonsingular, then S = L−1 M, where M, L are in SSF-1. Moreover, the factorization is unique. H S22 0 Proof. If S22 is invertible, then let Y1 = −1 , which is symplectic, 0 S22 and set H H S12 S22 S11 S22 ′ , (1.16) S = Y1 S = −1 S22 S21 In H In −S22 S12 , which is symplectic, which is also symplectic. Still, let Y2 = 0 In then set S ′′ = Y2 S ′ , we obtain H −1 S22 (S11 − S12 S22 S21 ) 0 ′′ , (1.17) S = −1 S22 S21 In or. . −1 H S22 (S11 − S12 S22 S21 ) 0 −1 S22 S21 In. . =. . −1 H H S12 S22 S22 −S22 −1 0 S22. . S,. or equivalently, −1 −1 In −S12 S22 S11 − S12 S22 S21 0 S, =Q Q −1 −1 0 S22 S22 S21 In H where Q = diag(S22 , In ). Hence, −1 A 0 S11 − S12 S22 S21 0 = M= −1 −H In S21 In S22. 4. (1.18).

(10) and L=. . −1 In −S12 S22 −1 0 S22. . =. . In G 0 AH. Since S ′′ is symplectic, from (1.17) and (1.8), we get. . .. −1 −H H −1 (A)H = (S11 − S12 S22 S21 )H = (S22 ) = S22 = AH ,. and, from (1.17) and (1.7), we have H H = H. From (1.18) and (1.6), we have GH = G. For the uniqueness, suppose that there are two SSF-1 pencil (M1 , L2 ) 6= −1 (M2 , L2 ) such that S = L−1 1 M1 and S = L1 M1 . Thus, through direct computation, we have  −H A1 + G1 A−H  1 H1 = A2 + G2 A2 H2   −H −A−H 1 H1 = −A2 H2 −G A−H = −G2 A−H  2   −H1 1 −H A1 = A2 And therefore (M1 , L2) = (M2 , L2).. Similar to the above theorem, one can have if S12 is invertible, then S = L−1 M with −1 S21 − S22 S12 S11 0 M= −1 −S12 S11 −In. and. L= which is in SSF-2.. 1.3. . −1 −S22 S12 In −1 −S12 0. . ,. Some further observations of symplectic matrices and pencils. Although we know the fact that S is invertible in last section, we still cannot conclude that S11 , S22 is nonsigular. Therefore, [1] gives an idea that how to make S11 , S22 to be nonsigular. Theorem 1.3 (The complementary bases theorem for symplectic matrices). Let S be 2n×2n symplectic matrix partitioned as in (1.5). Suppose that there are k linearly indepedent rows (cloumns) of Spq , where p, q ∈ {1, 2}, indexed by α ⊆ {1, ..., n}. Then, the rows (columns) of Sp′ q (Spq′ ) indexed by α′ , the complement of α, together with the rows (columns) of α of Spq forms a basis of C. That is, they constitude a nonsigular matrix. 5.

(11) Proof. Here, we claim the case that the linear indepedent rows of S21 and the complement rows of S21 in S22 forms a full rank matrix. First, choose an approciate permutation matrix P such that ′ ′ S11 S12 ′ S = (P ⊕ P )S = ′ ′ S21 S22 ′ with the first k rows S21 are linearly independent. ′ Let Y be a n-by-n nonsingular matrix such that Y S21 is in a row reduced −T echelon form. Note that Y ⊕ Y is symplectic. And, then let ′′ ′′ S11 S12 ′′ −T ′ , S = (Y ⊕ Y )S = ′′ ′′ S21 S22. where ′′ S21. and ′′ S22. =. =. . . X11 X12 X21 X22. Ik Z 0 0. . (1.19). . .. (1.20). So far the above multiplication does not change the relationship of the ′ ′ linear independence among the rows of [S21 , S22 ], we have only to show that ′′ rank(X22 ) = n − k. Since S is symplectic, by (1.9), we get ′′ ′′H ′′ ′′H 0 = (S ′′ S ′′−1 )21 = S21 S22 − S22 S21 .. (1.21). Combining (1.19), (1.20) and (1.21), we have X21 = −X22 Z H . Now, since S ′′ is nonsingular, the rows of S ′′ are linearly independent. In particular, [0, 0, X21 , X22 ] is linearly independent, and thus n − k = rank([0, 0, X21 , X22 ]) = rank([X21 , X22 ]) = rank(X22 [−Z H , In−k ]) = rank(X22 ).. Therefore, we can define a ”swap” matrix diag(v) diag(v ′ ) Π= , −diag(v ′ ) diag(v) 6. (1.22).

(12) where vi ∈ {0, 1}, and vi′ = 1 − vi , for i = 1, ...n. Notice that diag(v) and diag(v ′ ) are the product of permutation matrix and that Π is symplectic. Using the above results, we can easily know that if S is symplectic, then there is an invertible swap matrix Π such that Se = ΠS. is symplectic with Se22 or Se12 nonsigular and thus we may have −1 −1 e S = Π−1 1 S1 = Π1 L1 M1. and. −1 −1 e S = Π−1 2 S2 = Π2 L2 M2 ,. where (M1 , L1 ) and (M2 , L2 ) are in SSF-1 and SSF-2, respectively. This gives us that −1 −1 L−1 1 M1 = (Π1 Π2 )L2 M2. which means that SSF-1 and SSF-2 are equivalent through a symplectic matrix. That is, solving M1 x = λL1 x. can be considered as. −1 M2 x = λL2 (Π1 Π−1 2 ) x,. provided that L1 and L2 are invertible.. 1.4. The structure-preserving doubling algorithm. This section follows [10]. Given a symplectic pencil M − λL ∈ R2n×2n , we define L 2n×4n N (M, L) = [M∗ , L∗ ] ∈ R : rank[M∗ , L∗] = 2n, [M∗ , L∗ ] =0 . −M L ) ≤ 2n, N (M, L) 6= ∅. Since rank( −M Now, given [M∗ , L∗ ] ∈ N (M, L), let. ˆ = M∗ M and Lˆ = L∗ L, M. then ˆ − λLˆ M − λL → M is called a doubling transformation. The following two theorems in [10, Theorem 2.1 and 2.2] gives the ”doubling” and the ”structure-preserving” of the algorithm. 7.

(13) ˆ −λLˆ be the result of a doubling transformaTheorem 1.4. Let the pencil M ˆ ˆ tion of the symplectic M− λL is symplectic. pencil λL, then the pencil M − U U ˆ U = Lˆ U S 2 . S, then M =L Moreover, if M V V V V Theorem 1.5. Let M − λL be a symplectic pencil.. 1. If M − λL is in SSF-1, then there exists [M∗ , L∗] ∈ N (M, L) such ˆ − λLˆ in SSF-1. that M 2. If M − λL is in SSF-2, Q − P > 0 and Q − AT (Q − P )−1A ≥ 0, then ˆ − λLˆ in SSF-2. there exists [M∗ , L∗ ] ∈ N (M, L) such that M The aboves can be applied to solving the Riccati-type matrix equations, the continuous-time algebraic Riccati equation (CARE), −XGX + AT X + XA + H = 0, and the discrete-time algebraic Riccati equation (DARE), X = AT X(In + GX)−1 A + H, where A, G, H ∈ Rn×n with G, H ≥ 0, and the nonlinear matrix equation with plus sign (NME-P), X + AT X −1 A = Q, and the nonlinear matrix equation with minus sign (NME-M), X − AT X −1 A = Q, where A, Q ∈ Rn×n with Q > 0. The followings are the algorithms, for more detail, see [10]. A fact that X ≥ 0 solves DARE if and only if X satisfies In In G In A 0 S, = X 0 AT X −H In. (1.23). for some stable S ∈ Rn×n . Appling the above theorem, we have the following algorithms. Algorithm 1.1 (SDA-1). Set A0 = A, G0 = G and H0 = H. Iterates Ak+1 = Ak (In + Gk Hk )−1 Ak , Gk+1 = Gk + Ak Gk (In + Hk Gk )−1 ATk , Hk+1 = Hk + ATk (In + Hk Gk )−1 Hk Ak , 8.

(14) X > 0 solves the NME-P if and only if X satisfies In 0 In In A 0 = S, Q −In X AT 0 X. (1.24). for some S ∈ Rn×n . Appling to the doubling algorithm, we obtain the following. Algorithm 1.2 (SDA-2). Set A0 = A, Q0 = Q and P0 = 0. Iterates Ak+1 = Ak (Qk − Pk )−1 Ak ,. Qk+1 = Qk − ATk (Qk − Pk )−1 Ak , Pk+1 = Pk + Ak (Qk − Pk )−1 ATk ,. For the convergence of the algorithms are shown in [10, Theorem 3.1 and 4.1]. Theorem 1.6. Assume that X, Y ≥ 0 satisfies. X = AT X(I + GX)−1 A + H, Y = AY (I + HY )−1 AT + G. and let S = (I + GX)−1 A, T = (I + HY )−1 AT + G, then SDA-1 gives k. 1. Ak = (I + Gk X)S 2 , 2. H ≤ Hk ≤ Hk+1 ≤ X and k k k k X − Hk = (S T )2 (X + XGk X)(S T )2 ≤ (S T )2 (X + XY X)(S T )2 , 3. G ≤ Gk ≤ Gk+1 ≤ Y and k k k k T − Gk = (T T )2 (Y + Y Hk Y )(T T )2 ≤ (T T )2 (Y + Y XY )(T T )2 .. Moreover, if ρ(S) < 1, then kAk k2 , kX − Hk k2 and kY − Gk k2 tends to 0 as k→∞ Theorem 1.7. Suppose that X > 0 satisfies the NME-P and let S = X −1 A, then SDA-2 gives k. 1. Ak = (X − Pk )S 2 , 2. 0 ≤ Pk ≤ Pk+1 < X and Qk − Pk = (X − Pk ) + ATk (X − Pk )−1 Ak > 0, 3. X ≤ Qk+1 ≤ Qk < X and k k k k Qk − X = (S T )2 (X − Pk )S 2 ≤ (S T )2 XS 2 .. Under ρ(S) < 1, we have kAk k2 and kX − Qk k2 tends to 0 as k → ∞. 9.

(15) 2. A canonical form for Hamiltonian matrices and pencils. The results in this section follow the results in [9], and it is reproved in a direct way. We would first consider the Jordan canonical form of a Hamiltonian matrix and then derive the Kronecker canonical form of a Hamiltonian pencil through considering the Jordan canonical form for a Hamiltonian matrix.. 2.1. A canonical form for Hamiltonian matrices. In this section, we want to get a symplectic similarity of a Hamiltonian matrix. That is, we want to find a symplectic U such that HU = UΛ, where Λ is in Jordan canonical form. ¯ would also By similarity, we know that if λ is an eigenvalue of H, then −λ be an eigenvalue of H. But this does not guarantee the eigenvalues always be pairwise. For example, 0 1 J1 = −1 0 is Hamiltonian which has eigenvalues i and −i and thus it is difficult to deal with pure imaginary eigenvalue of a Hamiltonian matrix. Before considering the eigenvalues and eigenvectors of a Hamiltonian matrix, we see the following lamma. Lemma 2.1. Let A be n-by-n matrix over C, rows of U H span the left invariant subspace of A corresponding to eigenvalue λ1 and columns of V span the right invariant subspace of A corresponding to eigenvalue λ2 . If λ1 6= λ2 , then U H V = 0 and det(U H V ) 6= 0. Proof. Let columns of X1 , X2 forms the right and left invariant subspace of A, respectively. Consider X2H AX1 , we have ΛX2H X1 = X2H X1 Λ, where Λ = diag(J ’s are in Jordan form, for i = 1, 2, ..., k. Let i λ1 , Jλ2 , . . . , Jλk ) and Jλ X11 X12 · · · X1k  X21 X22 · · · X2k    X2H X1 =  .. .. .. .. , and then we get (Xij Jλj )pq = (Jλi Xij )pq .  . . . .  Xk1 Xk2 · · · Xkk Therefore, if λi 6= λj then Xij = 0 and det(Xii )6= 0, because det(X2H X1 )6= 0, for i, j = 1, ..., k. 10.

(16) Let V+ , V− , V0 span right invariant subspaces of H corresponding to positive, negative and zero real part eigenvalue respectively. The following lemma gives the left invariant subspaces of H. Lemma 2.2. Let V+ , V− , V0 be defined as above. Then V−H J , V+H J , V0H J sapn left invariant subspaces of H corresponding to positive, negative and zero real part eigenvalue respectively. Proof. Pre-multiplying J and then taking Hermitian of both sides into the H equation HV+ = V+ Λ+ , we get V+H (J H) = −ΛH + (V+ J ). Our goal is to find a symplectic  U such that HU = UΛ, and sofar we 0 V+H J V− 0 H H  . 0 0 have, let V = [V+ , V− , V0 ], V J V = V− J V+ H 0 0 V0 J V0. Lemma 2.3. The columns of V− (V+H J V− )−1 also span the right invariant subspace of H corresponding to Λ− . Proof. Consider HV− = V− Λ− . Pre-multiplying V+H J of both sides gives H H V+H J HV− = V+H J V− Λ− or −ΛH + V+ J V− = V+ J V− Λ− , so we obtain that H −1 (V+H J V− )−1 (−ΛH + ) = Λ− (V+ J V− ) . Hence, HV− (V+H J V− )−1 = V− Λ− (V+H J V− )−1. = V− (V+H J V− )−1 (−ΛH + ).. According to the lamma, we can [V+ , Ve− , V0 ], then  0 H e e  V J V = −In 0. let Ve− = V− (V+H J V− )−1 and Ve =  In 0 . 0 0 H 0 V0 J V0. (2.1). The following we focus on the invariant subspace of a zero real part eigen(k) value of H. Let columns of V0 form the invariant subspace of H coresponding to λk = ia(k) , a(k) ∈ R. From (k). (k). (k). (k). (V0 )H J H(V0 ) = −(V0 )H HH J (V0 ) or. (k). (k). (k). (k). (k). (k). (V0 )H J (V0 )Λ0 = (−Λ0 )H (V0 )H J (V0 ). 11.

(17) (k). Let Λ0 = iα(k) I + N (k) where α(k) ∈ R, k = 1, ..., l and the Jordan N (k) = diag(Nr1 (0), ..., Nrs (0)) (we call it nilpotent matrix in the following), we get K (k) N (k) = −(N (k) )H K (k) , (k). (2.2). (k). where K (k) = (V0 )H J (V0 ) which is skew-Hermitian.  −1   (−1)2    ∈ Rn×n and then we may  · Define Pn =     · n (−1) −1 H have Pn Nn (0)Pn = −Nn (0). Applying the fact, pre-multiplying P −1 = ) into (2.2) leads diag(Pr−1 , ..., Pr−1 s 1 P −1 K (k) N (k) = −P −1 (N (k) )H P P −1K (k) = N (k) P −1 K (k) .. (2.3). Notice that, in (2.3), P −1 K (k) commutes with nilpotent matrix. Definition 2.1. A matrix T ∈ R(n+1)×(n+1) is called a Toeplitz matrix if   a0 a1 a2 · · · an .   .. . ..   a−1 a0 a1   . .. .. T = . . a2  .  a−2 . .   . .. .. ..  .. . . . a1  a−n · · · a−2 a−1 a0. A square matrix is upper triangular and Toeplitz we called upper triangular Toeplitz. And a p-by-q upper triangular Toeplitz matrix takes the form T if p > q, 0 or 0 T if p < q, where T is upper triangular Toeplitz.. Proposition 2.4. The matrix A that commutes with nilpotent matrix if and only if A is upper triangular Toeplitz. Proof. See [7] Lemma 4.4.11. Corollary 2.5. Let N = diag(Nr1 , ..., Nrs ), where Nrj ∈ Rrj ×rj ’s are nilpotent matrices for j = 1, ..., s. Then a matrix E that commutes with N if and only if each (i, j) block of E is upper triangular Toeplitz for i, j = 1, ..., s. 12.

(18) P −1K (k) = E is upper triangular Toeplitz and K (k) is skew-Hermitian, so = Prp Epq ∈ Crp ×rq which take the form. (k) Kpq. .    where W =   . W, if rp = rq , 0 W , if rp < rq , 0 , if rp > rq , W. −a1 .. . (−1)rp a1 (−1)rp a2 (−1)rp a3. j = 1, ..., rp .. (2.4). −a1 a2 −a3 .. .. a1 −a2 .. . ···. (−1)rp arp. .     , aj ∈ C,  . In particularly, the main diagonal blocks Kpp of K (k) takes the form   −b1  b1 ib2     −b1 −ib2 −b3  (2.5)   , if rp is even,  .. .. ..   . . .  b1 ib2 b3 · · · ibrp   −ib1  ib1 b2      −ib −b −ib 1 2 3  , if rp is odd, (2.6)   .. .. ..   . . .  −ib1 −b2 −ib3 · · · −ibrp. where each brj ∈ R. To ensure the followings work, we need the next proposition. Proposition 2.6. Let. .   K (k) =  . K11 H −K12 .. .. K12 K22 .. .. H H −K2s −K1s. · · · K1s · · · K2s .. .. . . · · · Kss. .   , . (2.7). where Kpq takes the forms in (2.4), (2.5), and (2.6) for p, q = 1, ..., s, be defined as above. Then there is Q that commutes with nilpotent matrix such that the blocks of main diagonal of QH K (k) Q is invertible. 13.

(19) Proof. Let K. (k). .   = . K11 H −K12 .. .. K12 K22 .. .. H H −K1s −K2s. · · · K1s · · · K2s .. .. . . · · · Kss. .   , where Kpq takes the form in . K11 K12 (2.4), (2.5) and (2.6) for p, q = 1, ..., s. If K = with r1 < r2 H −K12 K22 and K11 is singular, then it contradicts the fact that K (k) is nonsingular. Without loss of generality, we assume that each Kpp ∈ Crp ×rp is singular and that has the same size, p = 1, ..., s, that is, r1 = · · · = rs . Suppose that K11 is singular and K1m is invertible, then Kmm is also singular and setting   Ir 1   Ir 2   ..   .   Q1 =  , I βI  rm  rm   ..   . Ir s (k). . ¯ H is nonsigular, and thus we get where β ∈ {1, i} such that βK1m − βK 1m H (k) Q1 K Q1 is also nonsingular with (1,1)-block nonsingular. Similarly, select appropriate Q2 , ..., Qs and set Q = Q1 · · · Qs , we have done.. Lemma 2.7. Let K (k) be defined in (2.7). Then there is a nonsingular Y (k) that commutes with N (k) such that (Y (k) )H K (k) Y (k) = diag(β1 Pr1 , ..., βs Prs ), where βj = ±1, ±i for j = 1, ..., s. Proof. From the previous proposition, we may assume that Kpp is invertible and a fact in (2.2), Kpp Nrp = −NrHp Kpp and Kpq Nrq = −NrHp Kpq , we have −1 H −1 −1 Nrp Kpq = Nrp Kpp Kpq Kpq Nrq = −Kpp Kpp. for p, q = 1, ...s. Setting . −1 I −K11 K12  I   I X1 =   ..  .. 14.  I.    ,  .

(20) which is commute with N (k) , we have  K11 0 K13  0 e e 23 K22 K   H e H K33 X1H K (k) X1 =  −K13 −K 23  ..  . H H H e −K1s −K2s −K3s. · · · K1s e 2s ··· K · · · K3s · · · Kss. .    ,  . e 2j = K2j + K H K −1 K1j for j = 2, ..., s. Similar method for selecting where K 12 11  −1 I −K11 K13   I     I X2 =   , X3 , ..., Xt , and we finally get   . .   . I e 22 , ..., K b ss ), X H K (k) X = diag(K11 , K. where X = X1 · · · Xt , which commutes with N (k) . The following, weconsider the elimination of[Kpp ]rp ×rp for rp is even. −a1  a ia2  1     −a −ia −a 1 2 3 Assume that Kpp =   , and a1 > 0. Let E0 =  .. .. ..   . . .  a1 ia2 a3 · · · iarp   −1  1 ib2     −1 −ib −b √1 Irp , then E0H Kpp E0 =  2 3   . Set E1 = Irp + |a1 |  .. .. ..   . . .  1 ib2 b3 · · · ibrp   −1  1 0      H H −1 0 −c 3 (−ib2 /2)Nrp , then E1 E0 Kpp E0 E1 =   , where Nrp  .. .. ..   . . .  1 0 c3 · · · icrp is nilpotent. Similarly, choose E2 = Irp + (−ic3 /2)Nr2p ,...,Ek and let Fp = E0 · · · Ek , we obtain FpH Kpp Fp = βp Prp with βp = 1. Note that if a1 < 0, then FpH Kpp Fp = βp Prp for βp = −1. The algorithm also works for rp is odd, we would have βp = ±i. Therefore, Y (k) = X1 · · · Xt F , where F = diag(F1 , ..., Fs ). 15.

(21) So far, we have (k). (k). (k). (k). (k). HV0 Y (k) = V0 Λ0 Y (k) = V0 Y (k) Λ0 , but the congruence (k). (k). (V0 Y (k) )H J (V0 Y (k) ) = diag(β1 Pr1 , ..., βs Prs ), and so (V0 Y )H J (V0 Y ) 6= J . The followings give a method for the above problem. Lemma 2.8. Let Pr ∈ Rr×r be defined above. −1 H 1. For βP2r , β = ±1, let r , βPr ), then we get Z (βP2r )Z = Z = diag(I Nr (−1)r βer eH r Jr and Z −1 N2r Z = . −NrH −1 2. For iβP2r+1 , β =  ±1, let Z = diag(Ir+1 , −iβPr ), then we have Ir r+1 H  and the nilpotent matrix will  Z (iβP2r+1 )Z = iβ(−1) −Ir   Nr er . 0 (−1)r+1 iβeH be Z −1 N2r+1 Z =  r H −Nr. Proof. Direct calculation gives the followings. 1. First, H. Z (βP2r )Z = =. . . Ir βPr (−1)r Pr2. . (−1)r βPr Ir .. βPr. . Ir βPr−1. . If r is even, then Pr−1 = −Pr and hence (−1)r Pr2 = −Ir . And if r is odd, then Pr−1 = Pr which gives (−1)r Pr2 = −Ir . As for Ir Nr er eH Ir −1 1 Z N2r Z = βPr−1 Nr βPr Nr (−1)r βer eH r . = −NrH 16.

(22) 2. Similarly, . Z H (βP2r+1 )Z =  . =. and Z −1 N2r+1 Z =. . . =. Ir+1 iβPr. Ir iβ(−1)r+1 (−1). r. Pr2 Ir iβ(−1)r+1. −Ir . .  . ,. .  Nr er Ir+1 H   0 e1 −iβPr−1 Nr . Nr er . 0 (−1)r+1 iβeH r H −Nr. Thus, the results hold. Remark 2.1. Let H1 =. . Nn (iα) en eH n 0 −Nn (iα)H. . H2 =. . Nn (iα) −en eH n 0 −Nn (iα)H. . and. .. There does not exist symplectic S such that H1 = S −1 H2 S. Proof. Suppose that there is a symplectic S such that H1 = S −1 H2 S. According to the last Lemma, we can find Z1 and Z2 such that, say, Z1−H J Z1−1 = P2n , Z2−H J Z2−1 = −P2n and Z1 H1 Z1−1 = N2n = Z2 H2 Z2−1 . Since H1 and H2 are symplectic similar, we have Z1 = Z2 S. And thus, J = Z1H P2n Z1. = S H Z2H P2n Z2 S = −S H J S = −J ,. a contradiction. The following lemma gives the congruence of two odd size matrices. 17.

(23) Lemma 2.9. Given two matrix pairs     Nr1 (iα1 ) er1 Ir 1 e1 ) = ( ) , iα1 (−1)r1 +1 iβ1 eH iβ1 (−1)r1 +1 (C1 , N r1 H −Nr1 (iα1 ) −Ir1 and. . e2 ) = ( (C2 , N. iβ2 (−1)r2 +1. Ir 2. −Ir2.  Nr2 (iα2 ) er2 ), , iα2 (−1)r2 +1 iβ2 eH r2 −Nr2 (iα2 )H  . where Nrk (iαk ) = iαk Irk + Nrk , αk ∈ R and k ∈ {1, 2}. Let √ 2 −1 (−1)r1 +1 β1 i v11 v12 V = = −1 −(−1)r1 +1 β1 i v21 v22 2. and. .    Z=   . Then. Ir 1 0 0 0 0 v11 0 0 0 0 Ir 2 0 0 0 v21 0 0 0 .    H Z diag(C1 , C2 )Z =    . and. Z −1. ". e1 N. e2 N. #. .    Z=   . 0 0 0 0 0 v12 Ir 1 0 0 0 0 0 0 0 v22 0 Ir 2 0. 0 0 0 0 0 0 0 0 w11 −Ir1 0 0 0 −Ir2 0 0 0 w21 √.    .   . Ir 1 0 0 0 Ir 2 0 0 0 w12 0 0 0 0 0 0 0 0 w22. Nr1 (iα1 ) 0 − √22 er1 0 Nr2 (iα2 ) − 22 er2 0 0 z11 0 0 0 0 0 0 0 0 z21. 18. .        .

(24) √. 0 0 −√ 22 (−1)r1 β1 ier1 2 0 0 (−1)r1 β1 ier2 2 √ √ 2 2 β (−1)r1 ieH β (−1)r2 ieH z12 r1 r2 2 1 2 2 H −Nr1 (iα1 ) 0 0 H 0 −N (iα ) 0 r2 2 √ √ 2 H 2 r1 +r2 H e − β β (−1) er2 z22 2 r1 2 1 2 where. .     ,   . 1 −[β1 (−1)r1 + β2 (−1)r2 ]i −[β1 (−1)r1 − β2 (−1)r2 ] w11 w12 = β1 (−1)r1 − β2 (−1)r2 −[β1 (−1)r1 + β2 (−1)r2 ]i w21 w22 2 i(α1 + α2 ) (−1)r1 +1 β1 (α1 − α2 ) z11 z12 1 . =2 and −(−1)r1 +1 β1 (α1 − α2 ) i(α1 + α2 ) z21 z22 . Proof. Direct calculation. w11 w12 Note that if β1 (−1) = −β2 (−1) , then = ±J1 . And for a w21 w22 Hermitian matrix, we know that all eigenvalues of it are all real and so there is a special relation between Hermitian matrices. r1. r2. . Definition 2.2. Let A ∈ Cn×n be Hermitian. The inertia of A is the ordered triple iner(A) = (i+ (A), i− (A), i0 (A)), where i+ (A), i− (A) and i0 (A) are the number of positive, negative, zero eigenvalues of A respectively. Theorem 2.10. Let A, B ∈ Cn×n be Hermitian matrices. Then there is a nonsingular matrix S ∈ Cn×n such that A = SBS H if and only if A and B have the same inertia. Proof. See [6] Theorem 4.5.8. Such a relation in Theorem 2.10, we called the congruent relation. A fact that if A is Hermitian then iA will be skew-Hermitian, and thus the eigenvalues of a skew-Hermitian matirx are all pure imaginary. From (2.1), we see that the i+ (iK) = i− (iK) and therfore K = V0H J V0 can be congruent to J . Applying the above theorems, we have the following theorem.. 19.

(25) Theorem 2.11 (Hamiltonian Jordan Canonical form). Let H be Hamiltonian. Then there is a symplectic U such that   Rr 0   Re De     R D c c     R D d d −1 , (2.8) U HU =  H  0  −Rr     0 −ReH   H   0 −Rc H Gd −Rd. where the blocks of Rr , Re , Rc , Rd are all in Jordan canonical forms and the sub-indices r, e, c and d means the blocks are non-zero eigenvalues, zero real part eigenvalues of even number, combination of two same zero real part eigenvalues of odd number and combination of two different zero real part eigenvalues of odd number, respectively, and the parts of sub-indices with e, c and d take the forms in Lemma 2.8 and Lemma 2.9. Proof. Consider Lemma 2.7, Lemma 2.8 and Lemma 2.9.. Corollary 2.12. A Hamiltonian matrix H ∈ C2n×2n is similar to a Hamiltonian triangular Jordan canonical form if and only if the algebraic multiplicities of all pure imaginary eigenvalues are even. Moreover, if H ∈ R2n×2n , then it is similar to a real Hamiltonian triangular Jordan canonical form if and only if the algebraic multiplicities of all positive pure imaginary eigenvalues are even.. 2.2. Hamiltonian Kronecker canonical form. In this section, we consider the Kronecker canonical form1 for a Hamiltonian pencil and we give a brief conclusion of the canonical form. Definition 2.3. A pencil A − λB ∈ Mm×n (C) is said to be regular if m = n and det(A − λB)6= 0. Otherwise, m 6= n or m = n but det(A − λB)= 0, the pencil is called singular. Before consindering this problem, we recall the generalized eigenvalue problem Ax = λBx. If A, B are both square and B is nonsingular, then the generalized eigenvalue value problem Ax = λBx can be considered as the eigenvalue problem B −1 Ax = λx. Hence, for a Hamiltonian pencil M − λL, 1. Also called the Weierstraβ canonical form.. 20.

(26) if L is invertible, then it can be considered as a Hamiltonian matrix L−1 M. So, in the following, we consider the singular Hamiltonian pencils and note that M, L of the singular Hamiltonian pencil M − λL here are square. Just like matrices have a canonical form, pencils also have a canonical form which is given by [2]. Here we give the theorem without proof. Theorem 2.13. For a pencil A − λB ∈ Mm×n (C), there are nonsingular matrices P ∈ Mm×m (C), Q ∈ Mn×n (C) such that P (A − λB)Q = diag(Lǫ1 , ..., Lǫq , LPη1 , ..., LPηq , I − λN, D − λI), where 1. Lǫ is the µ × (µ + 1) bidiagonal matrix  λ −1  .. ..  . . λ. .. −1.   . 2. LPη is the ”pertransposed” pencil . of dimension (µ + 1) × µ.. −1.   λ  . ... .    .. . −1  λ .. 3. N is a nilpotent Jordan matrix (infinite elementary divisor), and 4. D is in Jordan canonical form (finite elementary divisor). The relation in the above theorem for two pencils is called the strictly equivalent. That is, for any two pencils A1 − λB1 and A2 − λB2 of dimension m × n are said to be strictly equivalent if there are invertible matrices P and Q of order m and n respectively such that P (A1 − λB1 )Q = A2 − λB2 . Similar to previous section, we will discuss the Kronecker form of Hamiltonian pencils for Q is symplectic. First, check whether P (M − λL)Q is still Hamiltonian or not. Proposition 2.14. Let M − λL be Hamiltonian and P be nonsingular. If Q is symplectic, then P (M − λL)Q is also Hamiltonian. 21.

(27) Proof. (P MQ)J (P LQ)H = P MQJ QH LH P H = P MJ LH P H. = P (−LJ MH )P H. = −P LQJ QH MH P H. = −(P LQ)J (P MQ)H. In tihs section, we assume the Hamiltonian pencil is regular. From the definition, MJ LH = −LJ MH which is equivalent to P MQKQH LH P H = −P LQKQH MH P H ,. (2.9). for K = Q−1 J Q−H and invertible P and Q sucht that P (M − λL)Q is in the Kronecker form. Then H I D D I . = K K I O I NH K11 K12 Let K = , we have K21 K22  DK11 + K11 D H = 0    DK12 N H + K12 = 0 , NK21 D H + K21 = 0    K22 N H + NK22 = 0. and hence, from the second, third equation and K is skew-Hermitian, K12 = 0 and K21 = 0. For the first and last equation, it is dicussed in previous section. Moreover, let Q = [Q1 Q2 ], note that QH J Q = −K−1 , then −1 −1 H 2 −K11 = QH 1 J Q1 , −K22 = Q2 J Q2 and Q1 , Q2 are the deflating subspace to the finite and infinite eigenvalues, respectively for the reason that, from (2.9) without multipling P in the left hand side, H H I , = −LQK MQK I NH. For a pencil A − λB, if there are P ∈ Cn×k and Q ∈ Ck×k such that AP = BP Q, then columns of P spans the deflating subspace for A − λB. 2. 22.

(28) or or. M Q1 K11 Q2 K22 N H = −L Q1 K11 H H Q2 K22 , . −1 MQ1 = −LQ1 K11 D H K11 = LQ1 D . −1 H LQ2 = MQ2 K22 N K22 = MQ2 N. Theorem 2.15 (Hamiltonian Kronecker canonical form). Given a regular complex Hamiltonian pencil M − λL. Then there is a nonsingular P and a symplectic U such that L11 L12 M11 M12 −λ P (M − λL)U = L21 L22 M21 M22 with.   Rr − λI   Re − λI     Rc − λI ,  M11 − λL11 =   R − λI d     RM − λRL I − λR∞   0   De     Dc ,  M12 − λL12 =   D d     DM − λDL −λD∞   0   0     0 ,  M21 − λL21 =   G d     GM − λGL 0   H −Rr − λI   −ReH − λI   H   −Rc − λI ,  M22 −λL22 =  H  −R − λI d     HM − λHL H I + λR∞ where the subindices r, e, c, d are the same block in the Theorem of the Hamiltonian Jordan canonical form, ∞ is the block of infinite eigenvalue, and M, L is a combination of a block of pure imaginary eigenvalue and a block of eigenvalue infinite. 23.

(29) Note that if M − λL ∈ R2n×2n , the blocks of subindices M, L are invalid. Corollary 2.16. A regular Hamiltonian pencil M − λL has a Hamiltonian triangular Kronecker caninical form if and only if the algebraic multiplicities of all purely imaginary eigenvalues are even. Moreover, if M − λL is real, then it has a real Hamiltonian triangular Kronecker form if and only if the algebraic multiplicities of all positive pure imaginary eigenvalues are even.. 24.

(30) 3. Real Hamiltonian logarithm of a real symplectic matrices. Given a matrix A ∈ Rn×n , we define the matrix logarithm X of A by e = A, and it is well-known that A has a logarithm if and only if A is invertible and it has an even number of Jordan blocks of each size relative to every negative eigenvalue see [7]. In this section, we discuss when will a real symplectic has a real Hamiltonian logarithm. At first, we consider the following in [4]. X. Lemma 3.1. Given a m × m Jordan block Nm (λ) and a function f which is (m − 1)-times differentiable at λ, then   1 f (λ) f ′ (λ) 12 f ′′ (λ) · · · (m−1)! f (m−1) (λ)   .. 1  0 . (m−2)! f (λ) f ′ (λ) f (m−2) (λ)    . . . . . . f (Nm (λ)) =  (3.1) . . . . .  .  . . . .    0  ··· 0 f (λ) f ′ (λ) 0 ··· ··· 0 f (λ) −Ik 0 Nk (−1) 0 (1) and Let Dk (−1) = , Dk (−1) = 0 −Ik 0 Nk (−1)−T Nk (−1) ek eTk (2) . Dk (−1) = 0 Nk (−1)−T Theorem 3.2. [4, Theorem 2.6] Given a symplectic matrix S ∈ R2n×2n . 1. Suppose that −1 ∈ / λ(S), then there is a Hamiltonian H ∈ R2n×2n such that eH =S if and only if S has an even number of canonical blocks of Nk (λ) 0 relative to negative eigenvalue. the form 0 Nk (λ)−T 2. If −1 ∈ λ(S) and that the conditions relative to negative eigenvalue in 1. are satisfied. Then S has a real Hamiltonian logarithm if, relative to (1) −1, there are only blocks of Dk (−1), blocks of Dk (−1) with k is odd, (2) and an even number of blocks of Dk (−1) with k is even. It is also be told that a real symplectic S will not have a real Hamiltonian (2) logarithm H if S has odd number of blocks of Dk (−1). Here, we give a proof of this statement. Lemma 3.3. Given symplectic matrices S1 and S2 ∈ C2n×2n and S1 is similar to S2 . Let columns of U and V ∈ C2n×k be the generalized eigenvectors 25.

(31) corresponding to the same eigenvalue of S1 and S2 , respectively. That is, S1 = AS2 A−1 for some nonsingular A, and U = AV . If U H J U = 0 and V H J V 6= 0, then S1 and S2 are not a symplectic similarity. Proof. Suppose that A is symplectic,then 0 = U H J U = V H AH J AV = V H J V 6= 0, a contradiction. Note that the lemma gives us that if one generalized eigenspace of a symplectic matrix is J -orthogonal and the generalized eigenspace of the other symplectic matrix is not J -orthogonal, then the two symplectic matrices are not symplectically similar. Next, we give the theorem to the problem. N2n (−1) ∈ R4n×4n has no real HamilTheorem 3.4. S = N2n (−1)−T tonian logarithm. Proof. Suppose that there is a real Hamiltonian H such that eH = S, whose characteristic polynomial and the minimal polynomial is p(t) = (t2 + π 2 )2n . According to Theorem 2.11, there is a symplectic U such that J E −1 , UHU = −J H where J = Nn (πi)⊕Nn (−πi) and E = β1 en eTn +β2 e2n eT2n for β1 , β2 ∈ {−1, 1}, for the reason that H has two even Jordan blocks relative to pure imaginary eigenvalue. Thus, K F −1 H −1 , USU = Ue U = K −H where K = Nn (−1) ⊕ Nn (−1) and F ∈ C2n×2n . Since the generalized eigenspaces of S are span{e1 , ..., e2n } and span{e2n+1 , ..., e4n }, which are both J -orthogonal, while the generalized eigenspaces of eH , the same as H, are span{e1 , ..., en , e2n+1 , ..., e3n } and span{en+1 , ..., e2n , e3n+1 , ..., e4n }, which are not J -orthogonal individually. But, setting U = [e1 , ...e2n ] and V = [e1 , ..., en , e2n+1 , ..., e3n ], U = UV with U is symplectic, contradicting to the Lemma.. 26.

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