Port-Hamiltonian Realizations of Linear Time Invariant Systems

Transcription

1 Port-Hamiltonian Realizations of Linear Time Invariant Systems Christopher Beattie and Volker Mehrmann and Hongguo Xu December Abstract The question when a general linear time invariant control system is equivalent to a port-hamiltonian systems is answered Several equivalent characterizations are derived which extend the characterizations of 38 to the general non-minimal case An explicit construction of the transformation matrices is presented The methods are applied in the stability analysis of disc brake squeal Keywords: port-hamiltonian system passivity stability system transformation linear matrix inequality Lyapunov inequality even pencil quadratic eigenvalue problem AMS subject classification 93A3 93B17 93B11 1 Introduction The synthesis of system models that describe complex physical phenomena often involves the coupling of independently developed subsystems originating within different disciplines Systematic approaches to coupling such diversely generated subsystems prudently follows a system-theoretic network paradigm that focusses on the transfer of energy mass and other conserved quantities among the subsystems When the subsystem models themselves arise from variational principles then the aggregate system typically has structural features that reflects underlying conservation laws and often it may be characterized as a port-hamiltonian (PH) system see for some major references Although PH systems may be formulated in a more general framework we will restrict ourselves to input-state-output PH systems which have the form ẋ = (J R) x H(x) + (F P)u(t) (1) y(t) = (F + P) T x H(x) + (S + N)u(t) Department of Mathematics Virginia Tech Blacksburg VA 2461 USA beattie@vtedu Supported by Einstein Foundation Berlin through an Einstein Visiting Fellowship 2 Institut für Mathematik MA 4-5 TU Berlin Str des 17 Juni 136 D-1623 Berlin FRG mehrmann@mathtu-berlinde The authors gratefully acknowledge the support by the Deutsche Forschungsgemeinschaft (DFG) as part of the collaborative research center SFB 129 Sub- stantial efficiency increase in gas turbines through direct use of coupled unsteady combustion and flow dynamics project A2 Development of a reduced order model of pulsed detonation combuster and by Einstein Center ECMath in Berlin 3 Department of Mathematics University of Kansas Lawrence KS 6645 USA xu@mathkuedu Partially supported by Alexander von Humboldt Foundation and by Deutsche Forschungsgemeinschaft through the DFG Research Center Matheon Mathematics for Key Technologies 1

2 where x R n is the n-dimensional state vector; H : R n ) is the Hamiltonian a continuously differentiable scalar-valued vector function describing the distribution of internal energy among the energy storage elements of the system; J = J T R n n is the structure matrix describing the energy flux among energy storage elements within the system; R = R T R n n is the dissipation matrix describing energy dissipation/loss in the system; F±P R n m are the port matrices describing the manner in which energy enters and exits the system and S + N with S = S T R mm and N = N T R mm describing the direct feed-through of input to output The matrices R P and S must satisfy R P K = P T ; (2) S that is K is symmetric positive-semidefinite This implies in particular that R and S are also positive semidefinite R and S Port-Hamiltonian systems generalize the classical notion of Hamiltonian systems expressed in our notation as ẋ = J x H(x) The analog of the conservation of energy for Hamiltonian systems is for PH systems (1) the dissipation inequality: H(x(t 1 )) H(x(t )) t1 t y(t) T u(t) dt (3) which has a natural interpretation as an assertion that the increase in internal energy of the system as measured by H cannot exceed the total work done on the system H(x) is a storage function associated with the supply rate y(t) T u(t) In the language of system theory (3) constitutes the property that (1) is a passive system 1 One may verify with elementary manipulations that the inequality in (3) is an immediate consequence of the inequality in (2) and holds even when the coefficient matrices J R F P S and N depend on x or explicitly on time t (see 27) or indeed (with care taken to define suitable operator domains) when they are defined as linear operators acting on infinite dimensional spaces Notice that with a null input u(t) = the the dissipation inequality asserts that H(x) is non-increasing along any unforced system trajectory Thus H(x) defines a Lyapunov function for the unforced system so PH systems are implicitly Lyapunov stable 19 Similarly H(x) is non-increasing along any system trajectory that produces a null output y(t) = so PH systems also have Lyapunov stable zero dynamics 9 PH systems constitute a class of systems that is closed under power-conserving interconnection This means that port-connected PH systems produce an aggregate system that must also be port-hamiltonian This aggregate system hence will be guaranteed to be both stable and passive Modeling with PH systems thus represents physical properties in such a way as to facilitate automated modeling 23 and at same time encoding physical properties in the structure of equations This framework also provides compelling motivation to identify and preserve PH structure when producing reduced order surrogate models see Now consider a general linear time-invariant (LTI) system: ẋ = A x + B u y = C x + D u with A R n n B R n m C R m n and D R m m The system (4) is passive if there exists a continuously differentiable storage function H : R n ) such that (3) holds for all admissible inputs u 2 (4)

3 So the natural question arises: When is (4) equivalent to a port-hamiltonian system? Here specifically we consider equivalence to LTI port-hamiltonian systems taking the form ξ = (J R)Q ξ + (F P) ϕ (5) η = (F + P) T Q ξ + (S + N) ϕ with J = J T R = R T Q = Q T > S = S T N = N T where J R Q R n n F P R n m S N R m m and K as defined in (2) is positive semidefinite The notion of system equivalence that we consider here engages three invertible transformations connecting (4) and (5) one on each of the input output and state space: u(t) = Ṽϕ(t) η(t) = VT y(t) and x(t) = Tξ(t) (with Ṽ V T invertible) Within this context the supply rate associated with (4) is transformed as y(t) T u(t) = η(t) T V 1 Ṽϕ(t) We wish to constrain the permissible transformations characterizing system equivalence so as to be power conserving; that is so that supply rates remain invariant y(t) T u(t) = η(t) T ϕ(t) Thus we assume that Ṽ = V and we say that (4) is equivalent to (5) if there exist invertible matrices V and T such that u(t) = V ϕ(t) η(t) = V T y(t) and x(t) = T ξ(t) (6) Since PH systems are structurally passive our starting point is the following characterization of passivity introduced in 39 for minimal linear time invariant systems The system (4) is minimal if it is both controllable and observable The system (4) (and more specifically the pair of matrices (A B) with A R n n B R n m ) is controllable if rank si A B = n for all s C Similarly the system (4) (and the pair (A C) with A R n n C R m n is si A observable if rank = n for all s C C Theorem 1 (39) Assume that the LTI system (4) is minimal The matrix inequality A T Q + QA QB C T (7) B T Q C (D + D T ) has a solution Q = Q T > if and only if (4) is a passive system in which case: i) H(x) = 1 2 xt Qx defines a storage function for (4) associated with the supply rate y T u satisfying (3) ii) There exist maximum and minimum symmetric solutions to (7): < Q Q + such that for all symmetric solutions Q to (7) Q Q Q + This result yields an immediate consequence for PH realizations Corollary 2 Assume that the LTI system (4) is minimal Then (4) has a PH realization if and only if it is passive Moreover if (4) is passive then every equivalent system to (4) (as generated by transformations as in (6)) may also be directly expressed as a PH system 3

4 Proof If (4) has a PH realization then a fortiori it is passive Conversely if (4) is passive then (7) has a positive definite solution Q = Q T = T T T written in terms of a Cholesky or square root factor T Then we can define directly Q = I J = 1 2 (TAT 1 (TAT 1 ) T ) R = 1 2 (TAT 1 + (TAT 1 ) T ) F = 1 ( 2 TB + (CT 1 ) T ) P = 1 ( 2 TB + (CT 1 ) T ) S = 1 2 (D + DT ) N = 1 2 (D DT ) and (7) can be written in terms of these defined quantities as T T R P T 2 I P T S I which verifies (2) Thus J R Q F P S and N as defined in (8) indeed determine a PH system The matrix inequality (7) actually includes two properties the Lyapunov inequality in the upper left block A T Q + QA (9) via Lyapunov s theorem 26 guarantees that the uncontrolled system ẋ = Ax is stable ie A has all eigenvalues in the closed left half plane and those on the imaginary axis are semisimple while passivity is encoded in the solvability of the full matrix inequality (7) If the inequality (9) is strict then the system is asymptotically stable ie all eigenvalues of A are in the open left half plane and strictly passive ie also the dissipation inequality (3) is strict see 24 for a detailed analysis We will discuss the solution of these two types matrix inequalities in the more general situation of non-minimal systems in Section 2 Based on these results we will then construct explicit transformations from of a general LTI system (minimal or not) to port-hamiltonian form (5) in Section 3 2 Lyapunov and Riccati inequalities The solution of Lyapunov and Riccati inequalities as they arise in the characterizations (9) and (7) is usually addressed via the solution of semidefinite programming see 4 We discuss the more explicit characterization via invariant subspaces 21 Solution of Lyapunov inequalities The stability of A is a necessary condition for (9) and (7) which require that TAT 1 + T T A T T T or equivalently that the Lyapunov inequality (9) has a positive definite solution Q = T T T It is well known that the equality case in (9) always has a positive definite solution if A is stable see 26 In the following we recall see eg 4 a characterization of the complete set of solutions of the inequality case If A is stable but not asymptotically stable then due to the fact that the eigenvalues on the imaginary axis are all semi-simple using the real Jordan form of A see eg 2 there exist a nonsingular matrix M R n n such that MAM 1 = diag (A 1 α 2 J 2 α r J r ) (1) (8) 4

5 where A 1 R n 1 n 1 is asymptotically stable α 2 α r are real and distinct and J j = I nj j = 2 r To characterize the solution set of (9) we make the ansatz that I nj Q = M T diag (Q 1 ˆQ 2 ˆQ ) r M (11) and separately consider the determination of the block Q 1 and the other blocks Let W(n 1 ) be the set of symmetric positive semidefinite matrices Θ 1 R n 1 n 1 with the property that Θ 1 x for any eigenvector x of A 1 Then for any Θ 1 W(n 1 ) we define Q 1 to be the unique symmetric positive definite solution of the Lyapunov equation A T 1 Q 1 + Q 1 A 1 = Θ 1 see 26 The other matrices ˆQ j j = 2 r are chosen of the form Yj Z ˆQ j = j > Z j Y j with Z j = Z T j when α j > or an arbitrary ˆQ j > when α j = We have the following characterization of the solution set of (9) Lemma 3 Let A R n n Then the Lyapunov inequality (9) has a symmetric positive definite solution Q R n n if and only if A is stable If A is asymptotically stable then the solution set is given by the set of all symmetric positive definite solutions of the Lyapunov equation A T Q + QA = Θ where Θ is any symmetric positive semidefinite matrix with the property that Θx for any eigenvector x of A or in other words (A Θ) is observable If A is stable but not asymptotically stable then with the transformation (1) any solution of (9) has the form (11) solving the Lyapunov equation with Θ 1 W(n 1 ) A T Q + QA = M T diag(θ 1 )M (12) Proof Consider a transformation to the form (1) and set for a symmetric matrix Q M T QM 1 = Q ij partitioned accordingly Form M T (A T Q + QA)M 1 A T 1 Q 11 + Q 11 A 1 A T 1 Q 12 + α 2 Q 12 J 2 A T 1 Q 1r + α r Q 1r J r α 2 J T 2 = QT 12 + QT 12 A 1 α 2 (J T 2 Q 22 + Q 22 J 2 ) α 2 J T 2 Q 2r + α r Q 2r J r α r J T r Q T 1r + QT 1r A 1 α r J T r Q T 2r + α 2Q T 2r J 2 α r (J T r Q rr + Q rr J r ) again partitioned accordingly For any j = 2 r with α r partition Qj11 Q Q jj = j12 Q T j12 Q j22 according to the block structure in J j so that J T Qj12 Q T j12 Q j11 Q j22 j Q jj + Q jj J j = Q j11 Q j22 Q j12 + Q T j12 5

6 Then α j (J T j Q jj +Q jj J j ) if and only if Q j12 +Q T j12 = and it follows that Q j11 Q j22 = Yj Z Thus any matrix of the form Q jj = j > with Z Z j Y j = Z T j is a positive definite j solution If α j = then clearly any Q jj > is a solution In both cases the resulting diagonal blocks Q jj > satisfy α j (J T j Q jj + Q jj J j ) = Thus for i j the blocks Q ij have to satisfy one of the two matrix equations A T 1 Q 1j + α j Q 1j J j = α i J T i Q ij + α j Q ij J j = which implies that Q ij = for all i j since the spectra of the respective pairs A 1 α j J j and α i J i α j J j are different and thus the corresponding Sylvester equations only have as solution 26 For any positive semidefinite matrix Θ 1 satisfying Θ 1 x for any eigenvector x of A 1 by Lyapunov s Theorem 26 there exists a symmetric positive definite matrix Q 1 satisfying A T 1 Q 1 + Q 1 A 1 = Θ 1 Thus the Lyapunov inequality (9) has a solution Q > and it has the form (11) satisfying (12) The results for an asymptotically stable A can be easily derived as a special case with all ˆQ 2 ˆQ r void in (11) Conversely if the Lyapunov inequality has a solution Q > and if x is an eigenvector of A ie Ax = λx then ( λ + λ)x Qx Since x Qx > it follows that λ + λ ie the real part of λ is non-positive Suppose that A has a purely imaginary eigenvalue iα Since a similarity transformation of A does not change the Lyapunov inequality (9) we may assume that A is in (complex) Jordan canonical form and we may consider each Jordan block separately If A k is a single real Jordan block of size k k k > 1 with eigenvalue iα then for any Hermitian matrix ˆQ k = q ij we have that q 11 q 1k 1 A k Q q 11 q 12 + q 12 q 1k + q 2k 1 k + Q k A k = q 1k 1 q 2k 1 + q 1k q k 1k + q k 1k This matrix cannot be negative semidefinite since otherwise q 11 = which would contradict that ˆQ k is positive definite Therefore A has to be stable ie if it has purely imaginary eigenvalues these eigenvalues must be semi-simple Remark 1 The construction in Lemma 3 relies on the computation of the real Jordan form of A which usually cannot be computed in finite precision arithmetic For the numerical computation of the solution it is better to use the real Schur form see Solution of Riccati inequalities By Corollary 2 we can characterize (at least in the minimal case) the existence of a transformation to PH form via the existence of a symmetric positive definite matrix Q solving the 6

7 linear matrix inequality A T Q QA C T QB C B T Q D + D T (13) In this subsection we will discuss the explicit solution of this matrix inequality in the general (non-minimal) case It is clear that for the inequality to hold S := D + D T has to positive semidefinite However we will see below that we can restrict ourselves to the case that S := D + D T is positive definite so we only consider this case here and using Schur complements (13) is equivalent to the Riccati inequality (A BS 1 C) T Q + Q(A BS 1 C) + QBS 1 B T Q + C T S 1 C (14) We first investigate the influence of the purely imaginary eigenvalues of A Clearly a necessary condition for (13) to be solvable is that Q satisfies A T Q + QA Following Lemma 3 if A has the form (1) then Q must have the form (11) and A T Q + QA has the form (12) Written in compact form we get where MAM 1 A1 = A 2 M T QM 1 Q1 = Q 2 A 2 = diag(α 2 J 2 α r J r ) = A T 2 Q 2 = diag( ˆQ 2 ˆQ r ) satisfying A T 2 Q 2 + Q 2 A 2 = and Setting M T (A T Q + QA)M 1 = MB = B1 B 2 A T 1 Q 1 + Q 1 A 1 CM 1 = C 1 C 2 and premultiplying M T and post-multiplying M 1 to the first block row and column of (13) respectively one has A T 1 Q 1 Q 1 A 1 C T 1 Q 1B 1 C T 2 Q 2B 2 C 1 B T 1 Q 1 C 2 B T 2 Q 2 S Therefore Q 2 must be positive definite satisfying B T 2 Q 2 = C 2 A T 2 Q 2 + Q 2 A 2 = (15) and A T 1 Q 1 Q 1 A 1 C T 1 Q 1B 1 C 1 B T 1 Q (16) 1 S or equivalently it has to satisfy the reduced Riccati inequality Ψ(Q 1 ) := (A 1 B 1 S 1 C 1 ) T Q 1 +Q 1 (A 1 B 1 S 1 C 1 )+Q 1 B 1 S 1 B T 1 Q 1 +C T 1 S 1 C 1 (17) To study the solvability of (17) we note that A 1 is asymptotically stable We claim that A 1 B 1 S 1 C 1 is necessarily asymptotically stable as well To show this suppose that the 7

8 inequality (17) has a solution Q 1 > Since Q 1 B 1 S 1 B T 1 Q 1 + C T 1 S 1 C 1 it follows from Lemma 3 that A 1 B 1 S 1 C 1 is stable and there exits an invertible matrix M 1 such that M 1 (A 1 B 1 S 1 C 1 )M 1 1 has real Jordan form (1) M T 1 Q 1 M 1 1 has the form (11) and following (12) we have M T 1 (Q 1 B 1 S 1 B T 1 Q 1 + C T 1 S 1 C 1 )M 1 1 = diag(θ ) Then due to the positive definiteness of S it follows that C 1 M 1 1 = C 11 and by making use of the block diagonal structure of M T 1 Q 1 M 1 1 we also have M 1B 1 = B T 11 T Thus it follows that M 1 A 1 M 1 1 = diag(a 11 α 2 J 2 α r J r ) Since A 1 is asymptotically stable all α j J j must be void which implies that A 1 B 1 S 1 C 1 must be asymptotically stable as well In order to characterize the solution of the reduced Riccati inequality (17) we use the following lemmas The first is the orthogonal version of the Kalman decomposition Lemma 4 Consider a general LTI system of the form (4) Then there exists a real orthogonal matrix U such that A 11 A 13 U T AU = A 21 A 22 A 23 Ã11 Ã =: 12 Ã A (18) B 1 U T B = B 2 B1 =: CU = C 1 C 3 =: C1 C2 where the pair (Ã11 B 1 ) is controllable and the pair (A 11 C 1 ) is observable Proof Applying first the controllability and then the observability staircase forms of 37 to the system there exists a real orthogonal matrix U such that U T AU is in the form (18) with (Ã11 B 1 ) controllable and (A 11 C 1 ) observable Note that by using the block structure in Lemma 4 we have that (A 11 B 1 ) is controllable as well The next lemma considers the Riccati inequality in the form (refstcf) Lemma 5 Consider the reduced Riccati inequality (17) and suppose that both A 1 and A 1 B 1 S 1 C 1 are asymptotically stable with S > Then there exists a transformation to the condensed form (18) of A 1 B 1 C 1 such that U T (A 1 B 1 S 1 Ã11 C 1 )U = B 1 S 1 C1 Ã 12 B 1 S 1 C2 Ã 22 = A 11 B 1 S 1 C 1 A 13 B 1 S 1 C 3 A 21 B 2 S 1 C 1 A 22 A 23 B 2 S 1 C 3 A 33 (19) where A 11 B 1 S 1 C 1 A 22 and A 33 are all asymptotically stable In addition also A 11 is asymptotically stable (Ã11 B 1 ) and (A 11 B 1 ) are controllable and (A 11 C 1 ) is observable 8

9 The proof is straightforward The following Lemma is also well-known see eg but we present a proof in our notation Lemma 6 Suppose that (A B) is controllable (A C) is observable A is asymptotically stable and S > Then the Riccati equation (A BS 1 C) T Q + Q(A S 1 C) + QBS 1 B T Q + C T S 1 C = (2) has a solution Q > if and only if the Hamiltonian matrix A BS H := 1 C BS 1 B T C T S 1 C (A BS 1 C) T (21) has a Lagrangian invariant subspace ie there exist square matrices W 1 W 2 E such that W1 T W 2 = W2 T W 1 W1 T W2 T T has full column rank and W1 H W 2 = W1 W 2 Moreover if (22) holds then Q := W 2 W 1 1 > solves (2) E (22) Proof Suppose that Q > solves (2) It is straightforward to prove that W1 T W2 T T I Q satisfies the required conditions and (22) with E = A + BS 1 B T Q T = Conversely suppose there exist W 1 and W 2 satisfying all the described conditions Without loss of generality we may assume W1 T W2 T T has orthogonal columns Then it is well known 6 7 that there exists a CS decomposition ie there exist orthogonal matrices U V and diagonal matrices Γ such that U T W 1 V = U T Γ W 2 V = I where is invertible and 2 + Γ 2 = I (Note that the zero block in U T W 1 V could be void Partition U T A11 A AU = 12 U T BS 1 B T G11 G U = 12 A 21 A 22 G T 12 G 22 U T C T S 1 H11 H CU = 12 H T V T E11 E EV = H 22 E 21 E 22 Then (22) takes the form A 11 A 12 G 11 G 12 A 21 A 22 G T 12 G 22 Y 11 Y 12 A T 11 A T 21 Y T 12 Y 22 A T 12 A 22 Γ I = Γ I E11 E 12 E 21 E 22 By comparing the (22) block on both sides one has G 22 = Due to the positive semidefiniteness of S then also G 12 = By comparing the (12) blocks on both sides of the above 9

10 equation and using the nonsingularity of it follows that E 12 = Then by comparing the (32) block on both sides one has A 21 = Hence U T A11 A AU = 12 A 22 U T BS 1 B T G11 U = implying that (A B) is not controllable if W 1 is not invertible Therefore W 1 must be invertible Similarly the observability of (A C) implies that W 2 is invertible Define Q = W 2 W1 1 which is then invertible and also symmetric because of the relation WT 1 W 2 = W2 T W 1 From (22) we then obtain I H Q = I Q (W 1 RW 1 1 ) which implies that Q solves (2) The positive definiteness of Q follows from the fact that Q is invertible A is asymptotically stable and QBS 1 B T Q + C T S 1 C Remark 2 The previous lemmas still hold if A is replaced by A BS 1 C provided that this matrix is also asymptotically stable Note that (A B) is controllable if and only if (A BS 1 C B) is controllable and (A C) is observable if and only if (A BS 1 C C) is observable Lemma 6 shows that the Riccati equation (2) has a solution Q > whenever the Hamiltonian matrix H has a Lagrangian invariant subspace The existence of such an invariant subspace depends only on the purely imaginary eigenvalues of H eg 14 When such an invariant subspace exists then there are many such invariant subspaces Note that the eigenvalues of H are symmetric with respect to the imaginary axis in the complex plane If (22) holds then the union of the eigenvalues of E and E T form the spectrum of H One particular choice is that the spectrum of E is in the closed left half complex plane another choice is that it is in the closed right half complex plane The two corresponding solutions of the Riccati equation (2) are the minimal solution Q and the maximal solution Q + and all other solution of the Riccati equation lie (in the Loewner ordering of symmetric matrices) between these extremal solutions Example 1 Consider the example B = S = 1 C = 1 A = 1 α where α > So A and A BS 1 C = α are both asymptotically stable The Riccati equation (2) is q 2 2αq + 1 = which does not have a real positive semidefinite solution when α ( 1) If α = 1 then it has a unique solution q = 1 associated with E = If α > 1 then it has two solutions α ± α 2 1 > with E = ± α 2 1 Suppose that A 1 B 1 C 1 from (17) have the condensed form (18) with an orthogonal matrix U Partition Q11 U T Q12 Q 1 U = Q 22 Q T 12 The reduced Riccati inequality (17) then is equivalent to U T Ψ(Q 1 )U = Ψ11 (Q 1 ) Ψ12 (Q 1 ) Ψ 12 (Q 1 ) T Ψ22 (Q 1 ) (23) 1

11 where (leaving off arguments) Ψ 11 = (Ã11 B 1 S 1 C1 ) T Q11 + Q 11 (Ã11 B 1 S 1 C1 ) + Q 11 B1 S 1 BT 1 Q11 + C T 1 S 1 C1 Ψ 12 = (Ã11 B 1 S 1 ( C 1 B T 1 Q 11 )) T Q12 + Q 12 Ã 22 + Q 11 (Ã12 B 1 S 1 C2 ) + C T 1 S 1 C2 Ψ 22 = ÃT 22 Q 22 + Q 22 Ã 22 + (Ã12 B 1 S 1 C2 ) T Q12 + Q T 12(Ã12 B 1 S 1 C2 ) + Q T 12 B 1 S 1 BT 1 Q12 + C T 2 S 1 C2 For (17) to have a solution Q 1 > it is necessary that Ψ 11 (Q 1 ) = Ψ 11 ( Q 11 ) has a positive definite solution Q 11 or equivalently the dual Riccati inequality has a solution (Ã11 B 1 S 1 C1 )Ỹ + Ỹ(Ã11 B 1 S 1 C1 ) T + Ỹ C T 1 S 1 C1 Ỹ + B 1 S 1 BT 1 1 Ỹ = Q 11 > Using the partitioning in (18) for Ỹ = Y11 Y 12 Y12 T Y 22 can write this as Φ11 (Ỹ) Φ12(Ỹ) Φ 12 (Ỹ)T Φ 22 (Ỹ) where then one Φ 11 (Ỹ) = (A 11 B 1 S 1 C 1 )Y 11 + Y 11 (A 11 B 1 S 1 C 1 ) T + Y 11 C T 1 S 1 C 1 Y 11 + B 1 S 1 B T 1 Again it is necessary that Φ 11 (Ỹ) = Φ 11(Y 11 ) has a solution Y 11 > or equivalently that the dual inequality (A 11 B 1 S 1 C 1 ) T Q 11 + Q 11 (A 11 B 1 S 1 C 1 ) + Q 11 B 1 S 1 B T 1 Q 11 + C T 1 S 1 C 1 (24) has a positive definite solution Since (A 11 B 1 ) is controllable and (A 11 C 1 ) is observable the inequality (24) is solvable if and only if the corresponding Riccati equation is solvable see 4 25 From Lemma 6 we have as necessary condition for the solvability of (13) or (14) that A 11 B 1 S 1 C 1 is asymptotically stable and that the Hamiltonian matrix A11 B H = 1 S 1 C 1 B 1 S 1 B T 1 C T 1 S 1 C 1 (A 11 B 1 S 1 C 1 ) T (25) has a Lagrangian invariant subspace We now show these two conditions as well as (15) are also sufficient for the existence of a positive definite solution by explicit constructions Clearly we only need to consider (16) or (17) We first show that Ψ 11 has a solution Q 11 > where Ψ 11 is defined in (23) Suppose the Hamiltonian matrix H has a Lagrangian invariant subspace By Lemma 6 we can determine a matrix Q 11 > that solves the equation (24) with the eigenvalues of A 11 B 1 S 1 (C 1 B T 1 Q 11 ) in the closed left half complex plane Then Q Q 11 =: 11 is a solution of Ψ 11 = but Q 11 is only positive semidefinite if A 22 is present If A 11 B 1 S 1 (C 1 B T 1 Q 11 ) 11

12 has no purely imaginary eigenvalues or equivalently H does not have purely imaginary eigenvalues then the Hamiltonian matrix Ã11 H = B 1 S 1 C1 B1 S 1 BT 1 C T 1 S 1 C1 (Ã11 B 1 S 1 C1 ) T does not have purely imaginary eigenvalues either since the spectrum of H is the union of the spectra of H A 22 and A 22 and A 22 is asymptotically stable Consider the Riccati equation Ψ 11 + Ξ = (26) with Ξ being chosen such that (Ã11 B 1 S 1 C1 C T 1 S 1 C1 + Ξ) is observable; such Ξ always exists Recall that (Ã11 B 1 S 1 C1 B 1 ) is controllable The Riccati equation (26) corresponds to the Hamiltonian matrix H Ξ For a sufficiently small (in norm) Ξ by continuity this new Hamiltonian matrix has a Lagrangian invariant subspace (when its eigenvalues are away from the imaginary axis) and by Lemma 6 Ψ = Ξ has a positive definite solution Q11 with all the eigenvalues of à 11 B 1 S 1 ( C 1 B T Q 1 11 ) in the closed left half complex plane If A 11 B 1 S 1 (C 1 B T 1 Q 11 ) has purely imaginary eigenvalues (including ) then there exists an invertible matrix L such that L(A 11 B 1 S 1 (C 1 B T 1 Q 11)L 1 Σ1 = Σ 2 where Σ 1 has only purely imaginary eigenvalues and Σ 2 is asymptotically stable So there exists an invertible matrix L such that Là L 11 1 = L A11 B 1 S 1 (C 1 B T 1 Q 11 ) Σ 1 A 21 B 2 S 1 (C 1 B T 1 Q 11 ) A L 1 = Σ 2 Σ1 =: 22 Σ 2 Σ 32 A 22 where à 11 = Ã11 B 1 S 1 ( C 1 B T 1 Q 11) and Σ 2 is asymptotically stable Since similarity transformations will not affect our analysis without loss of generality we may assume that both L and L are identity matrices Subtracting (Ã11 B 1 S 1 C1 ) T Q 11 + Q 11(Ã11 B 1 S 1 C1 ) + Q 11 B 1 S 1 BT 1 Q 11 + C T 1 S 1 C1 = from (26) yields the Riccati equation (à 11) T Ỹ + Ỹà 11 + Ỹ B 1 S 1 BT 1 Ỹ + Ξ = (27) where Ỹ = Q 11 Q 11 Repartition B B 1 = 1 12 B 2

13 according to à 11 and set Ξ = Ξ 22 Ỹ = Y 2 Then the Riccati equation (27) is reduced to (Σ 2) T Y 2 + Y 2 Σ 2 + Y 2 B 2S 1 (B 2) T Y 2 + Ξ 22 = Since (à 11 B 1 ) is controllable so is (Σ 2 B 2 ) Recall also that Σ 2 is asymptotically stable Analogous to the previous case one can choose (sufficiently small) Ξ 22 with (Σ 2 Ξ 22) observable and then the reduced Riccati equation has a positive definite solution Y 2 with the eigenvalues of Σ 2 + B 2 S 1 (B 2 )T Y 2 in the closed left half complex plane Then Q 11 = Q Q 11 +Ỹ = Q Q 12 + = (Q Y 12 )T Q 22 + Y 11 Y 12 > 2 Y12 T Y 22 where Q 11 has the same size of Σ 1 and Y 22 has the same size of A 22 and is a solution to Ψ 11 = Ξ 22 since à 11 B 1 S 1 ( C 1 B T Q 1 11 ) = à 11 + B 1 S 1 Σ1 B BT 1 Ỹ = 1 S 1 (B 2 )T Y 2 Σ 2 + B 2 S 1 (B 2 )T Y 2 has its eigenvalues is in the closed left half complex plane Once we have such a solution Q 11 we can solve Ψ 12 = for Q 12 This Sylvester equation has a unique solution because Ã22 = A 33 is asymptotically stable and the eigenvalues of à 11 B 1 S 1 ( C 1 B T Q 1 11 ) are in the closed left half complex plane Having (leaving off arguments) solved the equality Ψ 12 = we finally need to solve the inequality Ψ 22 Let Θ = I Q 1 11 Q 12 I From the equations Ψ 11 = Ξ and Ψ 12 = we obtain Ξ Θ T (U T Ψ(Q 1 )U)Θ 1 = Θ T Θ 1 = Ψ22 On the other hand Θ T (U T Q 1 U)Θ 1 = Q11 Π Q T 12 Ξ Ξ 1 Q 11 Ξ Ψ 22 Q T 12 Π = Q 22 Q T 1 12 Q Q Q Q Q 11 Ξ Q 1 11 Q 12 and Θ(U T A 1 U)Θ 1 = Θ(U T B 1 ) = U T B 1 = 1 Ã11 à 12 Ã11 Q Q à 22 B1 Q 1 11 Q 12 à 22 (C 1 U)Θ 1 = C1 C2 C 1 Q 1 11 Q 12 13

14 The (22) block of Θ T (U T Ψ(Q 1 )U)Θ 1 can be rewritten as so that à T 22Π + ΠÃ22 + ( C 2 C 1 Q 1 11 Q 12 ) T S 1 ( C 2 C 1 Q 1 11 Q 12 ) Ψ 22 = ÃT 22Π + ΠÃ22 + ( C 2 C 1 Q 1 11 Q 12 ) T S 1 ( C 2 C 1 Q 1 11 Q 12 ) + Q T 12 Since Ã22 is asymptotically stable and ( C 2 C 1 Q 1 11 Q 12 ) T S 1 ( C 2 C 1 Q 1 11 Q 12 ) + Q T 12 one can always select Ξ such that the Lyapunov equation Ψ 22 = Ξ 1 1 Q 11 F Q Q has a unique solution Π > Clearly for Q 22 = Π + Q T 1 12 Q Q the matrix Q11 Q12 Q 1 = U U Q T > 22 satisfies Q T 12 ξ Ψ(Q 1 ) = U Ξ U T 1 1 Q 11 F Q Q ie Q 1 > solves (16) and (17) We summarize the conditions for the existence of a positive definite solution of the matrix inequality (13) in the following theorem Theorem 7 Consider a general LTI system of the form (4) with A stable and S = D+D T > Let M be invertible such that MAM 1 A1 B1 = MB = CM 1 = C A 2 B 1 C 2 2 where A 2 is diagonalizable and contains all the purely imaginary eigenvalues of A Let A 1 B 1 C 1 have the condensed form (19) with an orthogonal matrix U Then the LMI (13) has a positive definite solution Q if and only if the following conditions hold (a) There exists a positive definite matrix Q 2 satisfying B T 2 Q 2 = C 2 A 2 Q 2 = Q 2 A 2 (b) The block A 11 B 1 S 1 C 1 is asymptotically stable (c) The Hamiltonian matrix H defined in (25) has a Lagrangian invariant subspace If these conditions hold then the LMI (13) has a positive definite solution of the form Q = M T Q1 M > where Q Q 1 > solves (17) and Q 2 is determined from condition 2 (a) and (A BS 1 C) T Q + Q(A BS 1 C) + QBS 1 B T Q + C T S 1 C Ξ = M T U U Ξ T M where Ξ Ξ as in above construction 14

15 Proof The proof follows from the given construction Remark 3 Relation (22) is equivalent to the property that the even matrix pencil λ I I A B A T C T B T C S (28) is regular and has index at most one ie the eigenvalues at are all semi-simple (this follows because S > ) and that it has a deflating subspace spanned by the columns of Q I Y T T with Y = S 1 (C B T Q) ie I I Q I Y (A BY) = A B A T C T B T C S Q I Y (29) A pencil λn M is called even if N = N T and M = M T Since even pencils have the Hamiltonian spectral symmetry in the finite eigenvalues this means that there are equally many eigenvalues in the open left and in the open right half plane Numerically to compute Q it is preferable to work with the even pencil (29) over the Hamiltonian matrix (21) since explicit inversion of S can be avoided Numerically stable structure preserving methods for this task are available see 2 3 Note that this procedure can be directly applied to the original data in (4) if S = D + D T is positive definite and we attempt to solve the Riccati equation of (14) The situation is more complicated if S is singular because then it is well-known that the pencil (28) has Jordan blocks of size larger than 1 (the index of (28) is larger than 1) associated with the eigenvalue which means that the number of finite eigenvalues is less than 2n In this case one has to first deflate the high index and possibly singular part of the system This can be done with the help of the staircase form 8 which has been implemented in a production code 5 The situation is also more difficult if S is indefinite since in this case both BS 1 B T and C T S 1 C cannot be guaranteed to be positive semidefinite and Lemma 6 may no longer be valid Remark 4 To show that the system (4) is passive it is sufficient that the LMI (13) has a positive semidefinite solution Q In this case the conditions can be relaxed First of all the condition B T 2 Q 2 = C 2 can be relaxed to Ker B 2 Ker C T 2 rank C 2B 2 = rank C 2 and C 2 B 2 Also A 2 may have purely imaginary eigenvalues with Jordan blocks For example in the extreme case when C 2 = Q 2 = always satisfies the conditions for any A 2 Secondly for (16) or (17) we still require that A 11 B 1 Ŝ 1 C 1 is asymptotically stable and that the Hamiltonian matrix Ĥ has a Lagrangian invariant subspace Since this only requires that Q 1 a solution can be determined in a simpler way We may simply set Q 11 = Q 11 and Ψ 11 = Then with the block structure the solution of Ψ 12 = has a form Q12 Q 12 = To solve Ψ 22 = for Q Q11 Q12 22 one can show Q 1 = U Q T U 12 Q T and 22 solves the Riccati equation Ψ(Q 1 ) = Also under certain conditions Ã22 = A 33 does not have to be asymptotically stable This shows that for non-minimal systems passivity is more general than the PH property 15

16 3 Construction of port-hamiltonian realizations In the last section we have seen that the existence of a port-hamiltonian realization for (4) reduces to the existence of a nonsingular matrix T R n n or a positive definite matrix Q = T T T such that the matrix inequalities (9) or (7) hold In this section we develop a constructive procedure to check these conditions In our previous considerations the matrix V that is used for basis change in the input space suffices to be invertible but to implement the transformations in a numerical stable algorithms we require V to be real orthogonal Consider V = V 1 V 2 where V 1 is chosen so that the columns of V 1 form an orthonormal basis of the kernel of D+D T To construct such a V we can use a singular value or rank-revealing QR decomposition 16 Then for the symmetric part S of the feedthrough matrix we have S = 1 2 (VT DV + V T DV ) = 1 ( 2 VT D + D T ) V = (3) S 2 where S 2 is symmetric and invertible Set B1 B 2 = BV C T 1 C T 2 := C T V (31) each partitioned compatibly with V as in (3) Scaling the second block row and column of the matrix inequality (7) with V T and V respectively the matrix inequality A T Q + QA QB 1 C T 1 QB 2 C T 2 B T 1 Q C 1 B T 2 Q C 2 S 2 (32) has a solution Q = Q T > if and only if the matrix inequality A T Q + QA QB 2 C T 2 B T 2 Q C (33) 2 S 2 has a positive definite solution Q satisfying the constraint B T 1 Q C 1 We will characterize conditions when this constraint is satisfied in the following subsections For the solution of the matrix inequality then we can restrict the input and output space to the invertible part of D + D T and once the transformation matrices have been constructed this transformation can be extended to the full system making sure that the constraint is satisfied 31 Construction of the transformation in the case D = D T To explicitly construct the transformation to port Hamiltonian form let us first discuss the special case that D = D T ie S = Considering the matrix K in (2) to have K we must have P = and the block V 2 in the transformation of the feedthrough term is void while V = V 1 is any orthogonal matrix Then we can express the conditions (32) and (33) in terms of the factor T in a factorization Q = TT T Corollary 8 For a state-space system of the form (4) with D + D T = the following two statements are equivalent: 16

17 1 There exists a change of basis x = Tz with an invertible matrix T R n n such that the resulting realization in the new basis has PH structure as in (5) 2 There exists an invertible matrix T such that a) (TB) T = CT 1 and b) (TAT 1 ) T + TAT 1 (34) Note that without the constraint (34a) if A is stable then by Lemma 3 the second condition (34b) can always be satisfied Adding the constraint (34a) however makes the question nontrivial We have the following characterization of the transformation matrices T that satisfy (34a) Lemma 9 Consider B C T R n m and assume that rank B = r a) There exists an invertible transformation T satisfying condition (34a) if and only if Ker C T = Ker B rank CB = r and CB or equivalently there exists an invertible (orthogonal) matrix V such that BV = B 1 C T V = C T 1 C 1 B 1 = YY T > where B 1 C T 1 Rn r have full column rank and Y R r r is invertible b) Let N B R n (n r) have columns that form a basis of Ker B T If Condition a) is satisfied then any T satisfying condition (34a) has the form T = UT Z T with T = N T B Y 1 C 1 Z T Z = I (35) where U R n n is an arbitrary orthogonal matrix and Z R (n r) (n r) is arbitrary nonsingular Proof Condition (34a) is equivalent to C = B T T T T Necessary for the existence of an invertible T with this property are the conditions Ker C T = Ker T T TB = Ker B CB = B T T T TB (36) rank CB = rank B T T T TB = rank B = r Conversely Ker C T = Ker B is equivalent to the existence of an orthogonal matrix V R m m such that BV = B 1 C T V = C T 1 (37) with B 1 C T 1 Rn r of full column rank The conditions rank CB = r and CB together are equivalent to C 1 B1 > So there exists an invertible matrix Y (eg a Cholesky factor or a positive square root) such that C 1 B 1 = YY T The matrix T as in (35) is then well defined Furthermore T is invertible since if T Q = then C 1 Q = and N T BQ = The latter statement implies that Q Ran(B) so Q = B 1 z for some z and furthermore C 1 B 1 z = This in turn implies that z = and Q = ; so T is injective and hence invertible 17

18 The invertibility of T implies that B T T T T = V T B T 1 = V T (Y 1 C 1 B 1 ) T NB C T 1 Y T N T B Y 1 C 1 = V T (C1 B 1 )(C 1 B 1 ) 1 C 1 N T B Y 1 C 1 = V T C1 = C Hence (34a) holds with T = T Now suppose that T is any invertible transformation satisfying (34a) Then B T T T (TT 1 ) = CT 1 = B T T T which is equivalent to B T 1 T T ( (TT 1 )T (TT 1 ) I) = Y ( (TT 1 )T (TT 1 ) I) = From this it follows that (TT 1 )T (TT 1 Z ) = T Z I for some invertible matrix Z and that TT 1 Z 1 = U must be real orthogonal I In order to explicitly construct a transformation matrix T as in (35) it will be useful to construct bi-orthogonal bases for the two subspaces Ker B T and Ker C Toward this end let N C contain columns that form a basis of Ker C so that Ran N C = Ker C Such a matrix is easily constructed in a numerically stable way via the singular value decomposition or a rank-revealing QR decomposition of C see 16 Since B and C T are assumed to satisfy Ker B = Ker C T we have singular value decompositions B = U B1 U B2 ΣB with Σ B Σ C R r r both invertible We then obtain V B T C = U C ΣC V T C1 V T C2 (38) N C = V C2 N B = U B2 (39) Observe that N T B N C is nonsingular since if N T B N CQ = and z = N C Q then N T B z = implies that z Ran B But then z = B 1 Y = N C Q implies C 1 B 1 Y = where B 1 and C 1 are defined in (37) and so Y = and hence z = Then since N C has full column rank we have Q = Thus N T B N C is injective hence invertible Performing another singular value decomposition N T B N C = Ũ ṼT with positive diagonal and Ũ Ṽ real orthogonal we can perform a change of basis ÑB = N B Ũ 1/2 and Ñ C = N C Ṽ 1/2 and obtain that the columns of ÑB form a basis for Ker B T the columns of ÑC form a basis for Ker C and these two bases are bi-orthogonal ie ÑT BÑC = I and we have T = Ñ T B Y 1 C 1 T 1 = Ñ C B 1 Y T (4) Using the formula (35) we can express the conditions for a transformation to PH form that we have obtained so far in a more concrete way 18

19 Corollary 1 Consider system (4) with D = D T and rank B = r Let the columns of ÑB and ÑC span the kernels of B T C and satisfy ÑT BÑC = I Then system (4) is equivalent to a PH system if and only if 1 Ker C T = Ker B rank CB = r CB and 2 there exists an invertible matrix Z such that ( Z T I AT 1 Z 1 Z + I I and T T 1 are defined in (4) T AT 1 Z 1 I ) T (41) Proof The condition follows from Corollary 8 and the representation (35) by setting U = I and T as in (4) 32 Construction of the transformation in the case of general D For the case that D is general we will present a recursive procedure The first step is to perform the transformations (3) (31) and to obtain the following characterization when a transformation to PH form exists Lemma 11 Consider system (4) transformed as in (3) and (31) Let the columns of ÑB 1 and ÑC 1 form the kernels of B T 1 C 1 respectively and satisfy ÑT B 1 Ñ C1 = I Then the system is equivalent to a PH system if and only if 1 Ker C T 1 = Ker B 1 rank C 1 B 1 = rank B 1 and C 1 B 1 and 2 there exists an invertible matrix Z such that with and T = Ỹ := Z I Ñ T B 1 Y 1 C 1 Ỹ + ỸT (42) T AT 1 T B 2 C 2 T 1 V2 T DV 2 Z 1 I (43) T 1 = Ñ C1 B 1 Y T C 1 B 1 = YY T (44) and B 1 C 1 are defined similar as in (37) just with B C being replaced by B 1 C 1 Proof Condition (34a) in this case has the form (TB 1 ) T = C 1 T 1 If this condition holds then applying the result of Lemma 9 to B 1 and C 1 one has the form (35) The result is proved by applying this formula to (34b) Note that Ỹ is obtained from the one given in (34b) via a congruence (and similarity) transformation with diag(u I) where U is orthogonal 19

20 We can repartition the middle factor of Ỹ in (43) as T AT 1 T B 2 C 2 T 1 V2 T DV 2 = =: ÑT B 1 AÑC ÑT 1 B 1 AB 1 Y T ÑT B 1 B 2 Y 1 C 1 AÑC 1 Y 1 C 1 AB 1 Y T Y 1 C 1 B 2 C 2 Ñ C1 C 2 B 1 Y T V2 T DV 2 Ã 1 B 1 (45) C 1 D1 Thus we have that Ỹ = Z I Ã 1 B 1 Z 1 C 1 D1 I and condition (42) is analogous to condition (9) just replacing T A B C D with Z Ã 1 B 1 C 1 D 1 Hence the existence of Z can be checked again by using Lemma 9 This implies that the procedure of checking the existence of a transformation from (4) to a PH system can be performed in a recursive way One first performs the transformation (3) and checks whether a condition as in Part 1 of Lemma 11 does not hold in which case a transformation does not exist or one has that D + D T in (45) is invertible In the latter case if the associated matrix inequality does not have a positive definite solution then a transformation does not exist Otherwise a transformation exists and a corresponding transformation matrix T can be constructed by computing Z satisfying Ỹ(Z) + ỸT (Z) and the matrix T is formed accordingly After this the process is repeated in a recursive manner To formalize the recursive procedure let A B G = C D and suppose that (34a) is satisfied Then form where T = G 1 := ṼT T G T 1 Ṽ T I I Ṽ = V V is the matrix in the decomposition (3) times a permutation that interchanges the last block columns and T is obtained from (44) In this way we obtain ÑT B 1 AÑC ÑT 1 B 1 AB 1 Y T ÑT B 1 B 2 Y G 1 = 1 C 1 AÑC 1 Y 1 C 1 AB 1 Y T Y 1 C 1 B 2 Y T C 2 Ñ C1 C 2 B 1 Y T V2 T DV 2 V2 T DV 1 Y V1 T DV 2 V1 T DV 1 by using the fact that C 1 T 1 = (T B 1 ) T Y = V 2

21 where V is defined in (37) for B 1 C 1 By (3) we have that V T 2 DV 1 = V T 1 DV 2 V T 1 DV 1 = V T 1 DT V 1 so that we can express G 1 as G 1 = Ã1 B 1 C 1 D1 Γ 1 Γ T 1 Φ 1 Ã If 1 B 1 satisfies condition (34a) then in an analogous way we construct C 1 D1 such that T 1 = diag(t 1 I I) Ṽ 1 = diag(i V 1 I) G 2 = ṼT T 1 1 G T Ṽ 1 = Ã2 B 2 C 2 D2 Γ 2 Γ 11 Γ T 2 Φ 2 Γ 12 Γ T 11 Γ T 12 Φ 1 and we continue Since the size of the matrices T j is monotonically decreasing this procedure will terminate after a finite number of k steps with Ãk B k G k = ṼT T k 1 k 1 ṼT T G T 1 1 Ṽ T k 1Ṽk 1 = C k Dk Γ k ΓT k Φk where D k + D T k is positive definite Φ k = Φ T k T j = diag(t j I I) Ṽ j = diag(i lj V j I) and l j is the size of T j for j = k 1 If there exists an invertible matrix T k such that Tk I Ã k B k C k Dk T 1 k I ( Tk + I Ã k B k C k Dk T 1 k I ) T see the solvability conditions in the previous section then with T k = diag(t k I I) we have T k G k T 1 k + ( T k G k T 1 k )T (46) Observe that for each i j with j > i the matrices Ṽi and ṼT i commute with T j and and thus setting T = T k T Ṽ = Ṽ Ṽk 1 then and inequality (46) implies that T k G k T 1 k = ṼT TG T 1 Ṽ TG T 1 + ( TG T 1 ) Then the desired transformation matrix T is positioned in the top diagonal block of T and the matrix V is positioned in the bottom diagonal block of Ṽ T 1 j 21

22 Remark 5 The recursive procedure described above requires at each step the computation of three singular value decompositions to check the ranks of the matrices B j C j and to construct the bi-orthogonal bases so that (44) holds While each step of this procedure can be implemented in a numerically stable way the consecutive rank decisions and the consequence of performing these subtle decisions are hard to analyze similar to the case of staircase algorithms In general the strategy should be adapted to the goal of the computation which in our case is to obtain a representation in PH form that is robust to small perturbations 33 Explicit solution of linear matrix inequalities via even pencils We have seen that to check the existence of the transformation to PH form and to explicitly construct the transformation matrices T V we can consider the solution of the linear matrix inequality (7) As we have discussed before the best way to do this is via the transformation of the even pencil (28) In this subsection we combine the recursive procedure with the construction of a staircase like form for this even pencil Let for a given real symmetric matrix Q denote by A Ψ (Q) := T Q QA C T QB C B T Q D + D T the corresponding block matrix which is supposed to be positive semidefinite Let V 1 B 1 C 1 be defined as in (3) (31) If B 1 C 1 satisfy Part 1 of Lemma 11 then since Ker C T 1 = Ker B 1 there exist real orthogonal matrices Ũ1 Ṽ1 (which can be obtained by performing a permuted singular value decomposition of B 1 ) such that Ũ T 1 B 1 Ṽ 1 = Σ B Ṽ1C T C11 C 1 Ũ 1 = 12 where Σ B is invertible and C 12 Σ B is real symmetric and positive definite Transforming the desired Q correspondingly as Q11 Q Q := ŨT 1 QŨ1 = 12 Q T 12 Q 22 then since the linear matrix inequality (7) implies QB 1 = C T 1 it follows in the transformed variables that Q 22 := C T 12Σ 1 B > Q 12 := C T 11Σ 1 B (48) and it remains to determine Q 11 so that Q > To achieve this we set Q12 Q 1 Q := 22 QT 12 Q 12 Q T = T T 12 Q T 1 22 Q T I = 22 Q 1 22 QT 12 I (47) and we clearly have that Q Then we can rewrite Q as ( ) Q1 Q = + Q = T T Q1 + T 1 Q = T T Q1 22 Q 22 T 1 22

23 partitioned analogously and we obtain that Q > if and only if Q 1 > By performing a congruence transformation on Ψ (Q) with T Z := V T = Ũ1T V = V Ṽ1 I and using the fact that Q(ŨT 1 B 1Ṽ1) = (ṼT 1 C 1Ũ1) T for any real symmetric Q 1 it follows that (T 1 Z T AT ) T T T QT T T QT (T 1 AT ) T T CT 2 TT QT (T 1 B 2) Ψ (Q)Z = C 2 T (T 1 B 2) T T T QT 2S 2 Partitioning T 1 AT =: A11 A 12 A 21 A 22 T 1 B B13 2 = B 23 C 2 T = C 31 C 32 and using that we obtain that Z T Ψ (Q)Z = T T Q1 QT = Q 22 A T 11 Q 1 Q 1 A 11 Q 1 A 12 A T 21 Q 22 C T 31 Q 1B 13 A T 12 Q 1 Q 22 A 21 A T 22 Q 22 Q 22 A 22 C T 32 Q 22B 23 C 31 B T 13 Q 1 C 32 B T 23 Q 22 2S 2 In this way we have that (7) holds for some Q > if and only if Ψ 1 (Q 1 ) := =: A T 11 Q 1 Q 1 A 11 Q 1 A 12 A T 21 Q 22 C T 31 Q 1B 13 A T 12 Q 1 Q 22 A 21 A T 22 Q 22 Q 22 A 22 C T 32 Q 22B 23 C 31 B T 13 Q 1 C 32 B T 23 Q 22 2S 2 A T 1 Q 1 Q 1 A 1 C T 1 Q 1B 1 C 1 B T 1 Q 1 D 1 + D T 1 holds for some real symmetric positive definite Q 1 where Q22 A A 1 = A 11 C 1 = 21 B 1 = A 12 B 13 and C 31 D 1 + D T A T 1 = 22 Q 22 Q 22 A 22 C T 32 Q 22B 23 C 32 B T 23 Q 22 2S 2 This construction has reduced the solution of the linear matrix inequality (7) to the solution of a smaller linear matrix inequality of the same form Thus we can again proceed in a recursive manner with the same reduction process until either the condition in Part 1 of Lemma 9 no longer holds (in which case no solution exists) or D k + D T k is positive definite for some k 23

24 This reduction process can be considered as the construction of a structured staircase form for the even pencil (28) By applying a congruence transformation to the pencil (28) with the matrix T T Y = T V it follows that I I I λ I A 11 A 12 B 13 A 21 A 22 Σ B B 23 A T 11 A T 21 A T 12 A T 22 C T 12 C T 31 C T 32 Σ T B C 12 B T 13 B T 23 C 31 C 32 S1 where S 1 := 2S 2 By performing another congruence transformation with the matrix the pencil becomes λ Ỹ = I I I I I I Q 22 I I I I I A 11 A 12 B 13 A 21 A 22 Σ B B 23 A T 11 A T 21 A T 21 Q 22 A T 12 A T 22 Q 22 A 21 A T 22 Q 22 Q 22 A 22 C T 31 C T 32 Q 22B 23 Σ T B B T 13 B T 23 C 31 C 32 B T 23 Q 22 S1 By further moving the last block row and column to the fifth position and then the 2nd block 24

25 row and column to the fifth position ie by performing a congruence permutation with one obtains λ I I Γ 1 Γ T 1 I I I I I I I A 1 B 1 A T 1 C T 1 A T 21 B T 1 C 1 D 1 + D T 1 1 A 21 T 1 Σ 1 Σ T 1 where Γ 1 = I A T 1 = 22 B T 23 Σ 1 := Σ B and A 1 B 1 C 1 D 1 + D T 1 are as defined before In this way we may repeat the reduction process on the (11) block which corresponds to Ψ 1 In order to exploit the block structures of the pencil we use a slightly different compression technique for D 1 + D T 1 Note that we may write D 1 + D T D11 D 1 = 12 D T 12 S 1 with S 1 symmetric positive definite Then we have D 1 + D T 1 = I S 1 1 DT 12 I T D11 D 12 S 1 1 DT 12 S 1 I S 1 1 DT 12 I Let D 11 D 12 S 1 1 DT 12 = Z 1 S 2 Z T 1 where S 2 is invertible and Z 1 is orthogonal Then Z 1 S Z 1 I T (D 1 + D T 1 ) Z 1 S 1 1 DT 12 Z 1 I = S 2 S 1 =: S 2 A necessary condition for the existence of a transformation to PH form is that S 2 > or equivalently S 2 > If this holds then using the fact Z 1 S 1 1 DT 12 Z 1 I T Γ 1 = Z T 1 25

26 by performing a congruence transformation on the 3rd block rows and columns with Z 1 S 1 1 DT 12 Z 1 I and another congruence transformation on the fourth block row and column with Z 1 we obtain the pencil λ I I Γ 11 Γ 12 Γ T 11 Γ T 12 A 1 B 11 B 12 A T 1 C T 11 C T B T 11 C B T 12 C 21 S 2 31 T 11 T 21 T 31 Σ 1 Σ T 1 where Γ11 Γ 12 = Γ = I I In order to proceed B 11 C T 11 must satisfy the same conditions as B 1 C T 1 If these conditions hold then we can perform a second set of congruence transformation and transform the pencil to λ I I Γ 2 Γ 11 Γ T 2 Γ 12 Γ 13 Γ T 11 Γ T 12 Γ T 13 A 2 B 2 A T 2 C T 2 Θ T 2 11 B T 2 C 2 D 2 + D T Θ 2 T 2 Σ 2 Σ T T 11 T 21 T 41 T 51 Σ 1 Σ T 1 where Γ 11 Γ 12 Γ 13 = Γ 12 Γ 11 Γ 2 = I 26