only nite eigenvalues. This is an extension of earlier results from [2]. Then we concentrate on the Riccati equation appearing in H 2 and linear quadr


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1 The discrete algebraic Riccati equation and linear matrix inequality nton. Stoorvogel y Department of Mathematics and Computing Science Eindhoven Univ. of Technology P.O. ox 53, 56 M Eindhoven The Netherlands Fax: li Saberi School of Electrical Engineering and Computer Science Washington State University Pullman, W U.S.. Fax: (59) February 4, 998 bstract In this paper we study the discrete time algebraic Riccati equation. In particular we show that even in the most general cases there exists a oneone correspondence between solutions of the algebraic Riccati equation and deating subspaces of a matrix pencil. We will also study the relationship between algebraic Riccati equation and the discrete time linear matrix inequality. We show that in general only a subset of the set of rankminimizing solutions of the linear matrix inequality correspond to the solutions of the associated algebraic Riccati equation, and study under what conditions these sets are equal. In this process we also derive very weak assumptions under which a Riccati equation has a solution. Keywords: discretetime, algebraic Riccati equation, linear matrix inequality, rankminimizing solutions Introduction The algebraic Riccati equation and linear matrix inequality are important tools in linear systems and control theory. Since their introduction in control theory in the early sixties, they have appeared in an impressive range of problems in control theory including H 2 and H optimal control theory. In this paper, we rst establish some properties of the general discrete algebraic Riccati equation. In particular, we show that an algebraic Riccati equation can be associated to a matrix pencil and we will show that the solutions of the algebraic Riccati equation have a one to one relationship with deating subspaces of the matrix pencil having The work of li Saberi is partially supported by the National Science Foundation under Grant ECS y The research of dr... Stoorvogel has been supported by the Royal Netherlands cademy of Sciences and rts and by the Netherlands Organization for Scientic research (NWO).
2 only nite eigenvalues. This is an extension of earlier results from [2]. Then we concentrate on the Riccati equation appearing in H 2 and linear quadratic control. Establishing a connection between the solutions of a linear matrix inequality and its associated algebraic Riccati equation has been a longstanding research problem. In continuoustime setting it is shown (see [8, 2, 28]) that the set of boundary solutions of the continuous linear matrix inequality coincides with the set of real symmetric solutions of an appropriately dened continuous algebraic Riccati equation. The boundary solutions of the continuous linear matrix inequality are those solutions which minimize the rank of the given matrix in the continuous linear matrix inequality and hence are known as rankminimizing solutions. The rankminimizing solutions of the continuous linear matrix inequality play a prominent role in H 2 optimal control theory, and their characterization in terms of an appropriately dened continuous algebraic Riccati equation is of signicant interest. In this paper we will examine the connections between rankminimizing solutions of the linear matrix inequality and solutions of the algebraic Riccati equation in a discrete time setting. Our rst observation is that, unlike the continuoustime case, in general the rankminimizing solutions of the discrete linear matrix inequality cannot be obtained from solutions of an appropriately dened discrete algebraic Riccati equation. However, we show that a subset of rankminimizing solutions of the discrete linear matrix inequality which we refer to as strong rankminimizing solutions, coincides with the set of real symmetric solutions of the associated discrete algebraic Riccati equation. This fact was noted in [9] and we will present it in a more general setting. We will show that the rankminimizing solutions of the linear matrix inequality have a one to one relationship with deating subspaces of the matrix pencil. Moreover, the strongly rankminimizing solutions are solutions of the algebraic Riccati equation and hence, as noted before, can be connected to deating subspaces of this matrix pencil having only nite eigenvalues. We will also discuss results like the existence of solutions to the H 2 algebraic Riccati equation and in so doing we improve the existing results in the literature. We will use a reduction technique to derive our results for the general case. This is presented in section 2. The reduction technique for the H 2 case which is slightly more powerful is presented in the appendix. lso some properties of matrix pencils, which we will need in our derivations, are presented in the appendix. 2 The discrete algebraic Riccati equation 2. asic properties and denitions The classical discrete time Riccati equation such as it appeared in Linear Quadratic (LQ) control and H 2 optimal control (see e.g. [2, 6]) is an equation for an unknown symmetric matrix X which should be such that T X + D T D is invertible and X = T X? ( T X + C T D)( T X + D T D)? ( T X + D T C) + C T C (2.) Moreover a special solution of this equation was often needed; the socalled stabilizing solution which additionally is such that the matrix? ( T X + D T D)? ( T X + D T C) (2.2) 2
3 is asymptotically stable. However, it has become more and more obvious over the years that this equation does not capture important cases and we therefore need a more general discrete time Riccati equation. To be more specic we will list some of these problems: (i) If the system characterized by the system parameters (; ; C; D) is not leftinvertible then the classical discrete time Riccati equation has no symmetric solutions (ii) In the indenite LQ problem (see [2]) and the H optimal control (see e.g. [3,, 26]) a Riccati equation appeared of a slightly dierent structure. (iii) The antistabilizing solution of a Riccati equation, where the matrix (2.2) has all its eigenvalues outside the unit circle, is in certain cases also of interest (for instance in spectral factorization). It can be shown that the eigenvalues of the stabilizing solution are mirror images with respect to the unit circle of the eigenvalues of the antistabilizing solutions. Therefore, if the stabilizing solution has eigenvalues in, there does not exist an antistabilizing solution of the classical discrete time Riccati equation since a matrix obviously only has nite eigenvalues. However, it will be shown in this paper that a dierent denition of stability not on the basis of the eigenvalues of the matrix (2.2) but on the basis of eigenvalues of a pencil will allow us to naturally extend the denition of antistabilizing solutions to these cases. The general discrete algebraic Riccati equation we will study in this paper is formally introduced in the following denition: Denition 2. Let 2 R nn, 2 R nm, Q 2 R nn, R 2 R mm and R being symmetric, be given. Then X = T X? ( T X + S T )( T X + R) y ( T X + S) + Q Ker( T X + R) Ker( T X + S) and S 2 R mn, with Q (2.3a) (2.3b) is called the general discrete algebraic Riccati equation where M y denotes a generalized inverse of the matrix M. generalized inverse is dened by the properties MM y M = M and M y MM y = M y and is in general not unique. On the other hand, due to the condition (2.3b), equality (2.3a) is independent of the specic generalized inverse we are using. It is obvious that the above denition is a generalization of equation (2.). First of all, we replace the inverse by a generalized inverse and secondly we replace the matrices C; D in a more general form with Q; R; S. Note that if Q; R; S are such that Q S > ; R S T which we will sometimes refer to as the positive semidenite case then there exist matrices C; D such that Q S T = C T C D ; (2.5) S R D T (2.4) 3
4 and then equation (2.3a) reduces to (2.) except that the inverse is replaced by a generalized inverse. We also tried to motivate that we need an extension of the denition of stabilizing and antistabilizing solutions. The formal denition is presented in the following denition: Denition 2.2 Consider the following matrix pencil Q + T X? X T X + S T C T X + S T X + R (2.6) X is called a stabilizing solution of the algebraic Riccati equation if X satises (2.3) and is such that number of innite zeros of (2.6) precisely equals rank( T X + R) and the remaining zeros are nite and are inside the unit circle. X is called a semistabilizing solution of the algebraic Riccati equation if X satises (2.3) and is such that number of innite zeros of (2.6) precisely equals rank( T X + R) and the remaining zeros are nite and are inside or on the unit circle. X is called a antistabilizing solution of the algebraic Riccati equation if X satises (2.3) and is such that all zeros of (2.6) are outside the unit circle. For the denition of zeros of a matrix pencil we refer to appendix. Note that if a solution of the general Riccati equation (2.3a) is such that T X + R is invertible then the nite zeros of the pencil (2.6) are exactly the eigenvalues of the matrix (2.2). In other words, this denition is consistent with the denition given before for the classical case. Next we would like to see under what conditions a solution of the algebraic Riccati equation will be such that R + T X is invertible. In the semidenite case where we have (2.5) it was already wellknown (see e.g. [24]) that a solution of the Riccati equation will be such that R + T X is invertible if and only if (; ; C; D) is leftinvertible, or, equivalently, the following matrix H has full rank for all but nitely many z 2 C : H(z) = T (z? I? T )? Q S T I S R (zi? )? This function is often called the Popov function and its relation to the Riccati equation has been studied in many places (see e.g. [, 3]). In the positive semidenite case, i.e. if there exist matrix C and D satisfying (2.5), we have H = G G where G is the transfer matrix of (; ; C; D) and G (z) = G T (z? ) and then it is obvious that H has full rank if and only if (; ; C; D) is leftinvertible. The amazing fact for this special case is that either all or none of the solutions will satisfy this extra invertibility assumption. We will see that this latter property holds in general. We will need the following auxiliary lemma which is an extension of a result in [9]. The latter is the discretetime analagon of a results found in [23]. I (2.7) 4
5 Lemma 2.3 For any symmetric matrix X we have: H(z) = H(z; X) := T (z? I? T )? I L(X) (zi? )? I Proof : The identity can be veried by writing down a realization in descriptor form for H and H(; X). Just as a reminder, the inertia of a matrix is dened as the triple of the number of eigenvalues in the open left half plane, the number of eigenvalues on the imaginary axis and the number of eigenvalues in the open right half plane. The following lemma is an extension of [3]. Theorem 2.4 ssume that a symmetric matrix X satisfying (2.3) exists. Then H has full normal rank if and only if T X + R is invertible. The inertia of T X + R are equal to the inertia of H(z) for all but nitely many z on the unit circle. T X + R > if and only if H(z) > for any point z on the unit circle. Proof : ssume that a symmetric matrix X satises the algebraic Riccati equation. Then it can be checked straightforwardly that V T (z? )L(X)V (z) = H(z) where L and V are given by L(X) := Q + T X? X T X + S T ; T X + S T X + R V (z) := (zi? )? I I?( T X + R) y ( T X + S) Note that V is square and invertible for almost all z. Hence, (2.8) implies that the rank of L(X) equals the normal rank of H. Moreover, for all but nitely many points on the unit circle, the inertia of L(X) is equal to the inertia of H together with an appropriate number of zero eigenvalues. Since the Schur complement of T X + R in L(X) is zero we nd that the inertia of T X + R equals the inertia of H(z) for all but nitely many points on the unit circle. This guarantees in particular that H has full normal rank if and only if T X + R is invertible. Moreover, we have that H(z) > on the unit circle if and only if T X + R >. (2.8) The normal rank of a rational matrix H is dened as the maximal rank of H(z) over all z 2 C attained for all but nitely many z, it will be denoted by normrank H. which is 5
6 The last point in the above lemma is basically a special case of the second point but listed separately since it will play an important role in the rest of this paper. Note that the above lemma implies that a necessary condition for the existence of a solution to the discrete algebraic Riccati equation is that the inertia of H(z) are independent of z except for possibly some singularities. necessary condition for the existence of a solution with T X + R > is that H(z) > for all z on the unit disc. This condition was already presented for the discrete time in [2] and in [28] for the continuous time. Finally, note that H being of full normal rank guarantees that the generalized inverse in (2.3a) is a normal inverse and that (2.3b) is automatically satised. In other words, in that case we can simply focus on the equation: X = T X? ( T X + S T )( T X + R)? ( T X + S) + Q (2.9) In the positive semidenite case we then obtain the classical case of equation (2.). 2.2 Reduction to the case that H has full normal rank The algebraic Riccati equation presented in the previous subsection is very general. In this subsection, we will present a technique to reduce problems where H, as dened in (2.7), does not have full rank to the case where H has full rank. The method presented here needs an a priori given solution of the Riccati equation and is hence not interesting for computational purposes but is very useful in deriving properties of the Riccati equation. reduction method for the special case where H is positive semidenite on the unit circle is presented in appendix and only relies on an a priori solution of a linear matrix inequality instead of an a priori solution of the Riccati equation. To present the reduction scheme, we need the controllability subspace (see e.g. [4]) of a linear system (; ; C; D): Denition 2.5 subspace R is called a controllability subspace for the system : ( _x = x + u y = Cx + Du if for any initial condition x() = x 2 R there exists an input u which steers the state x to in nite time while keeping the output y identically. Equivalently R is a controllability subspace if there exists a matrix F such that R is the smallest + F invariant subspace containing im \ R and contained in Ker(C + DF ). R () is dened as the largest controllability subspace of the system. n important property of the largest controllability subspace is expressed in the following lemma (see e.g. [22]): Lemma 2.6 system with realization (; ; C; D) is left invertible if and only if R () = fg and ( T D T ) T is injective. 6
7 We assume that there exists one solution X of the algebraic Riccati equation (2.3). We dene S, Q, and R by: Q S S T R := Q + T X? X T X + S T T X + S T X : (2.) + R Then it is easy to check that X satises (2.3) if and only if ~ X = X? X satises: ~X = T ~ X? ( T ~ X + S T )( T ~ X + R) y ( T ~ X + S) + Q (2.a) Ker( T ~ X + R) Ker( T ~ X + S) (2.b) It is straightforward to check that we have: H(z) = G T (z? ) R y G(z) (2.2) where G is the transfer matrix associated to the system X with realization(; ; S; R). It is also easy to see that the normal rank of H is equal to the normal rank of G. Therefore if H does not have full normal rank then the system X is not leftinvertible. We dene R ( X) according to denition 2.5. Hence there exists a matrix F such that R ( X) is ( + F ) invariant and contained in Ker( S + RF ). We dene the shifted algebraic Riccati equation as follows: Denition 2.7 The shifted discrete algebraic Riccati equation associated with the discrete algebraic Riccati equation (2.3) is dened as: ~X = ~ T ~ X ~? ( ~ T ~ X + ~ S T )( T ~ X + R) y ( T ~ X ~ + ~ S) + ~ Q (2.3a) Ker( T ~ X + R) Ker( ~ T ~ X + ~ S) (2.3b) where ~ = + F, ~ S = S + RF and ~ Q = Q + S T F + F S + F RF. Observation 2.8 Let X be a solution of the algebraic Riccati equation (2.3) and let F be such that R ( X) is ( + F )invariant and contained in Ker( S + RF ). (i) X is a solution of the discrete algebraic Riccati equation (2.3) if and only if ~ X = X? X is a solution of the associated shifted discrete linear matrix inequality (2.3). (ii) Let X be a solution of the algebraic Riccati equation (2.3). The zeros of the matrix pencil (2.6) are equal to the zeros of the following matrix pencil: zi? ~Q + ~ T X ~ ~? X ~ ~ T X ~ + S ~ T C T X ~ ~ + S ~ T X ~ + R where ~ X = X? X. 7
8 This observation is an immediate consequence of the fact that a matrix X satises (2.3) if and only if rank Q + T X? X T X + S T = rank T X + R (2.4) T X + S T X + R The above observation shows that without loss of generality we can focus on the shifted algebraic Riccati equation which has more structure and is hopefully easier to handle. In particular if we choose a basis in the state space X X 2 such that X = R ( X) and a basis in the input space U U 2 such that U =? R ( X) \ Ker R. In that basis we get that ; ; Q ~ and R ~ have a special form: ~ = 2 22 ; ~ = 2 22 ; Q ~ = ; R = ; (2.5) Q 22 R 22 such that ( ; ) is controllable. Note that since R 22 is invertible we can even make sure that F is such that ~ S = and, since ( ; ) is controllable, we can also guarantee that and 22 have no eigenvalues in common by a suitable choice of F. Consider H, as dened by (2.7) for the original Riccati equation, for the shifted Riccati equation, which we denote by H s. Then analogously to (2.2) we obtain: H s (z) = R (2.6) Using the particular basis for the input space chosen above we nd: H s (z) = R 22 Consider an arbitrary solution X ~ of the shifted Riccati equation which we decompose in the basis chosen before: X = X X 2 X T 2 X 22 Then we have : H s (z) = G T X (z? )( T ~ X + R) y GX(z) (2.7) where GX is the transfer matrix of the system with realization ( ~ ; ; T X ~ ; T X + R). Since rank H s (z) = rank GX(z) for almost all z we nd that in our new basis GX(z) = GX;(z) (2.8) This implies in particular that T X + R j U = or, equivalently, using the decompositions we presented above: T X = ; T 2 X T 2 = Next, if we consider (2.3b) in our new basis we obtain: T X = ; T 2 X T + T 22 X T 2 = 8
9 Then, (2.8) and the above equalities imply that the transfer matrix associated to the system ( ; ; T 2 X ; ) is equal to. Since ( ; ) is controllable, this implies T 2 X = Finally, since X satises the Riccati equation we must have that im( ~ Q + ~ X ~? X) im ~ T X which combined with the above equalities yields that T X? X = ; T 22 X T 2? X T 2 = To make the use of all the above equalities we will use the following technical lemma: Lemma 2.9 Let (; ) be controllable. Then X = is the unique symmetric solution of the following linear matrix inequality: T X? X T X T X > (2.9) T X Proof : Let X be an arbitrary solution of (2.9). For any matrix F we nd: ( + F ) T X( + F )? X ( + F ) T X = T X( + F ) T X I F T T X? X I T X T X T X I > (2.2) F I Since (; ) is controllable there exists a matrix F such that + F is stable in which case (2.2) tells us that: ( + F ) T X( + F )? X > (2.2) Standard theory for the discrete time Lyapunov equation (see e.g []) then tells us that X >. Conversely, if we choose F such that + F is antistable then we again obtain (2.2) but standard theory for the discrete time Lyapunov equation then tells us that X 6. We have that X must be positive and negative semidenite which clearly implies that X must be. We have T X? X T X T X T X = and the above lemma then obviously implies that X =. Moreover since we have chosen F such that and 22 have no eigenvalues in common, we nd that the Sylvester equation T 22 X 2? X 2 = 9
10 has a unique solution X 2 =. In other words, we only have to compute X 22. It is easy to see that the algebraic Riccati equation reduces to: X 22 = T 22 X Q 22? T 22 X ( T 22 X R)? T 22 X : (2.22) Using lemma 2.4 it is easy to see that in the above reduced order Riccati equation we can work with a regular inverse instead of generalized inverse since the rational function H associated to this Riccati equation has full rank. This enables us to rst derive results for the case that H has full rank and then use the above reduction step to derive results for the general case. The results of the above reduction scheme are put together in the following theorem: Theorem 2. Let X be a solution of the algebraic Riccati equation (2.3) and let F be such that S + RF = and R ( X) is ( + F )invariant where X has realization (; ; S; R) with S and R dened by (2.). Moreover, we assume that we have chosen the appropriate bases as described above. (i) X is a solution of the algebraic Riccati equation (2.3) if and only if X? X = X 22 and X 22 is a solution of the reduced algebraic Riccati equation (2.22). (ii) Let X be a solution of the linear matrix inequality. The zeros of the matrix pencil (2.6) are equal to the zeros of the following matrix pencil: zi? 22? T 22 X 22 22? X 22 + Q 22 T 22 X C : T 22 X T 22 X R 22 It is in general computationally not very attractive to use this method to determine solutions of the linear matrix inequality and Riccati equation since we rst have to nd an initial solution X of the linear matrix inequality. ut it does yield a straightforward method to derive properties of the linear matrix inequality and the algebraic Riccati equations since all important features of solutions of the linear matrix inequality are preserved in the reduction scheme. If an initial solution of the linear matrix inequality is available (e.g. in the positive semidenite case) then the above does yield a computationally attractive method to determine solutions of the Riccati equation. 2.3 Solutions of the algebraic Riccati equation and deating subspaces We rst consider the case that H, dened by (2.7), has full rank. From theorem 2.4, we know that in this case if there exists a solution to the Riccati equation then it must be such that T X + R is invertible. Hence we can study equation (2.9) instead of (2.3). Moreover, an obvious necessary condition for the existence of solutions of the Riccati equation is that we can nd a matrix Z such that R + T Z is invertible.
11 Using such a matrix Z it is easy to check that a matrix X is a solution of the algebraic Riccati equation (2.9 if and only if nd: where ~? ~ Q I I X = I? LZ L? ~ T Z ~ T L = (R + T Z)? T ~ =? (R + T Z)? ( T Z + S) I cl (2.23) X ~Q = Q + T Z? ( T Z + S T )(R + T Z)? ( T Z + S) cl =? (R + T X)? ( T X + S) In fact, (2.23) states that V = im I X (2.24) is a deating subspace of the matrix pencil: ~ I? LZ L? Q ~? z I? ~ T Z ~ T (2.25) such that the zeros of the pencil restricted to V are the eigenvalues of the matrix cl, which are therefore nite. Note that this is a symplectic pencil. Moreover, it is a regular pencil by noting that the pencil can be constructed as a Schurcomplement of R + T Z of the following matrix zi Z? Q? T Z I? z T? T Z? S T T Z + S z T R + T Z fter reordering of rows and columns we obtain the following matrix C? + Q + T Z T Z + S T z T? IC T Z + S R + T Z z T Using lemma 2.3 it is then easy to check that this matrix has full rank because we assumed H has full rank. In the above we have obtained the following lemma: Lemma 2. ssume that H, given by (2.7), has full normal rank. subspace V of the form (2.24) is a deating subspace of the regular matrix pencil (2.25) such that the matrix pencil restricted to V has only nite zeros if and only if X is a solution of the algebraic Riccati equation (2.3).
12 Proof : If X is a solution of the Riccati equation then it is easy to show that V is a deating subspace of (2.25) such that the matrix pencil restricted to V has only nite zeros. For the converse we note that if X satises (2.23) then the rst row implies: im(i + (R + T Z)? T (X? Z)) im(? (R + T Z)? ( T Z + S)) Some straightforward manipulations then yield (2.3b). This makes it is easy to see that X satises (2.3a) as well. Since H has full normal rank, theorem 2.4 then tells us that T X +R is invertible and hence X satises (2.9). The matrix Z plays an important role in the above formulation but can be chosen rather arbitrarily. We will transform our problem to get rid of the matrix Z. Note that this has the disadvantage that we loose the structure of a symplectic pencil (and its analogy with the continuoustime Hamiltonian pencil) and in an increase in dimension. Therefore the above characterization has some clear advantages but for a deeper understanding of the structure of the problem it is attractive to remove this matrix Z. We dene the following matrix pencil: I?S T C? T?S T C S R? T R (2.26) Note that this pencil is no longer symplectic. The following lemma establishes that H has full normal rank if and only if this pencil is regular: Lemma 2.2 There exists such that the matrix: Q S T?I + T C S R T (2.27) has full rank if and only if H has full normal rank. In that case (2.27) has full rank for all but nitely many. s a consequence, H has full normal rank if and only if the pencil (2.26) is regular. Proof : First choose such that has no eigenvalue in or?. We get: I?? I (? I? Q S T?I + T I C S R T ( T? I)? Q(?? )? + S T (I? ) I I Q I C? T (I? T )? I H(? ) 2
13 where denotes an element that is not important in this argument. Hence we see that (2.27) has full rank if and only if is such that H(? ) has full rank which immediately yields the rst part of the lemma. We choose 2 C such that 6=. Then we obtain: I?S T T?S T C S R? T R I I???? I Q S T?? I + T (? )IC I S R T I This implies that the pencil (2.26) is regular if the matrix (2.27) has full rank for =?. Therefore it is immediate that the pencil is regular if and only if H has full normal rank. For the pencil (2.26) we will study deating subspaces of the form I V = XC P (2.28) The paper [2] connected solutions of the algebraic Riccati equation to deating subspaces of (2.26) for =. In [7, theorem 5.2.2] a similar result was obtained also for = and an extra invertibility requirement on the invariant subspace. We generalize their results and have the following result: Theorem 2.3 ssume that the rational matrix H has full normal rank. Choose such that the matrix in (2.27) has full rank. Let the pencil (2.26) be given and dene L by L(X) := Q + T X? X T X + S T : (2.29) T X + S T X + R (i) If a symmetric matrix X is such that the rank of L(X) is equal to m then there exists P such that V dened by (2.28) is a deating subspace of (2.26). Conversely if (2.28) is a deating subspace of (2.26) then X is such that the rank of L(X) is equal to m. (ii) If a matrix X satises the Riccati equation (2.3) then there exists P such that V dened by (2.28) is a deating subspace of (2.26) and the zeros of the pencil restricted to V are nite. Conversely if (2.28) is a deating subspace of (2.26) and the eigenvalues of the pencil restricted to V are nite then X is a solution of the Riccati equation (2.3). Proof : ssume L(X) has rank m. Let L be square but not necessarily invertible and L 2 be such that: L(X) L L 2 = ; L L 2 injective. (2.3) 3
14 We know that (2.27) has full rank and therefore I?? Q S T?I + T L C 2 S R T L? L? L 2 must be injective and hence L? L? L 2 which is a square matrix must be invertible. We choose: V = L + L 2 ; V 2 = L and P = L 2 (L? L? L 2 )? : Then it is easily checked that: I I I?S T XC V 2 T?S T XC S R P? T R P C V (2.3) We know that V 2? V = (L? L? L 2 ) is invertible and hence V? zv 2 is a regular pencil. y lemma., V dened by (2.28) is then a deating subspace for (2.26). To prove the converse in part (i) we assume that V is a deating subspace of the pencil (2.26). ut in that case we know there exists matrices V and V 2 with (V T ) surjective such that I I I?S T XC V 2 T?S T XC S R P? T R P ut after premultiplication with the matrix I W T X I C T X I we obtain that L(X) Moreover, V 2 P (V 2? V ) V 2 P (V 2? V ) = V T 2 V (2.32) is an injective matrix. fter all if V 2 x = and P V x = for some x 6= then we obtain from (2.32) that V x = which is in contradiction with the fact that V? zv 2 is a regular pencil. Hence the rank of L(X) is less than or equal to m. However, lemma 2.3 together with the assumption that H is of full rank guarantees that the rank of L(X) is at least m. For part (ii) we note that X satises the Riccati equation if and only if the rank of L(X) is equal to the rank of T X + R which is then invertible. Moreover, this is equivalent to 4
15 the requirement that in (2.3) we can choose L = I. On the other hand, V is a deating subspace of the pencil (2.26) such that the zeros of the pencil restricted to V are nite if and only if (2.3) is satised with V 2 = I. The same steps as in the proof of part (i) but with V 2 = L = I then yield a proof of part (ii). Suppose we have the matrix pencil (2.26). We can ask ourselves whether we have a result equivalent to theorem 2.3 in case the rational matrix H has no longer full rank. y lemma 2.2, we know that in the general case we have to work with singular pencils. We obtain the following result: Theorem 2.4 Choose such that the matrix in (2.27) has rank equal to 2n + normrank H. ssume that a solution of the Riccati equation exists. Let the pencil (2.26) be given and dene L by (2.29). If a matrix X satises the Riccati equation (2.3) then there exists a matrix P 2 R mn and an injective matrix M 2 R m(m?) such that V dened by (2.28) is a deating subspace of (2.26) such that the matrix pencil (2.26) restricted to V has no zeros in? or innity. Conversely, if (2.28) is a deating subspace of (2.26) such that the pencil (2.26) restricted to V has no zero in? or innity, then X is a solution of the Riccati equation (2.3). Proof : Let X be a solution of the algebraic Riccati equation. y lemma., it is sucient to show that there exist P; V and V 2 satisfying: I I I?S T XC V 2 T?S T XC S R P? T R P V (2.33) Since X satises the algebraic Riccati equation, we can apply the reduction technique described in subsection 2.2 with X = X. We get + F = 2 22 ; = 2 22 ; S + RF = ; Q + S T F + F S + F RF = Q 22 ; R = R 22 using a suitable feedback F and suitable bases. Moreover, Q; S and R are dened by (2.). y theorem 2.3, we can nd P 22 ; W and W 2 satisfying I I 22 T C I?C T D C W 2 T 22?C T D D T C D T D P 22? T 22 D T D P 22 C W (2.34) and such that W 2 is invertible since X 22 = is a solution of the reduced algebraic Riccati equation (2.22). 5
16 We can now easily construct suitable P; V and V 2 in this basis: P = + F; V = 2 W P 22 (W 2? W ) ; V 2 = I P 22 W W 2 and we see that the deating subspace indeed satises all the requirements of the theorem. Conversely suppose that we have a deating subspace of the pencil (2.26) of the form (2.28) such that (2.33) is satised for a regular pencil (V ; V 2 ) with no zero in? or. fter some algebraic manipulations we nd: L(X) V 2 P (V 2? V ) = with V 2 invertible. This implies that X satises the Riccati equation. One can determine deating subspaces of a matrix pencil using numerical tools which are not very well developed for the singular case but it can still be done (see e.g. [25, 27]). We know that each rank minimizing solution of the linear matrix inequality is associated to a deating subspace of the matrix pencil (2.26). If the pencil is regular (i.e. H has full normal rank) then the zeros of the matrix pencil restricted to a deating subspace are also zeros of the original (unrestricted) pencil. lthough this is in general not true for singular pencils, for deating subspaces with the special structure as in (3.2), we still have that the zeros of the matrix pencil restricted to a deating subspace are also zeros of the original (unrestricted) pencil. It can be checked that is an eigenvalue of the matrix pencil restricted to V if only if the matrix (2.6) has a zero for z =. Moreover is an eigenvalue of the matrix pencil restricted to V if and only if the matrix (2.6) has more than innite zeros. If X is a strongly rankminimizing solution and H has full normal rank then T X + R is invertible and the zeros of (2.6) are precisely the eigenvalues of the matrix? ( T X + R)? ( T X + S) In the case where H has no longer full normal rank then the matrix pencil (2.26) restricted to the deating subspace (3.2) has less than n zeros (where n is the dimension of ). These zeros are exactly the uncontrollable eigenvalues of the pair h? T y ( T X + S); (I? T y i T ) (2.35) where T = T X + R. On the other hand if we look for a deating subspace of the form (2.28)) as in theorem 2.4 then the matrix pencil (2.26) restricted to this deating subspace has exactly n zeros. Moreover there exists a matrix F such that these zeros are exactly the eigenvalues of the matrix? T y ( T X + S)? (I? T y T )F (2.36) In the singular case the matrix P in a deating subspace of the form (2.28) is not uniquely determined given X. This freedom translates directly in the freedom in the location of the zeros determined by the matrix F. We are also interested in (semi)stabilizing solutions of the algebraic Riccati equation as dened in denition 2.2: 6
17 Theorem 2.5 ssume that H has full normal rank. stabilizing solution, if it exists, is unique. Moreover, if a semistabilizing solution exists, it is actually a stabilizing solution if and only if Q S T z? I? S R? T zi?? has full rank for all z on the unit circle. C (2.37) Proof : stabilizing solution of the algebraic Riccati equation is clearly unique since solutions of the algebraic Riccati equation have a one to one relation with deating subspaces of the symplectic pencil (2.25). Since the symplectic pencil has at most n stable eigenvalues, a stable ndimensional subspace of the pencil is unique and hence also the associated solution to the algebraic Riccati equation is unique. The matrix pencil (2.26) is regular. Therefore, a semistabilizing solution of the algebraic Riccati equation is necessarily stabilizing if the matrix pencil (2.26) has no zeros on the unit circle. It is easy to see that if (2.26) has a zero then (2.37) has a nonempty kernel for z =. Hence if (2.37) has full rank for all z on the unit circle then the matrix pencil (2.26) has no zeros on the unit circle. 3 The linear matrix inequality and its associated algebraic Riccati equation The algebraic Riccati equation studied in section 2 is very general and includes the Riccati equation studied in H control (see e.g. [3,, 26]) as well as the Riccati equation studied in linear quadratic control (see e.g. [2,6,2]). In the rest of the paper we will concentrate on the Riccati equation used in linear quadratic control and the linear matrix inequality associated to it. We therefore require that the solution of the Riccati equation satises the additional requirement that T X + R > (3.) We know from section 2 that either all or none of the solutions of the algebraic Riccati equation satisfy this additional property. asically, we need to assume that H(z) > for all z on the unit circle, where H is dened by (2.7). Then we know that all solutions of the discrete algebraic Riccati equation satisfy (3.). However, we would like to stress that this assumption (3.) does not need to imply that we are dealing with the positive semidenite case where: Q S > R S T is satised. This is a sucient but certainly not a necessary condition. 7
18 Denition 3. Let 2 R nn, 2 R nm, Q 2 R nn, R 2 R mm and S 2 R mn with Q and R being symmetric be given. The matrix inequality for an unknown n n matrix X of the form L(X) := Q + T X? X T X + S T > (3.2) T X + S T X + R is called the discrete linear matrix inequality. Moreover a matrix X which satises (3.2) is referred to as a solution of the discrete linear matrix inequality. This discrete linear matrix inequality has been used earlier in [9, 4]. We denote the set of real symmetric solutions of the discrete linear matrix inequality (3.2) as?, i.e.? := X 2 R nn X = X T and L(X) > (3.3) Next we dene the notion of rankminimizing solutions for the discrete linear matrix inequality: Denition 3.2 solution X 2? is said to be rankminimizing if rank L(X) = =: min rank L(Y ) Y 2? Moreover, we denote the set of rankminimizing solutions of the discrete linear matrix inequality as? min, i.e.? min := f X 2? j rank L(X) = g (3.4) Finally we need the concept of strongly rankminimizing solutions. Denition 3.3 solution X 2? is said to be a strongly rankminimizing solution of the linear matrix inequality if: rank L(X) = rank( T X + R) (3.5) Moreover, we denote the set of strongly rankminimizing solutions of the linear matrix inequality as: L min := f X 2? j rank L(X) = rank( T X + R) g The name suggests that strongly rankminimizing solutions are also rankminimizing. This is indeed true as will be shown later. Moreover, whether a rankminimizing solution is strongly rank minimizing is completely determined by the zeros at innity of the following matrix pencil: Q + T X? X T X + S T C T X + S T X + R This property is stated in the following theorem: (3.6) 8
19 Theorem 3.4 ssume = normrank H. Let a rankminimizing solution of the linear matrix inequality be given. Then the matrix pencil (3.6) has at least zeros at innity. Moreover, X is a strongly rankminimizing solution if and only if the matrix pencil has exactly zeros at innity. Proof : Let C x and D x be such that: C T x D T x C x D x = L(X) and such that C x D x is surjective. Then the matrix pencil is strongly equivalent to the following matrix pencil: C x D C x The zeros at innity of this pencil are equal to the zeros of the pencil I? s?s sc x sd x at the origin. Since for small s, the matrix I? s is invertible we can transform this matrix using a standard Schur complement in the form: I? s s[d x + sc x (I? s)? ] It is easy to show that = normrank H = normrank[d x + sc x (I? s)? ]. Then it is to see that we have at least zeros at the origin. Moreover, we have more than zeros if and only if D x + sc x (I? s)? loses rank at the origin. Clearly D x + sc x (I? s)? loses rank at the origin is equivalent to the requirement rank D x <. Therefore we have zeros at the origin if and only if rank T X + R = rank D T x D x =. y denition, the latter shows that X is a strongly rank minimizing solution of the linear matrix inequality. We can also dene a stabilizing solution of the linear matrix inequality: Denition 3.5 solution X 2? is said to be stabilizing if all the nite zeros of the matrix pencil (3.6) are inside the unit circle and the number of zeros at innity is equal to the rank of L(X). X 2? is called a semistabilizing solution if all the nite zeros of the matrix pencil (3.6) are inside or on the unit circle and the number of zeros at innity is equal to the rank of L(X). Note that theorem 3.4 guarantees that stabilizing and semistabilizing rankminimizing solutions are also strongly rankminimizing. 9
20 If H has full normal rank and = normrank H. then the above denition of stabilizing solutions is for rankminimizing solutions equivalent to the standard denition that the matrix cl :=? ( T X + R)? ( T X + S) is stable. This is easy to see by noting that theorem 3.4 guarantees that T X + R is invertible. It is a very natural extension to the case where T X + R is no longer invertible in which case we can no longer use cl since the inverse does not exist. The condition that cl is stable is motivated by optimal control problems for the case that = normrank H. Hence the fact that the condition that cl is stable is not consistent with our denition based on the matrix pencil (3.5) if 6= normrank H is in our view acceptable. We dene the discretetime algebraic Riccati equation associated with the discrete linear matrix inequality (3.2) as follows: Denition 3.6 The H 2 algebraic Riccati equation associated with the discrete linear matrix inequality (3.2) is dened as (2.3) with the additional requirement (3.). The rest of this section will be devoted to the existence of solutions, rankminimizing solutions, strongly rankminimizing solutions and (semi)stabilizing solutions of the linear matrix inequality. Moreover, we will derive relationships between the dierent kind of solutions to the linear matrix inequality as well as the relation with the H 2 algebraic Riccati equation. We are interested in those solutions of the discrete linear matrix inequality which can be associated to solutions of the discrete algebraic Riccati equation. We show that the set L min in fact coincides with the set of real symmetric solutions of the discrete algebraic Riccati equation associated with the discrete linear matrix inequality. Obviously X satises L(X) > if and only if Ker(R + T X) Ker( T X + S T ), T X + R >, T X? X? ( T X + S T )( T X + R) y ( T X + S) + Q >. We have the following straightforward lemma: Lemma 3.7 The set of strongly rankminimizing solutions of the linear matrix inequality coincides with the set of real symmetric solutions of the H 2 algebraic Riccati equation associated with the linear matrix inequality. In other words, any symmetric real matrix X satisfying the H 2 algebraic Riccati equation (2.3) and the additional condition (3.) belongs to the set L min. Conversely any X 2 L min satises the discrete algebraic Riccati equation (2.3) and the additional condition (3.). In particular: L min? min (3.7) 2
21 Proof : Let X be any strongly rankminimizing solution. Then it is easy to check that X satises the algebraic Riccati equation (2.3) and hence, according to theorem 2.4 the rank of T X + R equals the normal rank of H. On the other hand by lemma 3.9 we have rank T X + D T D = rank L(X) is larger than or equal to the normal rank of H. The real symmetric solutions of the algebraic Riccati equation are a subset of all the rankminimizing solutions of the linear matrix inequality. Within the set of rankminimizing solutions they are the ones that maximize the rank of T X + R. This result is a generalization of a result from [9]. The following lemma shows sucient conditions for the existence of strongly rankminimizing solutions to the linear matrix inequality, i.e. it shows existence of real symmetric solutions of the algebraic Riccati equation. The result can be found in [5]. Lemma 3.8 ssume that H(z) > for all z on the unit circle, and (; ) is controllable. Then there exist a real symmetric solution of the algebraic Riccati equation and hence there exists a solution of the linear matrix inequality. The controllability condition in lemma 3.8 cannot be weakened unless other assumptions are imposed. This can be seen by the following example: = ; Q = ; S = ; R = ; = The linear matrix inequality has no solution. We have H(z) > for all z on the unit circle but (; ) is not controllable. On the other hand (; ) is stabilizable and hence we cannot even weaken our controllability assumption in the above lemma to stabilizability. We will now have a closer look at rankminimizing solutions of the linear matrix inequality. We obtain the following lemma: Lemma 3.9 We have: rank L(X) > normrank H 8 X 2?: (3.8) Proof : This follows from lemma 2.3. Obviously, since the normal rank of H(; X) equals the normal rank of H, we must have that the rank of L(X) is larger than normal rank of H. Note that to connect the linear matrix inequality with the Riccati equation, we have to require that rank L(X) = normrank H. In general there need not be a solution of the linear matrix inequality which achieves equality in (3.8). For some of our results we will explicitly assume there are solutions achieving equality in (3.8) or equivalently, using the notation of denition 3.2, we will assume = normrank H. For the particular case where (; ) is controllable and H has full rank, we can obtain a very explicit characterization of a set of solutions of the Riccati equation. This characterization is 2
22 given in the following lemma which is in a certain a generalization of a result from [2] which showed, under some additional assumptions, that there exists a oneone connection between the stable subspaces of the symplectic pencil and solutions of the Riccati equation. Lemma 3. Let (; ) be controllable. Moreover assume that H has full normal rank and H(z) > for all z on the unit circle. Choose such that (2.27) has full rank. Let ; : : : ; j be the nite zeros unequal to? outside the unit circle, and j+ ; : : : ; r be the zeros unequal to? on the closed unit disc of the matrix pencil (2.26), all of them without counting multiplicity. Choose ; : : : j such that for i = ; : : : ; j we have either i = i or i =? i. Then there exists a symmetric solution to the algebraic H 2 Riccati equation (2.9) with the additional requirement (3.) such that the eigenvalues of? (R + T X)? ( T X + S T ) (3.9) are ; : : : j ; j+ ; : : : r. Proof : y lemma 3.8 we know there exists at least one solution X of the linear matrix inequality. Let C and D satisfy (.). We can try to nd a solution of the algebraic Riccati equation: ~X = C T C + T ~ X? ( T ~ X + C T D)( T ~ X + D T D)? ( T ~ X + D T C) (3.) It is easy to check that X satises the Riccati equation (2.9) if and only if ~ X = X? X satises the Riccati equation (3.). Moreover, the eigenvalues of (3.9) are equal to the eigenvalues of? (D T D + T ~ X)? ( T ~ X + D T C) Finally the eigenvalues of the matrix pencil (2.26) are equal to the eigenvalues of the following pencil: T C I?C T DC? T?C T DC D T C D T D? T D T D (3.) Next, we know there exists a feedback F and a suitable basis in the state space such that: + F = 2 22 ; = 2 ; C + DF = C 2 ; where the eigenvalues of are precisely the invariant zeros of (; ; C; D) on the unit circle. We will try to nd a solution of the algebraic Riccati equation of the form: ~X = X 22 We nd that X 22 should satisfy the following algebraic Riccati equation: X 22 = T 22 X C T 2 C 2? ( T 22 X C T 2 D)(T 2 X D T D)? ( T 2 X D T C 2 ) (3.2) 22
23 such that the matrix 22? 2 ( T 2 X D T D)? ( T 2 X D T C 2 ) (3.3) has eigenvalues ; : : : ; j. Moreover, the matrix pencil associated to this Riccati equation is: 22 2 I T 2 C 2 I?C T 2 D C? T 22?C T 2 D C D T C 2 D T D? T 2 D T D (3.4) and it is easy to check that this matrix pencil has no eigenvalues on the unit circle. Moreover, the eigenvalues of the matrix pencil (3.) are precisely the eigenvalues of the reduced matrix pencil (3.4) together with the eigenvalues of. This implies in particular that the eigenvalues of are j+ ; : : : ; r. Let 22 be an n r n r matrix and 2 be a n r m r matrix. Choose a n r dimensional deating subspace V = X X 2 X 3 C of the pencil (3.4) associated with the eigenvalues ; : : : ; j. In other words, we have: 22 T 2 C 2 I?C T 2 D D T C 2 D T D X X 2 X 3 I 2 C V 2 T 22?C T 2 D T 2 D T D X X 2 X 3 C V Since j 6= for all j we have that V 2? V is invertible. Then it is easy to check that (X T X T 2 )T must be surjective. Moreover V r = im X X 2? ZX = im X is a deating subspace of the symplectic pencil [L ; L 2 ] with L = L 2 = 22? ~ 2 R? S ~?C T 2 C 2 + T 22 Z 22? S ~ T R ~? S ~ ;? Z I I ~ 2 R? T 2 T 22? ~ S T ~ R? 2 where ~ S = T 2 Z 22 + D T C 2, ~ R = T 2 Z 2 + D T D and Z is a symmetric matrix chosen such that ~ R is invertible. We have L XV 2 = L 2 XV. Moreover, [L ; L 2 ] has no eigenvalues on the unit circle and hence L? L 2 and V? V 2 are both invertible. We nd: (T? T 2 )? (T + T 2 )X = X(V + V 2 )(V? V 2 )? Using some algebraic manipulations we nd that: T = (T? T 2 )? (T + T 2 ) =? R? Q? T 23
24 with ( ; R) controllable, R > and Q symmetric. Moreover, we nd: T I?I = I T T?I We can then use the argument from [6, p. 87] to show that X is invertible and that X T 2 X is symmetric. In this way, we nd that I X 2 X? C X 3 X? is a deating subspace of (3.4). This guarantees that X 22 = X 2 X? satises (3.2). Moreover, since X T 2 X is symmetric we nd that X 22 is symmetric. Finally, the eigenvalues of (3.3) are the required ; : : : ; j. Then X = X + X 22 satises the requirements of the lemma. The above can be used to obtain the following characterization of the set? min : Lemma 3. ssume that H has full normal rank m and H(z) > on the unit circle. Moreover, assume that the uncontrollable eigenvalues ; 2 ; : : : J of (; ) are such that i j 6= for any i; j. Then the H 2 Riccati equation (2.9) with the additional requirement (3.) has at least one symmetric solution. Moreover: min X2? rank L(X) = m (3.5) Proof : We write and in Kalman canonical form: = 2 22 ; = (3.6) Next decompose Q, S and a potential solution of the algebraic Riccati equation X compatibly: Q = Q Q 2 Q T 2 Q 22 ; S = S S 2 ; X = X X 2 : (3.7) X T 2 X 22 We rst note that the ; block of the equation (2.9) reduces to : X = T X? ( T X + S )( T X + R)? ( T X + S ) + Q (3.8) We see that X must be the solution of the discrete algebraic Riccati equation associated to the controllable subsystem. Our assumptions with respect to the uncontrollable eigenvalues of 24
25 (; ) guarantees that there are no uncontrollable eigenvalues on the unit circle and if is an uncontrollable eigenvalue then? is not an uncontrollable eigenvalue. From lemma 3. we know the existence of a solution X of the Riccati equation (3.8) such that the eigenvalues ; 2 ; : : : ; n of x :=? ( T X + D T D)? T X (3.9) are such that i j 6= for i = ; : : : ; n and j = ; : : : ; J. Note that ; : : : ; J are precisely the eigenvalues of 22. Next we study the 2, block of the Riccati equation (2.9). It can be written in the following form: X T 2 = T 22 X T 2 x + [ T 2 X x + Q 2 ] Due to our condition on the relation between the eigenvalues of x and 22 we know this equation is uniquely solvable for X 2 (see e.g. [8]). Finally, we have the 2,2 block of the Riccati equation (2.9). It can be written in the following form: X 22 = T 22 X M where M depends on X and X 2 but is independent of X 22. We know 22 has no two eigenvalues which are the inverse of each other and hence this equation has a symmetric solution X 22. It is easily checked that the soconstructed X is a solution of the algebraic Riccati equation (2.9). We know that X is a solution of the discrete linear matrix inequality. Moreover, it is easy to check that the rank of L(X) is equal to m. ecause of (3.8) we nd that X is a rankminimizing solution of the discrete linear matrix inequality. (3.5) is then an immediate consequence. Remark Note that (3.5) does not hold in general. For instance if = ; Q = ; S = ; R = ; = then (3.5) is not true. ll matrices X yield a matrix L(X) with rank larger than or equal to 2. On the other hand m =. We are of course also interested in an analogous result as of the above theorem for the case H has no longer full normal rank. In [3] we nd necessary and sucient conditions for the existence of a stabilizing solution. We have the following sucient conditions for the existence of a, not necessarily stabilizing, solution. Lemma 3.2 ssume that H(z) > on the unit circle and there exists a solution of the linear matrix inequality. Moreover, assume that the uncontrollable eigenvalues ; 2 ; : : : J of (; ) are such that i j 6= for any i; j. Then the H 2 Riccati equation (2.3) with the additional requirement (3.) has at least one symmetric solution. Moreover:? min = f X 2? j rank L(X) = g (3.2) 25
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