Rankone LMIs and Lyapunov's Inequality. Gjerrit Meinsma 4. Abstract. We describe a new proof of the wellknown Lyapunov's matrix inequality about


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1 Rankone LMIs and Lyapunov's Inequality Didier Henrion 1;; Gjerrit Meinsma Abstract We describe a new proof of the wellknown Lyapunov's matrix inequality about the location of the eigenvalues of a matrix in some region of the complex plane. The proof makes use of standard facts from quadratic and semidenite programming. Links are established between the Lyapunov matrix, rankone LMIs and the Lagrange multiplier arising in duality theory. Keywords Linear Systems, Stability, LMI. 1 Introduction Let A C nn be a given complex matrix and let 1 a D = fs C : s b b c 1 s < 0g denote a given open region of the complex plane, where Hermitian matrix a b b C c has one strictly negative eigenvalue and one strictly positive eigenvalue, and the star denotes transpose conjugate. In the sequel, the notation P 0 or P 0 (resp. P 0 or P 0) means that matrix P is positive denite (resp. semidenite). The location of the eigenvalues of A can be characterized as follows. 1 Corresponding author. FAX: Laboratoire d'architecture et d'analyse des Systemes, Centre National de la Recherche Scientique, Avenue du Colonel Roche, Toulouse, cedex, France. Institut National des Sciences Appliquees, Complexe Scientique de Rangueil, Toulouse, cedex, France. Faculty of Applied Mathematics, University of Twente, P.O.Box 1, Enschede, 00 AE, The Netherlands. 1
2 Theorem 1 (Lyapunov's Inequality) Matrix A has all its eigenvalues in region D if and only if there is a matrix P = P 0 C nn such that I ap bp I A b 0: (1) P cp A Matrix inequality (1) is referred to as Lyapunov's inequality. Lyapunov's proof of the above theorem { originally developed in the case that D is the open left halfplane, i.e. a = c = 0, b = 1 and inequality (1) becomes A P + P A 0 { relies on the construction of a positive quadratic function whose derivative is negative along the trajectories of an associated dynamical system, see e.g. []. It can be extended to arbitrary regions D via a conformal mapping. Another proof of Theorem 1 can be found in [1, Theorem.19]. Eigenvectors of matrix A are used to show that existence of P implies stability of A, whereas the converse statement is shown via properties of the matrix exponential function. The aim of this note is to give a new, alternative proof of Lyapunov's inequality without referring to stability of the trajectories of a dynamical system or to matrix exponentials. We use elementary concepts from linear algebra, quadratic and semidenite programming. Links are established between the Lyapunov matrix and the Lagrange multiplier arising in duality theory. Relationships with rankone LMIs, the KalmanYakubovich Popov Lemma and (D; G)scaling in analysis are also pointed out. The proof relies on the following important result, proved e.g. as in [9, Lemma ]. Lemma 1 Two column vectors p; q C n with q nonzero satisfy a b [q p] [q p] 0 () b c if and only if p = sq for some s D C. Rankone LMI Problem First we show the equivalence between location of the eigenvalues of A in region D and a rankone LMI optimization problem or a rankone LMI feasibility problem. If s C is an eigenvalue of A, then there exists a nonzero vector q C n such that (A si)q = 0: () Pursuing an idea proposed in [, Chapter 1], it follows that all the eigenvalues of A belong to D if and only if the optimal value of the quadratic optimization problem = min q (A si) (A si)q s:t: s D C q q = 1 ()
3 is strictly positive, where 1 D C = fs C : s a b b c 1 s 0g is the closed region complementary to D in C. Dene the rankone positive semidenite matrix q q X = xx = 0 p p and use the notations A(s)q = [ A I ] x = Ax q = [ I 0 ] x = Qx p = [ 0 I ] x = Px to write inequality () as an LMI in rankone matrix X, namely ax b F (X) = Q P X bx cx Q P 0 () where F is a linear map from C nn to C nn. Using these notations, an alternative formulation of quadratic optimization problem () is given by the following lemma. Lemma The eigenvalues of matrix A belong to region D if and only if > 0 in rankone LMI optimization problem = min Trace A AX s:t: F (X) 0 X = X 0 Trace Q QX = 1 Rank X = 1: () The above rankone LMI problem is an optimization problem. It turns out that we can equivalently state this result via a feasibility problem, following an idea exposed in []. To see this, note that Lemma 1 and equation () imply that A has no eigenvalue in D C if and only if there is no nonzero vector q for which a b [q Aq] [q Aq] 0: () b c The lefthand side of inequality () can alternatively be expressed as aqq + baqq + b qq A + caqq A = I A aqq b qq Now dene the linear map bqq cqq I A : aq b G(Q) = Q I A bq cq I A 0 (8) from C nn to C nn. With Q denoting the nonzero rankone matrix Q = qq we arrive at the following result which is equivalent to Lemma.
4 Lemma The eigenvalues of matrix A belong to region D if and only if there is no solution to the rankone LMI feasibility problem G(Q) 0 Q = Q 0 Trace Q = 1 Rank Q = 1: (9) LMI Problem Now we show that the nonconvex rank constraints in LMI problems () and (9) are actually irrelevant. Let I N = A denote a matrix whose columns span the ndimensional right nullspace of full rowrank matrix A. If s k C is a nondefective eigenvalue of A (i.e. its algebraic multiplicity is equal to its geometric multiplicity) and q k C n is the corresponding eigenvector, then the vector qk x k = s k q k belongs to the right nullspace of matrix A. Similarly, if s k is a defective eigenvalue of A (i.e. its algebraic multiplicity is greater than its geometric multiplicity), then the corresponding chain of linearly independent generalized eigenvectors q k ; q k+1 ; q k+ ; : : : gives rise to vectors x k = qk s k q k x k+1 = q k+1 s k q k+1 + q k also belonging to the right nullspace of A. Let x k+ = V = [x 1 x n ] q k+ s k q k+ + q k+1 (10) denote a matrix built up from all the vectors x i associated with all the eigenvalues s i of A. It follows from the above discussion that the columns of N and V span the same vector space. By denition, vectors q i are linearly independent, thus we can dene linearly independent vectors q i C n such that [q 1 q n ] [q 1 q n ] = I: (11) Following these preliminaries, consider now the following relaxation of rankone LMI problem () = min Trace A AX s:t: F (X) 0 (1) X = X 0 Trace Q QX = 1 where the nonconvex rank constraint has been dropped. Since the nonconvex feasible set in problem () is a subset of the convex feasible set in problem (1), LMI optimization problem (1) is referred to as a convex relaxation of the nonconvex rankone LMI problem (). In relation to the above problem, we can state the following central result.
5 Lemma > 0 in rankone LMI optimization problem () if and only if > 0 in LMI optimization problem (1). Proof The inner product of positive semidenite matrices A A and X is always nonnegative, hence 0. Moreover, the fact that > 0 implies > 0 is trivial since the feasible set in problem () is a subset of the feasible set in problem (1), i.e. it holds. Consequently, in order to show that > 0 implies > 0, the remainder of the proof will consist in proving that = 0 implies = 0. So suppose that X is a positive semidenite matrix such that = 0 in problem (1). Let W be a n r full column rank matrix such that X = W W. By putting matrix A A into Schur form, it can easily be shown that Trace A AW W = 0 implies A AW W = 0. Consequently, the columns of W span a subspace that belongs to the right nullspace of A. In view of the above denition of matrix V, there exists a matrix M such that W = V M. Let m ij denote the entries of positive semidenite matrix MM C nn. For a given index k, it holds either m kk > 0 or m ik = m ki = 0 for all i = 1; : : : ; n. Matrix X is feasible for problem (1) thus F (X) = F (V MM V ) = nx i=1 nx j=1 m ij F (x i x j) 0: (1) Since matrix X cannot be zero by assumption, matrix MM is also nonzero and there exists at least one index k such that m kk > 0. Let x k+l be the last eigenvector in the chain of generalized eigenvectors with eigenvalue s k for which m (k+l)(k+l) is nonzero (note that l = 0 if s k is nondefective). From relations (), (10), (11) and (1) it follows that q k+l F (X)q k+l = m (k+l)(k+l) (a + bs k + b s k + cs k s k) 0: Since m (k+l)(k+l) > 0 we see that s k D C, hence vector x k in equation (10) is such that Trace A Ax k x k = 0 and F (x kx k ) 0 in virtue of Lemma 1. Consequently, matrix x kx k is a solution to rankone LMI problem () such that = 0 and the lemma is proved. The following result is then a straightforward corollary to Lemma : Lemma The eigenvalues of matrix A belong to region D if and only if > 0 in LMI optimization problem (1). Now consider the following relaxation to rankone LMI feasibility problem (9): G(Q) 0 Q = Q 0 Trace; Q = 1 (1) where the rank constraint has been dropped. Using the same kind of arguments as above, we can show the following counterpart to Lemma : Lemma The eigenvalues of matrix A belong to region D if and only if there is no solution to LMI feasibility problem (1).
6 Dual LMI Problem Now we use standard semidenite programming duality results [10] to come up with a more compact formulation of the stability conditions of Lemmas and and prove the Lyapunov's inequality of Theorem 1. Dene the linear map Q F D (P ) = P ap b P bp Q cp P dual to the map introduced in (). It is easy to show that ap = b P Trace F D (P )X = Trace F (X)P: Using standard duality arguments, we now prove that the LMI feasibility problem A A F D (P ) P = P 0 is dual to LMI optimization problem (1). To see this, build the Lagrangian bp cp L(P; X; Y ) = Trace (A A F D (P ))X Trace P Y = Trace A AX + Trace (F (X) Y )P of problem (1) where X = X 0 and Y = Y 0 are Lagrange multiplier matrices. The dual function associated with the Lagrangian reads Trace A AX iff (X) = Y 0 g(x; Y ) = min L(P; X; Y ) = P 1 otherwise: The dual optimization problem, obtained by maximizing dual function g(x; Y ) is therefore LMI optimization problem (1), where the equality constraint Trace Q QX = 1 ensures compactness of the feasible set. The matrix inequalities in problem (1) are strict, hence there is no duality gap and > 0 in LMI optimization problem (1) if and only if LMI problem (1) is feasible. Recall that N denotes a matrix whose columns span the right nullspace of A. Then it follows from the Elimination Lemma [] that feasibility problem (1) can equivalently be written as This is exactly the statement of Theorem 1. Similarly, we can dene G D (P ) = N F D (P )N = N F D (P )N 0 P = P 0: I A ap b P bp cp I A as the linear map from C nn to C nn dual to the linear map G(Q) introduced in (8). It is easy to show that Trace G D (P )Q = Trace G(Q)P: It now follows that nonexistence of a nonzero Q = Q 0 for which G(Q) 0 is equivalent to the existence of P = P 0 for which G D (P ) 0. In other words we proved Theorem 1. (1) (1) (1)
7 Numerical Examples.1 First Example Let A = 1 be a constant matrix with eigenvalues 1 and  and let be the stability region. Primal LMI problem (1) reads = min Trace s:t: D = fs C : s + s < 0g X X = X 0 Trace X = 1: X X 0 With a relative accuracy of 10 8, the LMI Control Toolbox 1.0. for Matlab. [] returns and X = = 0:191 0: : : : : :18180: : : : : : : : : : as the optimum of the above problem. In virtue of Theorem, is strictly positive hence all the eigenvalues of matrix A belong to region D.
8 Dual LMI problem (1) reads 1 P = P 0: 0 P + P 1 C A 1 0 With the help of the LMI Toolbox, we obtained the matrix P = 1:8810 0:108 0:108 0:09190 as a feasible solution for the above problem. On the other hand, LMI problem (1) reads 1 0 Q Q Q = Q 0; Trace Q = 1: This problem is infeasible, which is consistent with the above results and Theorem 1.. Second Example Now let A = 1 be a constant matrix with eigenvalues  and and let be the stability region. D = fs C : s + s < 0g With a relative accuracy of 10 8, the LMI Toolbox returns and X = = 0: : : : : : : : : : : : : : : : : as the optimum of primal problem (1). In virtue of Theorem, some eigenvalues of matrix A do not belong to region D. One can check that X = xx = 0:10 0: : : :10 0: : :190
9 is actually a rankone solution to LMI problem (1). Vector x can be written as q x = sq where q is an eigenvector of matrix A corresponding to the eigenvalue s = D C. One can check that positive semidenite matrix 0: : Q = qq = 0: : is a feasible solution for LMI problems (9) or (1). On the other hand, dual LMI problem (1) is found infeasible, which is consistent with the above results and Theorem 1. Conclusion We have proposed a new proof of Lyapunov's matrix inequality that relies on elementary optimization techniques and linear algebra. Following ideas proposed in [] and [, Chapter 1], we consider the eigenvalue location problem as a mere quadratic optimization problem. Then, the quadratic problem can be formulated as an LMI problem with a nonconvex rank constraint. The Lyapunov matrix can be viewed as a Lagrange multiplier matrix arising when dualizing this problem. In [, Chapter 1, x1..], it is shown that removing the nonconvex rankone constraint leads to a sucient LMI stability condition. Our contribution is in showing in Lemmas and that the LMI conditions are also necessary. In other words, the rank constraint in problems () and (9) are irrelevant as far as eigenvalue location is concerned. In a similar fashion, the eigenvalue location problem can be viewed as a frequencydependent analysis problem with one repeated scalar block si corresponding to the Laplace variable s. The Lyapunov matrix P plays the role of a Dscaling matrix associated with the repeated scalar block, and the irrelevance of the nonconvex rank constraint readily follows from the losslessness of the (D; G)scaling as pointed out in [8]. Equivalence of primal problem (1) and dual problem (1) can also be shown via geometric arguments similar to that used in the proof of the KalmanYakubovichPopov (KYP) Lemma in [9, Theorem 1], in the proof of losslessness of (D; G)scaling [, Lemma.1], in the Sprocedure [11] or also in the generalized Sprocedure proposed in [, Theorem 1]. Our approach is also very similar in spirit to the one pursued in [9] to provide an alternative proof of the KYP Lemma. Note however that in this reference the author considers a version of the KYP Lemma where the Laplace variable s varies on the imaginary axis or the unit circle. This result has been extended to other onedimensional curves of the complex plane such as the real axis [] or a segment on the imaginary axis []. These curves are boundaries of the twodimensional stability regions D considered in the present note. It is therefore expected that we can similarly derive more general versions of the KYP Lemma in twodimensional stability regions. 9
10 Finally, we are currently investigating the application of these techniques to the study of stability of polynomial matrices, twoindeterminate polynomial matrices and uncertain polynomial matrices. Related results will be reported elsewhere. References [1] S. Barnett \Polynomials and Linear Control Systems", Marcel Dekker, New York, 198. [] S. Boyd, L. El Ghaoui, E. Feron and V. Balakrishnan \Linear Matrix Inequalities in System and Control Theory", SIAM Studies in Applied Mathematics, Philadelphia, Pennsylvania, 199. [] L. El Ghaoui and S. I. Niculescu (Editors) \Advances in Linear Matrix Inequality Methods in Control", SIAM Advances in Control and Design, Philadelphia, Pennsylvania, [] T. Iwasaki, G. Meinsma and M. Fu \Generalized Sprocedure and Finite Frequency KYP Lemma", preprint, [] T. Kailath \Linear Systems", Prentice Hall, Englewood Clis, New Jersey, [] The MathWorks, Inc. \LMI Control Toolbox for Matlab", Release 1.0., See the home page [] G. Meinsma, Y. Shrivastava and M. Fu \A Dual Formulation of Mixed and the Losslessness of (D; G)scaling", IEEE Transactions on Automatic Control, Vol., No., pp. 10{10, 199. [8] A. Packard and J. Doyle \The Complex Singular Value", Automatica, Vol. 9, No. 1, pp. 1{109, 199. [9] A. Rantzer \On the KalmanYakubovichPopov Lemma", Systems and Control Letters, Vol. 8, pp. {10, 199. [10] L. Vandenberghe and S. Boyd \Semidenite Programming", SIAM Review, Vol. 8, pp. 9{9, 199. [11] V. A. Yakubovich \The Sprocedure in Nonlinear Control Theory", Vestnik Leningrad University of Mathematics, Vol., pp. {9, 19. In Russian,
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