Stability radii for real linear Hamiltonian systems with perturbed dissipation

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1 Stability radii for real linear amiltonian systems with perturbed dissipation Christian Mehl Volker Mehrmann Punit Sharma August 17, 2016 Abstract We study linear dissipative amiltonian (D) systems with real constant coefficients that arise in energy based modeling of dynamical systems. In this paper we analyze when such a system is on the boundary of the region of asymptotic stability, i.e., when it has purely imaginary eigenvalues, or how much the dissipation term has to be perturbed to be on this boundary. For unstructured systems the explicit construction of the real distance to instability (real stability radius) has been a challenging problem. In this paper, we analyze this real distance under different structured perturbations to the dissipation term that preserve the D structure and we derive explicit formulas for this distance in terms of low rank perturbations. We also show (via numerical examples) that under real structured perturbations to the dissipation the asymptotical stability of a D system is much more robust than for unstructured perturbations. Dissipative amiltonian system, port-amiltonian system, real distance to instability, real structured distance to instability, restricted real distance to instability. AMS subject classification. 93D20, 93D09, 65F15, 15A21, 65L80, 65L05, 34A30. 1 Introduction In this paper we study linear time-invariant systems with real coefficients. When a physical system is energy preserving, then a mathematical model should reflect this property and this is characterized by the property of the system being amiltonian. If energy dissipates from the system then it becomes a dissipative amiltonian (D) system, which in the linear time-invariant case can be expressed as ẋ = (J R) Qx, (1.1) where the function x x T Qx, with Q = Q T R n,n positive definite, describes the energy of the system, J = J T R n,n is the structure matrix that describes the energy flux among energy storage elements, and R R n,n with R = R T 0 is the dissipation matrix that describes energy dissipation in the system. Dissipative amiltonian systems are special cases of port-amiltonian systems, which recently have received at lot attention in energy based modeling, see, e.g., 3, 8, 19, 21, 22, Institut für Mathematik, MA 4-5, TU Berlin, Str. des 17. Juni 136, D Berlin, FRG. {mehl,mehrmann,sharma@math.tu-berlin.de. Supported by Einstein Stiftung Berlin through the Research Center Matheon Mathematics for key technologies in Berlin. 1

2 26, 25, 27, 28, 29. An important property of D systems is that they are stable, which in the time-invariant case can be characterized by the property that all eigenvalues of A = (J R)Q are contained in the closed left half complex plane and all eigenvalues on the imaginary axis are semisimple, see, e.g. 20 for a simple proof which immediately carries over to the real case discussed here. For general unstructured systems ẋ = Ax, if A has purely imaginary eigenvalues, then arbitrarily small perturbations (arising e.g. from linearization errors, discretization errors, or data uncertainties) may move eigenvalues into the right half plane and thus make the system unstable. So stability of the system can only be guaranteed when the system has a reasonable distance to instability, see 2, 5, 11, 12, 14, 31, and the discussion below. When the coefficients of the system are real and it is clear that the perturbations are real as well, then the complex distance to instability is not the right measure, since typically the distance to instability under real perturbations is larger than the complex distance. A formula for the real distance under general perturbations was derived in the ground-breaking paper 23. The computation of this real distance to instability (or real stability radius) is an important topic in many applications, see e.g. 10, 17, 18, 24, 30, 32. In this paper, we study the distance to instability under real (structure preserving) perturbations for D systems and we restrict ourselves to the case that only the dissipation matrix R is perturbed, because this is the part of the model that is usually most uncertain, due to the fact that modeling damping or friction is extremely difficult, see 10 and Example 1.1 below. Furthermore, the analysis of the perturbations in the matrices Q, J is not completely clear at this stage. Since D systems are stable, it is clear that perturbations that preserve the D structure also preserve the stability. owever, D systems may not be asymptotically stable, since they may have purely imaginary eigenvalues, which happens e.g. when the dissipation matrix R vanishes. So in the case of D systems, we discuss when the system is robustly asymptotically stable, i.e. small real and structure-preserving perturbations R to the coefficient matrix R keep the system asymptotically stable, and we determine the smallest perturbations that move the system to the boundary of the set of asymptotically stable systems. Motivated from an application in the area of disk brake squeal, we consider restricted structure-preserving perturbations of the form R = B B T, where B R n,r is a so-called restriction matrix of full column rank. This restriction allows the consideration of perturbations that only affect selected parts of the matrix R. (We mention that perturbations of the form B B T should be called structured perturbations by the convention following 13, but we prefer the term restricted here, because of the danger of confusion with the term structure-preserving for perturbations that preserve the D structure.) Example 1.1 The finite element analysis of disk brake squeal 10 leads to large scale differential equations of the form M q + (D + G) q + (K + N)q = f, where M = M T > 0 is the mass matrix, D = D T 0 models material and friction induced damping, G = G T models gyroscopic effects, K = K T > 0 models the stiffness, and N is a nonsymmetric matrix modeling circulatory effects. ere q denotes the derivative with respect to time. An appropriate first order formulation is associated with the linear system 2

3 ż = (J R)Qz, where J := G K N (K N ) 0, R := 1 D 2 N 1 2 N 0 M 0, Q := 0 K 1. (1.2) This system is in general not a D system, since for N 0 the matrix R is indefinite. But since brake squeal is associated with the eigenvalues in the right half plane, it is an important question for which perturbations the system is stable. This instability can be associated with the restricted indefinite perturbation matrix R + R := D N 1 2 N 0 and this matrix underlies further restrictions, since only a small part of the finite elements is associated with the brake pad and thus the perturbations in N are restricted to the coefficients associated with these finite elements. In the following denotes the spectral norm of a vector or a matrix while F denotes the Frobenius norm of a matrix. By Λ(A) we denote the spectrum of a matrix A R n,n, where R n,r is the set of real n r matrices, with the special case R n = R n,1. For A = A T R n,n, we use the notation A 0 and A 0 if A is positive or negative semidefinite, respectively, and A > 0 if A is positive definite. The Moore-Penrose of a matrix R n,r is denoted by A, see e.g. 9. The paper is organized as follows. In Section 2 we study mapping theorems that are needed to construct the minimal perturbations. The stability radii for D systems under perturbations in the dissipation matrix is studied in Section 3 and the results are illustrated with numerical examples in Section 4. 2 Mapping Theorems As in the complex case that was treated in 20, the main tool in the computation of stability radii for real D systems will be minimal norm solutions to structured mapping problems. To construct such mappings in the case of real matrices, and to deal with pairs of complex conjugate eigenvalues and their eigenvectors, we will need general mapping theorems for maps that map several real vectors x 1,..., x m to vectors y 1,..., y m. We will first study the complex case and then use these results to construct corresponding mappings in the real case. For given X C m,p, Y C n,p, Z C n,k, and W C m,k, the general mapping problem, see 16, asks under which conditions on X, Y, Z, and W the set S := { A C n,m AX = Y, A Z = W is non-empty, and if it is, then to construct the minimal spectral or Frobenius norm elements from S. To solve this mapping problem, we will need the following well-known result. Theorem Let A, B, C be given complex matrices of appropriate dimensions. Then for any positive number µ satisfying ( A µ max, A C ), (2.1) B 3,

4 there exists a complex matrix D of appropriate dimensions such that A C µ. (2.2) B D Furthermore, any matrix D satisfying (2.2) is of the form D = KA L + µ(i KK ) 1/2 Z(I LL ) 1/2, where K = (µ 2 I A A) 1/2 B, L = (µ 2 I AA ) 1/2 C, and Z is an arbitrary contraction, i.e., Z 1. Using Theorem 2.1 we obtain the following solution to the general mapping problem. Theorem 2.2 Let X C m,p, Y C n,p, Z C n,k, and W C m,k. Then the set S := { A C n,m AX = Y, A Z = W, is non-empty if and only if X W = Y Z, Y X X = Y, and W Z Z = W. If this is the case, then { S = Y X + (W Z ) (W Z ) XX + (I ZZ )R(I XX ) R C n,m. (2.3) Furthermore, we have the following minimal-norm solutions to the mapping problem AX = Y and A Z = W : 1) The matrix G := Y X + (W Z ) (W Z ) XX (2.4) is the unique matrix from S of minimal Frobenius norm G F = Y X 2 F + W Z 2 F trace((w Z )(W Z ) XX ) = inf A F. A S 2) The minimal spectral norm of elements from S is given by { µ := max Y X, W Z = inf A. (2.5) A S Moreover, assume that rank(x) = r 1 and rank(z) = r 2. If X = UΣV and Z = Û Σ V are singular value decompositions of X and Z, respectively, where U = U 1, U 2 with U 1 C m,r 1 and Û = Û1, Û2 with Û1 C n,r 2, then the infimum in (2.5) is attained by the matrix where F := Y X + (W Z ) (W Z ) XX + (I ZZ )Û2CU 2 (I XX ), (2.6) C = K(U1 (Y X )Û1)L + µ(i KK ) 1 2 P (I L L) 1 2, K = (µ 2 I U1 (Y X ) ZZ (Y X )U 1 ) 1 2 ( Û2 Y X U 1 ), ) L = (µ 2 I Û 1 (Y X )(Y X ) 1 ( ) 2 Û 1 Û1 (W Z ) U 2, and P is an arbitrary contraction. 4

5 Proof. Suppose that S is non-empty. Then there exists A C n,m such that AX = Y and A Z = W. This implies that X W = Y Z. Using the properties of the Moore-Penrose pseudoinverse, we have that Y X X = AXX X = AX = Y and W Z Z = A ZZ Z = A Z = W. Conversely, suppose that X W = Y Z, Y X X = Y, and W Z Z = W. Then S is non-empty, since for any C C n,m we have A = Y X + (W Z ) (W Z ) XX + (I ZZ )C(I XX ) S. In particular, this proves the inclusion in (2.3). To prove the other direction, let X = UΣV and Z = Û Σ V be singular value decompositions of X and Z, respectively, partitioned as in 2), with Σ 1 R r 1,r 1 and Σ 1 R r 2,r 2 being the leading principle submatrices of Σ and Σ, respectively. Let A S and set  := Û A11 A AU = 12, (2.7) A 21 A 22 where A 11 C r 1,r 2, A 12 C r 1,m r 2, A 21 C n r 1,r 2 and A 22 C n r 1,m r 2. Then A F =  F and A = Â. Multiplying ÛÂU X = AX = Y by Û from the left, we obtain A11 A 12 Σ1 V 1 A11 A = 12 U 1 Û A 21 A 22 0 A 21 A 22 U2 X = 1 Û Û2 Y = 1 Y Û2 Y, which implies that A 11 = Û 1 Y V 1 Σ 1 1 and A 21 = Û 2 Y V 1 Σ 1 1. (2.8) Analogously, multiplying UÂ Û Z = A Z = W from the left by U, we obtain that A 11 A 21 Σ1 V 1 A A 12 A = 11 A 21 Û 22 0 A 12 A 1 U Z = 1 U 22 U2 W = 1 W U2 W. This implies that Û 2 A 11 = ( Σ 1 1 ) V 1 W U 1 and A 12 = ( Σ 1 1 ) V 1 W U 2. (2.9) Equating the two expressions for A 11 from (2.8) and (2.9), we obtain Û 1 Y V 1 Σ 1 = ( Σ 1 ) V 1 W U 1 which shows that X W = Y Z. Furthermore, we get Û Â = 1 Y V 1 Σ 1 1 ( Σ 1 1 ) V 1 S U 2 Û Û2 Y V 1Σ 1 = 1 Y X U 1 Û1 (W Z ) U 2 1 A 22 Û2 Y X U 1 A 22. (2.10) Thus, using X U 1 U 1 = X, Z Û 1 Û 1 = Z, and U 1 U 1 = XX, we obtain A = ÛÂU = Û1Û 1 Y X U 1 U1 + Û2Û 2 Y X U 1 U1 + Û1Û 1 (W Z ) U 2 U2 + Û2A 22 U2 = Û1Û 1 Y X U 1 U1 + (I Û1Û 1 )Y X U 1 U1 + Û1Û 1 (SZ ) (I U 1 U1 ) + Û2A 22 U2 = Y X + (W Z ) (W Z ) XX + (I ZZ )Û2A 22 U2 (I XX ). (2.11) 5

6 1) In view of (2.10) and using that X U 2 = 0 and Û 2 (Z ) = 0, we obtain that A 2 F = Û Â 2 F = 1 Y X 2 U 1 Û2 Y + Û X 1 (W Z ) U 2 2 U 1 F + A 22 2 F F = Û Y X U 2 F + Û(W Z ) U 2 F Û 1 (W Z ) U 1 2 F + A 22 2 F = Y X 2 F + W Z 2 F trace ((W Z )(W Z ) XX ) + A 22 2 F. Thus, by setting A 22 = 0 in (2.11), we obtain a unique element of S which minimizes the Frobenius norm giving ) inf A F Y = X 2F + W A S Z ((W 2F trace Z )(W Z ) XX. 2) By definition of µ, we have { µ = max Y X, W Z { Û = max 1 Y X U 1 Û2 Y X U 1, Û 1 Y X U 1 Û1 (W Z ) U 2. Then it follows that for any A S, we have A =  µ with  as in (2.7). By Theorem 2.1 there exists matrices A S with A µ, i.e., inf A S A = µ. Furthermore, by Theorem 2.1 this infimum is attained with a matrix A as in (2.11), where ) A 22 = K (U 1 (Y X ) Û 1 L + µ(i KK ) 1 2 P (I L L) 1 2, K = L = ( ) µ 2 I U1 (Y X ) 1 ZZ (Y X 2 )U 1 (Û 2 Y X U 1 ), (µ 2 I Û 1 (Y X )(Y X ) Û 1 ) 1 2 ( Û 1 (W Z ) U 2 ), and P is an arbitrary contraction. ence the assertion follows by setting C = A 22. The special case p = k = 1 of Theorem 2.2 was originally obtained in 15, Theorem 2. The next result characterizes the ermitian positive semidefinite matrices that map a given X C n,m to a given Y C n,m, and we include the solutions that are minimal with respect to the spectral or Frobenius norm. This generalizes 20, Theorem 2.3 which only covers the case m = 1. Theorem 2.3 Let X, Y C n,m be such that rank(y ) = m. Define S := { C n,n = 0, X = Y. Then there exists = X Y > 0. If this holds, then 0 such that X = Y if and only if X Y = Y X and := Y (Y X) 1 Y (2.12) is well-defined and ermitian positive-semidefinite, and { ( S = + I n XX ) ( K K I n XX ) K C n,n. (2.13) Furthermore, we have the following minimal norm solutions to the mapping problem X = Y : 6

7 1) The matrix from (2.12) is the unique matrix from S with minimal Frobenius norm min { F S = Y (X Y ) 1 Y F. 2) The minimal spectral norm of elements from S is given by min { S = Y (X Y ) 1 Y and the minimum is attained for the matrix from (2.12). Proof. If S, then X Y = X X = (X) X = Y X and X Y = X X 0, since = 0. If X X were singular, then there would exist a vector v C m \ {0 such that v X Xv = 0 and, hence, Y v = Xv = 0 (as 0) in contradiction to the assumption that Y is of full rank. Thus, X Y > 0. Conversely, let X Y = Y X > 0 (which implies that also (X Y ) 1 > 0). Then in (2.12) is well-defined. Clearly X = Y and = Y (Y X) 1 Y 0, which implies that S. Let = + (I XX )K K(I XX ) C n,n be as in (2.13) for some K C n,n. Then clearly =, X = Y and also 0, as it is the sum of two positive semidefinite matrices. This proves the inclusion in (2.13). For the converse inclusion, let S. Since 0, we have that = A A for some A C n,n. Therefore, X = Y implies that (A A)X = Y, and setting Z = AX, we have AX = Z and A Z = Y. Since rank(y ) = m, we necessarily also have that rank(z) = m and rank(x) = m. Therefore, by Theorem 2.2, A can be written as for some C C n,n. Note that A = ZX + (Y Z ) (Y Z ) XX + (I ZZ )C(I XX ) (2.14) (Y Z ) XX = (Z ) Y XX = (Z ) Z ZX = (ZZ ) ZX = ZX, (2.15) since (ZZ ) = ZZ and Y X = Z Z. By inserting (2.15) in (2.14), we obtain that and thus, A = (Y Z ) + (I ZZ )C(I XX ), = A A = (Y Z )(Y Z ) + (I XX )C (I ZZ )(Y Z ) +(Y Z )(I ZZ )C(I XX ) + (I XX )C (I ZZ )(I ZZ )C(I XX ) = (Y Z )(Y Z ) + (I XX )C (I ZZ )(I ZZ )C(I XX ), (2.16) where the last equality follows since (Y Z )(I ZZ ) = Y (Z Z ZZ ) = Y (Z Z ) = 0. Setting K = (I ZZ )C in (2.16) and using Z (Z ) = (Z Z) 1 Z ( (Z Z) 1 Z ) = (Z Z) 1 = (Y X) 1 7

8 proves the inclusion in (2.13). 1) If S, then X Y = Y X > 0 and we obtain = Y (Y X) 1 Y + (I XX )K K(I XX ), (2.17) for some K C n,n. By using the formula BB + DD 2 F = BB 2 F +2 DB 2 F + D D 2 F for B = Y (Y X) 1/2 and D = K(I XX ), we obtain 2 F = Y (Y X) 1 Y 2 F +2 K(I XX )Y (Y X) F + (I XX )K K(I XX ) 2 F. ence, setting K = 0, we obtain = Y (Y X) 1 Y as the unique minimal Frobenius norm solution. 2) Let S be of the form (2.17) for some K C n,n. Since Y (Y X) 1 Y 0 and (I XX )K K(I XX ) 0, we have This implies that Y (Y X) 1 Y Y (Y X) 1 Y + (I XX )K K(I XX ). Y (Y X) 1 Y inf Y (Y X) 1 Y + (I XX )K K(I XX ) = inf. K C n,n S (2.18) One possible choice for obtaining equality in (2.18) is K = 0 which gives = Y (Y X) 1 Y S and = Y (Y X) 1 Y = min. S aving presented the complex versions of the mapping theorems, we now adapt these for the case of real perturbations. ere, real refers to the mappings, but not to the vectors that are mapped, because we need to apply the mapping theorems to eigenvectors which may be complex even if the matrix under consideration is real. Remark 2.4 If X C m,p, Y C n,p, Z C n,k, W C m,k are such that rank(x X) = 2p and rank(z Z) = 2k (or, equivalently, rank(re X Im X) = 2p and rank(re Z Im Z) = 2k), then minimal norm solutions to the mapping problem AX = Y and A Z = W with A R n,m can easily be obtained from Theorem 2.2. This follows from the observation that with AX = Y and A Z = W we also have AX = Y and A Z = W and thus ARe X Im X = Re Y Im Y and A Re Z Im Z = Re W Im W. We can then apply Theorem 2.2 to the real matrices X = Re X Im X, Y = Re Y Im Y, Z = Re Z Im Z, and W = Re W Im W. Indeed, whenever there exists a complex matrix A C n,m satisfying AX = Y and A Z = W, then there also exists a real one, because it is easily checked that the minimal norm solutions in Theorem 2.2 are real. (ere, we assume that for the case of the spectral norm the real singular value decompositions of X and Z are taken, and the contraction P is also chosen to be real.) A similar observation holds for solution of the real version of the ermitian positive semidefinite mapping problem in Theorem 2.3. Since the real version of Theorem 2.2 (and similarly of Theorem 2.3) is straightforward in view of Remark 2.4, we refrain from an explicit statement. The situation, however, changes considerably if the assumptions rank(re X Im X) = 2p and rank(re Z Im Z) = 2k as in 8

9 Remark 2.4 are dropped. In this case, it seems that a full characterization of real solutions to the mapping problems is highly challenging and very complicated. Therefore, we only consider the generalization of Theorem 2.3 to real mappings for the special case m = 1 which is in fact the case needed for the computation of the stability radii. We obtain the following two results. Theorem 2.5 Let x, y C n be such that rank(y ȳ) = 2. Then the set S := { R n,n, T = 0, x = y is non-empty if and only if x y > x T y (which includes the condition x y R). In this case let X := Re x Im x and Y := Re y Im y. Then the matrix := Y ( Y X ) 1 Y (2.19) is well defined and real symmetric positive semidefinite, and { S = + (I XX )K(I XX ) K R n,n, K T = K 0. (2.20) 1) The minimal Frobenius norm of an element in S is given by min { Y ( F S = Y T X ) 1 Y T and this minimal norm is uniquely attained by the matrix in (2.19). 2) The minimal spectral norm of an element in S is given by min { S = Y ( Y T X ) 1 Y T and the matrix in (2.19) is a matrix that attains this minimum. Proof. Let S, i.e. T = 0 and x = y. Since is real, we also have X = Y. Thus, by Theorem 2.3, X and Y satisfy X Y = Y X and Y X > 0. The first condition is equivalent to (Re x) T Im y = (Re y) T Im x which in turn is equivalent to x y R and the second condition is easily seen to be equivalent to x y > x T y. Conversely, if x y = y x and x y > x T y, then Y T X > 0 and hence := Y ( Y T X ) 1 Y T is positive semidefinite. Moreover, we obviously have X = Y and thus x = y. This implies that S. The inclusion in (2.20) is straightforward. For the other inclusion let S, i.e., T = 0 and x = y, and thus X = Y. Then, by Theorem 2.3, there exist L C n,n such that = + (I XX )L L(I XX ) = + (I XX ) ( Re(L ) Re(L) + Im(L) Im(L) ) (I XX ), F 9

10 where for the last identity we made use of the fact that is real. Thus, by setting K = (Re(L ) Re(L)+Im(L ) Im(L)), we get the inclusion in (2.20). The norm minimality in 1) and 2) follows immediately from Theorem 2.3, because any real map S also satisfies X = Y. Theorem 2.5 does not consider the case that y and ȳ are linearly dependent. In that case, we obtain the following result. Theorem 2.6 Let x, y C n, y 0 be such that ranky ȳ = 1. Then the set S = { R n,n, T = 0, x = y is non-empty if and only if x y > 0. In that case := yy x y is well-defined and real symmetric positive semidefinited. Furthermore, we have: 1) The minimal Frobenius norm of an element in S is given by (2.21) min { F S = y 2 x y and this minimal norm is uniquely attained by the matrix in (2.21). 2) The minimal spectral norm of an element in S is given by min { S = y 2 x y and the matrix in (2.19) is a matrix that attains this minimum. Proof. If S, i.e., T = 0 and x = y, then by Theorem 2.3 (for the case m = 1), we have that x y > 0. Conversely, assume that x and y satisfy x y > 0. Since y and ȳ are linearly dependent, there exists a unimodular α C such that αy ist real. But then also yy = (αy)(αy) is real and hence the matrix in (2.21) is well defined and real. By Theorem 2.3, it is the unique element from S with minimal Frobenius norm and also an element from S of minimal spectral norm. Remark 2.7 Note that results similar to Theorem 2.5 and Theorem 2.6 can also be obtained for real negative semidefinite maps. Indeed, for x, y C n such that ranky ȳ = 2, there exist a real negative semidefinite matrix R n,n such that x = y if and only if x y = y x and x y > x T y. Furthermore, it follows immediately from Theorem 2.5 by replacing y with y and with that a minimal solution in spectral and Frobenius norm is given by = Re y Im y ( Re y Im y Re x Im x ) 1 Re y Im y. An analogous argument holds for Theorem 2.6. Therefore, we will refer to Theorem 2.5 and Theorem 2.6 also in the case that we are seeking solutions for the negative semidefinite mapping problem. The minimal norm solutions for the real symmetric mapping problem with respect to both the spectral norm and Frobenius norm are well known, see 1, Theorem We do not restate this result in its full generality, but in terms of the following two theorems that are formulated in such a way that they allow a direct application in the remainder of this paper. 10

11 Theorem 2.8 Let x, y C n \ {0 be such that rank(x x) = 2. Then S := { R n,n T =, x = y is nonempty if and only if x y = y x. Furthermore, define X := Re x Im x, Y := Re y Im y, := Y X + (Y X ) T (XX ) T Y X. (2.22) 1) The minimal Frobenius norm of an element in S is given by min F = F = S and the minimum is uniquely attained by in (2.22). 2 Y X 2 F trace (Y X (Y X ) T XX ) 2) To characterize the minimal spectral norm, consider the singular value decomposition X = UΣV T and let U = U 1 U 2 where U 1 R n,2. Then and the minimum is attained by min = Y S X, Ĥ = (I n XX )KU T 1 Y X U 1 K T (I n XX ), (2.23) where K = Y X U 1 (µ 2 I 2 U T 1 Y X Y X U 1 ) 1/2 and µ := Y X. Proof. Observe that for T = R n,n the identity x = y is equivalent to X = Y. Thus, the result follows immediately from 1, Theorem applied to the mapping problem X = Y. Theorem 2.9 Let x, y C n \ {0 be such that rank(x x) = 1. Then S := { R n,n T =, x = y is nonempty if and only if x y = y x. Furthermore, we have: 1) The minimal Frobenius norm of an element in S is given by min F = y S x and the minimum is uniquely attained by the real matrix (If x and y are linearly dependent, then = yx x x.) := yx x 2 + xy x 2 (x y)xx x 4. (2.24) 2) The minimal spectral norm of an element in S is given by y min = S x, and the minimum is attained by the real matrix Ĥ := y y y x y x x x 1 x y 1 x y x y 1 y y x x (2.25) if x and y are linearly independent and for 11 Ĥ := yx x x otherwise.

12 Proof. By 1, Theorem (see also 20, Theorem 2.1 and 16) the matrices and Ĥ are the minimal Frobenius resp. spectral norm solutions to the complex ermitian mapping problem x = y. Thus, it only remains to show that and Ĥ are real. Since x and x are linearly dependent, there exists a unimodular α C such that αx is real. But then also αy = (αx) and thus xx = (αx)(αx) and yx = (αy)(αx) are real which implies the realness of. Analogously, Ĥ can be shown to be real. Obviously, the minimal Frobenius or spectral norm solutions from Theorem 2.9 have either rank one or two. The following lemma characterizes the rank of the minimal Frobenius or spectral norm solutions from Theorem 2.8 as well as the number of their negative and their positive eigenvalues, respectively. Lemma 2.10 Let x, y C n \ {0 be such that x y = y x and rank(x x) = 2. If and Ĥ are defined as in (2.22) and (2.23), respectively, then rank( ), rank(ĥ) 4 and both and Ĥ have at most two negative eigenvalues and at most two positive eigenvalues. Proof. Recall from Theorem 2.8 that X = Re x Im x, Y = Re y Im y and consider the singular value decomposition X = UΣV T = U Σ 0 V T with U = U 1 U 2 R n,n, U 1 R n,2, Σ R 2,2, and V R 2,2. If we set Y1 Y = U V T, where Y 1 R 2,2 and Y 2 R n 2,2, then = Y X + (Y X ) T (XX ) T Y X Y = U Σ Σ 1 U T Y + U 1 T Σ 1 Y2 Y Σ Σ 1 Y = U 1 T Σ 1 Y2 T Y Σ Y 2 U T Y1 U Σ U T U T. (2.26) Thus, is of rank at most four, and also has a n 2 dimensional neutral subspace, i.e., a subspace V R n satisfying z T z = 0 for all z, z V. This means that the restriction of to its range still has a neutral subspace of dimension at least 2 if rank( ) = 4, max{0, rank( ) 2 = 1 if rank( ) = 3, 0 if rank( ) 2. Thus, it follows by applying 7, Theorem that has at most two negative and at most two positive eigenvalues. On the other hand, we have Ĥ = (I n XX )KU T 1 Y X U 1 K T (I n XX ), (2.27) 12

13 where K = Y X U 1 W, W := (µ 2 I 2 U1 T Y X Y X U 1 ) 1/2 and µ := Y X. Then we obtain K = Y X Y U 1 W = U Σ U T Y U Y Σ 1 1 W = U Σ 1 1 W. 2 0 Y Σ 1 2 Also, := (I n XX )KU1 T Y X U 1 K T (I n XX ) 0 = U Y Σ 1 W U1 T Y U Σ U T U 2 Y Σ 1 1 W T = U Y 2 Σ 1 W Y 1 Σ 1 W T Σ 1 Y T 2 Σ 1 Y T 2 U T U T. (2.28) Inserting (2.26) and (2.28) into (2.27), we get Ĥ = Σ 1 Y = U 1 T Σ 1 Y2 T Y Σ 1 2 Y Σ 1 2 W Y Σ 1 1 W T Σ 1 Y2 T Z L T = U L LW Z T W T L T U T, U T where Z := Σ 1 Y1 T and L := Y 2 Σ 1. This implies that U T Z L T ĤU = L LW Z T W T L T and Z are real symmetric. Let Û Rn 2,n 2 be invertible such that ÛL = l l 12 l is in row echelon form, where L l11 l = 12 0 l 22 I2 0 0 Û Z L T L LW ZW T L T. Then T = I2 0 0 Û T L 0 = Z , where Z 11 = Z L LT LW ZW T LT R 4,4. This implies that rank(ĥ) = rank(z 11) 4. Furthermore, by Sylvester s law of inertia, Ĥ and Z 11 have the same number of negative eigenvalues. Thus, the assertion follows if we show that Z 11 has at most two negative eigenvalues. For this, first suppose that L is singular, i.e., l 22 = 0, which would imply that rank(z 11 ) 3. If rank(z 11 ) < 3, then clearly Z 11 can have at most two negative eigenvalues and at most two positive eigenvalues. If rank(z 11 ) = 3, then we have that Z 11 is indefinite. Indeed, if Z is indefinite then so is Z 11. If Z is positive (negative) semidefinite then LW ZW T LT is negative (positive) semidefinite. In this case Z 11 13

14 has two real symmetric matrices of opposite definiteness as block matrices on the diagonal. This shows that Z 11 is indefinite and thus can have at most two negative eigenvalues and at most two positive eigenvalues. If L is invertible, then I2 0 Z LT 0 W 1 L 1 L LW ZW T LT I2 0 0 L T W T = Z W 1 W T is a real symmetric amiltonian matrix. Therefore, by using the amiltonian spectral symmetry with respect to the imaginary axis, it follows that Z 11 has two positive and two negative eigenvalues. In the next section we will discuss real stability radii under restricted perturbations, where the restrictions will be expressed with the help of a restriction matrix B R n,r. To deal with those, we will need the following lemmas. Lemma 2.11 (20) Let B R n,r with rank(b) = r, let y C r \ {0, and let z C n \ {0. Then for all A R r,r we have BAy = z if and only if Ay = B z and BB z = z. Lemma 2.12 Let B R n,r with rank(b) = r, let y C r \ {0 and z C n \ {0. 1) If rank(z z) = 1, then there exists a positive semidefinite A = A T R r,r satisfying BAy = z if and only if BB z = z and y B z > 0. 2) If rank(z z) = 2. then there exists a positive semidefinite A = A T R r,r satisfying BAy = z if and only if BB z = z and y B z > y T B z. Proof. Let A R r,r, then by Lemma 2.11 we have that BAy = z if and only if Z BB z = z and Ay = B z. (2.29) If rank(z z) = 1, then by Theorem 2.6 the identity (2.29) is equivalent to BB z = z and y B z > 0. If rank(z z) = 2 then (2.29) is equivalent to BB z z = z z and Ay ȳ = B z B z, (2.30) because A is real. Note that rank(b z B z) = 2, because otherwise there would exist α C 2 \ {0 such that B z B zα = 0. But this implies that z zα = BB z BB zα = 0 in contradiction to the fact that rank(z z) = 2. Thus, by Theorem 2.5, there exists 0 A R r,r satisfying (2.30) if and only if BB z = z and y B z > y T B z. The following is a version of Lemma 2.12 without semidefiniteness. Lemma 2.13 Let B R n,r with rank(b) = r, let y C r \ {0 and z C n \ {0. Then there exist a real symmetric matrix A R r,r satisfying BAy = z if and only if BB z = z and y B z R. Proof. The proof is analogous to that of Lemma 2.12, using Theorem 2.8 and Theorem 2.9 instead of Theorem 2.5 and Theorem 2.6. In this section we have presented several mapping theorems, in particular for the real case. These will be used in the next section to determine formulas for the real stability radii of D systems. 14

15 3 Real stability radii for D systems Consider a real linear time-invariant dissipative amiltonian (D) system ẋ = (J R)Qx, (3.1) where J, R, Q R n,n are such that J T = J, R T = R 0 and Q T = Q > 0. For perturbations in the dissipation matrix we define the following real distances to instability. Definition 3.1 Consider a real D system of the form (3.1) and let B R n,r and C R q,n be given restrictions matrices. Then for p {2, F we define the following stability radii. 1) The stability radius r (R; B, C) of system (3.1) with respect to real general perturbations to R under the restriction (B, C) is defined by { r (R; B, C) := inf p R r,q, Λ ( (J R)Q (B C)Q ) ir. 2) The stability radius r S d (R; B) of system (3.1) with respect to real structure-preservig semidefinite perturbations from the set S d (R, B) := { R r,r T = 0 and (R + B B T ) 0 (3.2) under the restriction (B, B T ) is defined by r S d { (R; B) := inf p Sd (R, B), Λ ( (J R)Q (B B T )Q ) ir. 3) The stability radius r S i (R; B) of system (3.1) with respect to structure-preserving indefinite perturbations from the set S i (R, B) := { R r,r T = and (R + B B T ) 0 (3.3) under the restriction (B, B T ) is defined by r S i { (R; B) := inf p Si (R, B), Λ ( (J R)Q (B B T )Q ) ir. The formula for the stability radius r R,2 (R; B, C) is a direct consequence of the following well-known result. Theorem 3.2 (23) For a given M C p,m, define Then µ R (M) := ( inf { R m,p, det(i m M) = 0 ) 1. ( µ R (M) = inf σ 2 γ (0,1 Re M γ 1 Im M γ Im M Re M where σ 2 (A) is the second largest singular value of a matrix A. Furthermore, an optimal that attains the value of µ R (M) can be chosen of rank at most two. ), Applying this theorem to D systems we obtain the following corollary. 15

16 Corollary 3.3 Consider an asymptotically stable D system of the form (3.1) and let B R n,r and C R q,n be given restriction matrices. Then and ( ( r R,2 (R; B, C) = inf inf σ 2 ω R γ (0,1 Re M(ω) γ 1 Im M(ω) )) 1 γ Im M(ω) (3.4) Re M(ω) r R,2 (R; B, C) r R,F (R; B, C) 2 r R,2 (R; B, C), (3.5) where M(ω) := CQ ( (J R)Q iωi n ) 1B. Proof. By definition, we have { r R,2 (R; B, C) = inf R r,q, Λ ( (J R)Q (B C)Q ) ir = inf { R r,q, ω R, det ( iωi n (J R)Q + B CQ ) = 0 = inf { R r,q, ω R, det ( I n B CQ((J R)Q iωi n ) 1) = 0, where the last equality follows, since (J R)Q is asymptotically stable so that the inverse of (J R)Q iωi n exists for all ω R. Thus we have r R,2 (R; B, C) = inf { R r,q, ω R, det ( I n CQ((J R)Q iωi n ) 1 B ) = 0 ( ( ) ) 1 = inf µ R M(ω), ω R where M(ω) := CQ((J R)Q iωi n ) 1 B. Therefore, (3.4) follows from Theorem 3.2, and if is of rank at most two such that = r R,2 (R; B, C), then (3.5) follows by using the definition of r R,F (R; B, C) and by the fact that = r R,2 (R; B, C) r R,F (R; B, C) F rank( ) 2 = 2 r R,2 (R; B, C). To derive the real structured stability radii, we follow the strategy in 20 for the complex case, to reformulate the problem of computing r S d (R; B) or rs i (R; B) in terms of real structured mapping problems. The following lemma of 20 (expressed here for real matrices) gives a characterization when a D system of the form (3.1) has eigenvalues on the imaginary axis. Lemma , Lemma 3.1 Let J, R, Q R n,n be such that J T = J, R T = R 0 and Q T = Q > 0. Then (J R)Q has an eigenvalue on the imaginary axis if and only if RQx = 0 for some eigenvector x of JQ. Furthermore, all purely imaginary eigenvalues of (J R)Q are semisimple. After these preliminaries we obtain the following formulas for the stability radii. 3.1 The stability radius r S d (R; B) To derive formulas for the stability radius under real structure-preserving restricted and semidefinite perturbations we need the following two lemmas. Lemma 3.5 Let T = R n,n be positive semidefinite. Then x x x T x for all x C n, and equality holds if and only if x and x are linearly dependent. 16

17 Proof. Let S R n,n be a symmetric positive semidefinite square root of, i.e. S 2 =. Then using the Cauchy-Schwarz inequality we obtain that x T x = S x, Sx S x Sx = x x x x = x x, because x x = x x = x x as is real. In particular equality holds if and only if Sx and S x are linearly dependent which is easily seen to be equivalent to the linear dependence of x and x. Lemma 3.6 Let R, W R n,n be such that R T = R 0 and W T = W > 0. If x C n is such that rank(x x) = 2 and x W T RW x > x T W T RW x, set R := RW Re x Im x ( Re x Im x W RW Re x Im x ) 1 Re x Im x W R. Then R + R is symmetric positive semidefinite. Proof. Since R and W are real symmetric and x W T RW x > x T W T RW x, the matrix Re x Im x W RW Re x Im x is real symmetric and positive definite and, therefore, R is well defined and we have R = T R 0. We prove that R + R 0 by showing that all its eigenvalues are nonnegative. Since W is nonsingular, we have that W x 0. Also R is a real matrix of rank two satisfying R W Re x Im x = RW Re x Im x. This implies that (R + R )W Re x Im x = 0. (3.6) Since rankre x Im x = rankx x = 2 and since W is nonsingular, we have that W Re x and W Im x are linearly independent eigenvectors of R + R corresponding to the eigenvalue zero. Let λ 1,..., λ n be the eigenvalues of R and let η 1,..., η n be the eigenvalues of R + R, where both lists are arranged in nondecreasing order, i.e., 0 λ 1 λ n and η 1 η n. Since R is of rank two, by the Cauchy interlacing theorem 6, λ k η k+2 and η k λ k+2 for k = 1,..., n 2. (3.7) This implies that 0 η 3 η n, and thus the assertion follows once we show that η 1 = 0 and η 2 = 0. If R is positive definite, then λ 1,..., λ n satisfy 0 < λ 1 λ n and, therefore, 0 < η 3 η n. Therefore we must have η 1 = 0 and η 2 = 0 by (3.6). If R is positive semidefinite but singular, then let k be the dimension of the kernel of R. We then have k < n, because R 0. Letting l be the dimension of kernel of R + R, then using (3.7) we have that k 2 l k + 2, and we have η 1 = 0 and η 2 = 0 if we show that l = k + 2. Since W is nonsingular, the kernels of R and RW have the same dimension k. Let x 1,..., x k be linearly independent eigenvectors of RW associated with the eigenvalue zero, i.e., we have RW x i = 0 for all i = 1,..., k. Then R W x i = 0 for all i = 1,..., k, and hence (R + R )W x i = 0 for all i = 1,..., k. The linear independence of x 1,..., x k together with the nonsingularity of W implies that W x 1,..., W x k are linearly independent. By (3.6) we have (R + R )W Re x = 0 and (R + R )W Im x = 0. 17

18 Moreover, the vectors W Re x, W Im x, W x 1,..., W x k are linearly independent. Indeed, let α, β, α 1,..., α k R be such that αw Re x + βw Im x + α 1 W x α k W x k = 0. Then we have R(αW Re x + βw Im x) = 0 as RW x i = 0 for all i = 1,..., k. This implies that α = 0 and β = 0, because RW Re x and RW Im x are linearly independent. The linear independence of W x 1,..., W x k then implies that α i = 0 for all i = 1,..., k, and hence W Re x, W Im x, W x 1,..., W x k are linearly independent eigenvectors of R+ R corresponding to the eigenvalue zero. Thus, the dimension of the kernel of R + R is at least k + 2 and hence we must have η 1 = 0 and η 2 = 0. Using these lemmas, we obtain a formula for the structured real stability radius of D systems. Theorem 3.7 Consider an asymptotically stable D system of the form (3.1). Let B R n,r with rank(b) = r, and let p {2, F. Furthermore, for j {1, 2 let Ω j denote the set of all eigenvectors x of JQ such that (I n BB )RQx = 0 and rank ( RQx RQ x ) = j, and let Ω := Ω 1 Ω 2. Then r S d (R; B) is finite if and only if Ω is non-empty. If this is the case, then { r S d (R; B) = min inf (B RQx)(B RQx) x Ω 1 x QRQx, inf Y (Y X) 1 Y p, (3.8) p x Ω 2 where X = B T QRe x Im x and Y = B RQRe x Im x. Proof. By definition, we have r S d { (R; B) := inf p Sd (R, B), Λ ( (J R)Q (B B T )Q ) ir, where S d (R, B) := { R r,r T = 0 and (R + B B T ) 0. By using Lemma 3.4, we obtain that r S d { (R; B) = inf p Sd (R, B), (R + B B T )Qx = 0 for some eigenvector x of JQ { = inf p Sd (R, B), B B T Qx = RQx for some eigenvector x of JQ { = inf p Sd (R, B), B T Qx = B RQx for some x Ω, (3.9) since by Lemma 2.11 we have B B T Qx = RQx if and only if B T Qx = B RQx and BB RQx = RQx, and thus x Ω. From (3.9) and S d (R, B) { R r,r T = 0, we obtain that r S d (R; B) inf { p = T R r,r, 0, B T Qx = B RQx for some x Ω =: ϱ. (3.10) The infimum on the right hand side of (3.10) is finite if and only if Ω is non-empty. The same will also hold for r S d (R; B) if we show equality in (3.10). To this end we will use the abbreviations ϱ j := inf { p = T R r,r, 0, B T Qx = B RQx for some x Ω j 18

19 for j {1, 2, i.e., we have ϱ = min{ϱ 1, ϱ 2, and we consider two cases. Case (1): ϱ = ϱ 1. If x Ω 1, then BB RQx = RQx, and RQx and RQ x are linearly dependent. But then also B RQx and B RQ x are linearly dependent and hence, by Lemma 2.12 there exists R r,r such that 0 and B T Qx = B RQx (and thus ( ) 0 and ( )B T Qx = B RQx) if and only x QRQx > 0. This condition is satisfied for all x Ω 1. Indeed, since R is positive semidefinite, we find that x QRQx 0, and x QRQx = 0 would imply RQx = 0 and thus (J R)Qx = JQx which means that x is an eigenvector of (J R)Qx associated with an eigenvalue on the imaginary axis in contradiction to the assumption that (J R)Q only has eigenvalues in the open left half plane. Using minimal norm mappings from Theorem 2.6 we thus have r S d (R; B) ϱ 1 = inf { p = T R r,r, 0, B T Qx = B RQx, x Ω 1 = inf x Ω 1 { (B RQx)(B RQx) x QBB RQx = inf x Ω 1 { (B RQx)(B RQx) x QRQx. (3.11) As the expression in (3.11) is invariant under scaling of x, it is sufficient to take the infimum over all x Ω 1 of norm one. Then a compactness argument shows that the infimum is actually attained for some x Ω 1 and by 20, Theorem 4.2, the matrix := (B RQ x)(b RQ x) x QRQ x is the unique (resp. a) complex matrix of minimal Frobenius (resp. spectral) norm satisfying T = 0 and B T Qx = B RQx. Also, by Theorem 2.6 the matrix is real and by 20, Lemma 4.1 we have R + B B T 0. Thus S d (R, B). But this means that r S d (R; B) = ϱ 1 = ϱ. Case (2): ϱ = ϱ 2. If x Ω 2, then BB RQx = RQx, and Re RQx and Im RQx are linearly independent. But then, also Re B RQx and Im B RQx are linearly independent, because otherwise, also Re RQx = Re BB RQx and Im RQ x = Im BB RQ x would be linearly dependent. Thus, by Lemma 2.12 there exists R r,r such that 0 and B T Qx = B RQx if and only if x QRQx > x T QRQx. By Lemma 3.5 this condition is satisfied for all x Ω 2, because R and thus QRQ is positive semidefinite. Using minimal norm mappings from Theorem 2.5 we then obtain r S d (R; B) ϱ 2 = inf { p = T R r,r, 0, B T Qx = B RQx, x Ω 2 (3.12) = inf { Y (Y X) 1 Y p X = B T Q Re x Im x, Y = B RQ Re x Im x, x Ω 2 = Ỹ (Ỹ X) 1 Ỹ p, for some x Ω 2, where X = B T Q Re x Im x and Ỹ = B RQ Re x Im x. Indeed, the matrix Y (Y X) 1 Y is invariant under scaling of x, so that it is sufficient to take the infimum over all x Ω 2 having norm one. Then a compactness argument shows that the infimum is actually a minimum and attained for some x Ω 2. Then setting R := Ỹ (Ỹ X) 1 Ỹ, 19

20 we have equality in (3.12) if we show that R + B R B T is positive semidefinite, because this would imply that R S d (R, B). But this follows from Lemma 3.6 by noting that BB RQ Re x Im x = RQ Re x Im x. Indeed, by the definition of Ω 2, the vectors Re RQ x and Im RQ x are linearly independent, and R + B R B T = R BỸ (Ỹ X) 1 Ỹ B T = R BB RQRe x Im x ( Re x Im x QRQRe x Im x ) 1 Re x Im x QR(B ) T B T = R RQRe x Im x ( Re x Im x QRQRe x Im x ) 1 Re x Im x QR. is positive semidefinite by Lemma 3.6. This proves that r S d (R; B) = ϱ 2 = ϱ. Remark 3.8 In the case that R > 0 and J are invertible, the set Ω 1 from Theorem 3.7 is empty and hence Ω = Ω 2, because if R > 0 and if x is an eigenvector of JQ with rank ( RQx RQ x ) = 1 then, x is necessarily an eigenvector associated with the eigenvalue zero of JQ. Indeed, if RQx and RQ x are linearly dependent, then x and x are linearly independent, because RQ is nonsingular as R > 0 and Q > 0. This is only possible if x is associated with a real eigenvalue, and since the eigenvalues of JQ are on the imaginary axis, this eigenvalue must be zero. Remark 3.9 We mention that in the case that R and J are invertible, r S d (R; B) is also the minimal norm of a structure-preserving perturbation of rank at most two that moves an eigenvalue of (J R)Q to the imaginary axis. To be more precise, if S 2 (R, B) := { R r,r T =, rank 2, and (R + B B T ) 0 (3.13) and r S 2 { (R; B) := inf p S2 (R, B), Λ ( (J R)Q (B B T )Q ) ir, then we have r S 2 (R; B) = rs d (R; B). Indeed, assume that S 2(R, B) is such that (J R)Q (B B T )Q has an eigenvalue on the imaginary axis. By Lemma 3.4 we then have (R B B T )Qx = 0 for some eigenvector x of JQ. Since R and J are invertible, it follows from Remark 3.8 that RQx and RQ x are linearly independent. Since has rank at most two, it follows that the kernel of has dimension at least n 2. Thus, let w 3,..., w n be linearly independent vectors from the kernel of. Then B T Qx, B T Q x, w 3,..., w n is a basis of C n. Indeed, let α 1,..., α n C be such that Then multiplication with B yields α 1 B T Qx + α 2 B T Qx + α 3 w α n w n = 0. (3.14) 0 = α 1 B B T Qx + α 2 B B T Q x = α 1 RQx + α 2 RQ x and we obtain α 1 = α 2 = 0, because RQx and RQ x are linearly independent. But then (3.14) and the linear independence of w 3,..., w n imply α 3 = = α n = 0. Thus, setting T := B T Qx, B T Q x, w 3,..., w n we obtain that T is invertible and T T = diag(d, 0), where D = x QB B T Qx x T QB B T Qx x T QB B T Qx x x QB B T = QRQx x T QRQx Qx x T QRQx x QRQx 20.

21 Since by Lemma 3.5 we have x QRQx > x T QRQx, it follows that D is positive definite which implies S d (R; B) and hence r S 2 (R; B) rs d (R; B). The inequality is trivial as minimal norm elements from S d (R; B) have rank at most two. In this subsection we have characterized the real structured restricted stability radius under positive semidefinite perturbations to the dissipation matrix R. In the next subsection we extend these results to indefinite perturbations that keep R semidefinite. 3.2 The stability radius r S i (R; B) If the perturbation matrices R are allowed to be indefinite then the perturbation analysis is more complicated. We start with the following lemma. Lemma 3.10 Let 0 < R = R T, R = T R Rn,n be such that R has at most two negative eigenvalues. If dim (Ker(R + R )) = 2, then R + R 0. Proof. Let λ 1,..., λ n be the eigenvalues of R. As R has at most two negative eigenvalues we may assume that λ 3, λ 4,..., λ n 0 and we have the spectral decomposition R = n λ i u i u i i=1 with unit norm vectors u 1,..., u n R n. Since R = R + n λ i u i u i > 0. i=3 and λ 1 u 1 u 1 +λ 2u 2 u 2 is of rank two, we can apply the Cauchy interlacing theorem and obtain that R + R = R + λ 1 u 1 u 1 + λ 2u 2 u 2 has at least n 2 positive eigenvalues. But then using the fact that dim(ker(r + R )) = 2 we get R + R 0. Using this lemma, we obtain the following results on the stability radius r S i (R; B). Theorem 3.11 Consider an asymptotically stable D system of the form (3.1). Let B R n,r with rank(b) = r and let p {2, F. Furthermore, for j {1, 2 let Ω j denote the set of all eigenvectors x of JQ such that BB RQx = RQx and rank ( B T Qx B T Q x ) = j, and let Ω = Ω 1 Ω 2. 1) If R > 0, then r S i (R; B) is finite if and only if Ω is non-empty. In that case we have { r S i R,2 (R; B) = min (B RQx) inf x Ω 1 B T Qx, inf Y X (3.15) x Ω 2 and { R,F (R; B) = min (B RQx) inf x Ω 1 B Qx, inf r S i x Ω 2 Y X 2 F trace (Y X (Y X ) XX ), where X = Re B T Qx Im B T Qx and Y = Re B RQx Im B RQx for x Ω 2. (3.16) 21

22 2) If R 0 is singular and if r S i (R; B) is finite, then Ω is non-empty and we have { r S i R,2 (R; B) min (B RQx) inf x Ω 1 B T Qx, inf Y X (3.17) x Ω 2 and { r S i R,F (R; B) min (B RQx) inf x Ω 1 B T Qx, inf Y X 2 F trace (Y x Ω X (Y X ) XX ), 2 (3.18) where X = Re B T Qx Im B T Qx and Y = Re B RQx Im B RQx for x Ω 2. Proof. By definition r S i { (R; B) = inf p Si (R, B), Λ ( (J R)Q (B B T )Q ) ir, where S i (R, B) := { = T R r,r (R + B B T ) 0. Using Lemma 3.4 and Lemma 2.11 and following the lines of the proof of Theorem 3.7, we get r S i (R; B) = inf { p S i (R, B), B T Qx = B RQx for some eigenvector x of JQ satisfying BB RQx = RQx. Since all elements of S i (R, B) are real symmetric, we obtain that r S i { (R; B) inf p = T R r,r, B T Qx = B RQx for some x Ω =: ϱ (p). (3.19) The infimum in the right hand side of (3.19) is finite if Ω is non-empty, as by Lemma 2.13 for x Ω there exist = T R r,r such that B T Qx = B RQx if and only if x QBB RQx R. This condition is satisfied because of the fact that BB RQx = RQx and R is real symmetric. If r S i (R; B) is finite, then Ω is non-empty because otherwise the right hand side of (3.19) would be infinite. To continue, we will use the abbreviations ϱ (p) j := inf { p = T R r,r, B T Qx = B RQx for some x Ω j for j {1, 2, i.e., ϱ (p) = min{ϱ (p) 1, ϱ(p) 2, and we consider two cases. Case (1): ϱ (p) = ϱ (p) 1. If x Ω 1, then BB RQx = RQx and B T Qx and B T Q x are linearly dependent. Then using mappings of minimal spectral resp. Frobenius norms (and again using compactness arguments), we obtain from Theorem 2.9 that r S i (R; B) ϱ(p) 1 = inf = inf x Ω 1 { p = T R r,r, B T Qx = B RQx, x Ω 1 { B RQx p B T Qx p = B RQ x p B T Q x p for some x := x Ω 1. This proves 2) and in (3.15) and (3.16) in the case ϱ (p) = ϱ (p) 1. Case (2): ϱ (p) = ϱ (p) 2. If x Ω 2, then BB RQx = RQx and B T Qx and B T Q x are linearly independent. Using mappings of minimal spectral resp. Frobenius norms (and once more 22

23 using compactness arguments), we obtain from Theorem 2.8 that r S i R,2 (R; B) ϱ(2) 2 = inf { = T R r,r, B T Qx = B RQx, x Ω 2 = inf { Y X X = Re B T Qx Im B T Qx, Y = Re B RQx Im B RQx, x Ω 2 = Ŷ X, for some x Ω 2, where X = Re B T Q x Im B T Q x and Ŷ = Re B RQ x Im B RQ x, and ( r S i R,F (R; B) ) 2 ( = inf ) 2 { = inf 2 F = T R r,r, B T Qx = B RQx, x Ω 2 ( {2 Y X 2 F trace Y X (Y X ) XX ) X = Re B T Qx Im B T Qx, Y = Re B RQx Im B RQx, x Ω 2 ϱ (F ) 2 = 2 Ỹ X 2 F trace (Ỹ X (Ỹ X ) X X ) for some x Ω 2, where X = Re B T Q x Im B T Q x and Ỹ = Re B RQ x Im B RQ x. This proves 2) and in (3.15) and (3.16) in the case ϱ (p) = ϱ (p) 2. In both cases (1) and (2), it remains to show that equality holds in (3.15) and (3.16) when R > 0. This would also prove that in the case R > 0 the non-emptyness of Ω implies the finiteness of r S i R,F (R; B). Thus, assume that R > 0 and let = T R r,r and = T R r,r be such that they satisfy B T Q x = B RQ x and B T Q x = B RQ x, (3.20) and such that they are mappings of minimal spectral or Frobenius norm, respectively, as in Theorem 2.8 or Theorem 2.9, respectively. The proof is complete if we show that (R + B B T ) 0 and (R + B B T ) 0, because this would imply that and belong to the set S i (R, B). In the case ϱ (p) = ϱ (p) 1 this follows exactly as in the proof of 20, Theorem 4.5 which is the corresponding result to Theorem 3.11 in the complex case. (In fact, in this case and coincide with the corresponding complex mappings of minimal norm.) In the case ϱ (p) = ϱ (p) 2, we obtain from Lemma 2.10, that the matrices and are of rank at most four with at most two negative eigenvalues. This implies that also the matrices B B T and B B T individually have at most two negative eigenvalues. Indeed, let B 1 R n,n r be such that B B 1 R n,n is invertible then we can write B B T 0 = B B B B 1 and by Sylvester s law of inertia also B B T has at most two negative eigenvalues. A similar argument proves the assertion for B B T. Furthermore, using (3.20), we obtain (R + B B T )Q x = RQ x BB RQ x = RQ x RQ x = 0 and (R + B B T )Q x = RQ x BB RQ x = RQ x RQ x = 0, 23

24 since x, x Ω, i.e., BB RQ x = RQ x and BB RQ x = RQ x. Also rank x x = 2 and rank x x = 2, respectively, imply that ( dim Ker(R + B B ) ( T ) = 2 and dim Ker(R + B B ) T ) = 2. Thus, Lemma 3.10 implies that (R + B B T ) 0 and (R + B B T ) 0. Remark 3.12 It follows from the proof of Theorem 3.11 that in the case R 0, the inequalities in (3.17) and (3.18) are actually equalities if the minimal norm mappings and from (3.20) satisfy (R + B B T ) 0 and (R + B B T ) 0, respectively. In our numerical experiments this was always the case, so that we conjecture that in general equality holds in (3.17) and (3.18). 4 Numerical experiments In this section, we present some numerical experiments to illustrate that the real structured stability radii are indeed larger than the real unstructured ones. To compute the distances, in all cases we used the function fminsearch in Matlab Version No (R2009a) to solve the associated optimization problems. We computed the real stability radii r R,2 (R; B, B T ), r S i R,2 (R; B) and rs d R,2 (R; B) with respect to real restricted perturbations to R, as obtained in Theorem 3.2, Theorem 3.7, and Theorem 3.11, respectively, and compared them to the corresponding complex distances r C,2 (R; B, B T ), r S i C,2 (R; B) and rs d C,2 (R; B) as obtained in 20, Theorem 3.3, 20, Theorem 4.5 and 20, Theorem 4.2, respectively. We chose random matrices J, R, Q, B R n,n for different values of n 14 with J T = J, R T = R 0 and B of full rank, such that (J R)Q is asymptotically stable and all restricted stability radii were finite. The results in Table 4.1 illustrate that the stability radius r R,2 (R, B, B T ) obtained under general restricted perturbations is significantly smaller than the real stability radii obtained under structure-preserving restricted perturbations, and it also illustrates that the real stability radii may bd significantly larger than the corresponding complex stability radii. size n r C,2 (R; B, B T ) r R,2 (R; B, B T ) r S i C,2 (R; B) rs i R,2 (R; B) rs d C,2 (R; B) rs d R,2 (R; B) Table 4.1: Comparison between complex and real stability radii. Example 4.1 As a second example consider the lumped parameter, mass-spring-damper dynamical system shown in Figure 4.1, see e.g. 32. It has two point masses m 1 and m 2, 24