Structured eigenvalue/eigenvector backward errors of matrix pencils arising in optimal control

Transcription

1 Electronic Journal of Linear Algebra Volume 34 Volume 34 08) Article Structured eigenvalue/eigenvector backward errors of matrix pencils arising in optimal control Christian Mehl Technische Universitaet Berlin, Volker Mehrmann Technische Universitaet Berlin, Punit Sharma Indian Institute of Technology Delhi, ollow this additional works at: Part of the Control Theory Commons, Dynamic Systems Commons, the Numerical Analysis Computation Commons Recommended Citation Mehl, Christian; Mehrmann, Volker; Sharma, Punit 08), "Structured eigenvalue/eigenvector backward errors of matrix pencils arising in optimal control", Electronic Journal of Linear Algebra, Volume 34, pp DOI: This Article is brought to you for free open access by Wyoming Scholars Repository It has been accepted for inclusion in Electronic Journal of Linear Algebra by an authorized editor of Wyoming Scholars Repository or more information, please contact

2 STRUCTURED EIGENVALUE/EIGENVECTOR BACKWARD ERRORS O MATRIX PENCILS ARISING IN OPTIMAL CONTROL CHRISTIAN MEHL, VOLKER MEHRMANN, AND PUNIT SHARMA Abstract Eigenvalue eigenpair backward errors are computed for matrix pencils arising in optimal control In particular, formulas for backward errors are developed that are obtained under block-structure-preserving symmetrystructure-preserving perturbations It is shown that these eigenvalue eigenpair backward errors are sometimes significantly larger than the corresponding backward errors that are obtained under perturbations that ignore the special structure of the pencil Key words Eigenvalue backward error, Eigenvector backward error, Structured matrix pencil, Dissipative Hamiltonian system, H control, Linear quadratic optimal control AMS subject classifications 93D0, 93D09, 655, 5A, 5A Introduction In this paper, we consider the perturbation theory, in particular the calculation of structured backward errors, for eigenvalues eigenvectors of structured matrix pencils Lz) of the form ) Lz) = M + zn := 0 J R B J R) H Q 0 B H 0 S + z 0 E 0 E H where J, R, E, Q C n,n, B C n,m S C m,m satisfy J H = J, R H = R E H = E, Q H = Q, S H = S > 0, ie, S is positive definite These pencils are special cases of so-called even pencils, ie, matrix pencils P z) satisfying P z) = P z) H ; see, eg, [] Even pencils with an additional block-structure as in ) arise in optimal control H control problems as well as in the passivity analysis of dynamical systems or instance, if one considers the optimal control problem of minimizing the cost functional subject to the constraint t 0 x H Qx + u H Su dt ) Eẋ = Ax + Bu, xt 0 ) = x 0, then it is well known see [0, 3]) that the optimal solution is associated with the deflating subspace of a pencil of the form ) associated with the finite eigenvalues in the open left half plane If there exist exactly Received by the editors on December, 07 Accepted for publication on September 9, 08 Hling Editor: rancoise Tisseur Corresponding Author: Volker Mehrmann Institut für Mathematik, MA 4-5 TU Berlin, Str d 7 Juni 36, D-063 Berlin, Germany mehl@mathtu-berlinde, mehrmann@mathtu-berlinde) C Mehl V Mehrmann gratefully acknowledge support from Einstein Center ECMath via project SE3: Stability analysis of power networks power network models V Mehrmann also acknowledges support Deutsche orschungsgemeinschaft through CRC 90 Control of Self-Organizing Nonlinear Systems via project A0: Analysis computation of stability exponents for delay differential-algebraic equations Department of Mathematics, Indian Institute of Technology Delhi, Hauz Khas, New Delhi-006, India punitsharma@mathsiitdacin),

3 57 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control n eigenvalues in the open left half plane then this deflating subspace is an extended Lagrangian subspace or other applications in passivity analysis robust control, see [7]) Note that for general descriptor systems we need not have that E = E H However, if this is not the case then we can just carry out the polar decomposition see [5]) to obtain E = UẼ with U unitary Ẽ = ẼH Multiplying equation ) from the left with U H we obtain a new system that has the desired property E = E H, so wlog we assume that E = E H then partition A = J R into its skew-symmetric symmetric part Note that the condition E = E H holds by assumption if ) is a port-hamiltonian descriptor system; see [5, 7] In this case, we furthermore have that R 0, ie, it is positive semidefinite The solution of the optimal control problem becomes highly ill-conditioned when eigenvalues are close to the imaginary axis the solution usually ceases to exist when the eigenvalues are on the imaginary axis [6, ] When eigenvalues on the imaginary axis exist then it is an important question to find small perturbations to the system ) or the pencil ) that remove the eigenvalues from the imaginary axis [4, 3] These questions motivate the principle aims of this paper to determine backward errors associated with eigenvalues on the imaginary axis of pencils of the form ) We will consider in this paper the special case of pencils with Q = 0, which arises in optimal control without state weighting, in the context of passivity analysis [, 3] Thus, we will consider a pencil of the form 3) Lz) = M + zn := 0 J R B J R) H 0 0 B H 0 S + z 0 E 0 E H Unfortunately many of our results do not carry over easily to the case Q 0 where the perturbation theory becomes much more involved highly technical In the following, denotes the spectral norm of a vector or a matrix A denotes the robenius norm of a matrix A Hermn) SHermn) respectively denote the set of Hermitian skew- Hermitian matrices of size n By ir we denote the set of nonzero purely imaginary numbers, ie, ir = iα α R \ 0, by I n the identity matrix of size n or a matrix A we write A = 0 if each entry of A is equal to zero The sensitivity analysis of eigenvalues eigenvalue/eigenvector pairs in the following, called eigenpairs) of matrix pencils matrix polynomials with various structures has recently received a lot of attention; see, eg, [,, 3, 6, 8, 5] In particular, backward error formulas for structured matrix pencils polynomials with respect to structure preserving perturbations have been obtained in [, ] in [8, 9], respectively or pencils of the form 3), if the structure of the pencil is ignored, then for a given pair λ, x) C C n+m \ 0) the eigenpair backward error is defined as ηl, λ, x) = inf [ M N ] M, N C n+m,n+m, M M ) + λn N ) ) x = 0 It can be interpreted as the robenius norm of the smallest perturbation that makes λ, x) being an eigenpair of the perturbed pencil Minimizing this expression over all λ, x) ir) C n+m ) we obtain the distance of Lz) to the next pencil having eigenvalues on the imaginary axis thus, the passivity radius of Lz); see [4, 4] If the even structure of the pencil is taken into account, then a structured eigenpair backward

4 Christian Mehl, Volker Mehrmann, Punit Sharma 58 error with respect to structure-preserving perturbations can be defined as η even L, λ, x) = inf [ M N ] M Hermn + m), N SHermn + m), M M ) + λn N ) ) x = 0 Clearly, we have ηl, λ, x) η even L, λ, x) In fact, for a given λ, x) C C n+m \ 0), it is well known by [3, Theorem 46] that 4) ηl, λ, x) = Lλ)x x + λ, by [, Theorem 337] that 5) η even x Lλ)x x L, λ, x) = H Lλ)x x 4 + λ ) However, both formulas ignore the special block-structure of the pencil Lz), in particular the zero structure the definiteness of the matrix S, as we will show in this paper, eigenpair backward errors with respect to perturbations that preserve the block-structure possibly also the symmetry-structure may be significantly larger than the more generally obtained backward errors ηl, λ, x) η even L, λ, x) This is, in particular, the case when only one or two blocks are perturbed, while others are unperturbed, a situation that arises in many applications, eg when E is the incidence matrix of a network [0], in the case of semi-explicit differential-algebraic equations [9] where E is block-diagonal with an identity a zero block, when J is the structure matrix in a port-hamiltonian system [7], or when the weight matrix is just a scalar multiple of the identity as is common in optimal control problems for partial differential equations [6] We will mainly consider complex backward errors In some situations, the corresponding minimal-norm perturbations turn out to be real if the original pencil was real to start with In those cases, we easily obtain a corresponding result on real backward errors which we will explicitly state In other situations, however, this is not the case the techniques developed in this paper cannot be used to compute the corresponding real backward errors In those cases, the development of real structure-preserving backward errors remains a challenging open problem The remainder of this paper is organized as follows In Section, we review some minimal norm mapping problems In Section 3, we introduce a terminology define block- symmetry-structurepreserving eigenpair or eigenvalue backward errors for pencils Lz) of the form 3) These backward errors are computed while perturbing any two, three or all of the blocks J, R, E or B in Sections 4, 5 6, respectively The significance of these block- symmetry-structure-preserving backward errors over ηl, λ, x) η even L, λ, x) is shown via some numerical examples in Section 7 Preliminaries An important tool for the computation of backward errors are minimal norm solutions to mapping problems In this section, we will review some of these results restate them in a form that we need in the following sections The solution to the skew-hermitian mapping problem, ie, to find SHermn) that maps a matrix X C n,k to Y C n,k, is well known; see, eg, [], where also solutions that are minimal with respect to

5 59 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control the spectral the robenius norms are characterized The following theorem is a particular case of [, Theorem 3] Theorem Let X, Y C n,k Then there exists SHermn) satisfying X = Y if only if Y X X = Y Y H X = X H Y If the latter conditions are satisfied, then min SHermn), X = Y = Y X trace Y X Y X ) H XX ) ) the unique minimum is attained for = Y X Y X ) H X ) H X H Y X The second mapping problem that we will need is the following; see [7, Theorem ], [, Theorem ] Theorem Let u C m \ 0, r C n, w C n \ 0 s C m Define S = C n,m u = r, H w = s Then S if only if u H s = r H w If the latter condition is satisfied, then = ruh u + wsh w sh u)wu H w u is the unique matrix such that u = r H w = s, inf S = = r u + s w sh u w u Moreover, r inf = max S u, s w The following result see [, Remark ]) gives a real minimal robenius norm solution of the mapping problem considered in Theorem Theorem 3 Let u C m, r C n, w C n s C m be such that rank[u ū]) = rank[w w]) = define S R = R n,m u = r, H w = s Then S R if only if u H s = r H w u T s = r T w If the latter conditions are satisfied, then inf =, S R where = [r r][u ū] + [s s][w w] ) H [s s][w w] ) H [u ū][u ū] We mention that the form of the minimal norm perturbation given in [, Remark 4] slightly differs from the one given here, because in [] it was presented using real imaginary parts rather than complex vectors their complex conjugates

6 Christian Mehl, Volker Mehrmann, Punit Sharma Structured eigenpair backward errors In this section, we consider structured matrix pencils Lz) of the form 3) We use the results on the mapping problems from the previous section to estimate structure-preserving backward errors for eigenvalues λ or eigenpairs λ, x) of Lz), while perturbing only certain block entries of Lz) for the case when λ is purely imaginary S is definite To distinguish between different cases, we introduce a terminology for perturbations M + z N of the pencil Lz) = M + zn that affect only some of the blocks J, R, E, B of Lz) or example, suppose that only the blocks J E in Lz) are subject to perturbations Then the corresponding perturbations to M N are given by 3) M = 0 J 0 H J N = where J, E C n,n or λ C x C n+m \ 0 we then define 0 E 0 H E ) the block-structure-preserving eigenpair backward error η B J, E, λ, x) with respect to perturbations only to J E by 3) η B J, E, λ, x) = inf [ J E ] M M ) + λn N ) ) x = 0, M + z N B, where B denotes the set of all pencils M + z N as in 3) with J, E C n,n ; ) the symmetry-structure-preserving eigenpair backward error η S J, E, λ, x) with respect to structurepreserving perturbations only to J E by 33) η S J, E, λ, x) = inf [ J E ] M M ) + λn N ) ) x = 0, M + z N S where S denotes the set of all pencils M +z N as in 3) with J SHermn) E Hermn) or a given λ C, we also define the block-structure-preserving symmetry-structure-preserving eigenvalue backward errors η B J, E, λ) η S J, E, λ), respectively, by η B J, E, λ) := inf η B J, E, λ, x) η S J, E, λ) := inf η S J, E, λ, x) x C n+m \0 x C n+m \0 or other combinations of perturbations to the blocks J, R, E, B in Lz), the corresponding sets B S as well as the block- symmetry-structure-preserving eigenpair or eigenvalue backward errors are defined analogously 4 Perturbation in any two of the blocks J, R, E B In this section, we compute block- symmetry-structure-preserving backward errors of λ ir x C n+m \ 0 as approximate eigenpair, resp eigenvalue of the pencil Lz) defined in 3) while perturbing any two of the blocks J, R, E, or B at a time As we have discussed in the introduction, it is a common situation in many applications that not all blocks are perturbed Although restrictions in the perturbation structure are more common in some blocks than in the others, for completeness we also discuss several other perturbation combinations 4 Perturbation only in J E Let Lz) be a pencil as in 3) furthermore let λ, x) C C n+m \ 0) Suppose that only the blocks J E of Lz) are subject to perturbations Then by Section 3, B is the set of all pencils Lz) = M + z N, where M N have the block structure as in 3), S is the set of all pencils from B where in addition we have H J = J H E = E for the blocks in 3),,

7 53 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control The corresponding block-structure- symmetry-structure-preserving eigenpair backward errors η B J, E, λ, x) η S J, E, λ, x) are defined by 3) 33), respectively We first discuss under which conditions these backward errors are finite Remark 4 Let Lz) be a pencil as in 3), let λ ir x = [x T x T x T 3 ] T be such that x, x C n x 3 C m Then for any J, E C n,n corresponding Lz) = M + z N B, we have Lλ) Lλ))x = 0 if only if 4) 4) 43) J + λ E )x = J R + λe)x + Bx 3, J + λ E ) H x = J R λe)x, 0 = B H x + Sx 3, ie, η B J, E, λ, x) is finite if only if there exist matrices J E such that these equations are satisfied In the next lemma, we present conditions that are equivalent to the existence of matrices J E that satisfy the first two of the three equations in Remark 4 Lemma 4 Let Lz) be a pencil as in 3), let λ ir x = [x T x T x T 3 ] T be such that x, x C n x 3 C m urthermore, set r := J R + λe)x + Bx 3 s := J R λe)x Then the following statements are equivalent ) There exist J C n,n E C n,n satisfying 4) 4) ) There exists C n,n such that x = r H x = s 3) The identity x H 3 B H x = 0 is satisfied Moreover, we have inf J + E J, E C n,n satisfy 4) 4) 44) = inf + λ Cn,n, x = r, H x = s Proof ) ) : Let J C n,n E C n,n be such that 4) 4) are satisfied Then by setting = J + λ E we get x = r, H x = s which shows ) urthermore, using the Cauchy-Schwarz inequality in R ), we obtain J + λ E ) + λ ) J + E ) This implies thus, 45) inf + λ J, E C n,n satisfy 4) 4), = J + λ E inf J + E J, E C n,n satisfy 4) 4), inf + λ Cn,n, x = r, H x = s inf J + E J, E C n,n satisfy 4) 4),

8 Christian Mehl, Volker Mehrmann, Punit Sharma 53 which yields in 44) ) ) : Conversely, suppose that C n,n satisfies x = r H x = s Then by setting J = + λ E = λ + λ we get J + λ E =, hence, J E satisfy 4) 4) which proves ) urthermore, we obtain This implies inf J + E = inf J + E = + λ ) + λ + λ ) = + λ Cn,n, x = r, H x = s, J = + λ C n,n, x = r, H x = s, + λ, E = λ + λ hence, inf J + E inf J, E C n,n satisfy 4) 4) + λ C n,n, x = r, H x = s, which proves in 44) ) 3) : This follows from Theorem, because there exists C n,n satisfying x = r H x = s if only if x H s = r H x Since λ is purely imaginary, this latter equation is equivalent to x H 3 B H x = 0 Theorem 43 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m set r = J R+λE)x +Bx 3 s = J +R+λE)x Then η B J, E, λ, x) is finite if only if x 3 = 0 B H x = 0 If the latter conditions hold then 46) η B J, E, λ, x) = + λ η B J, E, λ) = σ minj R + λe), + λ where is given by = rx H x if x = 0, x s H x if x = 0, ) rx H x + xsh x I n xxh x otherwise Proof Combining Remark 4 Lemma 4, we obtain that η B J, E, λ, x) is finite if only if x satisfies x H 3 B H x = 0 B H x + Sx 3 = 0, or equivalently, x 3 = 0 B H x = 0, since S is definite Thus,

9 533 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control assume that x satisfies x 3 = 0 B H x = 0 Then, we obtain η B J, E, λ, x) = inf [ J E ] M M ) + λn N ) ) x = 0, M + z N B J = inf [ J E ], E C n,n satisfy 4) 4) = inf + λ Cn,n, x = r, H x = s 47) = inf + λ C n,n, x = r, H x = s, where the second last equality is due to Lemma 4 Thus, the formula for η B J, E, λ, x) in 46) follows from Theorem for the case x, x 0 for the case x = 0 or x = 0 it is straightforward Indeed, in the case x = 0, any matrix with x = r satisfies r x = rxh x is a matrix for which equality is attained The case x = 0 is analogous) Next we will prove the formula for η B J, E, λ) in 46) To this end, let M := y = [ y T y T 0 ] C n+m y, y C n, y, y ) 0, 0), B H y = 0 Then, we obtain ) + λ η B J, E, λ) = + λ ) inf η B J, E, λ, y) y C n+m \0 = inf y M C n,n, y = J R + λe)y, H y = J + R + λe)y inf 48) C n,n, y = J R + λe)y, H y = J + R + λe)y, y M where the second equality is due to 47) the inequality in the last line follows from the fact that for any C n,n, we have Defining µ := inf inf C n,n, y = J R + λe)y, H y = J + R + λe)y, y M we get by applying Theorem for the case of the spectral norm that J R + λe) H µ = inf max y, J R + λe)y 49) y M y y J R + λe) H y J R + λe)y 40) = min inf, inf, y C n \0,B H y =0 y y C n \0 y where in 49) we interpret the undefined expressions 0 0 that occur in the cases y = 0 or y = 0 as being equal to zero Let the columns of U = [u,, u k ] C n,k form an orthonormal basis of nullb H ) Then 4) J R + λe) H y J R + λe) H y inf y C n \0,B H y =0 y = inf y nullb H )\0 y J R + λe) H Uα = inf α C k \0 α = σ min J R + λe) H U ) )

10 Christian Mehl, Volker Mehrmann, Punit Sharma 534 By inserting 4) in 40), we get 4) µ = min σ min J R + λe) H ), σ min J R + λe) H U ) = σ min J R + λe) H ) = σ min J R + λe ), Using the value of µ from 4), we show that equality holds in 48) by constructing such that = = µ or this, let u v, respectively, be unit left right singular vectors of J R + λe) ) corresponding to the singular value σ := σ min J R + λe consider := σ uv H Then, clearly = = σ as is of rank one, Thus, we have equality in 48), ie, v = σ u = J R + λe)v H u = σ v = J R + λe) H u η B J, E, λ) = µ = = σ minj R + λe) + λ + λ + λ which finishes the proof Next we aim to compute the symmetry-structure-preserving eigenpair error η S J, E, λ, x), ie, when we have H J = J H E = E in the pencils Lz) = M + z N S We start with a criterion for the finiteness of the eigenpair error, where we focus on the case that λ is on the imaginary axis Remark 44 Let Lz) be a pencil as in 3), let λ ir x = [x T x T x T 3 ] be such that x, x C n x 3 C m Then using H J = J H E = E also the fact that λ is purely imaginary, the equations 4) 43) take the form J + λ E )x = J R + λe)x + Bx 3, J λ E )x = J R λe)x, 0 = B H x + Sx 3 Thus, combining the first two of these equations, we find that η S J, E, λ, x) is finite if only if there exist J SHermn) E Hermn) such that the equations J + λ E ) [ x x ] = [ J R + λe)x + Bx 3 J + R + λe)x ], 43) 0 = B H x + Sx 3 are satisfied We start with a lemma that contains equivalent conditions for equation 43) to be satisfied Lemma 45 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m, define X = [ x x ] Then the following statements are equivalent Y = [ J R + λe)x + Bx 3 J + R + λe)x ] ) There exist J SHermn) E Hermn) satisfying 43) ) There exists SHermn) such that X = Y

11 535 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control 3) X Y satisfy Y H X = X H Y Y X X = Y Moreover, we have inf J + E J SHermn), E Hermn) satisfying 43) 44) = inf + λ SHermn), X = Y Proof ) ) : Let J SHermn) E Hermn) be such that they satisfy 43), then by setting = J + λ E we get X = Y, SHermn) as λ ir The inequality in 44) then follows by the same arguments as in ) ) in the proof of Lemma 4 ) ) : Conversely, let SHermn) be such that X = Y Then, setting J = + λ E = λ + λ, we obtain J + λ E )X = Y as well as J SHermn) E Hermn), since λ ir Again, the proof in 44) follows by arguments similar to those in the part ) ) in the proof of Lemma 4 ) 3) : This follows immediately by Theorem Theorem 46 Let Lz) be a pencil defined as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m set X = [ ] x x Y = [ ] J R + λe)x + Bx 3 J + R + λe)x Then η S J, E, λ, x) is finite if only if Y H X = X H Y, Y X X = Y B H x + Sx 3 = 0 If the three latter conditions are satisfied, then 45) η S J, E, λ, x) = + λ Y X trace Y X Y X ) H XX )) Proof Combining Remark 44 Lemma 45 it follows that η S J, E, λ, x) is finite if only if x satisfies Y H X = X H Y, Y X X = Y B H x + Sx 3 = 0 In the following, let us assume that these conditions on x are satisfied Then we obtain η S J, E, λ, x) = inf [ J E ] J SHermn), E Hermn) satisfy 43) = inf + λ SHermn), X = Y, where the last equality is due to Lemma 45 Hence, 45) follows by using Theorem 4 Perturbations only in R E In this section, we consider the case where only the blocks R E in a pencil Lz) as in 3) are perturbed Let λ C x C n+m \ 0 Then by the terminology outlined in Section 3, the block- symmetry-structure-preserving eigenpair backward errors η B R, E, λ, x) η S R, E, λ, x) are defined by 46) η B R, E, λ, x) = inf [ R E ] M M ) + λn N ) ) x = 0, M + z N B,

12 Christian Mehl, Volker Mehrmann, Punit Sharma ) η S R, E, λ, x) = inf [ R E ] M M ) + λn N ) ) x = 0, M + z N S, respectively, where B is the set of all pencils of the form Lz) = M + z N with the block-structure 48) M = 0 R 0 H R N = 0 E 0 H E R, E C n,n, while S is the corresponding set of pencils Lz) = M + z N as in 48) with R, E Hermn) We highlight that in the case that only the block-structure is preserved, the perturbation matrices in 48) have exactly the same structure as the ones in 3), hence, by following exactly the same lines as in the previous section, we obtain the following theorem which shows that the values of η B R, E, λ, x), also of η B R, E, λ) := inf x C n+m \0 η B R, E, λ, x) are equal to the corresponding values η B J, E, λ, x) η B J, E, λ) Theorem 47 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m set r = J R+λE)x +Bx 3 s = J +R+λE)x Then η B R, E, λ, x) is finite if only if x 3 = 0 B H x = 0 If the latter conditions hold then η B R, E, λ, x) = η B J, E, λ, x) = + λ, η B R, E, λ) = η B J, E, λ) = σ minj R + λe), + λ where is given by = rx H x if x = 0, x s H x if x = 0, ) rx H x + xsh x I n xxh x otherwise Proof The proof is similar to the proof of Theorem 43 Next, we turn to the eigenpair backward error η S R, E, λ, x) for purely imaginary λ ir x = [x T x T x T 3 ] T C n+m \0 Note that in this case, H R = R H E = E In particular, the perturbations now have a different symmetry structure than the corresponding ones from the previous section, so that we expect the backward error η S R, E, λ, x) to differ from η S J, E, λ, x) We start again with a criterion for the finiteness of η S R, E, λ, x) continue with a lemma giving equivalent conditions Remark 48 Let Lz) be a pencil as in 3), let λ ir x = [x T x T x T 3 ] be such that x, x C n x 3 C m Then using H R = R H E = E also the fact that λ is purely imaginary, we find that there exist R, E Hermn) correspondingly Lz) = M + z N S such

13 537 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control that Lλ) Lλ))x = 0 if only if 49) 40) 4) R + λ E )x = J R + λe)x + Bx 3 R + λ E ) H x = J R λe)x 0 = B H x + Sx 3 Thus, η S R, E, λ, x) is finite if only if there exist R, E Hermn) such that 49) 4) are satisfied Lemma 49 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m let r = J R+λE)x +Bx 3 s = J R λe)x Then the following statements are equivalent ) There exist R, E Hermn) satisfying 49) 40) ) There exists C n,n such that x = r H x = s 3) The identity x H 3 B H x = 0 is satisfied Moreover, we have inf R + E R, E Hermn) satisfy 49) 40) + H = inf + H 4) λ Cn,n, x = r, H x = s Proof ) ) : Let R, E Hermn) be such that they satisfy 49) 40) Setting = R + λ E, we get x = r, H x = s Also note that R λ E are the unique Hermitian skew-hermitian parts of, respectively, ie, R = + H )/ λ E = H )/ This implies inf R + E + H = inf R + E = + H + H λ R, E Hermn) satisfy 49) 40) + H λ = R + λ E, R, E Hermn) satisfy 49) 40) Thus, we obtain inf R + E 43) J, E Hermn) satisfy 49) 40) + H inf + H λ Cn,n, x = r, H x = s which gives in 4) ) ) : Suppose that C n,n is such that x = r H x = s Then, by setting R = + H E = λ λ H ), we get R, E Hermn) such that 49) 40) are satisfied, because

14 Christian Mehl, Volker Mehrmann, Punit Sharma 538 R + λ E = Also, we have which implies + H inf = inf R + E R + E = + H + H λ + H λ Cn,n, x = r, H x = s C n,n, x = r, H x = s, R = + H, E = λ λ ) H hence 44) inf R + E R, E Hermn) satisfying 49) 40) + H inf + H λ C n,n, u = r, H w = s which finishes the proof of 4) ) 3) : By Theorem, there exists C n,n satisfying x = r H x = s if only if x H s = r H x which in turn is equivalent to x H 3 B H x = 0 In contrast to Theorem 46 47, we only obtain bounds for the symmetry-structure-preserving eigenpair backward error η S R, E, λ, x) Theorem 40 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T so that x, x C n x 3 C m, set r = J R+λE)x +Bx 3 s = J +R+λE)x Then η S R, E, λ, x) is finite if only if x 3 = 0 B H x = 0 If the latter conditions are satisfied, then 45) η S R, E, λ, x) + H + λ H if λ 46) λ η S R, E, λ, x) + H + λ H if λ, where is given by = rx H x if x = 0, x s H x if x = 0, ) rx H x + xsh x I n xxh x otherwise Proof Combining Remark 48 Lemma 49, we obtain that η S R, E, λ, x) is finite if only if x H 3 B H x = 0 B H x + Sx 3 = 0 The latter conditions hold if only if x 3 = 0 B H x = 0, because

15 539 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control S is definite Thus, let x = [x T x T x T 3 ] T be such that x 3 = 0 B H x = 0 Then we obtain from 47) by using Lemma 49 that η S R, E, λ, x)) = inf [ R E ] R, E Hermn), satisfying 49) 40) + H = inf + H 47) λ Cn,n, x = r, H x = s, where the last equality is due to Lemma 49 Note that for any C n,n, the Hermitian skew-hermitian parts of satisfy = + H + H This implies 48) 49) λ + H + H + H λ + H λ if λ if λ for all C n,n Then taking the infimum over all satisfying x = r H x = s in 48) 49), by using the minimal robenius norm mapping from Theorem we obtain 45) 46) Example 4 The reason why we only obtain bounds in Theorem 40 is the fact that the infimum in 47) need not be attained by the matrix from Theorem 40 As an example, consider the pencil Lz) as in 3) with J = [ 0 0 ] [ 0 0, R = 0 ] [ 0 0, E = B = 0 0 ], S = I, let λ = 4 i x = [ ] T, ie, x = x 3 = 0 C x = [ ] T We then obtain s = J + R + λe)x = 0 as well as r = J R λe)x + Bx 3 = [ 0 ] rx H [ = x = 0 0 ] which by 45) gives the bounds = η S R, E, λ, x) [ ] + [ 0 λ ] = 375 On the other h, for the Hermitian matrix := [ we have x = r, thus, we obtain from 47) that η S R, E, λ, x) = It remains an open problem to determine the exact value for η S R, E, λ, x) for the same reason, also the computation of the eigenvalue backward error η S R, E, λ) is a challenging problem ],

16 Christian Mehl, Volker Mehrmann, Punit Sharma Perturbations only in J R Next, we consider perturbations that only effect the blocks J R in a pencil Lz) as in 3) Let λ C x C n+m \ 0 Then by the terminology outlined in Section 3, the block- symmetry-structure-preserving eigenpair backward errors η B J, R, λ, x) η S J, R, λ, x) are defined by 430) η B J, R, λ, x) = inf [ J R ] M M ) + λn N ) ) x = 0, M + z N B 43) η S J, R, λ, x) = inf [ J R ] M M ) + λn N ) ) x = 0, M + z N S, respectively, where B is the set of all pencils of the form Lz) = M + z N with the block-structure 0 J R 0 M = J R ) H 0 0, N = 0, J, R C n,n, while S is the corresponding set of pencils Lz) = M + z N as in 48) with J SHermn) R Hermn) If the perturbations are restricted to be real, then the above backward errors are denoted by η B R J, R, λ, x) η S R J, R, λ, x), respectively As usual, we first investigate conditions for the finiteness of η B J, R, λ, x) Remark 4 Let Lz) be a pencil as in 3), let λ C x = [x T x T x T 3 ] T be such that x, x C n x 3 C m Then for any J, R C n,n corresponding Lz) = M + z N B, we have Lλ) Lλ))x = 0 if only if 43) 433) 434) J R )x = J R + λe)x + Bx 3, J R ) H x = J R λe)x, 0 = B H x + Sx 3 Consequently, η B J, R, λ, x) is finite if only if 43) 434) are satisfied Lemma 43 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m set r = J R+λE)x +Bx 3 s = J R λe)x Then the following statements are equivalent ) There exist J, R C n,n satisfying 43) 433) ) There exists C n,n such that x = r H x = s 3) There exist J SHermn), R Hermn) satisfying 43) 433) 4) The identity x H 3 B H x = 0 is satisfied Moreover, 435) 436) inf = inf inf J + R J, R C n,n satisfy 43) 433) = inf Cn,n, x = r, H x = s, J + R J SHermn), R Hermn) satisfying 43) 433) C n,n, x = r, H x = s

17 54 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control Proof ) ) : Let J, R C n,n be such that they satisfy 43) 433) By setting = J R we get x = r H x = s urthermore, we have J + R ) J + R ), where the last inequality is an elementary application of the Cauchy Schwartz inequality in R ) But then the inequality in 435) can be easily shown by following the arguments in the proof of ) ) in Lemma 4 ) ) : Suppose that C n,n is such that x = r H x = s define J = R = Then J R satisfy 43) 433) Also, we obtain J + R = hence, in 435) can be easily shown by following the arguments of the proof of ) ) in Lemma 4 ) 3) : Let C n,n be such that u = r H w = s Then by setting J = H R = + H, we get J SHermn), R Hermn) such that 43) 433) hold urthermore, we have J + R = H + + H = Thus, arguments similar to those in the proof of ) ) in Lemma 4 give in 436) 3) ) : Let J SHermn) R Hermn) be such that they satisfy 43) 433) Define = J R then x = r H x = s Note that J R are, respectively, the unique skew-hermitian Hermitian parts of, ie, J = H R = + H This implies J + R = H + + H, = Then again arguments similar to those in the proof of ) ) in Lemma 4 give in 436) ) 4) : This follows immediately from Theorem The following theorem yields the values of η B J, R, λ, x), η S J, R, λ, x), also of their real counterparts if Lz) is real It also gives the values of η B J, R, λ) := inf x C n+m \0 η B J, R, λ, x) η S J, R, λ) := inf x C n+m \0 η S J, R, λ, x) Theorem 44 Let Lz) be a pencil defined by 3), λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m, set r = J R+λE)x +Bx 3 s = J +R+λE)x Then the following statements hold: ) η B J, R, λ, x) η S J, R, λ, x) are finite if only if x 3 = 0 B H x = 0 If the latter conditions are satisfied, then η B J, R, λ, x) = η S J, R, λ, x) =,

18 Christian Mehl, Volker Mehrmann, Punit Sharma 54 as well as η B J, R, λ) = σ minj R + λe) η S J, R, λ) = σ min J R + λe), where is given by = rx H x if x = 0, x s H x if x = 0, ) rx H x + xsh x I n xxh x otherwise ) Suppose that Lz) is real If rank [x x ]) = rank [x x ]) =, then η B R J, R, λ, x) η S R J, R, λ, x) are finite if only if x 3 = 0, B T x = 0 λx T Ex = 0 If the latter conditions are satisfied, then 437) η B R J, R, λ, x) = η S R J, R, λ, x) =, where R n,n is given by = [r r][x x ] + [s s][x x ] ) H [s s][x x ] ) H [x x ][x x ] ) Proof The proof of ) follows the same lines as that of Theorem 43 by using Lemma 43 Theorem Concerning the proof of ), recall that when Lz) is real, then η B R J, R, λ, x) is the eigenpair backward error obtained by allowing only real perturbations to the blocks J R of Lz) Now for any J, R R n,n corresponding real Lz) = M + z N B, we have Lλ) Lλ))x = 0 if only if 438) 439) 440) J R )x = J R + λe)x + Bx 3, J R ) T x = J R λe)x, 0 = B T x + Sx 3 Since J R are real, 438) 439) can be equivalently written as 44) J R )[x x ] = [r r] J R ) T [x x ] = [s s] ollowing the lines of the proof of Lemma 43, there exist real matrices J R satisfying 44) if only if there exists R n,n such that [x x ] = [r r] T [x x ] = [s s] Applying Theorem 3, we find that this is the case if only if x H s = r H x x T s = r T x which, using the definition of r s, is in turn equivalent to the conditions x H 3 B H x = 0 λx T Ex = x T 3 B T x The latter conditions together with B T x + Sx 3 = 0 give x 3 = 0, B T x = 0 λx T Ex = 0, because S is assumed to be positive definite Therefore, from 430), η B R J, R, λ, x) is finite if only if x satisfies

19 543 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control x 3 = 0, B T x = 0 λx T Ex = 0 If this is the case, then we find that η B R J, R, λ, x) = inf [ J R ] J, R R n,n satisfy 44) = inf R n,n, [x x ] = [r r] T [x x ] = [s s] Thus 437) follows for η B R J, R, λ, x) by using Theorem 3 η S R J, R, λ, x) Similarly, we can also establish 437) for 44 Perturbation only to J B, or R B, or E B In this section, we obtain blockstructure-preserving eigenpair or eigenvalue backward errors when only the blocks J B in a pencil Lz) as in 3) are perturbed Unfortunately, is seems that this approach cannot be generalized to obtain the correpsonding symmetry-structure-preserving backward errors Let λ C x C n+m \ 0, then by the terminology outlined in Section 3, the block-structurepreserving eigenpair backward error η B J, B, λ, x) is defined by 44) η B J J, B, λ, x) = inf [ J B ] C n,n, B C n,m, M + z N B, M M ) + λn N ) ) x = 0, where B is the set of all pencils of the form Lz) = M + z N with M = 0 J B H J 0 0 H B 0 0 N = 0 If the perturbations are restricted to be real then the above error is denoted by η B R J, B, λ, x) Remark 45 Let Lz) be a pencil as in 3), let λ C x = [x T x T x T 3 ] T be such that x, x C n x 3 C m Then for any J C n,n, B C n,m, corresponding Lz) = M + z N B, we have Lλ) Lλ))x = 0 if only if which in turn is equivalent to 443) 444) J x + B x 3 = J R + λe)x + Bx 3, H J x = J R λe)x, H B x = B H x + Sx 3, [ ] [ ] x J B = J R + λe)x + Bx 3, x 3 =r =u [ ] H J B x = =w [ J + R + λe)x ] B H x + Sx 3 =s In particular, η B J, B, λ, x) is finite if only if 443) 444) are satisfied

20 Christian Mehl, Volker Mehrmann, Punit Sharma 544 Lemma 46 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m, let u, w, r s be defined as in 443) 444) Then the following statements are equivalent ) There exist J C n,n B C n,m satisfying 443) 444) ) There exists C n,n+m such that u = r H w = s 3) x satisfies x 3 = 0 Moreover, we have inf J + B J C n,n, B C n,m satisfy 443) 444) = inf C n,n+m, u = r, H w = s Proof ) ) is obvious while ) 3) is implied by Theorem using the fact that S is definite The last part then follows from the observation that any C n,n+m can be written as = [ ], where C n,n C n,m such that = [ ] = + Theorem 47 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m set u = [x T x T 3 ] T, w = x, r = J R + λe)x + Bx 3 s = [ J + R + λe)x ) T B H x + Sx 3 ) T ] T Then the following statements hold ) η B J, B, λ, x) is finite if only if x 3 = 0 In that case, we have η B J, B, λ, x) = +, [ η B ] H ) J, B, λ) = min σ min J R + λe B, σ min J R + λe), where are given by [ ] = ru H u if x = 0, ws H w if x = 0, ) ru H u + wsh w I n+m uuh u otherwise ) Suppose that Lz) is real If rank [x x ]) = rank [x x ]) = then η B R J, B, λ, x) is finite if only if x 3 = 0 λx T Ex = 0 If the latter conditions are satisfied, then 445) η B R J, B, λ, x) = +, where R n,n R n,m are given by [ ] = [r r][u ū] + [s s][w w] ) H [s s][w w] ) H [u ū][u ū] )

21 545 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control Proof The proof is analogous to the one of Theorem 43 by using Lemma 46 as well as Theorem in the complex case Theorem 3 in the real case Remark 48 A result similar to Theorem 47 can be obtained for the complex real blockstructure-preserving eigenpair backward errors η B R, B, λ, x) η B R R, B, λ, x) of a pair λ, x) ir) C n+m \ 0) when only the blocks R B in a pencil Lz) as in 3) are subject to perturbation In fact, one easily obtains η B R, B, λ, x) = η B J, B, λ, x) η B R R, B, λ, x) = η B R J, B, λ, x) As a consequence, we also have η B R, B, λ) = η B J, B, λ) inally, also the backward errors η B E, B, λ, x) η B E, B, λ) with respect to perturbations only in the blocks E B of Lz) as in 3) can be obtained in a similar manner Since the actual result differs slightly from the previous formulas, we present it as a theorem, but we omit the proof, since it is similar to the one of Theorem 43 Theorem 49 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 x = [x T x T x T 3 ] T such that x, x C n x 3 C m, set u = [λx T x T 3 ] T, w = x, [ ] r = J R + λe)x + Bx 3 s = λ J + R + λe)x ) T B H x + Sx 3 ) T T Then the following statements hold ) η B E, B, λ, x) is finite if only if x 3 = 0 In that case, we have η B E, B, λ, x) = +, Partition [ η B E, B, λ) = min σ min J R+λE) λ B ] ) H, σ minj R + λe), λ where C n,n C n,m are given by [ ] = ru H u if x = 0, ws H w if x = 0, ) ru H u + wsh w I n+m uuh u otherwise ) Suppose that Lz) is real If rank [x x ]) = rank [x x ]) = then η B R E, B, λ, x) is finite if only if x 3 = 0 λx T Ex = 0 If the latter conditions are satisfied, then η B R E, B, λ, x) = +, where R n,n R n,m are given by [ ] = [r r][u u] + [s s][w w] ) H [s s][w w] ) H [u u][u u] )

22 Christian Mehl, Volker Mehrmann, Punit Sharma Perturbation in any three of the matrices J, R, E B In this section, we define compute block- symmetry-structure-preserving eigenpair or eigenvalue backward errors for pencils Lz) as in 3), while we consider perturbations in any three of the blocks J, R, E, B of Lz) 5 Perturbations in the blocks J, R B We first concentrate on the case that perturbations are allowed to affect only the blocks J, R, B of a pencil Lz) as in 3) If λ C x C n+m \ 0, then following the terminology of Section 3, the block- symmetry-structure-preserving eigenpair backward errors η B J, R, B, λ, x) η S J, R, B, λ, x), respectively, are defined by η B J, R, B, λ, x) = inf [ J R B ] M M ) + λn N ) ) x = 0, M + z N B, η S J, R, B, λ, x) = inf [ J R B ] M M ) + λn N ) ) x = 0, M + z N S, where B denotes the set of all pencils of the form Lz) = M + z N with 0 J R B M = J R ) H 0 0, N = 0, B ) H 0 0 J, R C n,n, B C n,m, while S denotes the corresponding set of pencils that satisfy in addition J SHermn) R Hermn) If the perturbations are restricted to be real then the above backward errors are denoted by η B R J, R, B, λ, x) η S R J, R, B, λ, x), respectively Remark 5 Let Lz) be a pencil as in 3), let λ C x = [x T x T x T 3 ] T such that x, x C n x 3 C m Then for any J, R C n,n B C n,m, corresponding Lz) = M + z N B, we have Lλ) Lλ))x = 0 if only if J R )x + B x 3 = J R + λe)x + Bx 3, J R ) H x = J R λe)x, B ) H x = B H x + Sx 3, if only if 5) 5) [ ] [ ] x J R B = J R + λe)x + Bx 3, x 3 =r =u [ ] H J R B x = =w [ J + R + λe)x ] B H x + Sx 3 =s Lemma 5 Let Lz) be a pencil defined by 3), λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m let u, w, r s be as defined in 5) 5) Then the following statements are equivalent ) There exist J, R C n,n B C n,m satisfying 5) 5) ) There existss C n,n+m such that u = r H w = s 3) There exist J SHermn), R Hermn) B C n,m satisfying 5) 5) 4) x satisfies x 3 = 0

23 547 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control Moreover, we have inf [ J R B ] J, R C n,n, B C n,m satisfy 5) 5) = inf + C n,n, C n,m, [ ]u = r, [ ] H w = s, inf [ J R B ] J SHermn), R Hermn), B C n,m satisfy 5) 5) = inf + C n,n, C n,m, [ ]u = r, [ ] H w = s Proof As seen in the proof of Lemma 46, any C n,n+m can be written as = [ ] where C n,n C n,m such that = [ ] = + With this key observation the proof is obtained by following exactly the same arguments as in the proof of Lemma 43 Theorem 53 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 x = [x T x T x T 3 ] T such that x, x C n x 3 C m define û = [ x T x T 3 ] T, w = x, r = J R + λe)x + Bx 3 ŝ = [ J + R + λe)x ) T B H x + Sx 3 ) T ] T Partition Then the following statements hold: ) η B J, R, B, λ, x) is finite if only if x 3 = 0 In that case, we have η B J, R, B, λ, x) = +, [ η B J R+λE) J, R, B, λ) = min σ min B ] ) H, σ minj R + λe), where are given by [ ] = rû H û if x = 0, wŝ H w if x = 0, ) rû H û + wŝh w I n+m ûûh û otherwise ) If Lz) is real, rank [x x ]) = rank [x x ]) =, then η B R J, R, B, λ, x) is finite if only if x 3 = 0 λx T Ex = 0 If the latter conditions are satisfied, then η B R J, R, B, λ, x) = where R n,n R n,m are given by +, [ ] = [r r][û û] + [ŝ ŝ][w w] ) H [ŝ ŝ][w w] ) H [û û][û û] )

24 Christian Mehl, Volker Mehrmann, Punit Sharma 548 Proof Observe that if u = [x T x T 3 ] T s = [ J + R + λe)x ) T B H x + Sx 3 ) T ] T, then inf = inf ˆ + ˆ +, C n,n, [ ]u = r, [ ] H w = s ˆ, ˆ C n,n, [ ˆ ˆ ]û = r, [ ˆ ˆ ] H w = ŝ Therefore, the proof is analogous to that of Theorem 43 by using first Lemma 5 then Theorem for ) Theorem 3 for ) The following theorem presents the value of η S J, R, B, λ, x) its real counterpart if the original pencil is real It also gives η S J, R, B, λ) := inf x C n+m \0 η S J, R, B, λ, x) Theorem 54 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 x = [x T x T x T 3 ] T such that x, x C n x 3 C m define u = [x T x T 3 ] T, w = x, Partition r = J R + λe)x + Bx 3 s = [ J + R + λe)x ) T B H x + Sx 3 ) T ] T Then the following statements hold ) η S J, R, B, λ, x) is finite if only if x 3 = 0 In such a case the following holds η S J, R, B, λ, x) = +, [ η S ] H ) J, R, B, λ) = min σ min J R + λe) B, σ min J R + λe), where C n,n C n,m are given by [ ] = ru H u if x = 0, ws H w if x = 0, ) ru H u + wsh w I n+m uuh u otherwise ) If Lz) is real, rank [x x ]) = rank [x x ]) = then η S R J, R, B, λ, x) is finite if only if x 3 = 0 λx T Ex = 0 If the latter conditions are satisfied, then η S R J, R, B, λ, x) = where R n,n R n,m are given by +, [ ] = [r r][u u] + [s s][w w] ) H [s s][w w] ) H [u u][u u] ) Proof The proof is similar to that of Theorem 43 by using first Lemma 5, then Theorem for ) Theorem 3 for )

25 549 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control 5 Perturbations to R, E B, or J, E B This section is devoted to the block- symmetry-structure-preserving eigenpair eigenvalue backward errors when only the blocks R, E B of a pencil Lz) as in 3) are subject to perturbations Let λ C x C n+m \ 0, then in view of Section 3, we have the definitions η B R, E, B, λ, x) = inf η S R, E, B, λ, x) = inf [ R E B ] M M ) + λn N ) ) x = 0, M + z N B [ R E B ] M M ) + λn N ) ) x = 0, M + z N S respectively, where B is the set of all pencils of the form Lz) = M + z N with 0 R B 0 E 0 M = H R 0 0 N = H E 0 0 H B R, E C n,n, B C n,m, S is the corresponding set of all such pencils that in addition satisfy R, E Hermn) Remark 55 Let Lz) be a pencil as in 3), let λ ir x = [x T x T x T 3 ] T be such that x, x C n x 3 C m Then for any R, E C n,n B C n,m, corresponding Lz) = M + z N B, we have Lλ) Lλ))x = 0 if only if R + λ E )x + B x 3 = J R + λe)x + Bx 3, R + λ E ) H x = J R λe)x, H B x = B H x + Sx 3, which, in turn, is equivalent to [ ] [ ] x 53) R + λ E B = J R + λe)x + Bx 3, x 3 =r =u 54) [ ] H R + λ E B x = =w [ J + R + λe)x ] B H x + Sx 3 =s Lemma 56 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m let u, w, r s be defined as in 53) 54) Then the following statements are equivalent ) There exist R, E C n,n B C n,m satisfying 53) 54) ) There exists C n,n+m such that u = r H w = s 3) There exist R, E Hermn) B C n,m satisfying 53) 54) 4) x satisfies x 3 = 0,, Moreover, inf R E B R, E C n,n, B C n,m satisfy 53) 54) = inf + λ + C n,n, C n,m, [ ]u = r, [ ] H w = s,

26 Christian Mehl, Volker Mehrmann, Punit Sharma 550 inf R E B R, E Hermn), B C n,m satisfy 53) 54) + H = inf + H λ + C n,n, C n,m, [ ]u = r, [ ] H w = s Proof Again, by using the fact that any C n,n+m can be written as = [ ] where C n,n C n,m such that = [ ] = +, the proof is obtained by arguments similar to those in the proof Lemma 4 Lemma 49 Theorem 57 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] so that x, x C n, x 3 C m, define w = x, u = [x T x T 3 ] T, û = [ + λ ) / x T x T 3 ] T, r = J R + λe)x + Bx 3, s = [ J + R + λe)x ) T B H x + Sx 3 ) T ] T, ŝ = [ + λ ) / J + R + λe)x ) T B H x + Sx 3 ) T ] T Then η B R, E, B, λ, x) η S R, E, B, λ, x) are finite if only if x 3 = 0 urthermore, the following statements hold ) If x 3 = 0, then 55) 56) η B R, E, B, λ) = min η B R, E, B, λ, x) = σ min [ J R+λE) + λ +, B ] ) H, σ minj R + λe), + λ where are given by ) If x 3 = 0, then 57) [ ] = + ηs R, E, B, λ, x) when λ, 58) + λ ηs R, E, B, λ, x) when λ, where are given by [ ] = rû H û if x = 0, wŝ H w if x = 0, ) rû H û + wŝh w I n+m ûûh û + H + H + λ + λ otherwise H H ru H u if x = 0, ws H w if x = 0, ) ru H u + wsh w I n+m uuh u otherwise +, +,

27 55 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control Proof In view of 53) 54), we have η B R, E, B, λ, x) ) = inf [ R E B ] R, E C n,n, B C n,m satisfy 53) 54) = inf + λ + C n,n, C n,m, [ ]u = r, [ ] H w = s, where the last equality follows from Lemma 56 Observe that if we set = = we obtain η B R, E, B, λ, x) ) = inf +, then + λ C n,n, C n,m, [ ]û = r, [ ] H w = ŝ Thus 55) follows from Theorem, arguments similar to those in the proof of Theorem 43 give 56) Similarly, by using Lemma 56 in the definition of η S R, E, B, λ, x), we can write η S R, E, B, λ, x) ) + H = inf + H λ +, C n,n, or any C n,n we have = + H 59) 50) + + H λ + + H + H + H λ + H λ [ ]u = r, [ ] H w = s This implies + if λ + if λ for all C n,n C n,m Taking the infimum over all C n,n, C n,m satisfying [ ]u = r [ ] H w = s in 59) 50) followed by applying Theorem yields 57) 58) Remark 58 We mention that a result similar to Theorem 57 can also be obtained for the blockstructure-preserving eigenpair eigenvalue backward errors η B J, E, B, λ, x) η B J, E, B, λ), respectively, when perturbations are restricted to affect only the blocks J, E B of a pencil Lz) as in 3) In fact, for λ ir x C n+m, using arguments analogous to those in this section, we obtain that η B J, E, B, λ, x) = η B R, E, B, λ, x) η B J, E, B, λ) = η B R, E, B, λ) 53 Perturbation to J, R E Let Lz) be a pencil as in 3), let λ C x C n+m \0 In this section, we allow perturbations in the blocks J, R E of Lz) The block- symmetry-structurepreserving eigenpair backward errors η B J, R, E, λ, x) η S J, R, E, λ, x) are defined by [ J η B J, R, E, λ, x) = inf R E ] M M ) + λn N ) ) x = 0, M + z N B, [ J η S J, R, E, λ, x) = inf R E ] M M ) + λn N ) ) x = 0, M + z N S,

28 Christian Mehl, Volker Mehrmann, Punit Sharma 55 respectively, where B is the set of all pencils of the form Lz) = M + z N with 0 J R 0 0 E 0 M = J R ) H 0 0 N = H E S is the corresponding set of pencils from B that satisfy in addition J SHermn) R, E Hermn) Remark 59 If λ ir x = [x T x T x T 3 ] T are such that x, x C n x 3 C m, then for any J, R, E C n,n corresponding Lz) B, we have Lλ) Lλ))x = 0 if only if 5) 5) 53) J R + λ E ) x = J R + λe)x + Bx 3, =u =r J R + λ E ) H x = J R λe)x, =w =s 0 = B H x + Sx 3 Lemma 50 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m let u, w, r s be defined as in 5) 5) Then the following statements are equivalent ) There exist J, R, E C n,n satisfying 5) 5) ) There exists C n,n such that u = r H w = s 3) There exist J SHermn) R, E Hermn) satisfying 5) 5) 4) x satisfies x H 3 B H x = 0 Moreover, we have [ ] inf J R E J, R, E C n,n satisfy 5) 5) 54) = inf + λ Cn,n, u = r, H w = s, 55) [ ] inf J R E J SHermn), E, R Hermn) satisfying 5) 5) + H = inf + H + λ Cn,n, u = r, H w = s Proof ) ) : Let J, R, E C n,n be such that they satisfy 5) 5) By setting = J R + λ E we get u = r H w = s Also, we obtain 56) J + R + λ E ) + λ ) J + R + E ), where the latter inequality follows from the Cauchy-Schwarz inequality in R 3 ) Then in 54) can be shown similarly as ) ) in the proof of Lemma 4 ) ) : Conversely, let C n,n such that u = r H w = s Define J = + λ, R = + λ, E = λ + λ

29 553 Structured Eigenvalue/Eigenvector Backward Errors of Matrix Pencils Arising in Optimal Control Then J, R E satisfy J R + λ E = hence 5) 5) urthermore, we have J + R + E = + λ Thus, we get in 54) by following arguments similar to those of ) ) in the proof of Lemma 4 ) 3) : To show this, let C n,n be such that u = r H w = s Then by setting R = + H, J = H + λ ), E = λ H ) + λ ), we have J SHermn) R, E Hermn) using λ ir), furthermore we obtain J R + λ E = H + λ ) + + H + λ H ) + λ = + H + H ) Thus, J, R, E satisfy 5) 5), also J + R + E = + H + H + λ Now in 55) can be shown by arguments similar to those of ) ) in the proof of Lemma 49 3) ) : Suppose that J SHermn) R, E Hermn) satisfy 5) 5) Define = J R + λ E, then u = r H w = s Note that J + λ E is skew-hermitian since λ ir, therefore, J + λ E R are respectively the unique skew-hermitian Hermitian parts of, ie, This implies R = + H H J + λ E = H = J + λ E J + λ E H + λ J + E, where the last inequality is obtained with the help of the Cauchy-Schwarz inequality in R ) urthermore, we have 57) H + λ + + H J + E + R Thus, arguments similar to those in ) ) in the proof of Lemma 4 give in 55) ) 4) : This follows immediately from Theorem Theorem 5 Let Lz) be a pencil as in 3), let λ ir x C n+m \ 0 Partition x = [x T x T x T 3 ] T such that x, x C n x 3 C m define r = J R + λe)x + Bx 3 s = J +R+λE)x Then η B J, R, E, λ, x) η S J, R, E, λ, x) are finite if only if x 3 = 0 B H x = 0 If the latter conditions are satisfied, then =