Structured Backward Error for Palindromic Polynomial Eigenvalue Problems

Transcription

1 Structured Backward Error for Palindromic Polynomial Eigenvalue Problems Ren-Cang Li Wen-Wei Lin Chern-Shuh Wang Technical Report

2 Structured Backward Error for Palindromic Polynomial Eigenvalue Problems Ren-Cang Li Wen-Wei Lin Chern-Shuh Wang September 5, 2008 Abstract A detailed structured backward error analysis for four kinds of Palindromic Polynomial Eigenvalue Problems PPEP) d A l λ )x l = 0, A d l = εa l for l = 0, 1,..., d/2, where is one of the two actions: transpose and conjugate transpose, and ε {±1}. Each of them has its application background with the case taking transpose and ε = 1 attracting a great deal of attention lately because of its application in the fast train modeling. Computable formulas and bounds for the optimal structured backward errors are obtained. The analysis reveals distinctive features of PPEP from the usual Polynomial Eigenvalue Problems PEP) as investigated by Tisseur Linear Algebra Appl., 309: , 2000) and by Liu and Wang Appl. Math. Comput., 165: , 2005). 1 Introduction In vibration analysis of fast trains arises the Palindromic Quadratic Eigenvalue Problem PQEP) [2, 9, 11]: find a scalar λ C and an n-dimensional vector x C n \ {0} such that Qλ)x A 0 λ 2 + A 1 λ + A 0 )x = 0, 1.1) where A 0 and A 1 are n n real or complex) matrices with A 1 = A 1, and the superscript takes matrix transpose. The scalar λ and the nonzero vector x in 1.1) are an Department of Mathematics, University of Texas at Arlington, Arlington, P.O. Box 19408, Arlington, TX 76019, USA; rcli@uta.edu. Supported in part by the National Science Foundation under Grant DMS and by the National Center for Theoretical Sciences at National Tsing-Hua University, Taiwan where this work was initiated while this author was visiting the center. Department of Applied Mathematics, National Chiao Tung University, Hsinchu 300, Taiwan; wwlin@math.nctu,edu.tw. Department of Mathematics, National Cheng Kung University, Tainan 701, Taiwan; cswang@math.ncku,edu.tw. 1

3 eigenvalue and its associated eigenvector of Qλ). The pair {λ, x} is called an eigenpair of Qλ). Presently, PQEP is solved by certain kinds of structurally preserving linearization, such as Jacobi-type method [6, 12], QR-like algorithm [13], Patel s algorithm [8] and U RV - like algorithm [14], as well as the structure-preserving doubling algorithm [2]. Experience shows that PQEP from the fast train application is notoriously difficult to solve accurately, partly because of its widely varying magnitudes in eigenvalues. How to solve it accurately and robustly is still an active research topic. PQEP is a particular case of the so-called Polynomial Eigenvalue Problem PEP) of degree d : d ) A l λ l x = 0, 1.2) where A l, l = 0,1,...,d are n n complex matrices. When d = 2, PEP 1.2) is called the Quadratic Eigenvalue Problem QEP) [16]. For a PEP in its generality there is no relation among coefficient matrices, unlike PQEP. Naturally one defines the Palindromic Polynomial Eigenvalue Problem PPEP) to be a PEP 1.2) with A d l = A l, l = 0,1,..., d/2, namely, d Pλ)x A l λ )x l = 0, A d l = A l, for l = 0,1,..., d/2, 1.3) where d/2 is the largest integer that is no bigger than d/2. Let {µ,z} be a computed eigenpair of 1.2) or 1.3). Ideally, we would like to have d A lµ l ) z = 0. But practically we have d A l µ )z l = r with residual r 0 but tiny. Backward error analysis concerns if the computed eigenpair {µ,z} is an exact eigenpair of a nearby PEP, namely, d A l + ρ l A l )µ )z l = 0 with backward error matrices A l l = 0,1,...,d) hopefully tiny in magnitude, where ρ l l = 0,1,...,d) are scaling parameters, usually taken to be some norms of A l, respectively. For a general PEP, no constraints on and/or among A l, l = 0,1,...,d are imposed, except that A l, l = 0,1,...,d should have as tiny magnitude as possible. But when a PEP has worthy structures like PPEP, naturally we would ask if the computed eigenpair {µ,z} is an exact eigenpair of a nearby PPEP. This is the so-called structured backward error analysis for PPEP. Tisseur [15] thoroughly developed a backward error analysis for PEP in its generality, where no structure upon on δa l is imposed because there is not any on A l assumed. When A l, l = 0,1,...,d are Hermitian, naturally one would like to enforcing A l Hermitian, 2

4 too. This is the structured backward error analysis for the case. In the same article, Tisseur obtained a result when µ is real. When µ is complex, the structured backward error analysis is completed by Liu and Wang [10] who also investigated the corresponding question for the generalized and multi-parameter eigenvalue problem. Other two related works are Higham and Higham [3] and Hochstenbach and Plestenjak [7]. None of the results from these papers apply to PPEP s structured backward error analysis, however. In this paper, we shall consider PPEP in a more broader sense, with 1.3) being one of the four different kinds. To present these PPEP compactly, we let the superscript be either or H which takes complex conjugate and transpose and let ε {±1}. Collectively by a PPEP, we mean a polynomial eigenvalue problem of the following form d A l λ )x l = 0, A d l = εa l for l = 0,1,..., d/2. 1.4) We have mentioned the fast train application [2, 9, 11] which yielded a problem of this form with d = 2, = T, and ε = 1. It was first raised in the study of the vibration in the structural analysis for fast trains [5, 6], and then of the behavior of periodic surface acoustic wave SAW) filters [18]. PQEP with = H was raised in the computation of the Crawford number by the bisection and level set methods in [4]. A typical even degree PPEP of 1.4) can be derived from solving linear quadratic discrete-time optimal control problem subject to the higher order discrete system [1, 17]. Let {µ,z} be a computed eigenpair and let be either the spectral norm 1 2 or the Frobenius norm F. We are interested in knowing the optimal structured backward error d/2 min A l 2 subject to d ) A l + ρ l A l )µ l z = 0, ρ d l = ρ l 0, A d l + ρ d l A d l = εa l + ρ l A l ), for l = 0,1,..., d/2. 1.5) In the usual PEP case, backward errors always exist. But as one might expect, forcing backward perturbations to be structured creates critical cases where backward errors fail to exist, i.e., there are no A l, l = 0,1,...,d, satisfying the constraints in 1.5). In fact, as we shall see that the critical cases are when µ = 1 if = H and when µ = ±1 if =. The rest of this paper is organized as follows. Section 2 significantly simplifies 1.5) to a problem of solving a 2-by-2 matrix equation, which is solved in Section 3 for = H and in Section 4 for = T. It turns out the case when = H is considerably more complicated and interesting than the case when = T. Numerical examples are given in Section 5 to illustrate our results, and finally we present a few concluding remarks in Section 6. Notation. Throughout this paper, C n m is the set of all n m complex matrices, C n = C n 1, and C = C 1. Similarly define R n m, R n, and R except replacing the word complex by real. I n or simply I if its dimension is clear from the context) is the n n identity matrix. x is the complex conjugate of a scalar or the entry-wise complex conjugate 1 Notation 2 is generic and on a vector it is its l 2-norm. 3

5 of a vector or matrix. Rα) and Iα) are the real and imaginary part of α, respectively, and ι is the imaginary unit. We shall also adopt MATLAB-like convention to access the entries of vectors and matrices. i : j is the set of integers from i to j inclusive and i : i = {i}. For a matrix X, X i,j) is X s i,j)th entry, and X k:l,i:j) is X s submatrices consisting of intersections of row k to row l and column i to column j. X is X s Moore-Penrose inverse. 2 Simplify the Problem The constraints in 1.5) require ρ d l = ρ l and A d l = ε A l, for l = 0,1,..., d/2, and d ) d ρ l A l µ l z = r def = A l µ )z, l 2.1) where A l C n n l = 0,1,...,d). We will be investigating if 2.1) has a solution { A l,l = 0,1,..., d/2 } and if it does, we ll seek A l such that d/2 A l 2 = min, 2.2) where is either 2 or F. The two trivial cases i) when all ρ l = 0, and ii) when r = 0 will be excluded from our consideration. Assume n 2. Later we shall comment on which part of our analysis will cover the case when n = 1, i.e., the scalar polynomial case. Let Q C n n be a unitary matrix, i.e., QQ H = I n, such that α γ 0 β Q H z r ) = 0 0, 2.3) where Such Q always exists, and necessarily 2.3) implies It follows from 2.1) that r = r if = H, and r = r if =. 2.4) α = z 2, γ = zh r ᾱ, 2.5) β = zh r r z 2 z r 2 = 2 z 2 2 zh r ) 2 2 z 2 Q d ρ l A l µ l )QQ H z = Q r, 2.7) 4

6 or equivalently d ) ρ l B l µ l y = w, ρ d l = ρ l and B d l = ε Bl, for l = 0,1,..., d/2, 2.8) where γ α B l = Q A l )Q, y = Q H 0 β z =., w = Q r = 0,. 0 0 β = β and γ = γ if = H, and β = β and γ = γ if =. This reduction technique of going from 2.1) to 2.8) through performing 2.3) is similar to that used in [10, p.407] which came to our attention after this paper of ours is very much in its final stage. Since Q is unitary, 2.1) and 2.8) have the same solvability property, i.e., if one is solvable, so is the other, and moreover d/2 B l 2 = d/2 A l 2. Thus the optimal solution { B l,l = 0,1,..., d/2 } to 2.8) in the sense of d/2 B l 2 = min 2.9) gives rise to one solution { A l,l = 0,1,..., d/2 } to 2.1) in the sense of 2.2), and vice versa. Set δ 1 = γ/α, δ 2 = β/α. 2.10) and p = min d/2 B l 2 p : 2.8) satisfied for p = 2, F. 2.11) p = means that 2.8) is not solvable. It follows from 2.5) and 2.6) that δ 1 = zh r z 2, δ 2 = 2 r 2 2 z 2 2 zh r 2 z 2, 2 δ1 2 + δ 2 2 = r 2 z ) In particular δ 1 = γ α = zh r ᾱα = zh r z 2 2 for = H. 2.13) 5

7 Theorem 2.1 Suppose that 2.8) has a solution so does 2.1)). The optimal solution { B l,l = 0,1,..., d/2 } to 2.8) in the sense of 2.9) for the Frobenius norm satisfies B l ) i,j) = 0 for i, j > 2 and for l = 0,1,..., d/ ) There is an optimal solution { B l,l = 0,1,..., d/2 } in the sense of 2.9) for the spectral norm satisfies 2.14). Finally 2 F ) Proof No proof is needed for B l with ρ l = 0. Suppose all ρ l are positive. Equating the corresponding entries at the both sides of 2.8) leads to n linear equations in the entries of B l,l = 0,1,..., d/2. The last n 2 equations are homogenous and contain only those entries B l ) i,j) = 0 for i, j > 2 and for l = 0,1,..., d/2 and at the same time they do not appear in the first 2 equations. Therefore the optimal solution { B l,l = 0,1,..., d/2 } in the sense of 2.9) for the Frobenius norm must satisfy 2.14). For any 2-by-2 block matrix ) X11 X X = 12, X 21 X 22 we have X 2 X Therefore if { B l,l = 0,1,..., d/2 } is an optimal solution to 2.8) for the spectral norm, resetting B l ) i,j) = 0 for i, j > 2 and for l = 0,1,..., d/2 and leaving B l ) i,j) for 1 i, j 2 and for l = 0,1,..., d/2 untouched yield a solution to 2.8) and at the same time do not make d/2 B l 2 2 bigger. Since before resetting, { B l,l = 0,1,..., d/2 } is optimal, d/2 B l 2 2 will not change before and after the resetting. Thus the resulted { B l,l = 0,1,..., d/2 } is an optimal solution satisfying 2.14) for the spectral norm. Let { B l,l = 0,1,..., d/2 } be the optimal solution in the Frobenius norm. Then 2 d/2 B l 2 2 d/2 B l 2 F = F. On the other hand, let { B l,l = 0,1,..., d/2 } be an optimal solution satisfying 2.14) in the spectral norm. Then rank B l ) 2 for l = 0,1,..., d/2. Thus F d/2 B l 2 F d/2 2 B l 2 2 = 2 2, as expected. 6

8 Theorem 2.1 tells us that it suffices to consider the first two equations from the first 2 entries at the both sides of 2.8). In the two equations, only the entries B l ) i,j) for 1 i, j 2 and for l = 0,1,..., d/2 appear. We shall do so in the next two sections for each different combination of the superscript {, H} and ε {±1}. For this purpose, let l) b 11 b l) ) 12 B l ) 1:2,1:2) =. 2.16) b l) 21 b l) 22 Constraints among b l) ij because of B d l = ε Bl, for l = 0,1,..., d/2 will be imposed at the time we consider each case. Recall in 2.8) ρ d l = ρ l and B d l = ε Bl for l = 0,1,..., d/2. For even d, we have d ρ l B l µ l = = = = and B d/2 = ε Bd/2. For odd d, d ρ l B l µ l = = = = ρ l B l µ l + ρ d/2 B d/2 µ d/2 + ρ l B l µ l + ρ d/2 B d/2 µ d/2 + d l=d/2+1 0 j= ρ l B l µ l + ρ d/2 B d/2 µ d/2 + j=0 ρ l B l µ l ρ d j B d j µ d j ρ j ε B j µ d j ρ l B l + ε B l µd 2l) µ l + ρ d/2 B d/2 µ d/2 2.17) d 1)/2 d 1)/2 d 1)/2 d 1)/2 ρ l B l µ l + ρ l B l µ l + d l=d 1)/2+1 0 j=d 1)/2 d 1)/2 ρ l B l µ l + j=0 ρ l B l µ l ρ d j B d j µ d j ρ j ε B j µ d j ρ l B l + ε B l µd 2l) µ l. 2.18) Our detailed analysis in the next two sections is for the Frobenius norm only. Results for the spectral norm can then be deduced straightforwardly by using 2.15). So in the next two sections, in 2.9) is always the Frobenius norm. Recall the assumption r 0 which implies δ 1 + δ 2 > ) 7

9 For the ease of analysis, we assume also all ρ l > ) This will remove many special cases caused by one or more ρ l possibly being zeros. It seems that doing so would limit the applicability of our analysis to cases where no backward errors are preferred to some of the A l s, which is usually realized by setting the associated ρ l s to zero. But we note that such cases can be practically dealt with satisfactorily by giving sufficiently tiny positive numbers to the ρ l s. The following lemma will be needed in the later sections. Lemma 2.1 i) For τ = τ e ιϑ C, ϑ R, and η C, we have ) ) ) ) ) ) Rτ η) cos ϑ sin ϑ Rη) Rτ η) cos ϑ sin ϑ Rη) = τ, = τ. Iτ η) sin ϑ cos ϑ Iη) Iτ η) sinϑ cos ϑ Iη) ii) For ξ R and ϑ R, we have [ )] 2 ) cos ϑ sin ϑ I 2 + ξ = 1 + ξ 2 cos ϑ sinϑ )I sin ϑ cos ϑ 2 + 2ξ, sin ϑ cos ϑ ) ) cos ϑ cos ϑ = 1 [ )] cos 2ϑ sin 2ϑ I +, sin ϑ sin ϑ 2 sin 2ϑ cos 2ϑ ) ) sin ϑ sin ϑ = 1 [ )] cos 2ϑ sin 2ϑ I. cos ϑ cos ϑ 2 sin 2ϑ cos 2ϑ 3 Solve 2.8) when = H We recall 2.16), 2.17), 2.18), and Theorem 2.1, and keep in mind that = H, ε = ±1, and = F in 2.9). For even d, we need to solve l) b 11 b l) ) 12 ρ l b l) 21 b l) 22 + ε bl) 11 bl) 12 d/2) b 11 b d/2) 12 +ρ d/2 ε b d/2) 12 b d/2) 22 bl) 21 bl) 22 )µ d 2l ) ) ] α ) µ d/2 = 0 µ l ) γ, 3.1) β where εb d/2) 11 and εb d/2) 22 are real which means they are real for ε = 1 and pure imaginary for ε = 1). Componentwise, it gives ρ l b l) l) 11 + ε b 11 µd 2l) µ l + ρ d/2 b d/2) 11 µ d/2 = δ 1, 3.2) ρ l b l) l) 21 + ε b 12 µd 2l) µ l + ερ d/2 bd/2) 12 µ d/2 = δ ) 8

10 For odd d, we need to solve d 1)/2 l) b 11 b l) 12 ρ l b l) 21 b l) 22 ) + ε bl) 11 bl) 12 bl) 21 bl) 22 )µ d 2l ) µ l ) α = 0 ) γ. 3.4) β Componentwise, it gives We observe that: 1. b l) 22 d 1)/2 d 1)/2 for all l do not show up. Thus must bl) 22 ρ l b l) l) 11 + ε b 11 µd 2l) µ l = δ 1, 3.5) ρ l b l) l) 21 + ε b 12 µd 2l) µ l = δ ) = 0 for all l by 2.9) ) and 3.3) are decoupled. Thus they can be solved separately for optimal solutions in the sense of d/2 b l) 11 2 = min, 2 b d/2) b l) b l) 21 2) = min, respectively, to arrive at optimal solutions to 3.1) in the sense of 2.9) ) and 3.6) are decoupled. Thus they can also be solved separately for optimal solutions in the sense of d 1)/2 b l) 11 2 = min, d 1)/2 b l) b l) 21 2) = min, respectively, to arrive at optimal solutions to 3.4) in the sense of 2.9). We shall now solve each of 3.2), 3.3), 3.5), and 3.6) in terms of four lemmas. For this purpose, write µ = µ e ιθ, c l = coslθ), s l = sinlθ). 3.7) Then by Lemma 2.1 i), we have [ ] ) R b l) l) 11 + ε b 11 µd 2l µ l ) [ ] ) I b l) = µ l cl s l Rb l) l) ) 11 + ε b 11 µd 2l ) l) 11 + ε b 11 µd 2l µ l s l c l Ib l) l) 11 + ε b 11 µd 2l ) )[ )] ) = µ l cl s l I s l c 2 + ε µ d 2l cd 2l s d 2l Rb l) l s d 2l c d 2l Ib l). 3.8) 9

11 Lemma 3.1 Equation 3.2) which is for even d has a solution if and only if either µ 1 or µ = 1 and z H r/ εµ d/2 ) R. Otherwise it is inconsistent. Let 2 ξ even = ρ2 d/2 µ d 2 ζ even = µ d Φ even φ even + ρ2 d/ def = ξ even + ζ even = def = ξ even ζ even = µ 2l + µ 2d l)), 3.9), 3.10) µ l + µ d l) 2 + ρ 2 d/2 µ d, 3.11) µ l µ d l) 2, 3.12) J ε = 1 ) 1 + ε ε ) 2 1 ε 1 + ε When 3.2) has a solution, its optimal solution in the sense of d/2 by for 0 l d/2 1 ) Rb l) ) µ l + ε µ d l cd/2 l s Ib l) = ρ d/2 l l ξ even + εζ even s d/2 l c d/2 l and 3 and the optimal value 4 d/2 cd/2 s d/2 s d/2 c d/2 ε εb d/2) 11 = b l) 11 2 = ) ) Rδ1 ) Iδ 1 ) ρ d/2 µ d/2 ξ even + ζ even c d/2 s d/2 )J ε bl) µ l ε µ d l ξ even εζ even 11 2 = min is given, 3.14) ) Rδ1 ), 3.15) Iδ 1 ) Re ιθd/2 δ 1 ) 2 Φ even + Ie ιθd/2 δ 1 ) 2 φ even for ε = 1, Re ιθd/2 δ 1 ) 2 φ even + Ie ιθd/2 δ 1 ) 2 Φ even for ε = ) δ ) φ even ) Making J 1 = I 2 and J 1 = is all that is necessary The quantity ε εb d/2) 11 looks strange. But it is in fact b d/2) 11 when ε = 1 for which b d/2) 11 is real), and it is Ib d/2) when ε = 1 for which b d/2) 11 is pure imaginary). ) d/2 µ 4 e ιθd/2 z H r δ 1 can also be expressed as. µ z

12 When µ = 1 and z H r/ εµ d/2 ) R, 3.14), 3.15), and 3.16) can be significantly simplified to for 0 l d/2 1 ) Rb l) cd/2 l and Ib l) = 2ρ l Φ even J ε s d/2 l ) δ1 εµ d/2, ε εb d/2) 11 = ρ d/2 Φ even δ 1 εµ d/2, 3.18) d/2 b l) 11 2 = δ 1 2, Φ even = ρ 2 d/2 Φ ) even Proof Equation 3.2) needs to be split into two equations according to the real and imaginary parts because εb d/2) 11 is real and both b l) l) 11 and b 11 for l = 0,1,...,d/2 1 appear. By 3.8), we see that 3.2) is equivalently to Rb 0) Ib 0) where for l = 0,1,...,d/2 1 and Z 0 Z Z d/2 ) Z l = ρ l µ l cl s l. Rb ) Ib ) ε εb d/2) 11 = )[ s l I 2 + ε µ d 2l cd 2l c l s d 2l ) Rδ1 ), 3.20) Iδ 1 ) s d 2l c d 2l )], 3.21) ) Z d/2 = ρ d/2 µ d/2 cd/2 J ε. 3.22) s d/2 Set Z = Z 0 Z Z d/2 ) R 2 d+1). It can be computed, by Lemma 2.1 ii), that )[ Z l Zl = ρ 2 cl s l l µ 2l I 2 + ε µ d 2l cd 2l s l c l s d 2l s d 2l c d 2l )] 2 cl s l [ = µ 2l + µ 2d l)) )] I 2 + 2ε µ d cd s d for 0 l d/2 1, s d c d s l c l ) Z d/2 Z d/2 Therefore = ρ 2 d/2 µ d J ε cd/2 = ρ2 d/2 µ d 2 s d/2 [ I 2 + ε d/2 ) cd/2 s d/2 cd s d s d ) J c d ε )]. ZZ = Z l Zl cd = ξ even I 2 + εζ even 11 s d ) s d. 3.23) c d

13 ) cd s The eigenvalues of d are ±1. In fact, its eigendecomposition can be explicitly s d c d written out as ) ) ) ) cd s d cd/2 s = d/2 1 0 cd/2 s d/ ) s d c d s d/2 c d/2 0 1 s d/2 c d/2 So the eigenvalues of ZZ are Φ even and φ even, and moreover, we have an explicit eigendecomposition for ZZ : ) ) ) ZZ cd/2 s = d/2 ξeven + εζ even cd/2 s d/ ) s d/2 c d/2 ξ even εζ even s d/2 c d/2 If µ 1, then the smaller eigenvalue φ even > 0 and thus ZZ is invertible. The optimal solution in the sense of d/2 bl) 11 2 = min is Rb 0) Ib 0). Rb ) Ib ) ε εb d/2) 11 ) ) = Z Rδ1 ) = Z ZZ ) 1 Rδ1 ). 3.26) Iδ 1 ) Iδ 1 ) To arrive at 3.14), 3.15), and 3.16), we notice from 3.26) that ) Rb l) Ib l) ) = Zl ZZ ) 1 Rδ1 ), ε ) εb d/2) Iδ 1 ) 11 = Zd/2 ZZ ) 1 Rδ1 ). 3.27) Iδ 1 ) It can be verified that for 0 l d/2 1 ) Z l = ρ l µ l cd/2 s d/2 1 + ε µ d 2l s d/2 c d/2 1 ε µ d 2l ) cd/2 l s d/2 l s d/2 l c d/2 l ). 3.28) Equations 3.14) and 3.15) are the consequences of 3.22), 3.25), 3.27), and 3.28), upon noticing ) ) cd/2 s d/2 Rδ1 ) Re = ιθd/2 ) δ 1 ) s d/2 c d/2 Iδ 1 ) Ie ιθd/2, δ 1 ) ) { c d/2 s d/2 )Jε Rδ1 ) Re = ιθd/2 δ 1 ) for ε = 1, Iδ 1 ) Ie ιθd/2 δ 1 ) for ε = 1. Finally we have, by 3.14) and 3.15), the optimal value d/2 b l) 11 2 = [ µ l + ε µ d l ) 2 Re ιθd/2 δ 1 ) 2 ξ even + εζ even 12

14 µ l ε µ d l ) 2 ] + Ie ιθd/2 δ 1 ) 2 ξ even εζ even + ρd/2 µ d/2 ξ even + ζ even ) 2 c d/2 s d/2 )J ε ) Rδ1 ) 2. Iδ 1 ) which leads to 3.16). If, however, µ = 1, then rankz) = 1 and 3.20) is not solvable unless Rδ 1 ) Iδ 1 )) is parallel to c d/2 s d/2 ) J ε, or equivalently δ 1 εµ d/2 = z H r εµ d/2 z 2 2 R. 3.29) When it does, the optimal solution can be gotten from either equation from the first or the second component in 3.20). We shall do so now. When µ = 1, we have for l = 0,1,...,d/2 1 )[ )] cl s Z l = ρ l cd 2l s l I s l c 2 + ε d 2l l s d 2l c d 2l [ ) )] cl s = ρ l cd l s l + ε d l and s l = 2ρ l J ε cd/2 s d/2 c l s d l ) cd/2 l s d/2 l ) cd/2 Z d/2 = ρ d/2 J ε. s d/2 c d l ) J ε, 3.30) To see 3.18) and 3.19), we take the first equation in 3.20) for example. With µ = 1, the first equation is For ε = 1, it is 2c d/2 [ ] ρ l c l + εc d l )Rb l) + s l + εs d l )Ib l) ρ [ ] d/2 d/2 1 + ε)cd/2 1 ε)s d/2 ε εb 11 = Rδ 1). 3.31) [ ] ρ l c d/2 l Rb l) + s d l/2ib l) + ρ d/2 c d/2 ε εb d/2 11 = Rδ 1) which, under the condition that c d/2 0, leads to 3.18) and 3.19) upon noticing that Rδ 1 )/c d/2 = δ 1 /µ d/2 because of 3.29). If c d/2 = 0, the second equation in 3.20) will lead to the same conclusion for the case. The case for ε = 1 is similar. 13

15 Remark 3.1 Inequality 3.17) becomes an equality when Rδ 1 ) Iδ 1 )) is parallel to ) cd s the eigenvector of d associated with its eigenvalue ε. s d c d Remark 3.2 Equation 3.16) reveals what contributes the most to the structured backward error, while 3.17) fails to do at the price of being the simplest. That is that not all of δ 1 contributes equally in proportional to 1/ φ even to the structured backward error. In fact only either Re ιθd/2 δ 1 ) or Ie ιθd/2 δ 1 ), depending on what ε is, does. Conceivably, there are circumstances where the extreme tininess of either Re ιθd/2 δ 1 ) or Ie ιθd/2 δ 1 ) may offset the tininess of φ even to render reasonably tiny structured backward error, much smaller than 3.17) would indicate. Similar comment applies to part of the development in Lemma 3.3, too. Lemma 3.2 Equation 3.3) which is for even d always has a solution. Its optimal solution in the sense of 2 b d/2) ) b l) b l) 21 2 = min is given by 2 b d/2) 12 = ε ρ d/2 µ d/2 δ2 / 2 Ψ even, 3.32) b l) 12 = ερ l µ d l δ2 Ψ even, b l) 21 = ρ l µ l δ 2 Ψ even for l = 0,1,...,d/ ) satisfying where 2 b d/2) def Ψ even = b l) b l) 21 2) = δ 2 2 Ψ even, 3.34) µ 2l + µ 2d l)) + ρ 2 d/2 µ d / ) Proof Ψ even > 0 always since by assumption all ρ l > 0. It can be verified that b l) ij s given by 3.32) and 3.33) satisfy 3.3). On the other hand, 3.3) implies, by the Cauchy-Schwarz inequality, δ 2 2 Ψ even 2 b d/2) b l) b l) 21 2) for any solution {b l) ij } to 3.3). Lemma 3.3 Equation 3.5) which is for odd d has a solution if and only if either µ 1 or µ = 1 and z H r/ εµ d/2 ) R. Otherwise it is inconsistent. Let ξ odd = d 1)/2 µ 2l + µ 2d l)), 3.36) 14

16 d 1)/2 ζ odd = 2 µ d, 3.37) Φ odd φ odd d 1)/2 def = ξ odd + ζ odd = d 1)/2 def = ξ odd + ζ odd = µ l + µ d l) 2, 3.38) µ l µ d l) ) When 3.5) has a solution, its optimal solution in the sense of d 1)/2 given by for 0 l d 1)/2 1 ) Rb l) ) µ l + ε µ d l cd/2 l s Ib l) = ρ d/2 l l ξ odd + εζ odd s d/2 l c d/2 l and the optimal value d 1)/2 b l) 11 2 = cd/2 s d/2 s d/2 c d/2 ) ) Rδ1 ) Iδ 1 ) b l) µ l ε µ d l ξ odd εζ odd 11 2 = min is, 3.40) Re ιθd/2 δ 1 ) 2 Φ odd + Ie ιθd/2 δ 1 ) 2 φ odd for ε = 1, Re ιθd/2 δ 1 ) 2 φ odd + Ie ιθd/2 δ 1 ) 2 Φ odd for ε = ) δ 1 2 φ odd. 3.42) When µ = 1 and z H r/ εµ d/2 ) R, 3.40) and 3.41) can be significantly simplified to for 0 l d 1)/2 1 ) Rb l) cd/2 l and d 1)/2 Ib l) = 2ρ l Φ odd J ε s d/2 l b l) 11 2 = δ 1 2 k 1)/2, Φ odd = 4 Φ odd Z 0 Z d 1)/2 ) ) δ1 εµ d/2, 3.43) Proof Similarly to the proof of Lemma 3.1 above, 3.5) is equivalently to Rb 0) Ib 0). Rb d 1)/2) Ib d 1)/2) 15 =. 3.44) ) Rδ1 ), 3.45) Iδ 1 )

17 where Z l are as in 3.21). Let Z = Z 0 Z d 1)/2 ). Then, as in the proof of Lemma 3.1, ) ZZ cd s = ξ odd I 2 + εζ d odd s d c d ) ) ) cd/2 s = d/2 ξodd + εζ odd cd/2 s d/2. s d/2 c d/2 ξ odd εζ odd s d/2 c d/2 The eigenvalues of ZZ are Φ odd and φ odd. If µ 1, then the smaller eigenvalue φ odd > 0 and thus ZZ is invertible. The optimal solution in the sense of d 1)/2 b l) 11 2 = min is for l = 0,1,...,d 1)/2 1 ) Rb l) which lead to 3.40) and 3.41). Ib l) ) = Zl ZZ ) 1 Rδ1 ) Iδ 1 ) 3.46) If, however, µ = 1, then rankz) = 1 and 3.45) is not solvable unless Rδ 1 ) Iδ 1 )) is parallel to c d/2 s d/2 ) J ε, or equivalently 3.29) holds. When it does, the optimal solution can be gotten similarly as in the proof of Lemma 3.1. Remark 3.3 Inequality 3.42) becomes an equality when Rδ 1 ) Iδ 1 )) is parallel to ) cd s the eigenvector of d associated with its eigenvalue ε. s d c d Lemma 3.4 Equation 3.6) which is for odd d always has a solution. Its optimal solution in the sense of ) d 1)/2 b l) b l) 21 2 = min is given by satisfying where b l) 12 = ερ l µ d l δ2 Ψ odd, b l) 21 = ρ l µ l δ 2 Ψ odd for l = 0,1,...,d 1)/2 3.47) d 1)/2 Ψ odd = b l) b l) 21 2) = δ 2 2 Ψ odd. 3.48) d 1)/2 µ 2l + µ 2d l)). 3.49) Proof Ψ odd > 0 always since by assumption all ρ l > 0. It can be verified that b l) ij s given by 3.47) satisfy 3.6). On the other hand, 3.6) implies, by the Cauchy-Schwarz inequality, d 1)/2 1 δ 2 2 Ψ odd b l) b l) 21 2) for any solution {b l) ij } to 3.6). 16

18 We summarize our findings in Lemmas into the following theorem. Theorem 3.1 Suppose = H and ε = ±1 in 2.1). Let φ = { φeven for even d, φ odd, for odd d, Φ = { Φeven for even d, Φ odd, for odd d, Ψ = { Ψeven for even d, Ψ odd, for odd d, where φ even, φ odd, Φ even, Φ odd, Ψ even, and Ψ odd are as defined in Lemmas and let δ 1 and δ 2 be given as in 2.12) with r = r. We have 1. If µ = 1 and z H r/ εµ d/2 ) R, then F = If µ = 1 and z H r/ εµ d/2 ) R, then δ1 F = 2 Φ + δ 2 2 Ψ. 3. If µ 1, then F δ 1 2 φ + δ 2 2 Ψ. 3.50) Remark 3.4 Inequality 3.50) in general cannot be improved because ) it becomes an equality when Rδ 1 ) Iδ 1 )) cd s is parallel to the eigenvector of d associated with s d c d its eigenvalue ε. Besides bounding F as in 3.50), a close formula for it for the case µ 1 at the price of more complicated expressions can be gotten by combining 3.16), 3.34), 3.41), and 3.48). We omit the detail. Remark 3.5 Our solutions to 3.2) and 3.5) by Lemmas 3.1 and 3.3 are all that are needed for dealing with the scalar polynomial case, i.e., when n = 1. 4 Solve 2.8) when = We again recall 2.16), 2.17), 2.18), and Theorem 2.1, and keep in mind that = T, ε = ±1, and = F in 2.9). For even d, we need to solve l) b 11 b l) 12 ρ l b l) 21 b l) 22 ) + ε σε b d/2) 11 b d/2) 12 +ρ d/2 εb d/2) 12 σ ε b d/2) 22 l) b 11 b l) 21 b l) 12 b l) 22 )µ d 2l ) ) ] α ) µ d/2 = 0 µ l γ β), 4.1) 17

19 where σ ε = 1+ε 2, i.e., σ ε equals to 1 for ε = 1 and 1 for ε = 1. Componentwise, it gives For odd d, we need to solve d 1)/2 Componentwise, it gives We observe that: 1. b l) 22 ρ l b l) 11 + εbl) 11 µd 2l) µ l + ρ d/2 σ ε b d/2) 11 µ d/2 = δ 1, 4.2) ρ l b l) 21 + εbl) 12 µd 2l) µ l + ερ d/2 b d/2) 12 µ d/2 = δ ) l) b 11 b l) 12 ρ l b l) 21 b l) 22 d 1)/2 d 1)/2 ) + ε for all l do not show up. Thus must bl) 22 l) b 11 b l) ) 21 )µ d 2l b l) 12 b l) 22 µ l ) γ β) α =. 4.4) 0 ρ l b l) 11 + εbl) 11 µd 2l) µ l = δ 1, 4.5) ρ l b l) 21 + εbl) 12 µd 2l) µ l = δ ) = 0 for all l by 2.9) ) and 4.3) are decoupled. Thus they can be solved separately for optimal solutions in the sense that d/2 b l) 11 2 = min, 2 b d/2) b l) b l) 21 2) = min, respectively, to arrive at optimal solutions to 4.1) in the sense of 2.9) ) and 4.6) are decoupled. Thus they can also be solved separately for optimal solutions in the sense that d 1)/2 b l) 11 2 = min, d 1)/2 b l) b l) 21 2) = min, respectively, to arrive at optimal solutions to 4.4) in the sense of 2.9). We shall now solve each of 4.2), 4.3), 4.5), and 4.6) in terms of four lemmas. 18

20 Lemma 4.1 Equation 4.2) which is for even d has a solution if and only if ε = 1, or ε = 1 and µ ±1, or ε = 1 and µ = ±1 and z H r = 0. For ε = 1 or for ε = 1 and µ ±1, the optimal solution to 4.2) in the sense of d/2 bl) 11 2 = min is given by satisfying where b d/2) ρ d/2 µ d/2 δ 1 11 = σ ε, 4.7) Φ even b l) 11 = ρ l µ l + ε µ d l) δ 1 for l = 0,1,...,d/ ) Φ even d/2 def Φ even = b l) 11 2 = δ 1 2 Φ even, 4.9) µ l + εµ d l 2 + σε ρ 2 d/2 µ d. 4.10) For ε = 1 and µ = ±1 and z H r = 0, the optimal solution is all b l) 11 = 0. Proof Note all ρ l > 0 by the assumption 2.20). For ε = 1, Φ even > 0 always; For ε = 1, Φ even > 0 if and only if µ ±1. It can be verified that b l) ij s given by 4.7) and 4.8) satisfy 4.2) whenever Φ even > 0. On the other hand, 4.2) implies, by the Cauchy-Schwarz inequality, d/2 δ 1 2 Φ even b l) for any solution {b l) ij } to 4.2). For ε = 1 if µ = ±1, then 4.2) is consistent if and only if δ 1 = 0 or equivalently z H r = 0) for which case the optimal solution is b l) 11 = 0 for all l. Lemma 4.2 Equation 4.3) which is for even d always has a solution. Its optimal solution in the sense of 2 b d/2) ) b l) b l) 21 2 = min is given by b d/2) 12 = ε ρ d/2 µ d/2 δ 2 / 2 Ψ even, 4.11) b l) 12 = ερ l µ d l δ 2 Ψ even, b l) 21 = ρ l µ l δ 2 Ψ even for l = 0,1,...,d/ ) satisfying 2 b d/2) b l) b l) 21 2) = δ 2 2 Ψ even, 4.13) 19

21 where def Ψ even = µ 2l + µ 2d l)) + ρ 2 d/2 µ d / ) Proof Ψ even > 0 always. It can be verified that b l) ij s given by 4.11) and 4.12) satisfy 4.3). On the other hand, 4.3) implies, by the Cauchy-Schwarz inequality, δ 2 2 Ψ even 2 b d/2) b l) b l) 21 2) for any solution {b l) ij } to 4.3). Lemma 4.3 Equation 4.5) which is for odd d has a solution if and only if either µ ε or µ = ε and z H r = 0. For µ ε, then the optimal solution to 4.5) in the sense of d 1)/2 b l) 11 2 = min is given by satisfying where b l) 11 = ρ l µ l + ε µ d l) δ 1 Φ even, for l = 0,1,...,d 1)/2 4.15) d 1)/2 d 1)/2 def Φ odd = For µ = ε and z H r = 0, the optimal solution is all b l) 11 = 0. b l) 11 2 = δ 1 2 Φ odd, 4.16) µ l + εµ d l ) Proof Since d is odd, Φ odd = 0 if and only if µ = ε. It can be verified that b l) ij s given by 4.15) satisfy 4.5) when Φ even > 0. On the other hand, 4.5) implies, by the Cauchy-Schwarz inequality, d 1)/2 δ 1 2 Φ odd b l) for any solution {b l) ij } to 4.5). For µ = ε, then 4.5) is consistent if and only if δ 1 = 0 or equivalently z H r = 0) for which case the optimal solution is b l) 11 = 0 for all l

22 Lemma 4.4 Equation 4.6) which is for odd d always has a solution. Its optimal solution in the sense of ) d 1)/2 b l) b l) 21 2 = min is given by b l) 12 = ερ l µ d l δ 2 Ψ odd, b l) 21 = ρ l µ l δ 2 Ψ odd for l = 0,1,...,d 1)/2 4.18) satisfying where d 1)/2 Ψ odd = b l) b l) 21 2) = δ 2 2 Ψ odd, 4.19) d 1)/2 µ 2l + µ 2d l)). 4.20) Proof Ψ odd > 0 always. It can be verified that b l) ij s given by 4.18) satisfy 4.6). On the other hand, 4.6) implies, by the Cauchy-Schwarz inequality, d 1)/2 δ 2 2 Ψ odd b l) b l) 21 2) for any solution {b l) ij } to 4.6). We summarize our findings in Lemmas into the following theorem. Theorem 4.1 Suppose = and ε = ±1 in 2.1). Let Φ = { Φeven for even d, Φ odd, for odd d, Ψ = { Ψeven for even d, Ψ odd, for odd d, where Φ even, Φ odd, Ψ even, and Ψ odd are given by 4.10), 4.17), 4.14), and 4.20), respectively, and let δ 1 and δ 2 be given as in 2.12) with r = r. We have 1. If d is odd, µ = ε, or if d is even, µ = ±1, and ε = 1, then { +, if zh r 0, F = δ 2 / Ψ, if z H r = If d is odd, µ ε, or if d is even, µ ±1, or if d is even, ε = 1, then δ1 F = 2 Φ + δ 2 2 Ψ. 4.21) Remark 4.1 Our solutions to 4.2) and 4.5) by Lemmas 4.1 and 4.3 are all that are needed for dealing with the scalar polynomial case, i.e., when n = 1. 21

23 µ 1 Bound µ F ηµ,z) 10 4 Bound F ηµ,z) Index of approximate eigenpairs ε=1) Index of approximate eigenpairs ε= 1) Figure 5.1: Backward errors for the case = H; Left: ε = 1 and Right: ε = 1. Plotted quantities are µ 1, bounds by 3.50), F by combining 3.16) and 3.34), and Tisseur s ηµ, z) in 5.1). Bounds by 3.50) and F are visually indistinguishable from the plotted lines. 5 Numerical Examples In this section we test some numerical examples to illustrate our structured backward error compared with the existing unstructured) backward errors. For convenience, we consider PQEP only: Qλ) ελ 2 A 0 + λa 1 + A 0 with also A 1 = εa 1, and set ρ i = A i 2. By the existing unstructured) backward error, we mean the one from Tisseur [15] who defined the unstructured) backward error of an approximate eigenpair {µ,z} of Qλ) as ηµ,z) = min {δ : [Qµ) + Qµ)]z = 0, A 2 2 δρ 0, A 1 2 δρ 1, A 0 2 δρ 0 }, where Qλ) = λ 2 A 2 ) + λ A 1 ) + A 0 with no constraints imposed on A i. An explicit expression for ηµ,z) in terms of r = Qµ)z is given by [15] ηµ,z) = r 2 µ 2 ρ 0 + µ ρ 1 + ρ 0 ) z ) Take n = 10 and choose A 0, A 1 C randomly as by this piece of MATLAB code for = H: A0 = randnn)+i*randnn); A2 = varepsilon * A0 ; A1 = randnn)+i*randnn)); A1 = A1+varepsilon*A1 ; For = T, all needed is to replace A0 and A1 by A0. and A1., respectively. We compute 2n approximate eigenpairs {µ, z} of Qλ) by MATLAB s eig ) after linearizing Qλ). With {µ, z} running through all 2n approximate eigenpairs, we plots various quantities of interest in Figures 5.1 and 5.2. Figure 5.1 is for = H. It is worth pointing out that in our many numerical runs with different random matrices constructed as such and different n as well), there are always 22

24 F ηµ,z) F ηµ,z) Index of approximate eigenpairs ε=1) Index of approximate eigenpairs ε= 1) Figure 5.2: Backward errors for the case = T; Left: ε = 1 and Right: ε = 1. Plotted quantities are F by 4.21), and Tisseur s ηµ, z) in 5.1). some eigenvalues with µ being 1 in the working IEEE double) precision. As our analysis in Section 3 shows, if µ = 1, the matrix Z is singular and there may not exist a structured backward error unless z H r/ εµ d/2 ) is real. But this condition is nearly impossible to verify when the approximate eigenpair is so accurate in the working precision that r may have no correct significant digits. So in the numerical results presented in Figure 5.1, we purposely perturb the computed µ relatively by errors up to 10 8 to make µ 1 away from 0 by no less than about Doing so, however, does not prevent us from making our point with the numerical results, namely F is very sensitive with respect to µ 1 being near 0 or not. Figure 5.1 clearly shows when µ 1 is away from 0, F is not much worse than ηµ,z), and it appears that F is proportional to the reciprocal of µ 1. Figure 5.2 is for = T. We did not encounter the critical cases µ = ±1 as revealed in our analysis in Section 4 during our many numerical runs. The numerical results presented in this figure do not include any critical case, unfortunately. Because of the lack of critical cases, F is not much worse than ηµ,z) here. 6 Concluding Remarks We have presented a detailed structured backward error analysis for an approximate eigenpair {µ, z} of PPEP 1.4) by solving problem 1.5). Computable formulas and bounds for the optimal structured backward errors are obtained. These formulas and bounds show distinctive features of PPEP from the usual PEP as investigated by Tisseur [15] and Liu and Wang [10], namely there are critical cases µ = 1 for = H and µ = ±1 for = T) for which structured backward errors may not exist. In the PEP case, Tisseur [15] asked and solved what the optimal backward error for an approximated eigenvalue is by minimizing the backward errors for an approximate eigenpair {µ, z} over all possible vectors z. We also attempted to solve a similar question, namely what the optimal structured backward error for an approximated eigenvalue in 23

25 the PPEP case is. But it appears that our question is not as simple as in the PEP case because our formulas for structured backward errors for PPEP depend on the approximate eigenvector z in a more complicated way than Tisseur s. Also in the PEP case, Tisseur [15] asked and solved what the optimal backward errors for an approximate eigen-triplet an eigenvalue and its associated left and right eigenvectors). We attempted to solve a similar question, too, but did not succeed. References [1] R. Byers, D. S. Mackey, V. Mehrmann, and H. Xu. Symplectic, BVD, and palindromic approaches to discrete-time control problems. Technical report, Preprint , Institute of Mathematics, Technische Universität Berlin, [2] E. K.-W. Chu, T.-M. Hwang, W.-W. Lin, and C.-T. Wu. Vibration of fast trains, palindromic eigenvalue problems and structure-preserving doubling algorithms. J. Comput. Appl. Math., 229: , [3] D. J. Higham and N. J. Higham. Structured backward error and condition of generalized eigenvalue problems. SIAM J. Matrix Anal. Appl., 202): , [4] N. J. Higham, F. Tisseur, and P. M. Van Dooren. Detecting a definite Hermitian pair and a hyperbolic or elliptic quadratic eigenvalue problem, and associated nearness problems. Linear Algebra Appl., : , [5] A. Hilliges. Numerische Lösung von quadratischen Eigenwertproblemen mit Anwendungen in der Schiendynamik. Master s thesis, Technical University Berlin, Germany, July [6] A Hilliges, C. Mehl, and V. Mehrmann. On the solution of palindramic eigenvalue problems. In Proceedings of 4th European Congress on Computational Methods in Applied Sciences and Engineering ECCOMAS), Jyväskylä, Finland, [7] M. E. Hochstenbach and B. Plestenjak. Backward error, condition numbers, and pseudospectra for the multiparameter eigenvalue problem. Linear Algebra Appl., 375:63 81, [8] T.-M. Hwang, W.-W. Lin, and J. Qian. Numerically stable, structure-preserving algorithms for palindromic quadratic eigenvalue problems. Technical report, National Center for Theoretical Sciences, National Tsing Hua University, Taiwan, Preprints in Mathematics , [9] I. C. F. Ipsen. Accurate eigenvalues for fast trains. SIAM News, 37, [10] X.-G. Liu and Z.-X. Wang. A note on the backward errors for Hermite eigenvalue problems. Appl. Math. Comput., 1652): ,

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