Structured Backward Error for Palindromic Polynomial Eigenvalue Problems

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Structured Backward Error for Palindromic Polynomial

Eigenvalue Problems

Ren-Cang Li∗ Wen-Wei Lin† Chern-Shuh Wang‡

1 Introduction

In vibration analysis of fast trains arises the Quadratic Palindromic Eigenvalue Problem (QPEP) [1, 2, 3]: ﬁnd a scalar λ and an n-dimensional vector x such that

(AT₁λ2+ A0λ + A1)x = 0, (1.1)

where A0 and A1 are n× n (complex) matrices, AT0 = A0, and the superscript “·T”

takes matrix transpose. Presently, it is solved by certain kinds of structurally preserving linearization [1, 3]. Experience shows that QPEP from the fast train application is no-toriously 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.

QPEP is a particular case of the so-called Polynomial Eigenvalue Problem (PEP) Ã _k X `=0 A`λ` ! x = 0, (1.2)

where A`, ` = 0, 1, . . . , k are n× n matrices. When k = 2, PEP (1.2) is called Quadratic

Eigenvalue Problems (QEP) [5]. For a PEP in its generality there is no relation among co-eﬃcient matrices, unlike QPEP. Naturally one deﬁnes the Palindromic Polynomial Eigen-value Problem (PPEP) to be a PEP (1.2) with Ak−`= AT_`, ` = 0, 1, . . . ,bk/2c, namely,

Ã _k X `=0 A`λ` ! x = 0, A_k−` = AT_` for ` = 0, 1, . . . ,bk/2c (1.3)

wherebk/2c is the largest integer that is no bigger than k/2.

∗_{Department of Mathematics, University of Texas at Arlington, Arlington, P.O. Box 19408, Arlington,}

USA, TX 76019; [email protected]

†_{Department of Applied Mathematics, National Chiao Tung University, Hsinchu 300, Taiwan;}

[email protected],edu.tw

‡_Department _of _Mathematics, _National _Cheng _Kung _University, _Tainan _701, _Taiwan;

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Let {µ, z} be a computed eigenpair of (1.2) or (1.3). Ideally, we would like to have ³Pk

`=0A`µ`

´

z = 0. But practically we have Ã _k X `=0 A`µ` ! z =−r

with the residual r 6= 0 but tiny. Backward error analysis concerns if the computed eigenpair{µ, z} is an exact eigenpair of a nearby PEP, namely,

Ã _k X `=0 (A`+ ρ`∆A`)µ` ! z = 0

with ∆A`, ` = 0, 1, . . . , k hopefully tiny in magnitude, relative to A`, ` = 0, 1, . . . , k,

where ρ`, ` = 0, 1, . . . , k are scaling parameters, usually taken to be some norms of A`,

respectively. For a general PEP, no constraints on and/or among ∆A`, ` = 0, 1, . . . , k are

imposed, except that ∆A`, ` = 0, 1, . . . , k 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.

Tisseur [4] thoroughly developed a backward error analysis for PEP in its generality or when A`, ` = 0, 1, . . . , k are Hermitian. Her results do not 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 “·X” be either “·T” or “·H” which takes complex conjugate and transpose and let ε ∈ {±1}. Collectively by PPEP, we mean a polynomial eigenvalue problem of the following form

Ã _k X `=0 A`λ` ! x = 0, Ak−` = εAX` for ` = 0, 1, . . . ,bk/2c. (1.4)

Let{µ, z} be a computed eigenpair and let k · k be either the spectral norm1 k · k2 or the

Frobenius norm k · kF. We are interested in knowing the optimal structured backward

error

minqPbk/2c_`=0 k∆A`k2

subject to ³Pk_`=0(A`+ ρ`∆A`)µ`

´

z = 0, ρk−`= ρ`≥ 0,

Ak−`+ ρk−`∆Ak−`= ε(A`+ ρ`∆A`)X for ` = 0, 1, . . . ,bk/2c.

(1.5)

Notation. Throughout this paper, Cn×m is the set of all n× m complex matrices, Cn ₌ _Cn×1_{, and} _{C = C}1_{. Similarly deﬁne} _Rn×m_, _Rn_{, and} _{R except replacing the word}

complex by real. In (or simply I if its dimension is clear from the context) is the n× n

1_Notation_{k · k}

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identity matrix. ¯x is the complex conjugate of a scalar or the entry-wise complex conjugate of a vector or matrix. <(α) and =(α) 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:`,i:j) is X’s submatrices

X(k:`,i:j) consisting of intersections of row k to row ` and column i to column j. X† is X’s

Moore-Penrose inverse.

2 Simplify the Problem

The constraining equation in (1.5) gives ρk−`= ρ`and ∆Ak−` = ε∆AX` for ` = 0, 1, . . . ,bk/2c,

and _Ã k X `=0 ρ`∆A`µ` ! z = rdef= − Ã _k X `=0 A`µ` ! z, (2.1)

where ∆A` ∈ Cn×n (` = 0, 1, . . . , k). We will be investigating if (2.1) has a solution

{∆A`, ` = 0, 1, . . . ,bk/2c} and if it does, we’ll seek ∆A` such that bk/2c_X

`=0

k∆A`k2 = min, (2.2)

wherek · k is either k · k2 ork · kF. The two trivial cases 1) when all ρ` = 0, and 2) when

r = 0 will be excluded from our consideration.

Let Q∈ Cn×n be a unitary matrix, i.e., QQH= In, such that

QH(z br) =        α γ 0 β 0 0 .. . ... 0 0        , (2.3) where br = r if X=H, and br = ¯r ifX=T. (2.4)

Such Q always exists. Moreover we can take α =kzk2, γ = zHbr kzk2 , β =°°°°br− z H_br kzk2 2 z°°°° 2 = p kzk2 2kbrk22− |zHbr|2 kzk2 . (2.5)

These will be assumed in what follows. It follows from (2.1) that

QX Ã _k X `=0 ρ`∆A`µ` ! QQHz = QXr, (2.6)

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or equivalently Ã _k X `=0 ρ`∆B`µ` ! y = w, ρk−`= ρ` and ∆Bk−`= ε∆B_`X for ` = 0, 1, . . . ,bk/2c, (2.7) where ∆B`= QX(∆A`)Q, y = QHx =      α 0 .. . 0     , w = QXr =        bγ b β 0 .. . 0        , b

β = β andbγ = γ if X=H, and bβ = ¯β and bγ = ¯γ if X=T.

Since Q is unitary, (2.1) and (2.7) have the same solvability property, i.e., if one is solvable, so is the other, and moreover

bk/2c_X `=0 k∆B`k2 = bk/2c_X `=0 k∆A`k2.

Thus the optimal solution {∆B`, ` = 0, 1, . . . ,bk/2c} to (2.7) in the sense of bk/2c_X

`=0

k∆B`k2 = min (2.8)

gives rise to one solution {∆A`, ` = 0, 1, . . . ,bk/2c} to (2.1) in the sense of (2.2), and vice

versa. Set δ1 =bγ/α, δ2 = bβ/α. (2.9) and ∆p = min      v u u tbk/2cX `=0 k∆B`k2p : (2.7) satisﬁed      for p = 2,F. (2.10) ∆p=∞ means that (2.7) is not solvable. The assumption r 6= 0 implies |δ1| + |δ2| > 0.

Theorem 2.1 Suppose that (2.7) has a solution (so does (2.1)). The optimal solution {∆B`, ` = 0, 1, . . . ,bk/2c} to (2.7) in the sense of (2.8) for the Frobenius norm satisﬁes

(∆B`)(i,j)= 0 for i, j > 2 and for ` = 0, 1, . . . ,bk/2c. (2.11)

There is an optimal solution{∆B`, ` = 0, 1, . . . ,bk/2c} in the sense of (2.8) for the spectral

norm satisﬁes (2.11). Finally

∆2 ≤ ∆F ≤

√

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Proof No proof is needed for ∆B` with ρ` = 0. Suppose all ρ` are positive.

Equating the corresponding entries at the both sides of (2.7) leads to n linear equations in the entries of ∆B`, ` = 0, 1, . . . ,bk/2c. The last n − 2 equations are homogenous and

contain only those entries

(∆B`)(i,j)= 0 for i, j > 2 and for ` = 0, 1, . . . ,bk/2c

and at the same time they do not appear in the ﬁrst 2 equations. Therefore the optimal solution{∆B`, ` = 0, 1, . . . ,bk/2c} in the sense of (2.8) for the Frobenius norm must satisfy

(2.11).

For any 2-by-2 block matrix

Z = µ Z11 Z12 Z21 Z22 ¶ ,

we havekZk2 ≥ kZ11k2. Therefore if {∆B`, ` = 0, 1, . . . ,bk/2c} is an optimal solution to

(2.7) for the spectral norm, resetting (∆B`)(i,j)= 0 for i, j > 2 and for ` = 0, 1, . . . ,bk/2c

and leaving (∆B`)(i,j) for 1 ≤ i, j ≤ 2 and for ` = 0, 1, . . . , bk/2c untouched yield a

solution to (2.7) and at the same time do not make Pbk/2c_`=0 k∆B`k22 bigger. Since before

resetting,{∆B`, ` = 0, 1, . . . ,bk/2c} is optimal,

P_bk/2c

`=0 k∆B`k22 will not change before and

after the resetting. Thus the resulted {∆B`, ` = 0, 1, . . . ,bk/2c} is an optimal solution

satisfying (2.11) for the spectral norm.

Let{∆B`, ` = 0, 1, . . . ,bk/2c} be the optimal solution in the Frobenius norm. Then

∆2≤ v u u tbk/2cX `=0 k∆B`k22≤ v u u tbk/2cX `=0 k∆B`k2_F= ∆F.

On the other hand, let{∆B`, ` = 0, 1, . . . ,bk/2c} be an optimal solution satisfying (2.11)

in the spectral norm. Then rank(∆B`)≤ 2 for ` = 0, 1, . . . , bk/2c. Thus

∆F≤ v u u tbk/2cX `=0 k∆B`k2_F ≤ v u u tbk/2cX `=0 2k∆B`k22 = √ 2 ∆2, as expected.

Theorem 2.1 tells us that it suffices to consider the first two equations from the first 2 entries at the both sides of (2.7). In the two equations, only the entries (∆B`)(i,j) for

1≤ i, j ≤ 2 and for ` = 0, 1, . . . , bk/2c appear. We shall do so in the next four sections for each diﬀerent combination of the superscript X∈ {T,H} and ε ∈ {±1}. For this purpose,

let (∆B`)(1:2,1:2)= Ã b(`)₁₁ b(`)₁₂ b(`)₂₁ b(`)₂₂ ! . (2.13)

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Constraints among b(`)_ij because of ∆Bk−`= ε∆B`X for ` = 0, 1, . . . ,bk/2c will be imposed

at the time we consider each case.

Recall in (2.7), ρk−` = ρ` and ∆Bk−`= ε∆BX` for ` = 0, 1, . . . ,bk/2c. For even k, we

have k X `=0 ρ`∆B`µ` = k/2_X−1 `=0 ρ`∆B`µ`+ ρk/2∆Bk/2µk/2+ k X `=k/2+1 ρ`∆B`µ` = k/2_X−1 `=0 ρ`∆B`µ`+ ρk/2∆Bk/2µk/2+ 0 X j=k/2−1 ρk−j∆Bk−jµk−j = k/2_X−1 `=0 ρ`∆B`µ`+ ρk/2∆Bk/2µk/2+ k/2_X−1 j=0 ρjε∆BjXµk−j = k/2_X−1 `=0 ρ` ³ ∆B`+ ε∆BX` µk−2` ´ µ`+ ρk/2∆Bk/2µk/2 (2.14) and ∆B_k/2= ε∆BX k/2. For odd k, k X `=0 ρ`∆B`µ` = (k−1)/2_X `=0 ρ`∆B`µ`+ k X `=(k−1)/2+1 ρ`∆B`µ` = (k_X−1)/2 `=0 ρ`∆B`µ`+ 0 X j=(k−1)/2 ρk−j∆Bk−jµk−j = (k_X−1)/2 `=0 ρ`∆B`µ`+ (k_X−1)/2 j=0 ρjε∆BjXµk−j = (k_X−1)/2 `=0 ρ` ³ ∆B`+ ε∆B`Xµk−2` ´ µ`. (2.15)

Our detailed analysis in the next four sections is for the Frobenius norm only. Results for the spectral norm can then be deduced straightforwardly by using (2.12). So in the next four section,k · k in (2.8) is always the Frobenius norm. Recall also

|δ1| + |δ2| > 0.

For the ease of analysis, we assume

all ρ` > 0.

This will remove many special cases caused by one of more ρ` possibly being zeros. This

assumption seems to limit the applicability of our analysis to cases where no backward errors are preferred to some A` which is usually realized by setting the associated ρ` to

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zero. But we note that such cases can be practically dealt with quite satisfactorily by giving an extremely tiny positive numbers to the associated ρ`.

The following lemma will be needed in the later sections.

Lemma 2.1 For τ =|τ|eιφ∈ C, φ ∈ R, and η ∈ C, we have µ <(τ η) =(τ η) ¶ =|τ| µ cos φ − sin φ sin φ cos φ ¶ µ <(η) =(η) ¶ , µ <(τ ¯η) =(τ ¯η) ¶ =|τ| µ cos φ sin φ sin φ − cos φ ¶ µ <(η) =(η) ¶ . For ξ∈ R and ψ ∈ R, we have

· I2+ ξ µ cos ψ sin ψ sin ψ − cos ψ ¶¸2 = (1 + ξ2)I2+ 2ξ µ cos ψ sin ψ sin ψ − cos ψ ¶ , µ cos ψ sin ψ ¶ µ cos ψ sin ψ ¶T = 1 2 · I + µ cos 2ψ sin 2ψ sin 2ψ − cos 2ψ ¶¸ , µ − sin ψ cos ψ ¶ µ − sin ψ cos ψ ¶T = 1 2 · I− µ cos 2ψ sin 2ψ sin 2ψ − cos 2ψ ¶¸ .

3

X

=

H

and ε = 1

Recall (2.7), (2.14), and (2.15). For even k, we need to solve  k/2X−1 `=0 ρ` ÃÃ b(`)₁₁ b(`)₁₂ b(`)₂₁ b(`)₂₂ ! + Ã_¯ b(`)₁₁ ¯b(`)₂₁ ¯_b(`) 12 ¯b (`) 22 ! µk−2` ! µ`+ ρk/2 Ã b(k/2)₁₁ b(k/2)₁₂ ¯_b(k/2) 12 b (k/2) 22 ! µk/2  µα 0 ¶ = µ γ β ¶ ,

where b(k/2)₁₁ and b(k/2)₂₂ are real. Componentwise, it gives

k/2_X−1 `=0 ρ` ³ b(`)₁₁ + ¯b(`)₁₁µk−2` ´ µ`+ ρk/2b (k/2) 11 µ k/2 ₌ _δ 1, (3.1) k/2_X−1 `=0 ρ` ³ b(`)₂₁ + ¯b(`)₁₂µk−2` ´ µ`+ ρk/2¯b(k/2)12 µ k/2 ₌ _δ 2. (3.2)

For odd k, we need to solve  (kX−1)/2 `=0 ρ` ÃÃ b(`)₁₁ b(`)₁₂ b(`)₂₁ b(`)₂₂ ! + Ã_¯ b(`)₁₁ ¯b(`)₂₁ ¯ b(`)₁₂ ¯b(`)₂₂ ! µk−2` ! µ`  µα 0 ¶ = µ γ β ¶ . Componentwise, it gives (k_X−1)/2 `=0 ρ` ³ b(`)₁₁ + ¯b(`)₁₁µk−2` ´ µ` = δ1, (3.3) (k−1)/2_X `=0 ρ` ³ b(`)₂₁ + ¯b(`)₁₂µk−2` ´ µ` = δ2. (3.4)

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We observe that:

1. b(`)₂₂ for all ` do not show up. Thus must b(`)₂₂ = 0 for all ` by (2.8).

2. (3.1) and (3.2) are decoupled. Thus they can be solved separately for optimal solutions in the sense of k/2 X `=0 |b(`) 11|2 = min, 2|b (k/2) 12 |2+ k/2_X−1 `=0 ³ |b(`) 12|2+|b (`) 21|2 ´ = min, respectively, in view of (2.8).

3. (3.3) and (3.4) are decoupled. Thus they can also be solved separately for optimal solutions in the sense of

(k_X−1)/2 `=0 |b(`) 11|2 = min, (k_X−1)/2 `=0 ³ |b(`) 12| 2₊_|b(`) 21| 2´_{= min,} respectively, in view of (2.8).

4. Equation (3.1) is for even k and needs to be split into two equations according to the real and imaginary parts because b(k/2)₁₁ is real and both b(`)₁₁ and ¯b(`)₁₁ for ` = 0, 1, . . . , k/2−1 appear. For this purpose, write

µ =|µ|eιθ, c` = cos(`θ), s` = sin(`θ).

By Lemma 2.1, we have  < ³h b(`)₁₁ + ¯b(`)₁₁µk−2` i µ` ´ =³hb(`)₁₁ + ¯b(`)₁₁µk−2` i µ` ´   = |µ|` µ c` −s` s` c` ¶ Ã<(b(`) 11 + ¯b (`) 11µk−2`) =(b(`) 11 + ¯b (`) 11µk−2`) ! = |µ|` µ c` −s` s` c` ¶ · I2+|µ|k−2` µ ck−2` sk−2` sk−2` −ck−2` ¶¸ Ã <(b(`) 11) =(b(`) 11) ! . (3.5)

Therefore equivalently to (3.1), we have

(Z0 · · · Zk/2−1 Zk/2)            <(b(0) 11) =(b(0) 11) .. . <(b(k/2−1) 11 ) =(b(k/2−1) 11 ) b(k/2)₁₁            = µ <(δ1) =(δ1) ¶ , (3.6) where for ` = 0, 1, . . . , k/2− 1 Z` = ρ`|µ|` µ c` −s` s` c` ¶ · I2+|µ|k−2` µ ck−2` sk−2` sk−2` −ck−2` ¶¸ , (3.7)

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and Z_k/2= ρ_k/2|µ|k/2 µ ck/2 s_k/2 ¶ . (3.8)

Set Z = (Z0 · · · Zk/2−1 Zk/2)∈ R2×(k+1). It can be computed, again by Lemma 2.1, that

Z`Z`T = ρ2`|µ|2` µ c` −s` s` c` ¶ · I2+|µ|k−2` µ ck−2` sk−2` sk−2` −ck−2` ¶¸2µ c` −s` s` c` ¶T = ρ2_` ·³ |µ|2`₊_|µ|2(k−`)´_I 2+ 2|µ|k µ ck sk sk −ck ¶¸ for 0≤ ` ≤ k/2 − 1, Zk/2Zk/2T = ρ 2 k/2|µ| k µ c_k/2 sk/2 ¶ µ c_k/2 sk/2 ¶T = ρ 2 k/2|µ| k 2 · I + µ ck sk sk −ck ¶¸ , Therefore ZZT = k/2 X `=0 Z`Z`T = k/2_X−1 `=0 ρ2_` ·³ |µ|2`₊_|µ|2(k−`)´_I 2+ 2|µ|k µ ck sk sk −ck ¶¸ +ρ 2 k/2|µ|k 2 · I + µ ck sk sk −ck ¶¸ =  ρ2k/2|µ| k 2 + k/2_X−1 `=0 ρ2_` ³ |µ|2`₊_|µ|2(k−`)´   I2+|µ|k  ρ2k/2 2 + 2 k/2_X−1 `=0 ρ2_`  µck sk sk −ck ¶ . The eigenvalues of µ ck sk sk −ck ¶

are±1. So the eigenvalues of ZZT are

k/2_X−1 `=0 ρ2_` ³ |µ|`₊_|µ|k−`´2_{+ ρ}2 k/2|µ|k, Φeven def = k/2−1_X `=0 ρ2_` ³ |µ|`_{− |µ|}k−`´2_. _(3.9)

Thus if |µ| 6= 1, then the smaller eigenvalue Φeven > 0 and thus ZZT is invertible. The

optimal solution in the sense of Pk/2_`=0|b(`)₁₁|2= min is            <(b(0) 11) =(b(0) 11) .. . <(b(k/2−1) 11 ) =(b(k/2−1) 11 ) b(k/2)₁₁            = Z† µ <(δ1) =(δ1) ¶ = ZT(ZZT)−1 µ <(δ1) =(δ1) ¶ , (3.10)

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and k/2 X `=0 |b(`) 11|2≤ |δ1|2 Φeven . (3.11)

This inequality becomes an equality when (<(δ1) =(δ1)) T

is parallel to the eigenvector of µ

ck sk

sk −ck

¶

associated with its eigenvalue−1.

If, however,|µ| = 1, then rank(Z) = 1 and (3.6) is not be solvable unless (<(δ1) =(δ1)) T is parallel to (ck/2 sk/2) T , or equivalently δ1 µk/2 = zHr µk/2kzk2 2 ∈ R. (3.12)

When it does, the optimal solution can be gotten from either equation from the ﬁrst or the second component in (3.6). We shall do so now. When |µ| = 1, we have for ` = 0, 1, . . . , k/2− 1 Z` = ρ` µ c` −s` s` c` ¶ · I2+ µ ck−2` sk−2` s_k−2` −c_k−2` ¶¸ = ρ` ·µ c` −s` s` c` ¶ + µ ck−` sk−` sk−` −ck−` ¶¸ = 2ρ` µ ck/2ck/2−` ck/2sk/2−` s_k/2c_k/2_−` s_k/2s_k/2_−` ¶ , (3.13) and Z_k/2= ρ_k/2 µ ck/2 s_k/2 ¶ .

Thus the optimal solution for when|µ| = 1 and (3.12) holds is            <(b(0) 11) =(b(0) 11) .. . <(b(k/2−1) 11 ) =(b(k/2−1) 11 ) b(k/2)₁₁            = 1 ρ2 k/2+ 4 Pk/2−1 `=0 ρ2`         .. . 2ρ`ck/2−` 2ρ`sk/2−` .. . ρk/2         δ1 µk/2 (3.14) and k/2 X `=0 |b(`) 11|2= |δ1|2 Φ0_even, Φ 0 even def = ρ2_k/2+ 4 k/2_X−1 `=0 ρ2_`. (3.15)

5. Equation (3.2) is for even k. Set

Ψeven def = k/2_X−1 `=0 ρ2_` ³ |µ|2`₊_|µ|2(k−`)´_{+ ρ}2 k/2|µ|k/2. (3.16)

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Ψeven> 0 always. The optimal solution to (3.2) is √ 2 b(k/2)₁₂ = µ k/2_ρ k/2δ¯2/ √ 2 Ψeven , (3.17) b(`)₁₂ = µ k−`_ρ `δ¯2 Ψeven , b(`)₂₁ = µ¯ `_ρ `δ2 Ψeven for ` = 0, 1, . . . , k/2− 1 (3.18) satisfying 2|b(k/2)₁₂ |2+ k/2_X−1 `=0 ³ |b(`) 12| 2₊_|b(`) 21| 2´₌ |δ2|2 Ψeven . (3.19)

6. Equation (3.3) is for odd k. Similarly to Item 4 above, (3.3) is equivalently to

(Z0 · · · Z(k−1)/2)         <(b(0) 11) =(b(0) 11) .. . <(b((k−1)/2) 11 ) =(b((k−1)/2) 11 )         = µ <(δ1) =(δ1) ¶ , (3.20)

where Z` are as in (3.7). Let Z = (Z0 · · · Z(k−1)/2). Then

ZZT= (k_X−1)/2 `=0 ρ2_` ·³ |µ|2`₊_|µ|2(k−`)´_I 2+ 2|µ|k µ ck sk sk −ck ¶¸

whose eigenvalues are

(k_X−1)/2 `=0 ρ2_` ³ |µ|`₊_|µ|k−`´2_, _Φ odd def = (k−1)/2_X `=0 ρ2_` ³ |µ|`_{− |µ|}k−`´2_. _(3.21)

Thus if |µ| 6= 1, then the smaller eigenvalue Φodd > 0 and thus ZZT is invertible. The

optimal solution in the sense of P(k_`=0−1)/2|b(`)₁₁|2 = min is         <(b(0) 11) =(b(0) 11) .. . <(b((k−1)/2) 11 ) =(b((k−1)/2) 11 )         = Z† µ <(δ1) =(δ1) ¶ = ZT(ZZT)−1 µ <(δ1) =(δ1) ¶ (3.22) and (k_X−1)/2 `=0 |b(`) 11|2≤ |δ1|2 Φodd . (3.23)

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ck sk

sk −ck

¶

If, however,|µ| = 1, then rank(Z) = 1 and (3.20) is not be solvable unless (<(δ1) =(δ1)) T

is parallel to (ck/2 sk/2) T

, or equivalently (3.12) holds. When it does, the optimal solution can be gotten, as in Item 4, as

        <(b(0) 11) =(b(0) 11) .. . <(b((k−1)/2) 11 ) =(b((k−1)/2) 11 )         = 1 4P(k_`=0−1)/2ρ2_`       .. . 2ρ`ck/2−` 2ρ`sk/2−` .. .       δ1 µk/2 (3.24) and (k−1)/2_X `=0 |b(`) 11| 2 ₌ |δ1|2 Φ0_odd, Φ 0 odd def = 4 (k_X−1)/2 `=0 ρ2_`. (3.25)

7. Equation (3.4) is for odd k. Set

Ψodd= (k−1)/2_X `=0 ρ2_` ³ |µ|2`₊_|µ|2(k−`)´_. _(3.26)

Ψodd> 0 always. The optimal solution to (3.4) is

b(`)₁₂ = µ k−`_ρ `δ¯2 Ψodd , b(`)₂₁ = µ¯ `_ρ `δ2 Ψodd for ` = 0, 1, . . . , (k− 1)/2 (3.27) satisfying (k_X−1)/2 `=0 ³ |b(`) 12| 2₊_|b(`) 21| 2´₌ |δ2|2 Ψodd . (3.28)

We summarize our ﬁndings into the following theorem.

Theorem 3.1 Suppose X=Hand ε = 1 in (2.7). Let

Φ = ½

Φeven for even k,

Φodd, for odd k,

Φ0 = ½

Φ0_even for even k, Φ0_odd, for odd k, Ψ =

½

Ψeven for even k,

Ψodd, for odd k.

We have 1. If |µ| = 1 and zH_r/µk/2_{6∈ R, then ∆} F = +∞. 2. If |µ| = 1 and zHr/µk/2∈ R, then ∆F = r |δ1|2 Φ0 + |δ2|2 Ψ .

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3. If |µ| 6= 1, then ∆F ≤ r |δ1|2 Φ + |δ2|2 Ψ . (3.29)

Remark 3.1 One may work out the right-hand sides of (3.10) and (3.22) to write down

∆F exactly in the case when |µ| 6= 1. But it is rather complicated to do so. Inequality

(3.29) in general cannot be improved because it becomes an equality when (<(δ1) =(δ1)) T

ck sk

sk −ck

¶

4

X

=

H

and ε =

−1

For even k, we need to solve  k/2−1X `=0 ρ` ÃÃ b(`)₁₁ b(`)₁₂ b(`)₂₁ b(`)₂₂ ! − Ã_¯ b(`)₁₁ ¯b(`)₂₁ ¯_b(`) 12 ¯b (`) 22 ! µk−2` ! µ`+ ρ_k/2 Ã b(k/2)₁₁ b(k/2)₁₂ −¯b(k/2) 12 b (k/2) 22 ! µk/2  µα 0 ¶ = µ γ β ¶ ,

where b(k/2)₁₁ and b(k/2)₂₂ are imaginary. Componentwise, it gives

k/2_X−1 `=0 ρ` ³ b(`)₁₁ − ¯b(`)₁₁µk−2` ´ µ`+ ρk/2b (k/2) 11 µk/2 = δ1, (4.1) k/2_X−1 `=0 ρ` ³ b(`)₂₁ − ¯b(`)₁₂µk−2` ´ µ`− ρk/2¯b (k/2) 12 µ k/2 ₌ _δ 2. (4.2)

For odd k, we need to solve  (kX−1)/2 `=0 ρ` ÃÃ b(`)₁₁ b(`)₁₂ b(`)₂₁ b(`)₂₂ ! − Ã_¯ b(`)₁₁ ¯b(`)₂₁ ¯ b(`)₁₂ ¯b(`)₂₂ ! µk−2` ! µ`  µα 0 ¶ = µ γ β ¶ . Componentwise, it gives (k_X−1)/2 `=0 ρ` ³ b(`)₁₁ − ¯b(`)₁₁µk−2` ´ µ` = δ1, (4.3) (k−1)/2_X `=0 ρ` ³ b(`)₂₁ − ¯b(`)₁₂µk−2` ´ µ` = δ2. (4.4) We observe that:

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2. (4.1) and (4.2) are decoupled. Thus they can be solved separately for optimal solutions in the sense that

k/2 X `=0 |b(`) 11|2 = min, 2|b (k/2) 12 |2+ k/2−1_X `=0 ³ |b(`) 12|2+|b (`) 21|2 ´ = min, respectively, in view of (2.8).

3. (4.3) and (4.4) are decoupled. Thus they can also be solved separately for optimal solutions in the sense that

(k_X−1)/2 `=0 |b(`) 11| 2 _{= min,} (k_X−1)/2 `=0 ³ |b(`) 12| 2₊_|b(`) 21| 2´_{= min,} respectively, in view of (2.8).

4. Equation (4.1) is for even k and needs to be split into two equations according to the real and imaginary parts because b(k/2)₁₁ is real and both b(`)₁₁ and ¯b(`)₁₁ for ` = 0, 1, . . . , k/2−1 appear. For this purpose, write

µ =|µ|eιθ, c` = cos(`θ), s` = sin(`θ).

By Lemma 2.1, we have  < ³h b(`)₁₁ − ¯b(`)₁₁µk−2` i µ` ´ =³hb(`)₁₁ − ¯b(`)₁₁µk−2` i µ` ´   = |µ|` µ c` −s` s` c` ¶ Ã<(b(`) 11 − ¯b (`) 11µk−2`) =(b(`) 11 − ¯b (`) 11µk−2`) ! = |µ|` µ c` −s` s` c` ¶ · I2− |µ|k−2` µ ck−2` sk−2` sk−2` −ck−2` ¶¸ Ã <(b(`) 11) =(b(`) 11) ! . (4.5)

Therefore equivalently to (4.1), we have

(Z0 · · · Zk/2−1 Zk/2)            <(b(0) 11) =(b(0) 11) .. . <(b(k/2−1) 11 ) =(b(k/2−1) 11 ) =(b(k/2) 11 )            = µ <(δ1) =(δ1) ¶ , (4.6) where for ` = 0, 1, . . . , k/2− 1 Z` = ρ`|µ|` µ c` −s` s` c` ¶ · I2− |µ|k−2` µ ck−2` sk−2` sk−2` −ck−2` ¶¸ , (4.7) and Z_k/2= ρ_k/2|µ|k/2 µ −sk/2 ck/2 ¶ . (4.8)

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Set Z = (Z0 · · · Zk/2−1 Zk/2)∈ R2×(k+1). It can be computed, again by Lemma 2.1, that Z`Z`T = ρ2`|µ|2` µ c` −s` s` c` ¶ · I2− |µ|k−2` µ ck−2` sk−2` sk−2` −ck−2` ¶¸2µ c` −s` s` c` ¶T = ρ2_` ·³ |µ|2`₊_|µ|2(k−`)´_I 2− 2|µ|k µ ck sk sk −ck ¶¸ for 0≤ ` ≤ k/2 − 1, Zk/2Zk/2T = ρ 2 k/2|µ| k µ −sk/2 c_k/2 ¶ µ −sk/2 c_k/2 ¶T = ρ 2 k/2|µ|k 2 · I− µ ck sk sk −ck ¶¸ , Therefore ZZT = k/2 X `=0 Z`Z`T = k/2_X−1 `=0 ρ2_` ·³ |µ|2`₊_|µ|2(k−`)´_I 2− 2|µ|k µ ck sk sk −ck ¶¸ +ρ 2 k/2|µ| k 2 · I− µ ck sk sk −ck ¶¸ =  ρ2k/2|µ|k 2 + k/2_X−1 `=0 ρ2_` ³ |µ|2`₊_|µ|2(k−`)´   I2− |µ|k  ρ2k/2 2 + 2 k/2_X−1 `=0 ρ2_`  µck sk sk −ck ¶ . The eigenvalues of µ ck sk sk −ck ¶

are±1. So the eigenvalues of ZZT are

k/2_X−1 `=0 ρ2_` ³ |µ|`₊_|µ|k−`´2_{+ ρ}2 k/2|µ|k, Φeven def = k/2_X−1 `=0 ρ2_` ³ |µ|`_{− |µ|}k−`´2_. _(4.9)

Thus if |µ| 6= 1, then the smaller eigenvalue Φeven > 0 and thus ZZT is invertible. The

optimal solution in the sense of Pk/2_`=0|b(`)₁₁|2= min is            <(b(0) 11) =(b(0) 11) .. . <(b(k/2−1) 11 ) =(b(k/2−1) 11 ) =(b(k/2) 11 )            = Z† µ <(δ1) =(δ1) ¶ = ZT(ZZT)−1 µ <(δ1) =(δ1) ¶ , (4.10) and k/2 X `=0 |b(`) 11|2≤ |δ1|2 Φeven . (4.11)

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ck sk

sk −ck

¶

associated with its eigenvalue 1.

If, however,|µ| = 1, then rank(Z) = 1 and (4.6) is not be solvable unless (<(δ1) =(δ1)) T is parallel to (−sk/2 ck/2) T , or equivalently δ1 ιµk/2 = zH_r ιµk/2kzk2 2 ∈ R. (4.12)

When it does, the optimal solution can be gotten from either equation from the ﬁrst or the second component in (4.6). We shall do so now. When |µ| = 1, we have for ` = 0, 1, . . . , k/2− 1 Z` = ρ` µ c` −s` s` c` ¶ · I2− µ ck−2` sk−2` sk−2` −ck−2` ¶¸ = ρ` ·µ c` −s` s` c` ¶ − µ ck−` sk−` sk−` −ck−` ¶¸ = 2ρ` µ s_k/2s_k/2_−` −s_k/2c_k/2_−` −ck/2sk/2−` ck/2ck/2−` ¶ , (4.13) and Zk/2= ρk/2 µ −sk/2 ck/2 ¶ . Thus the optimal solution for when|µ| = 1 and (4.12) holds is

           <(b(0) 11) =(b(0) 11) .. . <(b(k/2−1) 11 ) =(b(k/2−1) 11 ) =(b(k/2) 11 )            = 1 ρ2_k/2+ 4Pk/2_`=0−1ρ2_`         .. . −2ρ`sk/2−` 2ρ`ck/2−` .. . ρ_k/2         δ1 ιµk/2 (4.14) and k/2 X `=0 |b(`) 11|2= |δ1|2 Φ0_even, Φ 0 even def = ρ2_k/2+ 4 k/2−1_X `=0 ρ2_`. (4.15)

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Ψeven> 0 always. The optimal solution to (4.2) is √ 2 b(k/2)₁₂ =−µ k/2_ρ k/2¯δ2/ √ 2 Ψeven , (4.17) b(`)₁₂ =−µ k−`_ρ `δ¯2 Ψeven , b(`)₂₁ = µ¯ `_ρ `δ2 Ψeven for ` = 0, 1, . . . , k/2− 1 (4.18) satisfying 2|b(k/2)₁₂ |2+ k/2_X−1 `=0 ³ |b(`) 12| 2₊_|b(`) 21| 2´₌ |δ2|2 Ψeven . (4.19)

6. Equation (4.3) is for odd k. Similarly to Item 4 above, (4.3) is eqivalently to

(Z0 · · · Z(k−1)/2)         <(b(0) 11) =(b(0) 11) .. . <(b((k−1)/2) 11 ) =(b((k−1)/2) 11 )         = µ <(δ1) =(δ1) ¶ , (4.20)

where Z` are as in (4.7). Let Z = (Z0 · · · Z(k−1)/2). Then

ZZT= (k_X−1)/2 `=0 ρ2_` ·³ |µ|2`₊_|µ|2(k−`)´_I 2− 2|µ|k µ ck sk sk −ck ¶¸

whose eigenvalues are

(k_X−1)/2 `=0 ρ2_` ³ |µ|`₊_|µ|k−`´2_, _Φ odd def = (k−1)/2_X `=0 ρ2_` ³ |µ|`_{− |µ|}k−`´2_. _(4.21)

Thus if |µ| 6= 1, then the smaller eigenvalue Φodd > 0 and thus ZZT is invertible. The

optimal solution in the sense of P(k_`=0−1)/2|b(`)₁₁|2 = min is         <(b(0) 11) =(b(0) 11) .. . <(b((k−1)/2) 11 ) =(b((k−1)/2) 11 )         = Z† µ <(δ1) =(δ1) ¶ = ZT(ZZT)−1 µ <(δ1) =(δ1) ¶ (4.22) and (k_X−1)/2 `=0 |b(`) 11|2≤ |δ1|2 Φodd . (4.23)

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ck sk

sk −ck

¶

If, however,|µ| = 1, then rank(Z) = 1 and (4.20) is not be solvable unless (<(δ1) =(δ1)) T

is parallel to (ck/2 sk/2) T

, or equivalently (4.12) holds. When it does, the optimal solution can be gotten, as in Item 4, as

        <(b(0) 11) =(b(0) 11) .. . <(b((k−1)/2) 11 ) =(b((k−1)/2) 11 )         = 1 4P(k_`=0−1)/2ρ2_`       .. . −2ρ`sk/2−` 2ρ`ck/2−` .. .       δ1 ιµk/2 (4.24) and (k−1)/2_X `=0 |b(`) 11| 2 ₌ |δ1|2 Φ0_odd, Φ 0 odd def = 4 (k_X−1)/2 `=0 ρ2_`. (4.25)

Ψodd= (k−1)/2_X `=0 ρ2_` ³ |µ|2`₊_|µ|2(k−`)´_. _(4.26)

The optimal solution to (4.4) is b(`)₁₂ =−µ k−`_ρ `δ¯2 Ψodd , b(`)₂₁ = µ¯ `_ρ `δ2 Ψodd for ` = 0, 1, . . . , (k− 1)/2 (4.27) satisfying (k_X−1)/2 `=0 ³ |b(`) 12| 2₊_|b(`) 21| 2´₌ |δ2|2 Ψodd . (4.28)

Theorem 4.1 Suppose X=Hand ε =−1 in (2.7). Let

Φ = ½

Φodd, for odd k,

Φ0= ½

Φ0_even for even k, Φ0_odd, for odd k, Ψ =

½

Ψodd, for odd k.

We have 1. If |µ| = 1 and zH_r/(ιµk/2₎_{6∈ R, then ∆} F= +∞. 2. If |µ| = 1 and zHr/(ιµk/2)∈ R, then ∆F = r |δ1|2 Φ0 + |δ2|2 Ψ .

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3. If |µ| 6= 1, then ∆F ≤ r |δ1|2 Φ + |δ2|2 Ψ . (4.29)

Remark 4.1 One may work out the right-hand sides of (4.10) and (4.22) to write down

∆F exactly in the case when |µ| 6= 1. But it is rather complicated to do so. Inequality

(4.29) in general cannot be improved because it becomes an equality when (<(δ1) =(δ1)) T

ck sk

sk −ck

¶

5

X

=

T

and ε = 1

For even k, we need to solve   k/2_X−1 `=0 ρ` ÃÃ b(`)₁₁ b(`)₁₂ b(`)₂₁ b(`)₂₂ ! + Ã b(`)₁₁ b(`)₂₁ b(`)₁₂ b(`)₂₂ ! µk−2` ! µ`+ ρ_k/2 Ã b(k/2)₁₁ b(k/2)₁₂ b(k/2)₁₂ b(k/2)₂₂ ! µk/2  µα 0 ¶ = µ ¯ γ ¯ β ¶ . Componentwise, it gives k/2_X−1 `=0 ρ` ³ b(`)₁₁ + b(`)₁₁µk−2` ´ µ`+ ρ_k/2b(k/2)₁₁ µk/2 = δ1, (5.1) k/2_X−1 `=0 ρ` ³ b(`)₂₁ + b(`)₁₂µk−2` ´ µ`+ ρ_k/2b(k/2)₁₂ µk/2 = δ2. (5.2)

For odd k, we need to solve  (kX−1)/2 `=0 ρ` ÃÃ b(`)₁₁ b(`)₁₂ b(`)₂₁ b(`)₂₂ ! + Ã b(`)₁₁ b(`)₂₁ b(`)₁₂ b(`)₂₂ ! µk−2` ! µ`  µα 0 ¶ = µ ¯ γ ¯ β ¶ . Componentwise, it gives (k_X−1)/2 `=0 ρ` ³ b(`)₁₁ + b(`)₁₁µk−2` ´ µ` = δ1, (5.3) (k−1)/2_X `=0 ρ` ³ b(`)₂₁ + b(`)₁₂µk−2` ´ µ` = δ2. (5.4) We observe that:

k/2 X `=0 |b(`) 11|2 = min, 2|b (k/2) 12 |2+ k/2_X−1 `=0 ³ |b(`) 12|2+|b (`) 21|2 ´ = min,

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respectively, in view of (2.8).

(k_X−1)/2 `=0 |b(`) 11|2 = min, (k_X−1)/2 `=0 ³ |b(`) 12|2+|b (`) 21|2 ´ = min, respectively, in view of (2.8). 4. Equation (5.1) is for even k. Set

Φeven def = k/2_X−1 `=0 ρ2_`¯¯¯µ`+ µk−`¯¯¯2+ ρ2_k/2|µ|k/2. (5.5)

Φeven> 0 always. The optimal solution to (5.1) is

b(k/2)₁₁ = ρk/2µ¯ k/2_δ 1 Φeven , (5.6) b(`)₁₁ = ρ` ¡ ¯ µ`+ ¯µk−`¢δ1 Φeven for ` = 0, 1, . . . , k/2− 1. (5.7)

Ψeven def = k/2_X−1 `=0 ρ2_` ³ |µ|2`₊_|µ|2(k−`)´_{+ ρ}2 k/2|µ| k_/2. _(5.8)

The optimal solution to (5.2) is √ 2 b(k/2)₁₂ = ρk/2µ¯ k/2_δ 2/ √ 2 Ψeven , (5.9) b(`)₁₂ = ρ`µ¯ k−`_δ 2 Ψeven , b(`)₂₁ = ρ`µ¯ `_δ 2 Ψeven for ` = 0, 1, . . . , k/2− 1 (5.10) satisfying 2|b(k/2)₁₂ |2+ k/2_X−1 `=0 ³ |b(`) 12| 2₊_|b(`) 21| 2´₌ |δ2|2 Ψeven . (5.11)

Φodd def = (k_X−1)/2 `=0 ρ2_`¯¯¯µ`+ µk−`¯¯¯ 2 . (5.12)

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Since k is odd for the case, Φodd= 0 if µ =−1, and Φodd > 0 if µ6= −1. So when µ 6= −1,

the optimal solution to (5.3) is b(`)₁₁ = ρ` ¡ ¯ µ`+ ¯µk−`¢δ1 Φeven for ` = 0, 1, . . . , (k− 1)/2. (5.13)

If, however, µ =−1, then (5.3) is consistent if and only δ1 = 0 for which case the optimal

solution is b(`)₁₁ = 0 for all `.

Ψodd= (k_X−1)/2 `=0 ρ2_` ³ |µ|2`₊_|µ|2(k−`)´_. _(5.14)

b(`)₁₂ = ρ`µ¯ k−`_δ 2 Ψodd , b(`)₂₁ = ρ`µ¯ `_δ 2 Ψodd for ` = 0, 1, . . . , (k− 1)/2 (5.15) satisfying (k_X−1)/2 `=0 ³ |b(`) 12|2+|b (`) 21|2 ´ = |δ2| 2 Ψodd . (5.16)

Theorem 5.1 Suppose X=T and ε = 1 in (2.7). Let

Φ = ½

Φodd, for odd k,

Ψ = ½

Ψodd, for odd k.

We have

1. If µ =−1, k is odd, and zHr¯6= 0, then ∆F = +∞.

2. If µ =−1, k is odd, and zHr = 0, then ∆¯ F =|δ2|/

√ Ψodd. 3. If µ6= −1 or k is even, then ∆F = r |δ1|2 Φ + |δ2|2 Ψ .

6

X

=

T

and ε =

−1

For even k, we need to solve  k/2X−1 `=0 ρ` ÃÃ b(`)₁₁ b(`)₁₂ b(`)₂₁ b(`)₂₂ ! − Ã b(`)₁₁ b(`)₂₁ b(`)₁₂ b(`)₂₂ ! µk−2` ! µ`+ ρk/2 Ã 0 b(k/2)₁₂ −b(k/2) 12 0 ! µk/2  µα 0 ¶ = µ ¯ γ ¯ β ¶ .

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Componentwise, it gives k/2_X−1 `=0 ρ` ³ b(`)₁₁ − b(`)₁₁µk−2` ´ µ` = δ1, (6.1) k/2_X−1 `=0 ρ` ³ b(`)₂₁ − b(`)₁₂µk−2` ´ µ`− ρ_k/2b(k/2)₁₂ µk/2 = δ2. (6.2)

For odd k, we need to solve  (kX−1)/2 `=0 ρ` ÃÃ b(`)₁₁ b(`)₁₂ b(`)₂₁ b(`)₂₂ ! − Ã b(`)₁₁ b(`)₂₁ b(`)₁₂ b(`)₂₂ ! µk−2` ! µ`  µα 0 ¶ = µ ¯ γ ¯ β ¶ . Componentwise, it gives (k_X−1)/2 `=0 ρ` ³ b(`)₁₁ − b(`)₁₁µk−2` ´ µ` = δ1, (6.3) (k−1)/2_X `=0 ρ` ³ b(`)₂₁ − b(`)₁₂µk−2` ´ µ` = δ2. (6.4) We observe that:

k/2 X `=0 |b(`) 11|2 = min, 2|b (k/2) 12 |2+ k/2_X−1 `=0 ³ |b(`) 12|2+|b (`) 21|2 ´ = min, respectively, in view of (2.8).

(k_X−1)/2 `=0 |b(`) 11|2 = min, (k_X−1)/2 `=0 ³ |b(`) 12|2+|b (`) 21|2 ´ = min, respectively, in view of (2.8). 4. Equation (6.1) is for even k. Set

Φeven def = k/2_X−1 `=0 ρ2_`¯¯¯µ`− µk−`¯¯¯2. (6.5)

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Since k is even, Φeven= 0 if µ =±1, and Φeven> 0 if µ6= ±1. When µ 6= ±1, the optimal solution to (6.1) is b(`)₁₁ = ρ` ¡ ¯ µ`− ¯µk−`¢δ1 Φeven for ` = 0, 1, . . . , k/2− 1. (6.6)

If, however, µ =±1, then (6.1) is consistent if and only δ1 = 0 for which case the optimal

Ψeven> 0 always. The optimal solution to (6.2) is

√ 2 b(k/2)₁₂ =−ρk/2µ¯ k/2_δ 2/ √ 2 Ψeven , (6.8) b(`)₁₂ =−ρ`µ¯ k−`_δ 2 Ψeven , b(`)₂₁ = ρ`µ¯ `_δ 2 Ψeven for ` = 0, 1, . . . , k/2− 1 (6.9) satisfying 2|b(k/2)₁₂ |2+ k/2_X−1 `=0 ³ |b(`) 12| 2₊_|b(`) 21| 2´₌ |δ2|2 Ψeven . (6.10)

Φodd def = (k_X−1)/2 `=0 ρ2_`¯¯¯µ`− µk−`¯¯¯2. (6.11)

Since k is odd for the case, Φodd= 0 if µ = 1, and Φodd> 0 if µ6= 1. So when µ 6= 1, the

optimal solution to (6.3) is b(`)₁₁ = ρ` ¡ ¯ µ`− ¯µk−`¢δ1 Φeven for ` = 0, 1, . . . , (k− 1)/2. (6.12)

If, however, µ = 1, then (6.3) is consistent if and only δ1 = 0 for which case the optimal

Ψodd= (k−1)/2_X `=0 ρ2_` ³ |µ|2`₊_|µ|2(k−`)´_. _(6.13)

b(`)₁₂ =−ρ`µ¯ k−`_δ 2 Ψodd , b(`)₂₁ = ρ`µ¯ `_δ 2 Ψodd for ` = 0, 1, . . . , (k− 1)/2 (6.14)

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satisfying (k_X−1)/2 `=0 ³ |b(`) 12|2+|b (`) 21|2 ´ = |δ2| 2 Ψodd . (6.15)

Theorem 6.1 Suppose X=T and ε =−1 in (2.7). Let

Φ = ½

Φodd, for odd k,

Ψ = ½

Ψodd, for odd k.

We have

1. For even k and µ =±1 or for odd k and µ = 1, if zHr¯6= 0, then ∆F = +∞.

2. For even k and µ =±1 or for odd k and µ = 1, if zHr = 0, then ∆¯ F =|δ2|/

√ Ψ.

3. For even k if µ6= ±1 or for odd k if µ 6= 1, then ∆F= r |δ1|2 Φ + |δ2|2 Ψ .

References

[1] E. K.-W. Chu, T.-M. Hwang, W.-W. Lin, and C.-T. Wud, Vibration of fast trains, palindromic eigenvalue problems and structure-preserving doubling algorithms, J. Comput. Appl. Math., 219 (2008), pp. 237–252.

[2] I. C. F. Ipsen, Accurate eigenvalues for fast trains, SIAM News, 37 (2004).

[3] D. S. Mackey, N. Mackey, C. Mehl, and V. Mehrmann, Structured polynomial eigenvalue problems: Good vibrations from good linearizations, SIAM J. Matrix Anal. Appl., 28 (2006), pp. 1029–1051.

[4] F. Tisseur, Backward error and condition of polynomial eigenvalue problems, Linear Algebra Appl., 309 (2000), pp. 339–361.

[5] F. Tisseur and K. Meerbergen, The quadratic eigenvalue problem, SIAM Rev., 43 (2001), pp. 235–386.