Download Structured condition number for Palindromic polynomial eigenvalue

Structured condition number for Palindromic
polynomial eigenvalue problems
Yueh-Cheng Kuo∗
Abstract
In this paper, we investigate eigenvalue condition number for ?-Palindromic
polynomial eigenvalue problems (?-PPEP) :
!
d
X
`
λ A` x = 0, Ad−` = A?` , for ` = 0, 1, . . . , bd/2c,
`=0
where ? is either transpose > or conjugate transpose H. It is easy to see that
the eigenvalues of a ?-PPEP contain both eigenvalues λ and 1/λ? . Result
on unstructured condition number for a nonzero simple eigenvalue of polynomial eigenvalue problem (PEP) have been investigated by Tisseur (Linear
Algebra Appl., 309:339-361, 2000). In particular we show that under some
reasonable assumptions on the weights of perturbation, the condition numbers for λ and 1/λ? of ?-PPEP are equal. Particular attention is paid to
the effect of structure-preserving perturbations on the eigenvalue of ?-PPEP.
We give computable expressions of structured condition numbers for nonzero
simple eigenvalues of ?-PPEP and, it is difference between structured and unstructured condition numbers. In addition we also show that the structured
condition numbers of λ and 1/λ? are equal.
1
Introduction
Consider the matrix polynomial has the form
Q(λ) = λ2 A>
0 + λA1 + A0 ,
(1.1)
where A0 , A1 are square real or complex matrices with A>
1 = A1 and the super>
script “ · ” takes matrix transpose. We will assume that Q(λ) is regular, that is,
∗
Department of Applied Mathematics, National University of Kaohsiung, Kaohsiung, 811,
Taiwan ([email protected]). Research supported in part by the National Center for Theoretical
Sciences of Taiwan.
1
detQ(λ) ≡
/ 0. In vibration analysis of rail tracts of fast trains arises the Palindromic
Quadratic Eigenvalue Problem (PQEP)[2, 6]: find a scalar λ ∈ C and a nonzero
vector x such that
Q(λ)x = 0,
(1.2)
where λ and x are an eigenvalue and its associated eigenvector. PQEP is a particular case of the P olynomial Eigenvalue P roblem (PEP) of degree d:
!
d
X
λ` A` x = 0,
(1.3)
`=0
where A` , ` = 0, . . . , d are square real or complex matrices. For a PEP, there is
no relation among coefficient matrices, unlike PQEP. It is natural to define the
P alindromic P olynomial Eigenvalue P roblem (PPEP) to be a PEP (1.3) with
Ad−` = A>
` , ` = 0, 1, . . . , bd/2c, that is
!
d
X
λ` A` x = 0, Ad−` = A>
(1.4)
` , for ` = 0, 1, . . . , bd/2c,
`=0
where bd/2c is the largest integer that is less than or equal to d/2.
A standard way of solving the PPEP is to convert it into a linear polynomial.
Recently, some pioneering work [9, 10] proposed a good linearization of PPEP
which linearizes (1.4) into a palindromic linear pair λZ > + Z. Later, PPEP is
solved by certain kinds of linearizarion, such as Jacobi-type method [10, 4], QRlike algorithm [11], Patel’s algorithm [5] and URV like algorithm [12], as well as
the structure-preserving doubling algorithm [2].
Condition numbers and backward errors play an important role in numerical
linear algebra. Condition numbers measure the sensitivity of solution of a problem
to perturbations in the data and backward errors measure the stability of a numerical method. Tisseur [13] developed the condition number and backward error
for PEP (1.3). When we consider PPEP (1.4), naturally one would like to the
perturbation matrices 4A` , ` = 0, 1, . . . , d satisfying
4Ad−` = 4A>
` , for ` = 0, 1, . . . , bd/2c.
This is the structured perturbation analysis. There is a growing interest in structured perturbation analysis due to the development of algorithms for structured
problems. It is natural to define the structured condition number and backward
error for PPEP to be the condition number and backward error with particular
emphasis on respecting structure in the coefficient matrix, respectively. The related works on the structured condition number and backward error of generalized
2
eigenvalue problems have been investigated in [3, 7], and the structured backward
error of PPEP have been considered in [8]. However, none of the results from
these apply to PPEP’s structured condition number analysis. This paper we will
investigate the effect of eigenvalue for structured perturbation on PPEP.
In this paper, we shall consider PPEP in more broader sense. To present these
PPEP compactly, we let the superscript “.? ” be either “.> ” or “.H ” which takes
transpose or complex conjugate transpose. Then we say that a matrix polynomial
P (λ) is ?-P alindromic of degree d if P (λ) has the following form
P (λ) = λd Ad + λd−1 Ad−1 + · · · + A0 , Ad−` = A?` , for ` = 0, 1, . . . , bd/2c, (1.5)
and the corresponding ?-P alindromic P olynomial Eigenvalue P roblem (?-PPEP)
P (λ)x = 0, x 6= 0.
(1.6)
We have mentioned the fast train application which yielded a problem of this form
with d = 2 and ? = >. Take the ? of ?-PPEP (1.6) then we obtain
x? P (1/λ? ) = 0.
Thus the eigenvalues of a ?-PPEP contain both eigenvalues λ and 1/λ? . (Here, we
allow λ = 0 and interpret 1/λ? as the eigenvalue ∞.)
This paper is organized as follows. In Section 2, we consider the condition
number for a nonzero simple eigenvalue of ?-PPEP. We show that the condition
numbers for eigenvalues λ and 1/λ? are equal. The eigenvalue structured condition
number for ?-PPEP are considered in Section 3. We give computable expressions of
structured condition number for nonzero simple eigenvalues and we also show that
the structured condition numbers for eigenvalues λ and 1/λ? are equal. Numerical
results are given in Section 4 and the paper is concluded in Section 5.
Throughout this paper, Cn×m is the set of all n × m complex matrices and the
notations k · k2 and k · kF are the 2-norm and Frobenius norm, respectively. For a
matrix X, X(k:`,i:j) is X’s submatrix consisting of intersections of row k to row `
and column i to column j. In is the n × n identity matrix. For a vector x ∈ Cn ,
x̄ is the complex conjugate of entry-wise complex conjugate of x. For a complex
number z ∈ C, R(z) and I(z) are the real and imaginary part of z, respectively.
|z| is the modulus of z and Arg(z) is the principal
√ value of argument of z. (i.e.,
ιArg(z)
z = |z|e
where −π < Arg(z) ≤ π and ι = −1.)
2
Eigenvalue condition number
Let λ ∈ C be a nonzero simple eigenvalue of a regular matrix polynomial P (λ)
with corresponding right eigenvector x. A normwise condition number of λ can be
3
defined by
κν (λ, P ) = lim sup
→0
| 4 λ|
: (P (λ + 4λ) + 4P (λ + 4λ))(x + 4x) = 0,
|λ|
k 4 A` kν ≤ ω` , ` = 0, 1, . . . , d ,
(2.7)
where ν = 2, F and perturbation matrix polynomial
4P (λ) = λd 4 Ad + λd−1 4 Ad−1 + · · · + 4A0 .
(2.8)
The ω` are nonnegative weights that represent tolerances against which the perturbations are measure.
By expanding the first constraint in (2.7) and multiplying y H from the left then
we obtain
4λy H P 0 (λ)x = −y H 4 P (λ)x + O(2 ).
(2.9)
Since λ is a simple eigenvalue, y H P 0 (λ)x 6= 0 [1]. Thus
4λ = −
y H 4 P (λ)x
+ O(2 ).
y H P 0 (λ)x
(2.10)
An explicit formula for this condition number with ν = 2 is given in [13], the
perturbations 4A` for ` = 0, 1, . . . , d attain the condition number are of rank 1,
we then have k 4 A` k2 = k 4 A` kF for ` = 0, 1, . . . , d. Hence the following result is
a trivial extension of [13].
Theorem 2.1. [13] The normwise condition κν (λ, P ) is given by
P
d
`
`=0 |λ| ω` kxk2 kyk2
, ν = 2, F.
κν (λ, P ) =
|λ||y H P 0 (λ)x|
A ?-Palindromic P (λ) in (1.5) is a matrix polynomial with additional constraints
Ad−` = A?` for ` = 0, . . . , bd/2c, but the weights ω` (` = 0, . . . , bd/2c) given in the
definition of condition number κν (λ, P ) (2.7) are independent. It is natural to
consider the condition number of ?-PPEP to be the condition number of PEP with
ωd−` = ω` for ` = 0, 1, . . . , bd/2c, that is
| 4 λ|
κ̄ν (λ, P ) = lim sup
: (P (λ + 4λ) + 4P (λ + 4λ))(x + 4x) = 0,
|λ|
→0
k 4 Ad−` kν ≤ ω` , k 4 A` kν ≤ ω` ` = 0, . . . , bd/2c . (2.11)
4
Table 2.1: Eigenvalues and corresponding right and left eigenvectors of ?Palindromic.
Structure of P (λ) eigenvalues
>-Palindromic
λ
1/λ
H-Palindromic
λ
1/λ̄
right eigenvector
x
ȳ
x
y
left eigenvector
y
x̄
y
x
Let P (λ) be a regular ?-Palindromic of degree d and λ ∈ C be a nonzero simple
eigenvalue of P (λ) with corresponding right eigenvector x, then we have
P (λ)x = 0 ⇔ x? P ? (λ) = 0 ⇔ (λ? )d x? P (1/λ? ) = 0
(2.12)
showing that 1/λ? is a eigenvalue of P (λ) with corresponding left eigenvector x? H .
Similarly, if y is left eigenvector of P (λ) corresponding to λ then y ? H is right
eigenvector of P (λ) corresponding 1/λ? . We summarize those relations together
with their ?-variants in Table 2.1.
In the following theorem, we show that the condition numbers for λ and 1/λ?
defined by (2.11) are equal.
Theorem 2.2. Let P (λ) be a ?-Palindromic. Suppose λ ∈ C is a nonzero simple
eigenvalue of P (λ), so that 1/λ? is a simple eigenvalue of P (λ). Then
κ̄ν (λ, P ) = κ̄ν (1/λ? , P ), ν = 2, F,
(2.13)
where κ̄ν (λ, P ) is given by (2.11).
Proof. From Table 2.1, we obtain that if (λ, x, y) is an eigentriple for P (λ) then
(1/λ? , y ? H , x? H ) is also an eigentriple for P (λ). For convenience of the notation,
we suppose ωd−` = ω` for ` = 0, 1, . . . , bd/2c. By applying Theorem 2.1 to the
eigenvalue 1/λ? of P (λ), we have
P
d
? `
?H
k2 ky ? H k2
`=0 |1/λ | ω` kx
?
κ̄ν (1/λ , P ) =
|1/λ? ||x? P 0 (1/λ? )y ? H |
P
d
`
|1/λ|
ω
kxk2 kyk2
`
`=0
=
|1/λ||y H (P 0 (1/λ? ))? x|
P
d
`
|λ|
ω
` kxk2 kyk2
`=0
=
, ν = 2, F.
(2.14)
|λ|d−1 |y H (P 0 (1/λ? ))? x|
5
Recall that P (λ) is in (1.5), then
λP 0 (λ) = dλd Ad + (d − 1)λd−1 Ad−1 + · · · + 2λ2 A2 + λA1 ,
λd−1 (P 0 (1/λ? ))? = λd−1 A?1 + 2λd−2 A?2 + · · · + (d − 1)λA?d−1 + dA?d
= λd−1 Ad−1 + 2λd−2 Ad−2 + · · · + (d − 1)λA1 + dA0 ,
it yields
λd−1 (P 0 (1/λ? ))? + λP 0 (λ) = dP (λ).
(2.15)
Since y H P (λ)x = 0, by (2.15) it holds that
λd−1 y H (P 0 (1/λ? ))? x = −λy H P 0 (λ)x.
(2.16)
From (2.14) and (2.16), we have
P
κ̄ν (1/λ? , P ) =
d
`=0
|λ|` ω` kxk2 kyk2
|λ||y H P 0 (λ)x|
= κ̄ν (λ, P ), ν = 2, F.
Remark 2.1. From the definition of κ̄ν (λ, P ) in (2.11), we have
| 4 λ|
≤ κ̄ν (λ, P ) + O(2 ).
|λ|
(2.17)
Roughly speaking, (2.17) shows that if order ω` perturbations are made in A` and
Ad−` for each ` = 0, 1, . . . , bd/2c, then relative error of each eigenvalue λ may be
bounded by κ̄ν (λ, P ). Using Theorem 2.2, we conclude that the relative errors of
λ and 1/λ? have the same upper bound for a ?-PPEP.
3
Structured eigenvalue condition number
In Section 2 we consider a normwise eigenvalue condition number that respect
?-PPEP. However, since the matrix polynomial P (λ) that we care about is ?Palindromic, we may wish to preserve the ?-Palindromic structure when we perturb
A` in the definition of the condition number. This is structured perturbations
analysis. In this section we assume that the perturbation matrix polynomial 4P (λ)
in (2.8) is ?-Palindromic, i.e., 4P (λ) ∈ P where
P = {4P (λ) : 4P (λ) is ?-Palindromic of degree d
and k 4 A` kF ≤ ω` , for ` = 0, 1, . . . , bd/2c }.
6
(3.18)
For convenience of the notation, throughout this section we suppose that
ω`+d := ω` , for ` = 0, 1, . . . , bd/2c.
The structured condition number for a nonzero simple eigenvalue λ ∈ C can be
defined by
| 4 λ|
struct
κF (λ, P ) = lim sup
: (P (λ + 4λ) + 4P (λ + 4λ))(x + 4x) = 0,
|λ|
→0
4 P (λ) ∈ P .
(3.19)
From the definitions (2.11) and (3.19), it is clear that
κstruct
(λ, P ) ≤ κ̄F (λ, P ).
F
As for the unstructured case, from (2.10) we have
κstruct
(λ, P ) =
F
δ struct (λ, x, y)
,
|λ||y H P 0 (λ)x|
(3.20)
where
δ struct (λ, x, y) :=
sup |y H 4 P (λ)x|.
(3.21)
4P (λ)∈P
So, if we can construct an optimal solution of (3.21) then the structured eigenvalue
condition number can be obtained.
The following lemmas will be useful to construct a ?-Palindromic matrix polynomial for the solution of (3.21).
Lemma 3.1. Let x, y be the vectors in Cn with kxk2 =
exists a unitary matrix Q ∈ Cn×n such that

x? y
p
 1 − |x? y|2


0
QH x = e1 , ỹ := Q? y = 

..

.
0
where e1 denotes the first column of In .
7
kyk2 = 1. Then there




,


(3.22)
Proof. Since kxk2 = 1, there exists a unitary matrix Q1 ∈ Cn×n , i.e., QH
1 Q1 = In ,
such that
QH
1 x = e1 .
(3.23)
It is easy to see that the first column of Q1 is x and Q?1 is a unitary matrix. Since
kyk2 = 1,
? xy
?
(3.24)
Q1 y =
y1
p
where y1 ∈ Cn−1 with ky1 k2 = 1 − |x? y|2 . Furthermore, there exists an (n − 1) ×
b2 such that
(n − 1) unitary matrix Q
hp
i>
b?2 y1 =
Q
1 − |x? y|2 0 · · · 0 .
(3.25)
Denote
Q2 =
1 0
b2
0 Q
then Q2 is a unitary matrix. Let
Q = Q1 Q2 .
From (3.23)-(3.25), we obtain (3.22).
q
Lemma 3.2. Let α, β ∈ C and τ = |α|2 + 21 |β|2 . Then
τ=
|αw0 + βw1 | .
sup
(3.26)
w0 , w1 ∈ C,
|w0 |2 + 2|w1 |2 = 1
This bound can be attained when
|α| ιθ
|β| ι(θ−θ1 )
w0 =
e , w1 =
e
τ
2τ
where θ1 = Arg(β) − Arg(α) and θ is arbitrary.
Proof. Applying triangle inequality we observe that
sup
|αw0 + βw1 | ≤
w0 , w1 ∈ C,
|w0 |2 + 2|w1 |2 = 1
sup
w0 , w
√1 ∈ C,
|w0 |2 + | 2w1 |2 = 1
β √
|α||w0 | + √ | 2w1 |
2
r
=
1
|α|2 + |β|2 = τ.
2
(3.27)
Let w0 = |α|
eιθ and w1 = |β|
eι(θ−θ1 ) where θ1 = Arg(β) − Arg(α) and θ is arbitrary.
τ
2τ
Then |w0 |2 + 2|w1 |2 = 1 and it is easy to check that the equality in (3.27) holds
8
Let P (λ) be a regular ?-Palindromic of degree d and λ ∈ C be a nonzero simple
eigenvalue of P (λ). Suppose that x and y are normalized right and left eigenvectors
of P (λ) associated λ, respectively. For 4P (λ) ∈ P, we have from Lemma 3.1 that
!
d
X
y H 4 P (λ)x = y H
λ ` 4 A` x
`=0
d
X
= y H Q? H Q?
!
λ ` 4 A`
QQH x
`=0
i
p
?
?
2
= x y, 1 − |x y| , 0, · · · , 0 4 PQ (λ)e1
h
= ỹ H 4 PQ (λ)e1 ,
(3.28)
where
4PQ (λ) =
d
X
`
?
λ Q 4 A` Q :=
d
X
λ` 4 B`
(3.29)
`=0
`=0
is a ?-Palindromic. Since Frobenius norm is unitary invariant,
k 4 B` kF = k 4 A` kF ≤ ω` , for ` = 0, 1, . . . , bd/2c.
(3.30)
PQ ≡ Q? PQ = {Q? 4 P (λ)Q : 4P (λ) ∈ P}
(3.31)
Let
where P is defined in (3.18). From (3.29)-(3.31), we have
PQ = P.
(3.32)
Substituting (3.28) and (3.32) into (3.21), we obtain that for odd d
δ struct (λ, x, y) =
sup
|ỹ H 4 PQ (λ)e1 |
4PQ (λ)∈P
=
sup
k 4 B` kF ≤ ω`
(d−1)/2
X H `
d−`
?
ỹ λ 4 B` + λ 4 B` e1 `=0
(3.33a)
and for even d
δ struct (λ, x, y) =
sup
|ỹ H 4 PQ (λ)e1 |
4PQ (λ)∈P
=
sup
?
= 4Bd/2
4Bd/2
k 4 B` kF ≤ ω`
d/2−1
X H `
d−`
?
H
d/2
.
λ
4
B
+
λ
4
B
e
+
ỹ
λ
4
B
e
ỹ
`
1
1
d/2
`
`=0
(3.33b)
9
Since ỹ has form in (3.22), we only consider the first 2 × 2 block of 4B` at
(3.33). The optimal solutions of (3.33) have the form
(4B` )(1:2,1:2)
02×(n−2)
4B` =
for ` = 0, 1, . . . , bd/2c,
(3.34)
0(n−2)×2
0(n−2)×(n−2)
where 0k×m is the k × m zero block. Denote
b` ≡ (4B` )(1:2,1:2) =
4B
b`11 b`12
b`21 b`22
.
(3.35)
Solve δ struct (λ, x, y) when ? = >.
We see that for ` = 0, 1, . . . , b(d − 1)/2c
H
`
d−`
ỹ H (λ` 4 B` + λd−` 4 B`> )e1 = e>
4 B`> )
1 ⊗ ỹ vec(λ 4 B` + λ
H
d−` >
= λ` e>
e1 ⊗ ỹ H vec(4B`> )
1 ⊗ ỹ vec(4B` ) + λ
b` ),
= z H vec(4B
(3.36a)
`
where ⊗ denotes the Kronecker product, vec denotes the operator that stacks the
columns of a matrix into a long vector and
i
h
p
p
z`H = (λ` + λ(d−`) )x> y, λ` 1 − |x> y|2 , λ(d−`) 1 − |x> y|2 , 0 .
(3.36b)
b` kF =
When d is odd: from (3.33a) and (3.36) and using the equality k 4 B
b` )k2 we have
kvec(4B
(d−1)/2
(d−1)/2
X
X
struct
H
b
ω` kz` k2
δ
(λ, x, y) =
sup
z
vec(4
B
)
≤
`
`
b` )k2 ≤ω` kvec(4B
`=0
`=0
(d−1)/2
=
X
ω` ξ` ,
(3.37a)
`=0
where
q
ξ` := kz` k2 = |λ` |2 + |λd−` |2 + 2R(λ` λ̄d−` )|x> y|2 , 0 ≤ ` ≤ (d − 1)/2. (3.37b)
It is easy to check that if we choose
p
ω` (λ̄` + λ̄(d−`) )x> y λ̄(d−`) 1 − |x> y|2
b
p
4B` =
for ` = 0, 1, . . . , (d − 1)/2,
ξ`
λ̄` 1 − |x> y|2
0
10
then the equality in (3.37) holds, i.e.,
(d−1)/2
δ
struct
(λ, x, y) = X
ω` ξ` .
(3.38)
`=0
When d is even:
H
ỹ H (λd/2 4 Bd/2 )e1 = λd/2 e>
1 ⊗ ỹ vec(4Bd/2 )
bd/2 ).
= z H vec(4B
(3.39a)
h
i
p
H
zd/2
= λd/2 x> y, λd/2 1 − |x> y|2 , 0, 0 .
(3.39b)
d/2
where
From (3.37) and (3.39), we see that the δ struct (λ, x, y) in (3.33) can be rewritten as
X
d/2
H
b
,
δ struct (λ, x, y) =
sup
z
vec(4
B
)
`
`
b?
b
4B
=
4
B
`=0
d/2
d/2
b` )k2 ≤ ω`
kvec(4B
where z`H are given in (3.36b) and (3.39b). By triangle inequality and applying
Lemma 3.2 when ` = d/2, we have
δ
struct
(λ, x, y) ≤
sup
b > = 4B
b
4B
d/2
d/2
b` )kF ≤ ω`
kvec(4B
d/2 X
H
b
z` vec(4B` )
`=0
d/2−1
=
X
`=0
=
d/2
X
ω` kz` k2 + r
sup
d/2 2 d/2 2
b11 +2b21 =ωd/2
H
bd/2 )
zd/2 vec(4B
ω` ξ` ,
(3.40a)
`=0
where
( p
|λ` |2 + |λd−` |2 + 2R(λ` λ̄d−` )|x> y|2 ` 6= d/2,
ξ` = q 1 d
|λ |(1 + |x> y|2 )
` = d/2.
2
(3.40b)
It is easy to check that if we choose
p
ω` (λ̄` + λ̄(d−`) )x> y λ̄(d−`) 1 − |x> y|2
b
p
for ` = 0, 1, . . . , d/2 − 1
4B` =
ξ`
λ̄` 1 − |x> y|2
0
11
and
bd/2 = ωd/2
4B
ξd/2
d/2 >
λ̄p
x y
1 d/2
λ̄
1 − |x> y|2
2
1 d/2
λ̄
2
p
1 − |x> y|2
0
,
then the equality in (3.40a) holds, i.e.,
δ
struct
(λ, x, y) = d/2
X
ω` ξ` .
(3.41)
`=0
Solve δ struct (λ, x, y) when ? = H.
The following lemma will be need in this case.
Lemma 3.3. Let α, β ∈ C and γi ∈ C for i = 1, 2, . . . , m, and let τ =
Then
m
X
γi wi .
τ=
sup
αw0 + β w̄0 +
wi ∈ C,
P
m
i=0
p
(|α| + |β|)2 +
(3.42)
i=1
|wi |2 = 1
This bound can be attained when
w0 = (|α| + |β|)eιθ /τ and wi = |γi |eιθi /τ for i = 1, 2, . . . , m,
(3.43)
where θ = Arg(αβ)/2 − Arg(α) and θi = Arg(αβ)/2 − Arg(γi ). (Note that if
Arg(γi ) = Arg(αβ)/2 for some i then the wi of optimal solution can be chosen as
a real number.)
Proof. By triangle inequality we observe that
m
m
X
X
γi wi ≤
sup
|α||w0 | + |β||w̄0 | +
|γi ||wi |
sup
αw0 + β w̄0 +
w ∈ C,
wi ∈ C,
Pm i
P
i=1
i=1
2
2
m
i=0 |wi | = 1
i=0 |wi | = 1
v
u
m
X
u
= t(|α| + |β|)2 +
|γi |2 .
(3.44)
i=1
Hence the expression in (3.44) is an upper bound. To show that P
this bound can
2
be attained we consider wi as in (3.43) for i = 0, 1, . . . , m. Then m
i=1 |wi | = 1
and using the fact that θ = Arg(αβ)/2 − Arg(α) and θi = Arg(αβ)/2 − Arg(γi ), it
observes that
Arg(αβ)/2 = Arg(αw0 ) = Arg(β w̄0 ) = Arg(γi wi ),
(3.45)
for i = 1, 2, . . . , m. From (3.45), it is easy to check that the equality in (3.44) holds.
This completes the proof.
12
Pm
i=1
|γi |2 .
We see that for ` = 0, 1, . . . , b(d − 1)/2c
f` (vec(4B` )) : = ỹ H (λ` 4 B` + λd−` 4 B`H )e1
H
`
d−`
= e>
4 B`H )
1 ⊗ ỹ vec(λ 4 B` + λ
H
d−` >
= λ` e>
e1 ⊗ ỹ H vec(4B`H )
1 ⊗ ỹ vec(4B` ) + λ
= α` b`11 + β` b̄`11 + γ` b`21 + λd−2` γ` b̄`12 ,
(3.46a)
where
α` = λ` xH y, β` = λ(d−`) xH y, γ` = λ`
p
1 − |xH y|2 ,
(3.46b)
and b`11 , b`21 , b`12 ∈ C as in (3.35). When d is even and ` = d/2
H
fd/2 (vec(4Bd/2 )) : = ỹ H (λd/2 4 Bd/2 )e1 = λd/2 e>
1 ⊗ ỹ vec(4Bd/2 )
d/2
d/2
= αd/2 b11 + γd/2 b21 .
(3.46c)
where
αd/2 = λd/2 xH y, γd/2 = λd/2
d/2
p
1 − |xH y|2 ,
(3.46d)
d/2
H
).
and b11 ∈ R, b21 ∈ C as in (3.35) (since 4Bd/2 = 4Bd/2
Substituting (3.46) into (3.33) and applying the triangle inequality, we obtain
δ struct (λ, x, y)

P(d−1)/2
sup
|f` (vec(4B` ))|

 k4B` kF ≤ω` `=0
Pd/2−1
≤
sup
|f` (vec(4B` ))| +
`=0

 k4B` kF ≤ω`
d is odd,
sup
fd/2 (vec(4Bd/2 ))
d is even.
k 4 Bd/2 kF ≤ ωd/2
H
4Bd/2 = 4Bd/2
(3.47)
The equality holds if all principal argument of f` (vec(4B` )) are equal. By applying Lemma 3.3 to f` (vec(4B` )) for ` = 0, 1, . . . , b(d − 1)/2c and Lemma 3.2 to
fd/2 (vec(4B` )) when d is even, we obtain
( P
(d−1)/2
`=0
ωτ
d is odd,
δ struct (λ, x, y) ≤
Pd/2−1 ` `
`=0 ω` τ` + ωd/2 τd/2 d is even,
(3.48a)
( p
|λ` |2 + |λd−` |2 + 2|λd ||xH y|2 ` 6= d/2,
τ` = q 1 d
|λ |(1 + |xH y|2 )
` = d/2.
2
(3.48b)
where
13
From (3.45)-(3.46), we know that the argument of ỹ H λ` 4 B` + λd−` 4 B`? e1
at the optimal solution is Arg(α` β` )/2, where α` , β` are given in (3.46b). When
d/2
d is even, from (??) and b11 ∈ R, the argument of ỹ H λd/2 4 Bd/2 e1 at the
optimal
solution
is Arg(αd/2 ), where αd/2 is given in (??). Let ϑ = Arg(α0 β0 )/2 =
Arg λd (xH y)2 /2. Then for ` = 1, 2 . . . , b(d − 1)/2c,
Arg(α` β` )/2 = Arg λd (xH y)2 /2 = ϑ
and when d is even
Arg(αd/2 ) = Arg λd/2 xH y = ϑ.
Thus, the equality in (3.48a) holds; that is,
( P
(d−1)/2
`=0
ωτ
d is odd,
struct
δ
(λ, x, y) =
Pd/2−1 ` `
d/2
`=0 ω` τ` + ωd/2 |λ | d is even.
(3.49)
Theorem 3.1. Let P (λ) be a ?-Palindromic of degree d and λ ∈ C be a nonzero
simple eigenvalue of P (λ). Let x and y be normalized right and left eigenvectors
associated λ, respectively. The structured condition number of λ is given by
(i) When ? = >:
 P
(d−1)/2

ξ` ω`

`=0

d is odd,

H
0
|λ||y P (λ)x|
P
(3.50)
κstruct
(λ,
P
)
=
d/2−1
F

ξ` ω` + |λ|d/2 ωd/2

`=0

d is even,

|λ||y H P 0 (λ)x|
q
where ξ` = |λ` |2 + |λd−` |2 + 2R(λ` λd−` )|x> y|2 , for ` = 0, 1, . . . , b(d − 1)/2c.
(ii) When ? = H:
 P
(d−1)/2

τ` ω`

`=0

d is odd,

H
0
|λ||y P (λ)x|
struct
P
κF (λ, P ) =
(3.51)
d/2−1
d/2

ωd/2

`=0 τ` ω` + |λ|

d is even,

|λ||y H P 0 (λ)x|
p
where τ` = |λ` |2 + |λd−` |2 + 2|λd ||xH y|2 , for ` = 0, 1, . . . , b(d − 1)/2c.
Remark 3.1. When ? = H, if the normalized right and left eigenvectors associated
λ are linearly dependent then |xH y| = 1 and κstruct
(λ, P ) = κ̄F (λ, P ).
F
14
The next corollary considers the relation between κ̄F (λ, P ) and κstruct
(λ, P )
F
when λ is a simple eigenvalue of ?-Palindromic P (λ) with λ = 1/λ? .
Corollary 3.2. Let P (λ) be a ?-Palindromic. Suppose that λ̂ with λ̂ = 1/λ̂? is a
simple eigenvalue of P (λ). Then
a. If ? = >, then λ̂ = ±1 and κstruct
(1, P ) = κ̄F (1, P ), κstruct
(−1, P ) = 0.
F
F
b. If ? = H, then |λ̂| = 1 and κstruct
(λ̂, P ) = κ̄F (λ̂, P ).
F
Proof. Since λ̂ = 1/λ̂? is a simple eigenvalue of P (λ), we know from Table 2.1 that
if x is normalized right eigenvector associated λ̂ then
x̄ If ? = >,
y=
x If ? = H,
is normalized left eigenvector associated λ̂. Hence
|x> y| = |x> x̄| = 1
|xH y| = |xH x| = 1
If ? = >,
If ? = H.
Then it is straightforward to show the results of this corollary by Theorem 3.1.
Remark 3.2. Let P (λ) be a >-Palindromic. Suppose that d and n are odd, then
P (λ) has dn (dn is odd) eigenvalues. Since the eigenvalues of P (λ) arrear in pairs
(λ, 1/λ), then there exists an eigenvalue λ̂ = 1/λ̂, i.e., λ̂ = ±1.
In the following theorem, we show that the structured condition numbers of λ
and 1/λ? are equal.
Theorem 3.3. Let P (λ) be a ?-Palindromic. Suppose λ ∈ C is a nonzero simple
eigenvalue of P (λ), so that 1/λ? is a simple eigenvalue of P (λ). Then
κstruct
(λ, P ) = κstruct
(1/λ? , P ).
F
F
(3.52)
Proof. Table 2.1 obtains that if (λ, x, y) is an eigentriple for P (λ) then (1/λ? , y ? H , x? H )
is also an eigentriple for P (λ). Without loss of generality, we suppose kxk2 =
kyk2 = 1. Using equation (3.20), we obtain the relation
δ struct (1/λ? , y ? H , x? H )
|1/λ? ||x? P 0 (1/λ? )y ? H |
|λ? |d δ struct (1/λ? , y ? H , x? H )
=
.
|λ? |d−1 |x? P 0 (1/λ? )y ? H |
κstruct
(1/λ? , P ) =
F
15
(3.53)
Since (y ? H )? x? H = |y H x? H | = |x? y|, it is easy to obtain
|λ? |d δ struct (1/λ? , y ? H , x? H ) = δ struct (λ, x, y).
(3.54)
Taking the absolute of both sides of (2.16) leads to
|λ||y H P 0 (λ)x| = λd−1 y H (P 0 (1/λ? ))? x
? = λd−1 y H (P 0 (1/λ? ))? x = |λ? |d−1 |x? P 0 (1/λ? )y ? H |.
(3.55)
From (3.53)-(3.55), we have
δ struct (λ, x, y)
|λ||y H P 0 (λ)x|
= κstruct
(λ, P ).
F
κstruct
(1/λ? , P ) =
F
The proof is complete.
4
Numerical results
To illustrate our results we present some numerical examples. All computations
were carried out using Matlab 2007a with the machine precision eps ≈ 2.22 ×
10−16 . In each example we report all numbers in five significant digits only, thought
all calculations are carried out in full precision.
As mentioned before, theoretically, the eigenvalues of ?-PPEP appear in pairs
(λ, 1/λ? ). So, if we sort the eigenvalues in the ascending order by modulus, the
product of the ith and (dn + 1 − i)th sorted eigenvalues should be one. Therefore,
we define the reciprocities of computed eigenvalues by
r(λi ) = |λi λ?dn+1−i − 1|, i = 1, . . . , bdn/2c.
(4.56)
Example 1. Take n = 15 and ? = >. We generated random a complex matrix
A0 ∈ C15×15 by Matlab code:
A0= randn(n)+i*randn(n);
Consider >-Palindromic P (λ) = λA>
0 + A0 and compute its eigenpairs {Λ0 , X0 }
by QZ algorithm. We obtain that −1 is a simple eigenvalue of P (λ). From Table
2.1, it is well know that the left eigenvector associated with eigenvalue λ is the
conjugate of right eigenvector associated with 1/λ.
Figure 4.1 shows the corresponding condition numbers and structured condition numbers of P (λ) with ω1 = ω0 = kA0 kF (ω0 and ω1 are the weights that
16
300
Condition number
−14
1
250
Reciprocity of eigenvalues
0.8
κF (λ,P)
200
x 10
r(λ)
κstruct
(λ,P)
F
0.6
150
0.4
100
0.2
50
0
10−2
10 −1
10 0
|λ |
10 1
10 2
0
10−2
10 −1
10 0
|λ |
10 1
10 2
Figure 4.1: Condition number for P (λ); Left and reciprocity of eigenvalues; Right.
represent tolerances of perturbations are made in A>
0 and A0 , respectively) and
the reciprocity property of eigenvalues. The notations “o”, “∗” and “+” denotes
(λ, P ) and r(λ), respectively. Since the probthe results computed κ̄F (λ, P ), κstruct
F
lem is >-PPEP, the eigenvalues of P (λ) appear in pairs (λ, 1/λ). From Theorem
2.2 and Theorem 3.3, we know that κF (λ, P ) = κ̄F (1/λ, P ) and κstruct
(λ, P ) =
F
κstruct
(1/λ,
P
).
This
is
confirmed
by
Figure
4.1.
In
addition,
Corollary
3.2
shows
F
struct
that the κF (−1, P ) = 0, this behavior is seen in this figure.
Next, we consider two kinds of perturbation matrices.
• Unstructured perturbations: We generate randomly two complex matrices 4A, 4B ∈
e = A0 + 4A, B
e = A> + 4B
C15×15 with k4AkF = k4BkF = 10−9 . Let A
e + A.
e
and Pe(λ) = λB
• Structured perturbations: We generate randomly a complex matrix 4A0 ∈
e0 = A0 + 4A0 and PeS (λ) = λA
e> + A
e0 .
C15×15 with k4A0 k = 10−9 . Let A
0
e and Λ
e S of Pe(λ) and PeS (λ) by QZ algorithm, respecCompute eigenvalues Λ
e Λ
eS
tively. Suppose that the computed eigenvalues Λe of P (λ) are “exact” and Λ,
are approximate eigenvalues. We then compare the relative error and reciprocity of
eigenvalues of those two cases in Figure 4.2. We see that structured and unstructured perturbations have similar accuracy in relative error of eigenvalues since
κF (λ, P ) and κstruct
(λ, P ) have the same order. But structured perturbations esF
sentially no effect the reciprocity of eigenvalues that because the perturbed system
PeS (λ) is still a >-Palindromic.
17
10−8
Relative error of eigenvalues
10−8
10−10
Reciprocity of eigenvalues
10−10
10−12
Unstructured perturbations
Structured perturbations
Unstructured perturbations
Structured perturbations
−12
10
10−14
10−16
10−2
10 −1
10 0
|λ |
10 1
10 2
10−14
10−15
10−2
10−1
10 0
|λ |
10 1
10 2
Figure 4.2: Relative error of eigenvalues; Left and reciprocity of eigenvalues; Right.
Example 2. This example concerns ? = H. Take n = 10, d = 2 and generate
randomly Ao , A1 ∈ C10×10 with AH
1 = A1 and kA0 k2 = kA1 k2 = 1 by Matlab’s
2 H
code. Let P (λ) = λ A0 +λA1 +A0 be a H-Palindromic. We compute 2n eigenpairs
(λ, x) of P (λ) by QZ algorithm of linearizing P (λ), given by
H
λx
−A1 A0
A0 0
ζ, with ζ =
.
GEP :
ζ=λ
0 In
x
In
0
We obtain six simple eigenvalues with property that λ = 1/λ̄. Since kA0 k2 =
kA1 k2 = 1, [13] shows that solving GEP by QZ algorithm is backward stable for
the H-PPEP. From Table 2.1, it is well know that the left eigenvector associated
with eigenvalue λ is the right eigenvector associated with 1/λ̄. Figure 4.3 shows the
corresponding condition numbers and structured condition numbers of P (λ) with
ω2 = ω0 = kA0 kF and ω1 = kA1 kF and the reciprocity property of eigenvalues.
Since the problem is H-PPEP, the eigenvalues of P (λ) appear in pairs (λ, 1/λ̄).
From Theorem 2.2 and Theorem 3.3, we know that κF (λ, P ) = κ̄F (1/λ, P ) and
κstruct
(λ, P ) = κstruct
(1/λ, P ). This is confirmed by Figure 4.3. In addition, CorolF
F
lary 3.2 shows that the κstruct
(λ, P ) = κ̄F (λ, P ) for each simple eigenvalue of P (λ)
F
with λ = 1/λ̄, this behavior is seen very clearly in this figure.
5
Conclusions
We have derived directly computable structured conditions for ?-PPEP. These formulas show the structured condition number and condition number of PEP as
18
120
Condition number
−15
8
Reciprocity of eigenvalues
7
100
6
80
5
κF (λ,P)
60
r(λ)
4
κstruct
(λ,P)
F
3
40
2
20
0
x 10
10 −1
10 0
|λ |
10 1
1
0
10 −1
10 0
|λ |
10 1
Figure 4.3: Condition number for P (λ); Left and reciprocity of eigenvalues; Right.
investigated by Tisseur [13] are difference. In our numerical experiences, the ratio
of structured condition number and condition number of ?-PPEP close to 1. We
used our formulas of structured condition number to show that the structured condition numbers of λ and 1/λ? are equal. That is, the eigenvalues λ and 1/λ? have
the same sensitivity to structured perturbations for ?-PPEP. Similarly, we also
showed that the condition numbers of λ and 1/λ? are equal under some reasonable
assumptions.
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