Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 824641 8 pageshttpdxdoiorg1011552013824641

Research ArticleA Note on the ⊤-Stein Matrix Equation

Chun-Yueh Chiang

Center for General Education National Formosa University Huwei 632 Taiwan

Correspondence should be addressed to Chun-Yueh Chiang chiangnfuedutw

Received 16 May 2013 Revised 16 July 2013 Accepted 16 July 2013

Academic Editor Masoud Hajarian

Copyright copy 2013 Chun-Yueh ChiangThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

This note is concerned with the linear matrix equation 119883 = 119860119883⊤119861 + 119862 where the operator (sdot)⊤ denotes the transpose (⊤) of a

matrix The first part of this paper sets forth the necessary and sufficient conditions for the unique solvability of the solution 119883The second part of this paper aims to provide a comprehensive treatment of the relationship between the theory of the generalizedeigenvalue problem and the theory of the linear matrix equationThe final part of this paper starts with a brief review of numericalmethods for solving the linear matrix equation In relation to the computed methods knowledge of the residual is discussed Anexpression related to the backward error of an approximate solution is obtained it shows that a small backward error implies asmall residual Just like the discussion of linear matrix equations perturbation bounds for solving the linear matrix equation arealso proposed in this work

1 Introduction

Our purpose of this work is to study the so-called ⊤-Steinmatrix equation

119883 = 119860119883⊤119861 + 119862 (1)

where 119860 119861 119862 isin R119899times119899 are known matrices and 119883 isin R119899times119899

is an unknown matrix to be determined Our interest inthe ⊤-Stein equation originates from the study of completelyintegrable mechanical systems that is the analysis of the ⊤-Sylvester equation

119860119883 + 119883⊤119861 = 119862 (2)

where119860 119861 and119862 are matrices inR119899times119899 [1 2] By means of thegeneralized inverses or QZ decomposition [3] the solvabilityconditions of (2) are studied in [1 2 4] Suppose that thematrix pencil119860minus120582119861

⊤ is regular that is 119886119860+119887119861⊤ is invertible

for some scalars 119886 and 119887 The ⊤-Sylvester equation (2) can bewritten as

(119886119860 + 119887119861⊤)119883 + 119883

⊤(119886119861 + 119887119860

⊤) = 119886119862 + 119887119862

⊤ (3)

Premultiplying both sides of (3) by (119886119860+119887119861⊤)minus1 we have

119883 + 119880119883⊤119881 = 119863 (4)

where 119880 = (119886119860 + 119887119861⊤)minus1 119881 = 119886119861 + 119887119860

⊤ and 119863 =

(119886119860 + 119887119861⊤)minus1(119886119862 + 119887119862

⊤) This is of the form (1) In other

words numerical approaches for solving (2) can be obtainedby transforming (2) into the form of (1) and then applyingnumerical methods to (1) for the solution [4ndash6] With this inmind in this note we are interested in the study of ⊤-Steinmatrix equation (1)

Our major purpose in this work can be divided into threeparts First we determine necessary and sufficient conditionsfor the unique solvability of the solution to (1) In doing soZhou et al [7] transform (1) to the standard Stein equation

119882 = 119860119861⊤119882119860⊤119861 + 119860119862

⊤119861 + 119862 (5)

with respect to the unknown matrix 119882 isin R119898times119899 and give thefollowing necessary condition

120583] = 1 forall120583 ] isin 120590 (119860⊤119861) (6)

Here 120590(119860⊤119861) is the set of all eigenvalues of 119860

⊤119861 Zhou

and his coauthors show that if (5) has a unique solutionthen (1) has a unique solution However a counterexample isprovided in [7] to show that the relation (6) is only a necessarycondition for the unique solvability of (1)

2 Abstract and Applied Analysis

In [4 8] the periodic QZ (PQZ) decomposition [3] isapplied to consider the necessary and sufficient conditions ofthe unique solvability of (1) conditions given in [8] ignorethe possibility of the existence of the unique solution while 1is a simple root of 120590(119860⊤119861) This condition is included in oursubsequent discussion and the following remark is providedto support our observation

Remark 1 Let 119860 = minus1 and let 119861 = 1 that is 120590(119860119861⊤) = minus1

It is clear that the scalar equation 119883 = minus119883⊤

+ 119862 has aunique solution 119883 = 1198622 But condition (6) is not satisfiedby choosing 120583 = ] = minus1

It can also be observed from Remark 1 that even if (1) isuniquely solvable it does not imply that (5) (namely119883 = 119883+

119862 minus 119862) is uniquely solvable Conditions in [4 Equation(46)]provided those conditions for the unique solvability of thesolution to (1) via a structured algorithm In our workthrough a complete analysis of square coefficient matrices interms of the analysis of the spectra of thematrix119860⊤119861 the newapproach to the condition of unique solvability of the⊤-Steinequation (1) can be obtained

Second we present the invariant subspace method andmore generally the deflating subspace method to solve the⊤-Stein equation Our methods are based on the analysis ofthe eigeninformations for a matrix pencil We carry out athorough discussion to address the various eigeninformationencountered in the subspace methods These ideas can beimplemented in algorithms easily

Finally we take full account of the error analysis of(1) Expressions and implications such as the residual thebackward error and perturbation bounds are derived in thiswork Note that for an approximate solution 119884 of (1) thebackward error tells us how much the matrices 119860 119861 and 119862

must be perturbed An important point found in Section 5is that a small backward error indicates a small value for theresidual R = 119884 minus 119860119884

⊤119861 minus 119862 but the reverse is not usually

trueBeginning in Section 2 we formulate the necessary and

sufficient conditions for the existence of the solution of (1)directly by means of the spectrum analysis In Section 3we provide a deflating subspace method for computingthe solution of (1) Numerical methods for solving (1) andthe related residual analysis are discussed in Section 4 Theassociated error analysis of (1) is given in Section 5 andconcluding remarks are given in Section 6

2 Solvability Conditions ofthe Matrix Equation (1)

In order to formalize our discussion let 119860 otimes 119861 be theKronecker product of matrices 119860 and 119861 let 119868

119899be the 119899 times 119899

identity matrix and let sdot 119865denote the Frobenius norm

With the Kronecker product (1) can be written as theenlarged linear system

(1198681198992 minus (119861

⊤otimes 119860)P) vec (119883) = vec (119862) (7)

where vec(119883) stacks the columns of 119883 into a column vectorandP is the Kronecker permutation matrix [9] which mapsvec(119883) into vec(119883⊤) that is

P = sum

1le119894119895le1198992

119890119895119890⊤

119894otimes 119890119894119890⊤

119895 (8)

where 119890119894denotes the 119894th column of the 1198992 times 119899

2 identitymatrix1198681198992 Due to the specific structure of P it has been shown in

[10 Corollary 4310] that

P⊤(119861⊤otimes 119860)P = 119860 otimes 119861

⊤ (9)

It then follows that

((119861⊤otimes 119860)P)

2

= (119861⊤otimes 119860)PP

⊤(119860 otimes 119861

⊤) = 119861⊤119860 otimes 119860119861

⊤

(10)

since P2 = 1198681198992 and P = P⊤ Note that eigenvalues of

matrices119860⊤119861 and119860119861⊤ are the same By (10) and the property

of the Kronecker product [11 Theorem 48] we know that

120590 (((119861⊤otimes 119860)P)

2

)

= 120582119894120582119895| 120582119894 120582119895isin 120590 (119860

⊤119861) = 120582

1 120582

119899 1 le 119894 119895 le 119899

(11)

That is the eigenvalues of (119861⊤

otimes 119860)P are related to thesquare roots of the eigenvalues of 120590(119860

⊤119861) but from (10)

no more information can be used to decide the positivity ornonnegativity of the eigenvalues of (119861⊤ otimes 119860)P A questionimmediately arises as to whether it is possible to obtainthe explicit expression of the eigenvalues of (119861

⊤otimes 119860)P

provided the eigenvalues of 119860⊤119861 are given In the followingtwo lemmas we first review the periodic QZ decompositionfor two matrices and then apply it to discuss the eigenvaluesof (119861⊤ otimes 119860)P

Lemma2 (see [3]) Let119860 and119861 be twomatrices inR119899times119899Thenthere exist unitary matrices 119875119876 isin C119899times119899 such that 119880

119860= 119875119860119876

and 119880119861= 119876119867119861⊤119875119867 are two upper triangular matrices

Lemma 3 Let 119860 and 119861 be two matrices in R119898times119899 Then

(1) (119861⊤ otimes 119860)P = (119876 otimes 119875119867)(119880119860otimes 119880119861)P(119876

119867otimes 119875)

(2) 120590((119861⊤ otimes 119860)P) = 120582119894 plusmnradic120582

119894120582119895| 120582119894 120582119895isin 120590(119860

⊤119861) =

1205821 120582

119899 1 le 119894 lt 119895 le 119899

Hereradic119911 denotes the principal square root of a complex number119911

Proof Part 1 follows immediately from Lemma 2 since 119880119860=

119875119860119876 and 119880119861= 119876119867119861⊤119875119867 for some unitary matrices 119875 and 119876

that is

(119861⊤otimes 119860)P = (119876 otimes 119875

119867) (119880119861otimes 119880119860) (119875 otimes 119876

119867)P

= (119876 otimes 119875119867) (119880119860otimes 119880119861)P (119876

119867otimes 119875)

(12)

Abstract and Applied Analysis 3

Let the diagonal entries of 119880119860and 119880

119861be denoted by 119886

119894119894

and 119887119895119895 respectively Then 119880

119860otimes 119880119861is an upper triangular

matrix with given diagonal entries specified by 119886119894119894and 119887119895119895

After multiplying119880119860

otimes 119880119861byP from the right the position

of the entry 119886119894119894119887119895119895is changed to be in the 119895+119899(119894minus1) th row and

the 119894 + 119899(119895 minus 1) th column of the matrix (119880119860

otimes 119880119861)P They

are then reshuffled by a sequence of permutation matrices toform a block upper triangular matrix with diagonal entriesarranged in the following order

1198861111988711 [

0 1198861111988722

1198862211988711

0] [

0 11988611119887119899119899

11988611989911989911988711

0]

1198862211988722 [

0 1198862211988733

1198863311988722

0] [

0 11988611989911989911988722

11988622119887119899119899

0]

[0 119886

119899minus1119899minus1119887119899119899

119886119899119899119887119899minus1119899minus1

0] 119886119899119899119887119899119899

(13)

Note that the reshuffling process is not hard to see when 119899 =

2 119880119860= [1198861111988612

0 11988622] and 119880

119861= [1198871111988712

0 11988722

] we have

(119880119860otimes 119880119861)P =

[[[

[

1198861111988711

1198861211988711

1198861111988712

1198861211988712

0 0 1198861111988722

1198861211988722

0 1198862211988711

0 1198862211988712

0 0 0 1198862211988722

]]]

]

(14)

However it is conceptually simple and regular but opera-tionally tedious to reorder (119880

119860otimes 119880119861)P to show this result

even for 119899 = 3 and that will be left as an exerciseBy (13) it can be seen that

120590 ((119861⊤otimes 119860)P) = 119886

119894119894119887119894119894 plusmnradic119886119894119894119886119895119895119887119894119894119887119895119895 1 le 119894 119895 le 119899

= 120582119894 plusmnradic120582

119894120582119895 1 le 119894 119895 le 119899

(15)

where 120582119894= 119886119894119894119887119894119894isin 120590(119860

⊤119861) for 1 le 119894 le 119899

Before demonstrating the unique solvability conditionswe need to define that a subset Λ = 120582

1 120582

119899 of

complex numbers is said to be ⊤-reciprocal free if and onlyif whenever 119894 = 119895 120582

119894= 1120582119895 This definition also regards 0 and

infin as reciprocals of each other Then we have the followingsolvability conditions of (1)

Theorem 4 The ⊤-Stein matrix equation (1) is uniquelysolvable if and only if the following conditions are satisfied

(1) the set of 120590(119860⊤119861) minus1 is ⊤-reciprocal free(2) minus1 can be an eigenvalue of the matrix119860⊤119861 but must be

simple

Proof From (7) we know that the ⊤-Stein matrix equation(1) is uniquely solvable if and only if

1 notin 120590 ((119861⊤otimes 119860)P) (16)

By Lemma 3 if 120582 isin 120590(119860⊤119861) then 1120582 notin 120590(119860

⊤119861) Otherwise

1 = radic120582 sdot (1120582) isin ((119861⊤

otimes 119860)P) On the other hand ifminus1 isin 120590(119860

⊤119861) and minus1 is not a simple eigenvalue then 1 isin

120590((119861⊤otimes119860)P) This verifies (16) and the proof of the theorem

is complete

It is worthy noting that the condition (1) of Theorem 4 iscontained in the condition (6) (also appear in [7 Theorem1]) which is the necessary and sufficient conditions forthe solvability of the solution to the Stein equation (5)However as mentioned before in Remark 1 or [7 Example2] condition (1) is just a necessary condition for uniquesolvability of the solution to (1) The ⊤-Stein matrix equation(1) is uniquely solvable provided that both conditions (1) and(2) of Theorem 4 are satisfying

3 The Connection between DeflatingSubspace and (1)

The relationship between solution of matrix equations andthe matrix eigenvalue problems has been widely studied inmany applications It is famous that solution of Riccati andpolynomial matrix equations can be found by computinginvariant subspaces of matrices and deflating subspaces ofmatrix pencils [12] This reality leads us to find some algo-rithms for computing solution of (1) based on the numericalcomputation of invariant or deflating subspaces

Given a pair of 119899 times 119899 matrices 119860 and 119861 recall that thefunction119860minus120582119861 in the variable 120582 is said to be thematrix pencilrelated to the pair (119860 119861) For a 119896-dimensional subspaceX isin

C119899 is called a deflating subspace for the pencil119860minus120582119861 if thereexists a 119896-dimensional subspaceY isin C119899 such that

119860X sube Y 119861X sube Y (17)

that is

119860119883 = 1198841198791 119861119883 = 119884119879

2 (18)

where119883119884 isin C119899times119896 are two full rank matrices whose columnsspan the spacesX andY respectively and matrices 119879

1 1198792isin

C119896times119896 In particular if in (18) 119883 = 119884 and 119861 = 1198792= 119868 for an

119899 times 119899 identity matrix 119868 then we have the simplified formula

119860119883 = 1198831198791 (19)

Here the spaceX spanned by the columns of the matrix119883 iscalled an invariant subspace for 119860 and satisfies

119860X sube X (20)

One strategy to analyze the eigeninformation is to transformone matrix pencil to its simplified and equivalent form Thatis two matrix pencils 119860 minus 120582119861 and 119860 minus 120582119861 are said to beequivalent if and only if there exist two nonsingular matrices119875 and 119876 such that

119875 (119860 minus 120582119861)119876 = 119860 minus 120582119861 (21)

In the subsequent discussion we will use the notion sim todescribe this equivalence relation that is 119860 minus 120582119861 sim 119860 minus 120582119861

Our task in this section is to identify eigenvectors ofproblem (18) and then associate these eigenvectors (left andright) with the solution of (1) We begin this analysis bystudying the eigeninformation of two matrices 119860 and 119861where 119860 minus 120582119861 is a regular matrix pencil

4 Abstract and Applied Analysis

Note that for the ordinary eigenvalue problem if theeigenvalues are different then the eigenvectors are linearlyindependent This property is also true for every regularmatrix pencil and is demonstrated as follows For a detailedproof the reader is referred to [13 Theorem 73] and [14Theorem 42]

Theorem5 Given a pair of 119899 times 119899matrix119860 and119861 if thematrixpencil 119860 minus 120582119861 is regular then its Jordan chains correspondingto all finite and infinite eigenvalues carry the full spectralinformation about the matrix pencil and consist of 119899 linearlyindependent vectors

Lemma 6 Let 119860 minus 120582119861 isin C119899times119899 be a regular matrix pencilAssume that matrices 119883

119894 119884119894isin C119899times119899119894 119894 = 1 2 are full rank

and satisfy the following equations

119860119883119894= 119884119894119877119894 (22a)

119861119883119894= 119884119894119878119894 (22b)

where 119877119894and 119878

119894 119894 = 1 2 are square matrices of size 119899

119894times 119899119894

Then

(i) 119877119894minus 120582119878119894isin C119899119894times119899119894 are regular matrix pencils for 119894 = 1 2

(ii) if 120590(1198771minus 1205821198781) cap 120590(119877

2minus 1205821198782) = 120601 then the matrix

[1198831

1198832] isin R119899times(1198991+1198992) is full rank

We also need the following useful lemma

Lemma 7 Given two regular matrix pencils119860119894minus120582119861119894isin C119899119894times119899119894

1 le 119894 le 2 consider the following equations with respect to119880119881 isin C1198991times1198992

1198601119880 = 119881119860

2 (23a)

1198611119880 = 119881119861

2 (23b)

Then if 120590(1198601minus 1205821198611) cap 120590(119860

2minus 1205821198612) = 120601 (23a) has the unique

solution 119880 = 119881 = 0

Proof For 1198992= 1 we get

1198601119906 = 1198862V

1198611119906 = 1198872V

(24)

where 1198862 1198872

isin C 119906 V isin C1198991times1 We may without loss ofgenerality assume that 119887

2= 0 then119860

1119906 = (119886

21198872)1198611119906 and thus

119906 = V = 0 Now for any 1198992

gt 1 consider the generalizedSchur decomposition of 119860

2minus 1205821198612 We can assume that 119860

2=

[119886119894119895] and 119861

2= [119887119894119895] are upper triangular matrices (ie 119886

119894119895=

119887119894119895= 0 1 le 119895 lt 119894 le 119899

2) Denote that the 119894th columns of 119880 and

119881 are 119906119894and V119894 respectively Thus

1198601119906119894=

119894

sum

119896=1

119886119896119894V119896 (25a)

1198611119906119894=

119894

sum

119896=1

119887119896119894V119896 (25b)

for 119894 = 1 2 1198992

If 119894 = 1 we obtained 1199061

= V1

= 0 from the abovediscussion Given an integer 119894 such that 1 le 119894 lt 119899

2and assume

that 119906119896= V119896= 0 for 1 le 119896 le 119894 we claim 119906

119894+1= V119894+1

= 0indeed from (25a) and (25b) we have

1198601119906119894+1

= 119886119894+1119894+1

V119894+1

1198611119906119894+1

= 119887119894+1119894+1

V119894+1

(26)

Again the case 119896 = 119894 + 1 is following the special case 1198992= 1

By using mathematical induction we prove this lemma

Corollary 8 Given119860 isin C119899times119899 andΛ isin C119896times119896 if 120590(119860)cap120590(Λ) =

120601 then the equation with respect to 119880 isin C119899times119896

119860119880 = 119880Λ (27)

has the unique solution 119880 = 0

Now we have enough tools to analyze the solution of (1)associated with some deflating spaces We first establish animportant matrix pencil let the matrix pencil M minus 120582L bedefined as

M minus 120582L = [119861119860⊤

0

minus119862119860⊤

119868119899

] minus 120582 [119868119899

0

119860119862⊤

119860119861⊤] isin R

2119899times2119899

(28)

it is clear that

120590 (M minus 120582L) = 120590 (119861119860⊤) cup 120590 (119868

119899minus 120582119860119861

⊤) (29)

a direct calculation shows that119883 is a solution of the (1) if andonly if

M [119868119899

119883119860⊤] = [

119868119899

119860119883⊤]119861119860

⊤

L [119868119899

119883119860⊤] = [

119868119899

119860119883⊤]

(30)

or if and only if its dual form is

[minus119860119883⊤

119868119899]M = [minus119883119860

⊤119868119899]

[minus119860119883⊤

119868119899]L = 119860119861

⊤[minus119883119860

⊤119868119899]

(31)

Armed with the property given in Theorem 5 andLemma 7 we can now tackle the problemof determining howthe deflating subspace is related to the solution of (1)

Theorem 9 Let 119860 119861 and 119862 isin R119899times119899 be given in (1) and let

M [1198801

1198811

] = [1198802

1198812

]1198791 (32a)

L [1198801

1198811

] = [1198802

1198812

]1198792 (32b)

where [ 119880119894119881119894

] is full rank 119894 = 1 2 Assume that the set of 120590(119861119860⊤)is ⊤-reciprocal free Then one has

Abstract and Applied Analysis 5

(1) 1198801= 1198802= 0 if 120590(119879

1minus 1205821198792) = 120590(119868

119899minus 120582119860119861

⊤)

(2) 1198801and 119880

2are nonsingular if 119879

1minus 1205821198792sim 119861119860⊤minus 120582119868119899

Moreover if 119860 is nonsingular then 119883 = 1198811119880minus1

1119860minus⊤

=

119880minus⊤

2119881⊤

2119860minus⊤ is the unique solution of (1)

Proof From (32a) and (32b) we get

119861119860⊤1198801= 11988021198791 (33a)

minus119862119860⊤1198801+ 1198811= 11988121198791 (33b)

1198801= 11988021198792 (33c)

119860119862⊤1198801+ 119860119861⊤1198811= 11988121198792 (33d)

(i) It follows from (33a) and (33c) that since 120590(119861119860⊤

minus

120582119868119899) cap 120590(119879

1minus 1205821198792) = 120601 we have 119880

1= 1198802

= 0 byLemma 7

(ii) It can be seen that there exist two nonsingularmatrices 119880 and 119881 such that

M [0

119880] = [

0

119881]1198792

L [0

119880] = [

0

119881]1198791

(34)

Hence together with (32a) and (32b) we have

M [0 1198801

119880 1198811

] = [0 1198802

119881 1198812

] [1198792

0

0 1198791

]

L [0 1198801

119880 1198811

] = [0 1198802

119881 1198812

] [1198791

0

0 1198792

]

(35)

Since 120590(Mminus120582L) = 120590(119861119860⊤minus120582119868119899)cup120590(119868

119899minus120582119860119861

⊤) and 120590(119861119860

⊤minus

120582119868119899) cap 120590(119868

119899minus 120582119860119861

⊤) = 120601 by Theorem 5 and Lemma 6 the

matrix [0 1198801

119880 1198811

] is nonsingular Together with (33c)1198801and119880

2

are nonsingularLet119883

119894= 119881119894119880minus1

119894 119894 = 1 2 then form (33b) and (33d)

119860119862⊤+ 119860119861⊤1198831= 11988121198792119880minus1

1= 1198832

minus119862119860⊤+ 1198831= 11988121198791119880minus1

1= 1198832119861119860⊤

(36)

or

119860119862⊤+ 119860119861⊤1198831= 1198832

119860119862⊤+ 119860119861⊤119883⊤

2= 119883⊤

1

(37)

Since the set of 120590(119860119861⊤) = 120590(119861119860

⊤) is ⊤-reciprocal free

together with

119883⊤

1minus 1198832minus 119860119861⊤(119883⊤

1minus 1198832)⊤

= 0 (38)

we get 1198831= 119883⊤

2 If 119860 is nonsingular it is easy to verify that

two matrices 1198831119860minus⊤ and 119883

⊤

2119860minus⊤ are both satisfying ⊤-Stein

equation (1) The proof of part (ii) is complete

Remark 10 (1) It is easily seen that [ 119868119899119883119860⊤ ] and [

1198801

1198812

] both spanthe unique deflating subspace of M minus 120582L corresponding tothe set of 120590(119861119860⊤) Otherwise in part (ii) we know that 119879

2

is nonsingular We then are able to transform the formulaedefined in (32a) and (32b) into the generalized eigenvalueproblem as follows

M [1198801

1198811

] = L [1198801

1198811

]119861119860⊤ (39)

That is some numerical methods for the computation ofthe eigenspace ofMminus120582L corresponding to the set of 120590(119861119860⊤)can be designed and solved (1)

(2) Since the transport of the unique solution 119883 of (1)is equal to the unique solution 119884 of the following matrixequation

119884 = 119861⊤119884119860⊤+ 119862⊤ (40)

analogous to the consequences of Theorem 9 the similarresults can be obtainedwith respect to (40) if119861 is nonsingularHowever we point out that (1) can be solved by computingdeflating subspaces of other matrix pencils For instance welet

M1minus 120582L

1= [

119860⊤119861 0

minus119862 minus 119860119862⊤119861 119868119899

] minus 120582 [119868119899

0

0 119860119861⊤] (41)

Assume that the set of 120590(119861119860⊤) is ⊤-reciprocal free it can beshown thatM

1[119868119899

119883] = L

1[119868119899

119883] 119860⊤119861 and it has similar results

to the conclusion ofTheorem 9The unique solution119883 of (1)can be found by computing deflating subspaces of the matrixpencil M

1minus 120582L

1without the assumption of the singularity

of 119860 and 119861

4 Computational Methods for Solving (1)Numerical methods for solving (1) have received great atten-tion in theory and in practice and can be found in [5 6] forthe Krylov subspace methods and in [15ndash17] for the Smith-type iterative methods In particular Smith-type iterativemethods are only workable in the case 120588(119860119861

⊤) lt 1 where

120588(119860119861⊤) denotes the spectral radius of 119860119861

⊤ In the recentyears a structure algorithm has been studied for (1) [4] viaPQZ decomposition which consists of transforming intoSchur form by a PQZ decomposition and then solving theresulting triangular system by way of back-substitution Inthis section we revisit these numericalmethods and point outthe advantages and drawbacks of all algorithms

41 Krylov Subspace Methods Since the ⊤-Stein equation isessentially a linear system (7) we certainly can use the Krylovsubspace methods to solve (7) See for example [5 6] andthe reference cited therein The general idea for applying theKrylov subspace methods is defining the ⋆-Stein operatorTas T 119883 rarr 119883 minus 119860119883

⊤119861 and its adjoint liner operator T as

Tlowast 119884 rarr 119884 minus 119861119884⊤119860 such that ⟨T(119883) 119884⟩ = ⟨119883Tlowast(119884)⟩

Here 119883 119884 isin R119898times119899 and the notion ⟨sdot sdot⟩ is denoted as theFrobenius inner product Then the iterative method basedon the Krylov subspaces for (1) is as follows

6 Abstract and Applied Analysis

(i) The conjugate gradient (CG) method [5]

119883119896+1

= 119883119896+

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2119875119896

119877119896+1

= 119862 minusT (119883119896+1

) = 119877119896minus

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2T (119875119896)

119863119896+1

= Tlowast(119877119896+1

) +

1003817100381710038171003817119877119896+11003817100381710038171003817

2

10038171003817100381710038171198771198961003817100381710038171003817

2119863119896

(42)

with an initial matrix 1198830and the corresponding initial

conditions

1198770= 119862 minusT (119883

0) 119863

0= Tlowast(1198770) (43)

Note that when the solvability conditions of Theorem 4are met the CG method is guaranteed to converge in a finitenumber of iterations for any initial matrix119883

0

42 The Bartels-Stewart-Like Algorithm [18] In this sectionwe focus on the discussion of the Bartels-Stewart algorithmwhich is known to be a numerical stable algorithm to solve⊤-Stein equationsThis method is to solve (1) by means of thePQZ decomposition [18] Its approach has been discussed in[4] and can be summarized as follows From Lemma 3 weknow that there exist two unitary matrices 119875 and 119876 (see [3]for the computation procedure) such that

119875119883119876 minus 119875119860119876 sdot 119876119867119883⊤119875⊤sdot 119875119861119876 = 119875119862119876 (44)

With 119860 = 119875119860119876 and 119861⊤

= 119876119867119861⊤119875119867 being uppertriangular

the transformed equation looks like

[11988311

11990912

11990921

11990922

] minus [11986011

11988612

0 11988622

] [119883⊤

11119909⊤

21

119909⊤

12119909⊤

22

] [11986111

0

21

22

]

= [11986211

11988812

11988821

11988822

]

(45)

with119883 = [1111990912

1199092111990922

] We then have

11990922

minus 11988622119909⊤

2222

= 11988822 (46)

11990921

minus 11988622119909⊤

1211986111

= 11988821

+ 11988622119909⊤

2221 (47)

11990912

minus 11986011119909⊤

2122

= 11988812

+ 11988612119909⊤

2222 (48)

11988311

minus 11986011119883⊤

1111986111

= 11986211

+ 11988612119909⊤

1211986111

+ 11986011119909⊤

2121

+ 11988612119909⊤

2221

(49)

Thus the Bartels-Stewart algorithm can easily be constructedby first solving 119909

22from (46) using 119909

22to obtain 119909

12and 119909

21

from (47) and (48) and then repeating the same discussionas (46)ndash(48) by taking advantage of the property of 119860

11and

11986111being lower triangular matrices from (49)

43 Smith-Type Iterative Methods Recently a class methodreferred to as the Smith accelerative iteration has been studiedfor solving the Sylvester matrix equation [15] The Smithaccelerative iteration has attracted much interests in manypapers (see [7 17] and the references therein) because of itsnice numerical behavior a quadratic convergence rate lowcomputational cost and high numerical reliability despite thelack of a rigorous error analysis Since the Sylvester matrixequation can be transformed into the Stein matrix equationwith a suitable transformation Zhou et al proposed Smith-type iterative methods (including the Smith accelerativeiteration Smith (ℓ) iteration and 119903-Smith iteration) forsolving the Stein matrix equation [17] and applying Smith-type iterative methods to (1) [7] In this section we try toexplain the Smith accelerative iteration based on the invariantsubspace method and summarize the recent results from [7]

Originally the Smith-type iterative methods are devel-oped to solve the standard Stein equation

119883 = A119883B +C ABC isin R119899times119899

(50)

As mentioned before the unknown119883 is highly related to thegeneralized eigenspace problems

[B 0

minusC 119868] [

119868

119883] = [

119868 0

0 A] [

119868

119883]B (51a)

or

A [119883 119868] [B 0

0 119868] = [119883 119868] [

119868 0

minusC A] (51b)

Premultiplying (51a) by the matrix [ B 0minusAC 119868 ] and postmul-

tiplying (51b) by the matrix [119868 0

minusCB A ] we get

[B2 0

minusC minusACB 119868119899

] [119868119899

119883] = [

119868119899

0

0 A2] [

119868119899

119883]B2

A2[119883 119868119899] [

B2 0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minusC minusACB A2]

(52)

Then for any positive integer 119896 gt 0 we obtain

[B2119896minus1

0

minus119862119896

119868119899

] [119868119899

119883] = [

119868119899

0

0 A2119896minus1] [

119868119899

119883]B2119896minus1

A2119896minus1

[119883 119868119899] [

B2119896minus1

0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minus119862119896

A2119896minus1]

(53)

where the sequence C119896 is defined by

119862119896= 119862119896minus1

+A2119896minus1

119862119896minus1

B2119896minus1

119896 ge 1 (54a)

1198620= C (54b)

The explicit expression of 119862119896is given as follows

119862119896=

2119896minus1

sum

119894=1

A119894CB119894 (55)

Abstract and Applied Analysis 7

Under the condition 120588(A)120588(B) lt 1 it is easy to see that 119862119896

is convergent and

lim sup119896rarrinfin

2119896

radic1003817100381710038171003817119883 minus 119862

119896

1003817100381710038171003817 le 120588 (A) 120588 (B) (56)

that is 119862119896converges quadratically to 119883 as 119896 rarr infin

This iterative method (54a) and (54b) is called the Smithaccelerative iteration [15] In recent years some modifiediterative methods are so-called Smith-type iteration whichare based on Smith iteration and improve its speed ofconvergence See for example [17] and the references citedtherein

Since the condition 120588(A)120588(B) lt 1 implies that theassumptions of Theorem 4 hold (1) is equivalent to (5) Wecan apply the Smith iteration to the (1) with the substitution(ABC) = (119860119861

⊤ 119860⊤119861 119862+119860119862

⊤119861) One possible drawback

of the Smith-type iterativemethods is that they cannot alwayshandle the case when there exist eigenvalues 120582 120583 isin 120590(119860

⊤119861)

such that 120582120583 = minus1 even if the unique solution119883 exists Basedon the solvable conditions given in this work it is possible todevelop a specific technique working on the particular caseand it is a subject currently under investigation

5 Error Analysis

Error analysis is a way for testing the stability of a numericalalgorithm and evaluating the accuracy of an approximatedsolution In the subsequent discussion we want to considerthe backward error and perturbation bounds for solving (1)

As indicated in (44) matrices 119860 and 119861⊤ are both upper-

triangularWe can then apply the error analysis for triangularlinear systems in [19 Section 31] and [20] to obtain

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 119888119898119899

u (1 +1003817100381710038171003817100381711986010038171003817100381710038171003817119865

1003817100381710038171003817100381711986110038171003817100381710038171003817119865

)1003817100381710038171003817100381711988310038171003817100381710038171003817119865

(57)

where 119888119898119899

is a content depending on the dimensions 119898 and119899 and u is the unit roundoff Since the PQZ decomposition isa stable process it is true that

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 1198881015840

119898119899u (1 + 119860

119865119861119865) 119883119865 (58)

with a modest multiple 1198881015840119898119899

Note that the inequality of the form (58) can serve

as a stopping criterion for terminating iterations generatedfrom the Krylov subspace methods [5 6] and Smith-typeiterative methods [15ndash17] In what follows we will derivethe error associated with numerical algorithms following thedevelopment in [20 21]

51 Backward Error Like the discussion of the ordinarySylvester equations [20] the normwise backward error of anapproximate solution 119884 of (1) is defined by

120578 (119884) equiv min 120598 119884 = (119862 + 120575119862) + (119860 + 120575119860)119884⊤

(119861 + 120575119861)

120575119860119865le 120598120572 120575119861

119865le 120598120573 120575119862

119865le 120598120574

(59)

where 120572 equiv 119860119865 120573 equiv 119861

119865 and 120574 equiv 119862

119865 Let R equiv 120575119862 +

120575119860119884⊤119861 + 119860119884

⊤120575119861 + 120575119860119884

⊤120575119861 which implies that R = 119884 minus

119860119884⊤119861 minus 119862 It can be seen that the residualR satisfies

R119865le 120578 (119884) (120574 + 119884

119865120572120573 (2 + 120578 (119884))) (60)

From (60) we know that a small backward error indeedimplies a small relative residual R Since the coefficientmatrices in (1) include nonlinearity it appears to be an openproblem to obtain the theoretical backward error with respectto the residual Again similar to the Sylvester equationdiscussed in [20 Section 162] the conditions under whicha ⊤-Stein equation has a well-conditioned solution remainunknown

52 Perturbation Bounds Consider the perturbed equation

119883 + 120575119883 = (119860 + 120575119860) (119883 + 120575119883)⊤

(119861 + 120575119861) + (119862 + 120575119862)

(61)

Let 119878(119883) = 119883minus119860119883⊤119861 be the corresponding⊤-Stein operator

We then have 119878(120575119883) = 120575119862+119860(119883+120575119883)⊤120575119861+120575119860(119883+120575119883)

⊤(119861+

120575119861) With the application of norm it follows that

120575119883119865le10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

120575119862119865+ 120575119878

119865(119883119865+ 120575119883

119865) (62)

where 120575119878119865

equiv 119860119865120575119861119865+ 120575119860

119865(119861119865+ 120575119861

119865) When

120575119878119865is small enough so that 1 ge 119878

minus1119865sdot 120575119878

119865 we can

rearrange the previous result to be

120575119883119865

119883119865

le

10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

1 minus1003817100381710038171003817119878minus11003817100381710038171003817119865

sdot 120575119878119865

(120575119862119865

119883119865

+ 120575119878119865) (63)

With 119862119865

= 119878(119883)119865

le 119878119865sdot 119883119865and the condition

number 120581(119878) equiv 119878119865

sdot 119878minus1119865 we arrive at the standard

perturbation result120575119883119865

119883119865

le120581 (119878)

1 minus 120581 (119878) sdot 120575119878119865119878119865

(120575119862119865

119862119865

+120575119878119865

119878119865

)

(64)

Thus the relative error in119883 is controlled by those in119860 119861 and119862 magnified by the condition number 120581(119878)

On the other hand we can also drop the high-order termsin the perturbation to obtain

120575119883 minus 119860120575119883⊤119861 = 119860119883

⊤120575119861 + 120575119860119883

⊤119861 + 120575119862 (65)

We then rewrite the system in terms of

Q vec (120575119883) = [(119883⊤119861)⊤

otimes 119868119898119868119899otimes (119860119883

⊤) 119868119898119899

] [

[

vec (120575119860)

vec (120575119861)vec (120575119862)

]

]

(66)

where Q = 119868119898119899

minus (119861⊤

otimes 119860)P Let 120577 = max120575119860119865119860119865

120575119861119865119861119865 120575119862

119865119862119865 It can be shown that

120575119883119865

119883119865

le radic3Ψ120577 (67)

where Ψ=Qminus1[120572(119883⊤119861)⊤otimes 119868119898

120573119868119899otimes (119860119883

⊤)120574119868119898119899

]2119883119865

8 Abstract and Applied Analysis

A possible disadvantage of the perturbation bound (67)which ignores the consideration of the underlying structureof the problem is that it overestimates the effect of theperturbation on the data But this ldquouniversalrdquo perturbationbound is accessible to any given matrices 119860 119861 and 119862 of (1)

Unlike the perturbation bound (67) it is desirable toobtain a posteriori error bound by assuming 120575119860 = 120575119861 = 0

and 120575119862 = 119883 minus 119860119883⊤119861 minus 119862 in (61) This assumption gives rise

to

120575119883119865

119883119865

le

10038171003817100381710038171003817119875minus1100381710038171003817100381710038172119877119865

119883119865

(68)

It is true that while doing numerical computation this boundgiven in (68) provides a simpler way for estimating the errorof the solution of (1)

6 Conclusion

In this note we propose a novel approach to the necessaryand sufficient conditions for the unique solvability of thesolution 119883 of the ⊤-Stein equation for square coefficientmatrices in terms of the analysis of the spectra 120590(119860

⊤119861)

Solvability conditions have been derived and algorithmshave been proposed in [4 8] by using PQZ decompositionOn the other hand one common procedure to solve theStein-type equations is by means of the invariant subspacemethod We believe that our discussion is the first whichimplements the techniques of the deflating subspace forsolving ⊤-Stein matrix equation and might also give riseto the possibility of developing an advanced and effectivesolver in the future Also we obtain the theoretical residualanalysis backward error analysis and perturbation boundsfor measuring accurately the error in the computed solutionof (1)

Acknowledgments

The author wishes to thank Proffessor Eric King-wah Chu(Monash University Australia) Proffessor Matthew M Lin(National Chung Cheng University Taiwan) and two anony-mous referees for many interesting and valuable suggestionson the paperThis research work is partially supported by theNational Science Council (NSC101-2115-M-150-002) and theNational Center for Theoretical Sciences in Taiwan

References

[1] H W Braden ldquoThe equations 119860⊤119883 plusmn 119883⊤119860 = 119861rdquo SIAM Journal

on Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021998

[2] X Ma J Han and F Pu ldquoA study of solution existence formatrix equation 119860119883 + 119860

⊤119862 = 119861rdquo Journal of Shenyang Institute

of Aeronautical Engineering vol 4 pp 64ndash66 2003[3] A W Bojanczyk G H Golub and P Van Dooren ldquoPeri-

odic schur decomposition algorithms and applicationsrdquo inAdvanced Signal Processing Algorithms Architectures andImplementations vol 1770 of Proceedings of SPIE pp 31ndash42 SanDiego Calif USA July 1992

[4] C-Y Chiang E K-W Chu andW-W Lin ldquoOn the lowast-Sylvesterequation 119860119883 plusmn 119883

lowast119861lowastrdquo Applied Mathematics and Computation

vol 218 no 17 pp 8393ndash8407 2012[5] Y Su and G Chen ldquoIterative methods for solving linear matrix

equation and linear matrix systemrdquo International Journal ofComputer Mathematics vol 87 no 4 pp 763ndash774 2010

[6] M Wang X Cheng and M Wei ldquoIterative algorithms forsolving the matrix equation 119860119883119861 + 119862119883

⊤119863 = 119864rdquo Applied

Mathematics andComputation vol 187 no 2 pp 622ndash629 2007[7] B Zhou J Lam and G-R Duan ldquoToward solution of matrix

equation 119860119891(119883)119861 + 119862rdquo Linear Algebra and its Applications vol435 no 6 pp 1370ndash1398 2011

[8] K D Ikramov and O Yu Vorontsov ldquoThe matrix equation119883+

119860119883⊤119861 = 119862 conditions for unique solvability and a numerical

algorithm for its solutionrdquo Doklady Mathematics vol 85 pp265ndash267 2012

[9] D S Bernstein Matrix Mathematics Theory Facts and For-mulas Princeton University Press Princeton NJ USA 2ndedition 2009

[10] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1994 Correctedreprint of the 1991 original

[11] F ZhangMatrix Theory Basic Results and Techniques Univer-sitext Springer 1999

[12] D A Bini B Iannazzo and B Meini Numerical Solution ofAlgebraic Riccati Equations Society for Industrial and AppliedMathematics (SIAM) Philadelphia Pa USA 2012

[13] I Gohberg P Lancaster and L Rodman Matrix PolynomialsComputer Science and Applied Mathematics Academic PressNew York NY USA 1982

[14] C-Y Chiang andMM Lin ldquoDeflating subspaces and Sylvesterequationsrdquo Tech Rep NCTS Taiwan 2013

[15] R A Smith ldquoMatrix equation119883119860+ 119861119883 = 119862rdquo SIAM Journal onApplied Mathematics vol 16 pp 198ndash201 1968

[16] T Penzl ldquoA cyclic low-rank Smith method for large sparseLyapunov equationsrdquo SIAM Journal on Scientific Computingvol 21 no 4 pp 1401ndash1418 1999

[17] B Zhou J Lam and G-R Duan ldquoOn Smith-type iterativealgorithms for the Stein matrix equationrdquo Applied MathematicsLetters vol 22 no 7 pp 1038ndash1044 2009

[18] RH Bartels andGW Stewart ldquoSolution of thematrix equation119860119883+119883119861 = 119862 algorithm 432rdquoCommunications of the ACM vol15 no 9 pp 820ndash826 1972

[19] G H Golub and C F Van Loan Matrix Computations JohnsHopkins Studies in the Mathematical Sciences Johns HopkinsUniversity Press Baltimore Md USA 3rd edition 1996

[20] N J Higham Accuracy and Stability of Numerical Algo-rithms Society for Industrial andAppliedMathematics (SIAM)Philadelphia Pa USA 2nd edition 2002

[21] A R Ghavimi and A J Laub ldquoBackward error sensitivityand refinement of computed solutions of algebraic Riccatiequationsrdquo Numerical Linear Algebra with Applications vol 2no 1 pp 29ndash49 1995

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

In [4 8] the periodic QZ (PQZ) decomposition [3] isapplied to consider the necessary and sufficient conditions ofthe unique solvability of (1) conditions given in [8] ignorethe possibility of the existence of the unique solution while 1is a simple root of 120590(119860⊤119861) This condition is included in oursubsequent discussion and the following remark is providedto support our observation

Remark 1 Let 119860 = minus1 and let 119861 = 1 that is 120590(119860119861⊤) = minus1

It is clear that the scalar equation 119883 = minus119883⊤

+ 119862 has aunique solution 119883 = 1198622 But condition (6) is not satisfiedby choosing 120583 = ] = minus1

It can also be observed from Remark 1 that even if (1) isuniquely solvable it does not imply that (5) (namely119883 = 119883+

119862 minus 119862) is uniquely solvable Conditions in [4 Equation(46)]provided those conditions for the unique solvability of thesolution to (1) via a structured algorithm In our workthrough a complete analysis of square coefficient matrices interms of the analysis of the spectra of thematrix119860⊤119861 the newapproach to the condition of unique solvability of the⊤-Steinequation (1) can be obtained

Second we present the invariant subspace method andmore generally the deflating subspace method to solve the⊤-Stein equation Our methods are based on the analysis ofthe eigeninformations for a matrix pencil We carry out athorough discussion to address the various eigeninformationencountered in the subspace methods These ideas can beimplemented in algorithms easily

Finally we take full account of the error analysis of(1) Expressions and implications such as the residual thebackward error and perturbation bounds are derived in thiswork Note that for an approximate solution 119884 of (1) thebackward error tells us how much the matrices 119860 119861 and 119862

must be perturbed An important point found in Section 5is that a small backward error indicates a small value for theresidual R = 119884 minus 119860119884

⊤119861 minus 119862 but the reverse is not usually

trueBeginning in Section 2 we formulate the necessary and

sufficient conditions for the existence of the solution of (1)directly by means of the spectrum analysis In Section 3we provide a deflating subspace method for computingthe solution of (1) Numerical methods for solving (1) andthe related residual analysis are discussed in Section 4 Theassociated error analysis of (1) is given in Section 5 andconcluding remarks are given in Section 6

2 Solvability Conditions ofthe Matrix Equation (1)

In order to formalize our discussion let 119860 otimes 119861 be theKronecker product of matrices 119860 and 119861 let 119868

119899be the 119899 times 119899

identity matrix and let sdot 119865denote the Frobenius norm

With the Kronecker product (1) can be written as theenlarged linear system

(1198681198992 minus (119861

⊤otimes 119860)P) vec (119883) = vec (119862) (7)

where vec(119883) stacks the columns of 119883 into a column vectorandP is the Kronecker permutation matrix [9] which mapsvec(119883) into vec(119883⊤) that is

P = sum

1le119894119895le1198992

119890119895119890⊤

119894otimes 119890119894119890⊤

119895 (8)

where 119890119894denotes the 119894th column of the 1198992 times 119899

2 identitymatrix1198681198992 Due to the specific structure of P it has been shown in

[10 Corollary 4310] that

P⊤(119861⊤otimes 119860)P = 119860 otimes 119861

⊤ (9)

It then follows that

((119861⊤otimes 119860)P)

2

= (119861⊤otimes 119860)PP

⊤(119860 otimes 119861

⊤) = 119861⊤119860 otimes 119860119861

⊤

(10)

since P2 = 1198681198992 and P = P⊤ Note that eigenvalues of

matrices119860⊤119861 and119860119861⊤ are the same By (10) and the property

of the Kronecker product [11 Theorem 48] we know that

120590 (((119861⊤otimes 119860)P)

2

)

= 120582119894120582119895| 120582119894 120582119895isin 120590 (119860

⊤119861) = 120582

1 120582

119899 1 le 119894 119895 le 119899

(11)

That is the eigenvalues of (119861⊤

otimes 119860)P are related to thesquare roots of the eigenvalues of 120590(119860

⊤119861) but from (10)

no more information can be used to decide the positivity ornonnegativity of the eigenvalues of (119861⊤ otimes 119860)P A questionimmediately arises as to whether it is possible to obtainthe explicit expression of the eigenvalues of (119861

⊤otimes 119860)P

provided the eigenvalues of 119860⊤119861 are given In the followingtwo lemmas we first review the periodic QZ decompositionfor two matrices and then apply it to discuss the eigenvaluesof (119861⊤ otimes 119860)P

Lemma2 (see [3]) Let119860 and119861 be twomatrices inR119899times119899Thenthere exist unitary matrices 119875119876 isin C119899times119899 such that 119880

119860= 119875119860119876

and 119880119861= 119876119867119861⊤119875119867 are two upper triangular matrices

Lemma 3 Let 119860 and 119861 be two matrices in R119898times119899 Then

(1) (119861⊤ otimes 119860)P = (119876 otimes 119875119867)(119880119860otimes 119880119861)P(119876

119867otimes 119875)

(2) 120590((119861⊤ otimes 119860)P) = 120582119894 plusmnradic120582

119894120582119895| 120582119894 120582119895isin 120590(119860

⊤119861) =

1205821 120582

119899 1 le 119894 lt 119895 le 119899

Hereradic119911 denotes the principal square root of a complex number119911

Proof Part 1 follows immediately from Lemma 2 since 119880119860=

119875119860119876 and 119880119861= 119876119867119861⊤119875119867 for some unitary matrices 119875 and 119876

that is

(119861⊤otimes 119860)P = (119876 otimes 119875

119867) (119880119861otimes 119880119860) (119875 otimes 119876

119867)P

= (119876 otimes 119875119867) (119880119860otimes 119880119861)P (119876

119867otimes 119875)

(12)

Abstract and Applied Analysis 3

Let the diagonal entries of 119880119860and 119880

119861be denoted by 119886

119894119894

and 119887119895119895 respectively Then 119880

119860otimes 119880119861is an upper triangular

matrix with given diagonal entries specified by 119886119894119894and 119887119895119895

After multiplying119880119860

otimes 119880119861byP from the right the position

of the entry 119886119894119894119887119895119895is changed to be in the 119895+119899(119894minus1) th row and

the 119894 + 119899(119895 minus 1) th column of the matrix (119880119860

otimes 119880119861)P They

are then reshuffled by a sequence of permutation matrices toform a block upper triangular matrix with diagonal entriesarranged in the following order

1198861111988711 [

0 1198861111988722

1198862211988711

0] [

0 11988611119887119899119899

11988611989911989911988711

0]

1198862211988722 [

0 1198862211988733

1198863311988722

0] [

0 11988611989911989911988722

11988622119887119899119899

0]

[0 119886

119899minus1119899minus1119887119899119899

119886119899119899119887119899minus1119899minus1

0] 119886119899119899119887119899119899

(13)

Note that the reshuffling process is not hard to see when 119899 =

2 119880119860= [1198861111988612

0 11988622] and 119880

119861= [1198871111988712

0 11988722

] we have

(119880119860otimes 119880119861)P =

[[[

[

1198861111988711

1198861211988711

1198861111988712

1198861211988712

0 0 1198861111988722

1198861211988722

0 1198862211988711

0 1198862211988712

0 0 0 1198862211988722

]]]

]

(14)

However it is conceptually simple and regular but opera-tionally tedious to reorder (119880

119860otimes 119880119861)P to show this result

even for 119899 = 3 and that will be left as an exerciseBy (13) it can be seen that

120590 ((119861⊤otimes 119860)P) = 119886

119894119894119887119894119894 plusmnradic119886119894119894119886119895119895119887119894119894119887119895119895 1 le 119894 119895 le 119899

= 120582119894 plusmnradic120582

119894120582119895 1 le 119894 119895 le 119899

(15)

where 120582119894= 119886119894119894119887119894119894isin 120590(119860

⊤119861) for 1 le 119894 le 119899

Before demonstrating the unique solvability conditionswe need to define that a subset Λ = 120582

1 120582

119899 of

complex numbers is said to be ⊤-reciprocal free if and onlyif whenever 119894 = 119895 120582

119894= 1120582119895 This definition also regards 0 and

infin as reciprocals of each other Then we have the followingsolvability conditions of (1)

Theorem 4 The ⊤-Stein matrix equation (1) is uniquelysolvable if and only if the following conditions are satisfied

(1) the set of 120590(119860⊤119861) minus1 is ⊤-reciprocal free(2) minus1 can be an eigenvalue of the matrix119860⊤119861 but must be

simple

Proof From (7) we know that the ⊤-Stein matrix equation(1) is uniquely solvable if and only if

1 notin 120590 ((119861⊤otimes 119860)P) (16)

By Lemma 3 if 120582 isin 120590(119860⊤119861) then 1120582 notin 120590(119860

⊤119861) Otherwise

1 = radic120582 sdot (1120582) isin ((119861⊤

otimes 119860)P) On the other hand ifminus1 isin 120590(119860

⊤119861) and minus1 is not a simple eigenvalue then 1 isin

120590((119861⊤otimes119860)P) This verifies (16) and the proof of the theorem

is complete

It is worthy noting that the condition (1) of Theorem 4 iscontained in the condition (6) (also appear in [7 Theorem1]) which is the necessary and sufficient conditions forthe solvability of the solution to the Stein equation (5)However as mentioned before in Remark 1 or [7 Example2] condition (1) is just a necessary condition for uniquesolvability of the solution to (1) The ⊤-Stein matrix equation(1) is uniquely solvable provided that both conditions (1) and(2) of Theorem 4 are satisfying

3 The Connection between DeflatingSubspace and (1)

The relationship between solution of matrix equations andthe matrix eigenvalue problems has been widely studied inmany applications It is famous that solution of Riccati andpolynomial matrix equations can be found by computinginvariant subspaces of matrices and deflating subspaces ofmatrix pencils [12] This reality leads us to find some algo-rithms for computing solution of (1) based on the numericalcomputation of invariant or deflating subspaces

Given a pair of 119899 times 119899 matrices 119860 and 119861 recall that thefunction119860minus120582119861 in the variable 120582 is said to be thematrix pencilrelated to the pair (119860 119861) For a 119896-dimensional subspaceX isin

C119899 is called a deflating subspace for the pencil119860minus120582119861 if thereexists a 119896-dimensional subspaceY isin C119899 such that

119860X sube Y 119861X sube Y (17)

that is

119860119883 = 1198841198791 119861119883 = 119884119879

2 (18)

where119883119884 isin C119899times119896 are two full rank matrices whose columnsspan the spacesX andY respectively and matrices 119879

1 1198792isin

C119896times119896 In particular if in (18) 119883 = 119884 and 119861 = 1198792= 119868 for an

119899 times 119899 identity matrix 119868 then we have the simplified formula

119860119883 = 1198831198791 (19)

Here the spaceX spanned by the columns of the matrix119883 iscalled an invariant subspace for 119860 and satisfies

119860X sube X (20)

One strategy to analyze the eigeninformation is to transformone matrix pencil to its simplified and equivalent form Thatis two matrix pencils 119860 minus 120582119861 and 119860 minus 120582119861 are said to beequivalent if and only if there exist two nonsingular matrices119875 and 119876 such that

119875 (119860 minus 120582119861)119876 = 119860 minus 120582119861 (21)

In the subsequent discussion we will use the notion sim todescribe this equivalence relation that is 119860 minus 120582119861 sim 119860 minus 120582119861

Our task in this section is to identify eigenvectors ofproblem (18) and then associate these eigenvectors (left andright) with the solution of (1) We begin this analysis bystudying the eigeninformation of two matrices 119860 and 119861where 119860 minus 120582119861 is a regular matrix pencil

4 Abstract and Applied Analysis

Note that for the ordinary eigenvalue problem if theeigenvalues are different then the eigenvectors are linearlyindependent This property is also true for every regularmatrix pencil and is demonstrated as follows For a detailedproof the reader is referred to [13 Theorem 73] and [14Theorem 42]

Theorem5 Given a pair of 119899 times 119899matrix119860 and119861 if thematrixpencil 119860 minus 120582119861 is regular then its Jordan chains correspondingto all finite and infinite eigenvalues carry the full spectralinformation about the matrix pencil and consist of 119899 linearlyindependent vectors

Lemma 6 Let 119860 minus 120582119861 isin C119899times119899 be a regular matrix pencilAssume that matrices 119883

119894 119884119894isin C119899times119899119894 119894 = 1 2 are full rank

and satisfy the following equations

119860119883119894= 119884119894119877119894 (22a)

119861119883119894= 119884119894119878119894 (22b)

where 119877119894and 119878

119894 119894 = 1 2 are square matrices of size 119899

119894times 119899119894

Then

(i) 119877119894minus 120582119878119894isin C119899119894times119899119894 are regular matrix pencils for 119894 = 1 2

(ii) if 120590(1198771minus 1205821198781) cap 120590(119877

2minus 1205821198782) = 120601 then the matrix

[1198831

1198832] isin R119899times(1198991+1198992) is full rank

We also need the following useful lemma

Lemma 7 Given two regular matrix pencils119860119894minus120582119861119894isin C119899119894times119899119894

1 le 119894 le 2 consider the following equations with respect to119880119881 isin C1198991times1198992

1198601119880 = 119881119860

2 (23a)

1198611119880 = 119881119861

2 (23b)

Then if 120590(1198601minus 1205821198611) cap 120590(119860

2minus 1205821198612) = 120601 (23a) has the unique

solution 119880 = 119881 = 0

Proof For 1198992= 1 we get

1198601119906 = 1198862V

1198611119906 = 1198872V

(24)

where 1198862 1198872

isin C 119906 V isin C1198991times1 We may without loss ofgenerality assume that 119887

2= 0 then119860

1119906 = (119886

21198872)1198611119906 and thus

119906 = V = 0 Now for any 1198992

gt 1 consider the generalizedSchur decomposition of 119860

2minus 1205821198612 We can assume that 119860

2=

[119886119894119895] and 119861

2= [119887119894119895] are upper triangular matrices (ie 119886

119894119895=

119887119894119895= 0 1 le 119895 lt 119894 le 119899

2) Denote that the 119894th columns of 119880 and

119881 are 119906119894and V119894 respectively Thus

1198601119906119894=

119894

sum

119896=1

119886119896119894V119896 (25a)

1198611119906119894=

119894

sum

119896=1

119887119896119894V119896 (25b)

for 119894 = 1 2 1198992

If 119894 = 1 we obtained 1199061

= V1

= 0 from the abovediscussion Given an integer 119894 such that 1 le 119894 lt 119899

2and assume

that 119906119896= V119896= 0 for 1 le 119896 le 119894 we claim 119906

119894+1= V119894+1

= 0indeed from (25a) and (25b) we have

1198601119906119894+1

= 119886119894+1119894+1

V119894+1

1198611119906119894+1

= 119887119894+1119894+1

V119894+1

(26)

Again the case 119896 = 119894 + 1 is following the special case 1198992= 1

By using mathematical induction we prove this lemma

Corollary 8 Given119860 isin C119899times119899 andΛ isin C119896times119896 if 120590(119860)cap120590(Λ) =

120601 then the equation with respect to 119880 isin C119899times119896

119860119880 = 119880Λ (27)

has the unique solution 119880 = 0

Now we have enough tools to analyze the solution of (1)associated with some deflating spaces We first establish animportant matrix pencil let the matrix pencil M minus 120582L bedefined as

M minus 120582L = [119861119860⊤

0

minus119862119860⊤

119868119899

] minus 120582 [119868119899

0

119860119862⊤

119860119861⊤] isin R

2119899times2119899

(28)

it is clear that

120590 (M minus 120582L) = 120590 (119861119860⊤) cup 120590 (119868

119899minus 120582119860119861

⊤) (29)

a direct calculation shows that119883 is a solution of the (1) if andonly if

M [119868119899

119883119860⊤] = [

119868119899

119860119883⊤]119861119860

⊤

L [119868119899

119883119860⊤] = [

119868119899

119860119883⊤]

(30)

or if and only if its dual form is

[minus119860119883⊤

119868119899]M = [minus119883119860

⊤119868119899]

[minus119860119883⊤

119868119899]L = 119860119861

⊤[minus119883119860

⊤119868119899]

(31)

Armed with the property given in Theorem 5 andLemma 7 we can now tackle the problemof determining howthe deflating subspace is related to the solution of (1)

Theorem 9 Let 119860 119861 and 119862 isin R119899times119899 be given in (1) and let

M [1198801

1198811

] = [1198802

1198812

]1198791 (32a)

L [1198801

1198811

] = [1198802

1198812

]1198792 (32b)

where [ 119880119894119881119894

] is full rank 119894 = 1 2 Assume that the set of 120590(119861119860⊤)is ⊤-reciprocal free Then one has

Abstract and Applied Analysis 5

(1) 1198801= 1198802= 0 if 120590(119879

1minus 1205821198792) = 120590(119868

119899minus 120582119860119861

⊤)

(2) 1198801and 119880

2are nonsingular if 119879

1minus 1205821198792sim 119861119860⊤minus 120582119868119899

Moreover if 119860 is nonsingular then 119883 = 1198811119880minus1

1119860minus⊤

=

119880minus⊤

2119881⊤

2119860minus⊤ is the unique solution of (1)

Proof From (32a) and (32b) we get

119861119860⊤1198801= 11988021198791 (33a)

minus119862119860⊤1198801+ 1198811= 11988121198791 (33b)

1198801= 11988021198792 (33c)

119860119862⊤1198801+ 119860119861⊤1198811= 11988121198792 (33d)

(i) It follows from (33a) and (33c) that since 120590(119861119860⊤

minus

120582119868119899) cap 120590(119879

1minus 1205821198792) = 120601 we have 119880

1= 1198802

= 0 byLemma 7

(ii) It can be seen that there exist two nonsingularmatrices 119880 and 119881 such that

M [0

119880] = [

0

119881]1198792

L [0

119880] = [

0

119881]1198791

(34)

Hence together with (32a) and (32b) we have

M [0 1198801

119880 1198811

] = [0 1198802

119881 1198812

] [1198792

0

0 1198791

]

L [0 1198801

119880 1198811

] = [0 1198802

119881 1198812

] [1198791

0

0 1198792

]

(35)

Since 120590(Mminus120582L) = 120590(119861119860⊤minus120582119868119899)cup120590(119868

119899minus120582119860119861

⊤) and 120590(119861119860

⊤minus

120582119868119899) cap 120590(119868

119899minus 120582119860119861

⊤) = 120601 by Theorem 5 and Lemma 6 the

matrix [0 1198801

119880 1198811

] is nonsingular Together with (33c)1198801and119880

2

are nonsingularLet119883

119894= 119881119894119880minus1

119894 119894 = 1 2 then form (33b) and (33d)

119860119862⊤+ 119860119861⊤1198831= 11988121198792119880minus1

1= 1198832

minus119862119860⊤+ 1198831= 11988121198791119880minus1

1= 1198832119861119860⊤

(36)

or

119860119862⊤+ 119860119861⊤1198831= 1198832

119860119862⊤+ 119860119861⊤119883⊤

2= 119883⊤

1

(37)

Since the set of 120590(119860119861⊤) = 120590(119861119860

⊤) is ⊤-reciprocal free

together with

119883⊤

1minus 1198832minus 119860119861⊤(119883⊤

1minus 1198832)⊤

= 0 (38)

we get 1198831= 119883⊤

2 If 119860 is nonsingular it is easy to verify that

two matrices 1198831119860minus⊤ and 119883

⊤

2119860minus⊤ are both satisfying ⊤-Stein

equation (1) The proof of part (ii) is complete

Remark 10 (1) It is easily seen that [ 119868119899119883119860⊤ ] and [

1198801

1198812

] both spanthe unique deflating subspace of M minus 120582L corresponding tothe set of 120590(119861119860⊤) Otherwise in part (ii) we know that 119879

2

is nonsingular We then are able to transform the formulaedefined in (32a) and (32b) into the generalized eigenvalueproblem as follows

M [1198801

1198811

] = L [1198801

1198811

]119861119860⊤ (39)

That is some numerical methods for the computation ofthe eigenspace ofMminus120582L corresponding to the set of 120590(119861119860⊤)can be designed and solved (1)

(2) Since the transport of the unique solution 119883 of (1)is equal to the unique solution 119884 of the following matrixequation

119884 = 119861⊤119884119860⊤+ 119862⊤ (40)

analogous to the consequences of Theorem 9 the similarresults can be obtainedwith respect to (40) if119861 is nonsingularHowever we point out that (1) can be solved by computingdeflating subspaces of other matrix pencils For instance welet

M1minus 120582L

1= [

119860⊤119861 0

minus119862 minus 119860119862⊤119861 119868119899

] minus 120582 [119868119899

0

0 119860119861⊤] (41)

Assume that the set of 120590(119861119860⊤) is ⊤-reciprocal free it can beshown thatM

1[119868119899

119883] = L

1[119868119899

119883] 119860⊤119861 and it has similar results

to the conclusion ofTheorem 9The unique solution119883 of (1)can be found by computing deflating subspaces of the matrixpencil M

1minus 120582L

1without the assumption of the singularity

of 119860 and 119861

4 Computational Methods for Solving (1)Numerical methods for solving (1) have received great atten-tion in theory and in practice and can be found in [5 6] forthe Krylov subspace methods and in [15ndash17] for the Smith-type iterative methods In particular Smith-type iterativemethods are only workable in the case 120588(119860119861

⊤) lt 1 where

120588(119860119861⊤) denotes the spectral radius of 119860119861

⊤ In the recentyears a structure algorithm has been studied for (1) [4] viaPQZ decomposition which consists of transforming intoSchur form by a PQZ decomposition and then solving theresulting triangular system by way of back-substitution Inthis section we revisit these numericalmethods and point outthe advantages and drawbacks of all algorithms

41 Krylov Subspace Methods Since the ⊤-Stein equation isessentially a linear system (7) we certainly can use the Krylovsubspace methods to solve (7) See for example [5 6] andthe reference cited therein The general idea for applying theKrylov subspace methods is defining the ⋆-Stein operatorTas T 119883 rarr 119883 minus 119860119883

⊤119861 and its adjoint liner operator T as

Tlowast 119884 rarr 119884 minus 119861119884⊤119860 such that ⟨T(119883) 119884⟩ = ⟨119883Tlowast(119884)⟩

Here 119883 119884 isin R119898times119899 and the notion ⟨sdot sdot⟩ is denoted as theFrobenius inner product Then the iterative method basedon the Krylov subspaces for (1) is as follows

6 Abstract and Applied Analysis

(i) The conjugate gradient (CG) method [5]

119883119896+1

= 119883119896+

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2119875119896

119877119896+1

= 119862 minusT (119883119896+1

) = 119877119896minus

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2T (119875119896)

119863119896+1

= Tlowast(119877119896+1

) +

1003817100381710038171003817119877119896+11003817100381710038171003817

2

10038171003817100381710038171198771198961003817100381710038171003817

2119863119896

(42)

with an initial matrix 1198830and the corresponding initial

conditions

1198770= 119862 minusT (119883

0) 119863

0= Tlowast(1198770) (43)

Note that when the solvability conditions of Theorem 4are met the CG method is guaranteed to converge in a finitenumber of iterations for any initial matrix119883

0

42 The Bartels-Stewart-Like Algorithm [18] In this sectionwe focus on the discussion of the Bartels-Stewart algorithmwhich is known to be a numerical stable algorithm to solve⊤-Stein equationsThis method is to solve (1) by means of thePQZ decomposition [18] Its approach has been discussed in[4] and can be summarized as follows From Lemma 3 weknow that there exist two unitary matrices 119875 and 119876 (see [3]for the computation procedure) such that

119875119883119876 minus 119875119860119876 sdot 119876119867119883⊤119875⊤sdot 119875119861119876 = 119875119862119876 (44)

With 119860 = 119875119860119876 and 119861⊤

= 119876119867119861⊤119875119867 being uppertriangular

the transformed equation looks like

[11988311

11990912

11990921

11990922

] minus [11986011

11988612

0 11988622

] [119883⊤

11119909⊤

21

119909⊤

12119909⊤

22

] [11986111

0

21

22

]

= [11986211

11988812

11988821

11988822

]

(45)

with119883 = [1111990912

1199092111990922

] We then have

11990922

minus 11988622119909⊤

2222

= 11988822 (46)

11990921

minus 11988622119909⊤

1211986111

= 11988821

+ 11988622119909⊤

2221 (47)

11990912

minus 11986011119909⊤

2122

= 11988812

+ 11988612119909⊤

2222 (48)

11988311

minus 11986011119883⊤

1111986111

= 11986211

+ 11988612119909⊤

1211986111

+ 11986011119909⊤

2121

+ 11988612119909⊤

2221

(49)

Thus the Bartels-Stewart algorithm can easily be constructedby first solving 119909

22from (46) using 119909

22to obtain 119909

12and 119909

21

from (47) and (48) and then repeating the same discussionas (46)ndash(48) by taking advantage of the property of 119860

11and

11986111being lower triangular matrices from (49)

43 Smith-Type Iterative Methods Recently a class methodreferred to as the Smith accelerative iteration has been studiedfor solving the Sylvester matrix equation [15] The Smithaccelerative iteration has attracted much interests in manypapers (see [7 17] and the references therein) because of itsnice numerical behavior a quadratic convergence rate lowcomputational cost and high numerical reliability despite thelack of a rigorous error analysis Since the Sylvester matrixequation can be transformed into the Stein matrix equationwith a suitable transformation Zhou et al proposed Smith-type iterative methods (including the Smith accelerativeiteration Smith (ℓ) iteration and 119903-Smith iteration) forsolving the Stein matrix equation [17] and applying Smith-type iterative methods to (1) [7] In this section we try toexplain the Smith accelerative iteration based on the invariantsubspace method and summarize the recent results from [7]

Originally the Smith-type iterative methods are devel-oped to solve the standard Stein equation

119883 = A119883B +C ABC isin R119899times119899

(50)

As mentioned before the unknown119883 is highly related to thegeneralized eigenspace problems

[B 0

minusC 119868] [

119868

119883] = [

119868 0

0 A] [

119868

119883]B (51a)

or

A [119883 119868] [B 0

0 119868] = [119883 119868] [

119868 0

minusC A] (51b)

Premultiplying (51a) by the matrix [ B 0minusAC 119868 ] and postmul-

tiplying (51b) by the matrix [119868 0

minusCB A ] we get

[B2 0

minusC minusACB 119868119899

] [119868119899

119883] = [

119868119899

0

0 A2] [

119868119899

119883]B2

A2[119883 119868119899] [

B2 0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minusC minusACB A2]

(52)

Then for any positive integer 119896 gt 0 we obtain

[B2119896minus1

0

minus119862119896

119868119899

] [119868119899

119883] = [

119868119899

0

0 A2119896minus1] [

119868119899

119883]B2119896minus1

A2119896minus1

[119883 119868119899] [

B2119896minus1

0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minus119862119896

A2119896minus1]

(53)

where the sequence C119896 is defined by

119862119896= 119862119896minus1

+A2119896minus1

119862119896minus1

B2119896minus1

119896 ge 1 (54a)

1198620= C (54b)

The explicit expression of 119862119896is given as follows

119862119896=

2119896minus1

sum

119894=1

A119894CB119894 (55)

Abstract and Applied Analysis 7

Under the condition 120588(A)120588(B) lt 1 it is easy to see that 119862119896

is convergent and

lim sup119896rarrinfin

2119896

radic1003817100381710038171003817119883 minus 119862

119896

1003817100381710038171003817 le 120588 (A) 120588 (B) (56)

that is 119862119896converges quadratically to 119883 as 119896 rarr infin

This iterative method (54a) and (54b) is called the Smithaccelerative iteration [15] In recent years some modifiediterative methods are so-called Smith-type iteration whichare based on Smith iteration and improve its speed ofconvergence See for example [17] and the references citedtherein

Since the condition 120588(A)120588(B) lt 1 implies that theassumptions of Theorem 4 hold (1) is equivalent to (5) Wecan apply the Smith iteration to the (1) with the substitution(ABC) = (119860119861

⊤ 119860⊤119861 119862+119860119862

⊤119861) One possible drawback

of the Smith-type iterativemethods is that they cannot alwayshandle the case when there exist eigenvalues 120582 120583 isin 120590(119860

⊤119861)

such that 120582120583 = minus1 even if the unique solution119883 exists Basedon the solvable conditions given in this work it is possible todevelop a specific technique working on the particular caseand it is a subject currently under investigation

5 Error Analysis

Error analysis is a way for testing the stability of a numericalalgorithm and evaluating the accuracy of an approximatedsolution In the subsequent discussion we want to considerthe backward error and perturbation bounds for solving (1)

As indicated in (44) matrices 119860 and 119861⊤ are both upper-

triangularWe can then apply the error analysis for triangularlinear systems in [19 Section 31] and [20] to obtain

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 119888119898119899

u (1 +1003817100381710038171003817100381711986010038171003817100381710038171003817119865

1003817100381710038171003817100381711986110038171003817100381710038171003817119865

)1003817100381710038171003817100381711988310038171003817100381710038171003817119865

(57)

where 119888119898119899

is a content depending on the dimensions 119898 and119899 and u is the unit roundoff Since the PQZ decomposition isa stable process it is true that

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 1198881015840

119898119899u (1 + 119860

119865119861119865) 119883119865 (58)

with a modest multiple 1198881015840119898119899

Note that the inequality of the form (58) can serve

as a stopping criterion for terminating iterations generatedfrom the Krylov subspace methods [5 6] and Smith-typeiterative methods [15ndash17] In what follows we will derivethe error associated with numerical algorithms following thedevelopment in [20 21]

51 Backward Error Like the discussion of the ordinarySylvester equations [20] the normwise backward error of anapproximate solution 119884 of (1) is defined by

120578 (119884) equiv min 120598 119884 = (119862 + 120575119862) + (119860 + 120575119860)119884⊤

(119861 + 120575119861)

120575119860119865le 120598120572 120575119861

119865le 120598120573 120575119862

119865le 120598120574

(59)

where 120572 equiv 119860119865 120573 equiv 119861

119865 and 120574 equiv 119862

119865 Let R equiv 120575119862 +

120575119860119884⊤119861 + 119860119884

⊤120575119861 + 120575119860119884

⊤120575119861 which implies that R = 119884 minus

119860119884⊤119861 minus 119862 It can be seen that the residualR satisfies

R119865le 120578 (119884) (120574 + 119884

119865120572120573 (2 + 120578 (119884))) (60)

From (60) we know that a small backward error indeedimplies a small relative residual R Since the coefficientmatrices in (1) include nonlinearity it appears to be an openproblem to obtain the theoretical backward error with respectto the residual Again similar to the Sylvester equationdiscussed in [20 Section 162] the conditions under whicha ⊤-Stein equation has a well-conditioned solution remainunknown

52 Perturbation Bounds Consider the perturbed equation

119883 + 120575119883 = (119860 + 120575119860) (119883 + 120575119883)⊤

(119861 + 120575119861) + (119862 + 120575119862)

(61)

Let 119878(119883) = 119883minus119860119883⊤119861 be the corresponding⊤-Stein operator

We then have 119878(120575119883) = 120575119862+119860(119883+120575119883)⊤120575119861+120575119860(119883+120575119883)

⊤(119861+

120575119861) With the application of norm it follows that

120575119883119865le10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

120575119862119865+ 120575119878

119865(119883119865+ 120575119883

119865) (62)

where 120575119878119865

equiv 119860119865120575119861119865+ 120575119860

119865(119861119865+ 120575119861

119865) When

120575119878119865is small enough so that 1 ge 119878

minus1119865sdot 120575119878

119865 we can

rearrange the previous result to be

120575119883119865

119883119865

le

10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

1 minus1003817100381710038171003817119878minus11003817100381710038171003817119865

sdot 120575119878119865

(120575119862119865

119883119865

+ 120575119878119865) (63)

With 119862119865

= 119878(119883)119865

le 119878119865sdot 119883119865and the condition

number 120581(119878) equiv 119878119865

sdot 119878minus1119865 we arrive at the standard

perturbation result120575119883119865

119883119865

le120581 (119878)

1 minus 120581 (119878) sdot 120575119878119865119878119865

(120575119862119865

119862119865

+120575119878119865

119878119865

)

(64)

Thus the relative error in119883 is controlled by those in119860 119861 and119862 magnified by the condition number 120581(119878)

On the other hand we can also drop the high-order termsin the perturbation to obtain

120575119883 minus 119860120575119883⊤119861 = 119860119883

⊤120575119861 + 120575119860119883

⊤119861 + 120575119862 (65)

We then rewrite the system in terms of

Q vec (120575119883) = [(119883⊤119861)⊤

otimes 119868119898119868119899otimes (119860119883

⊤) 119868119898119899

] [

[

vec (120575119860)

vec (120575119861)vec (120575119862)

]

]

(66)

where Q = 119868119898119899

minus (119861⊤

otimes 119860)P Let 120577 = max120575119860119865119860119865

120575119861119865119861119865 120575119862

119865119862119865 It can be shown that

120575119883119865

119883119865

le radic3Ψ120577 (67)

where Ψ=Qminus1[120572(119883⊤119861)⊤otimes 119868119898

120573119868119899otimes (119860119883

⊤)120574119868119898119899

]2119883119865

8 Abstract and Applied Analysis

A possible disadvantage of the perturbation bound (67)which ignores the consideration of the underlying structureof the problem is that it overestimates the effect of theperturbation on the data But this ldquouniversalrdquo perturbationbound is accessible to any given matrices 119860 119861 and 119862 of (1)

Unlike the perturbation bound (67) it is desirable toobtain a posteriori error bound by assuming 120575119860 = 120575119861 = 0

and 120575119862 = 119883 minus 119860119883⊤119861 minus 119862 in (61) This assumption gives rise

to

120575119883119865

119883119865

le

10038171003817100381710038171003817119875minus1100381710038171003817100381710038172119877119865

119883119865

(68)

It is true that while doing numerical computation this boundgiven in (68) provides a simpler way for estimating the errorof the solution of (1)

6 Conclusion

In this note we propose a novel approach to the necessaryand sufficient conditions for the unique solvability of thesolution 119883 of the ⊤-Stein equation for square coefficientmatrices in terms of the analysis of the spectra 120590(119860

⊤119861)

Solvability conditions have been derived and algorithmshave been proposed in [4 8] by using PQZ decompositionOn the other hand one common procedure to solve theStein-type equations is by means of the invariant subspacemethod We believe that our discussion is the first whichimplements the techniques of the deflating subspace forsolving ⊤-Stein matrix equation and might also give riseto the possibility of developing an advanced and effectivesolver in the future Also we obtain the theoretical residualanalysis backward error analysis and perturbation boundsfor measuring accurately the error in the computed solutionof (1)

Acknowledgments

The author wishes to thank Proffessor Eric King-wah Chu(Monash University Australia) Proffessor Matthew M Lin(National Chung Cheng University Taiwan) and two anony-mous referees for many interesting and valuable suggestionson the paperThis research work is partially supported by theNational Science Council (NSC101-2115-M-150-002) and theNational Center for Theoretical Sciences in Taiwan

References

[1] H W Braden ldquoThe equations 119860⊤119883 plusmn 119883⊤119860 = 119861rdquo SIAM Journal

on Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021998

[2] X Ma J Han and F Pu ldquoA study of solution existence formatrix equation 119860119883 + 119860

⊤119862 = 119861rdquo Journal of Shenyang Institute

of Aeronautical Engineering vol 4 pp 64ndash66 2003[3] A W Bojanczyk G H Golub and P Van Dooren ldquoPeri-

odic schur decomposition algorithms and applicationsrdquo inAdvanced Signal Processing Algorithms Architectures andImplementations vol 1770 of Proceedings of SPIE pp 31ndash42 SanDiego Calif USA July 1992

[4] C-Y Chiang E K-W Chu andW-W Lin ldquoOn the lowast-Sylvesterequation 119860119883 plusmn 119883

lowast119861lowastrdquo Applied Mathematics and Computation

vol 218 no 17 pp 8393ndash8407 2012[5] Y Su and G Chen ldquoIterative methods for solving linear matrix

equation and linear matrix systemrdquo International Journal ofComputer Mathematics vol 87 no 4 pp 763ndash774 2010

[6] M Wang X Cheng and M Wei ldquoIterative algorithms forsolving the matrix equation 119860119883119861 + 119862119883

⊤119863 = 119864rdquo Applied

Mathematics andComputation vol 187 no 2 pp 622ndash629 2007[7] B Zhou J Lam and G-R Duan ldquoToward solution of matrix

equation 119860119891(119883)119861 + 119862rdquo Linear Algebra and its Applications vol435 no 6 pp 1370ndash1398 2011

[8] K D Ikramov and O Yu Vorontsov ldquoThe matrix equation119883+

119860119883⊤119861 = 119862 conditions for unique solvability and a numerical

algorithm for its solutionrdquo Doklady Mathematics vol 85 pp265ndash267 2012

[9] D S Bernstein Matrix Mathematics Theory Facts and For-mulas Princeton University Press Princeton NJ USA 2ndedition 2009

[10] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1994 Correctedreprint of the 1991 original

[11] F ZhangMatrix Theory Basic Results and Techniques Univer-sitext Springer 1999

[12] D A Bini B Iannazzo and B Meini Numerical Solution ofAlgebraic Riccati Equations Society for Industrial and AppliedMathematics (SIAM) Philadelphia Pa USA 2012

[13] I Gohberg P Lancaster and L Rodman Matrix PolynomialsComputer Science and Applied Mathematics Academic PressNew York NY USA 1982

[14] C-Y Chiang andMM Lin ldquoDeflating subspaces and Sylvesterequationsrdquo Tech Rep NCTS Taiwan 2013

[15] R A Smith ldquoMatrix equation119883119860+ 119861119883 = 119862rdquo SIAM Journal onApplied Mathematics vol 16 pp 198ndash201 1968

[16] T Penzl ldquoA cyclic low-rank Smith method for large sparseLyapunov equationsrdquo SIAM Journal on Scientific Computingvol 21 no 4 pp 1401ndash1418 1999

[17] B Zhou J Lam and G-R Duan ldquoOn Smith-type iterativealgorithms for the Stein matrix equationrdquo Applied MathematicsLetters vol 22 no 7 pp 1038ndash1044 2009

[18] RH Bartels andGW Stewart ldquoSolution of thematrix equation119860119883+119883119861 = 119862 algorithm 432rdquoCommunications of the ACM vol15 no 9 pp 820ndash826 1972

[19] G H Golub and C F Van Loan Matrix Computations JohnsHopkins Studies in the Mathematical Sciences Johns HopkinsUniversity Press Baltimore Md USA 3rd edition 1996

[20] N J Higham Accuracy and Stability of Numerical Algo-rithms Society for Industrial andAppliedMathematics (SIAM)Philadelphia Pa USA 2nd edition 2002

[21] A R Ghavimi and A J Laub ldquoBackward error sensitivityand refinement of computed solutions of algebraic Riccatiequationsrdquo Numerical Linear Algebra with Applications vol 2no 1 pp 29ndash49 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

Let the diagonal entries of 119880119860and 119880

119861be denoted by 119886

119894119894

and 119887119895119895 respectively Then 119880

119860otimes 119880119861is an upper triangular

matrix with given diagonal entries specified by 119886119894119894and 119887119895119895

After multiplying119880119860

otimes 119880119861byP from the right the position

of the entry 119886119894119894119887119895119895is changed to be in the 119895+119899(119894minus1) th row and

the 119894 + 119899(119895 minus 1) th column of the matrix (119880119860

otimes 119880119861)P They

are then reshuffled by a sequence of permutation matrices toform a block upper triangular matrix with diagonal entriesarranged in the following order

1198861111988711 [

0 1198861111988722

1198862211988711

0] [

0 11988611119887119899119899

11988611989911989911988711

0]

1198862211988722 [

0 1198862211988733

1198863311988722

0] [

0 11988611989911989911988722

11988622119887119899119899

0]

[0 119886

119899minus1119899minus1119887119899119899

119886119899119899119887119899minus1119899minus1

0] 119886119899119899119887119899119899

(13)

Note that the reshuffling process is not hard to see when 119899 =

2 119880119860= [1198861111988612

0 11988622] and 119880

119861= [1198871111988712

0 11988722

] we have

(119880119860otimes 119880119861)P =

[[[

[

1198861111988711

1198861211988711

1198861111988712

1198861211988712

0 0 1198861111988722

1198861211988722

0 1198862211988711

0 1198862211988712

0 0 0 1198862211988722

]]]

]

(14)

However it is conceptually simple and regular but opera-tionally tedious to reorder (119880

119860otimes 119880119861)P to show this result

even for 119899 = 3 and that will be left as an exerciseBy (13) it can be seen that

120590 ((119861⊤otimes 119860)P) = 119886

119894119894119887119894119894 plusmnradic119886119894119894119886119895119895119887119894119894119887119895119895 1 le 119894 119895 le 119899

= 120582119894 plusmnradic120582

119894120582119895 1 le 119894 119895 le 119899

(15)

where 120582119894= 119886119894119894119887119894119894isin 120590(119860

⊤119861) for 1 le 119894 le 119899

Before demonstrating the unique solvability conditionswe need to define that a subset Λ = 120582

1 120582

119899 of

complex numbers is said to be ⊤-reciprocal free if and onlyif whenever 119894 = 119895 120582

119894= 1120582119895 This definition also regards 0 and

infin as reciprocals of each other Then we have the followingsolvability conditions of (1)

Theorem 4 The ⊤-Stein matrix equation (1) is uniquelysolvable if and only if the following conditions are satisfied

(1) the set of 120590(119860⊤119861) minus1 is ⊤-reciprocal free(2) minus1 can be an eigenvalue of the matrix119860⊤119861 but must be

simple

Proof From (7) we know that the ⊤-Stein matrix equation(1) is uniquely solvable if and only if

1 notin 120590 ((119861⊤otimes 119860)P) (16)

By Lemma 3 if 120582 isin 120590(119860⊤119861) then 1120582 notin 120590(119860

⊤119861) Otherwise

1 = radic120582 sdot (1120582) isin ((119861⊤

otimes 119860)P) On the other hand ifminus1 isin 120590(119860

⊤119861) and minus1 is not a simple eigenvalue then 1 isin

120590((119861⊤otimes119860)P) This verifies (16) and the proof of the theorem

is complete

It is worthy noting that the condition (1) of Theorem 4 iscontained in the condition (6) (also appear in [7 Theorem1]) which is the necessary and sufficient conditions forthe solvability of the solution to the Stein equation (5)However as mentioned before in Remark 1 or [7 Example2] condition (1) is just a necessary condition for uniquesolvability of the solution to (1) The ⊤-Stein matrix equation(1) is uniquely solvable provided that both conditions (1) and(2) of Theorem 4 are satisfying

3 The Connection between DeflatingSubspace and (1)

The relationship between solution of matrix equations andthe matrix eigenvalue problems has been widely studied inmany applications It is famous that solution of Riccati andpolynomial matrix equations can be found by computinginvariant subspaces of matrices and deflating subspaces ofmatrix pencils [12] This reality leads us to find some algo-rithms for computing solution of (1) based on the numericalcomputation of invariant or deflating subspaces

Given a pair of 119899 times 119899 matrices 119860 and 119861 recall that thefunction119860minus120582119861 in the variable 120582 is said to be thematrix pencilrelated to the pair (119860 119861) For a 119896-dimensional subspaceX isin

C119899 is called a deflating subspace for the pencil119860minus120582119861 if thereexists a 119896-dimensional subspaceY isin C119899 such that

119860X sube Y 119861X sube Y (17)

that is

119860119883 = 1198841198791 119861119883 = 119884119879

2 (18)

where119883119884 isin C119899times119896 are two full rank matrices whose columnsspan the spacesX andY respectively and matrices 119879

1 1198792isin

C119896times119896 In particular if in (18) 119883 = 119884 and 119861 = 1198792= 119868 for an

119899 times 119899 identity matrix 119868 then we have the simplified formula

119860119883 = 1198831198791 (19)

Here the spaceX spanned by the columns of the matrix119883 iscalled an invariant subspace for 119860 and satisfies

119860X sube X (20)

One strategy to analyze the eigeninformation is to transformone matrix pencil to its simplified and equivalent form Thatis two matrix pencils 119860 minus 120582119861 and 119860 minus 120582119861 are said to beequivalent if and only if there exist two nonsingular matrices119875 and 119876 such that

119875 (119860 minus 120582119861)119876 = 119860 minus 120582119861 (21)

In the subsequent discussion we will use the notion sim todescribe this equivalence relation that is 119860 minus 120582119861 sim 119860 minus 120582119861

Our task in this section is to identify eigenvectors ofproblem (18) and then associate these eigenvectors (left andright) with the solution of (1) We begin this analysis bystudying the eigeninformation of two matrices 119860 and 119861where 119860 minus 120582119861 is a regular matrix pencil

4 Abstract and Applied Analysis

Note that for the ordinary eigenvalue problem if theeigenvalues are different then the eigenvectors are linearlyindependent This property is also true for every regularmatrix pencil and is demonstrated as follows For a detailedproof the reader is referred to [13 Theorem 73] and [14Theorem 42]

Theorem5 Given a pair of 119899 times 119899matrix119860 and119861 if thematrixpencil 119860 minus 120582119861 is regular then its Jordan chains correspondingto all finite and infinite eigenvalues carry the full spectralinformation about the matrix pencil and consist of 119899 linearlyindependent vectors

Lemma 6 Let 119860 minus 120582119861 isin C119899times119899 be a regular matrix pencilAssume that matrices 119883

119894 119884119894isin C119899times119899119894 119894 = 1 2 are full rank

and satisfy the following equations

119860119883119894= 119884119894119877119894 (22a)

119861119883119894= 119884119894119878119894 (22b)

where 119877119894and 119878

119894 119894 = 1 2 are square matrices of size 119899

119894times 119899119894

Then

(i) 119877119894minus 120582119878119894isin C119899119894times119899119894 are regular matrix pencils for 119894 = 1 2

(ii) if 120590(1198771minus 1205821198781) cap 120590(119877

2minus 1205821198782) = 120601 then the matrix

[1198831

1198832] isin R119899times(1198991+1198992) is full rank

We also need the following useful lemma

Lemma 7 Given two regular matrix pencils119860119894minus120582119861119894isin C119899119894times119899119894

1 le 119894 le 2 consider the following equations with respect to119880119881 isin C1198991times1198992

1198601119880 = 119881119860

2 (23a)

1198611119880 = 119881119861

2 (23b)

Then if 120590(1198601minus 1205821198611) cap 120590(119860

2minus 1205821198612) = 120601 (23a) has the unique

solution 119880 = 119881 = 0

Proof For 1198992= 1 we get

1198601119906 = 1198862V

1198611119906 = 1198872V

(24)

where 1198862 1198872

isin C 119906 V isin C1198991times1 We may without loss ofgenerality assume that 119887

2= 0 then119860

1119906 = (119886

21198872)1198611119906 and thus

119906 = V = 0 Now for any 1198992

gt 1 consider the generalizedSchur decomposition of 119860

2minus 1205821198612 We can assume that 119860

2=

[119886119894119895] and 119861

2= [119887119894119895] are upper triangular matrices (ie 119886

119894119895=

119887119894119895= 0 1 le 119895 lt 119894 le 119899

2) Denote that the 119894th columns of 119880 and

119881 are 119906119894and V119894 respectively Thus

1198601119906119894=

119894

sum

119896=1

119886119896119894V119896 (25a)

1198611119906119894=

119894

sum

119896=1

119887119896119894V119896 (25b)

for 119894 = 1 2 1198992

If 119894 = 1 we obtained 1199061

= V1

= 0 from the abovediscussion Given an integer 119894 such that 1 le 119894 lt 119899

2and assume

that 119906119896= V119896= 0 for 1 le 119896 le 119894 we claim 119906

119894+1= V119894+1

= 0indeed from (25a) and (25b) we have

1198601119906119894+1

= 119886119894+1119894+1

V119894+1

1198611119906119894+1

= 119887119894+1119894+1

V119894+1

(26)

Again the case 119896 = 119894 + 1 is following the special case 1198992= 1

By using mathematical induction we prove this lemma

Corollary 8 Given119860 isin C119899times119899 andΛ isin C119896times119896 if 120590(119860)cap120590(Λ) =

120601 then the equation with respect to 119880 isin C119899times119896

119860119880 = 119880Λ (27)

has the unique solution 119880 = 0

Now we have enough tools to analyze the solution of (1)associated with some deflating spaces We first establish animportant matrix pencil let the matrix pencil M minus 120582L bedefined as

M minus 120582L = [119861119860⊤

0

minus119862119860⊤

119868119899

] minus 120582 [119868119899

0

119860119862⊤

119860119861⊤] isin R

2119899times2119899

(28)

it is clear that

120590 (M minus 120582L) = 120590 (119861119860⊤) cup 120590 (119868

119899minus 120582119860119861

⊤) (29)

a direct calculation shows that119883 is a solution of the (1) if andonly if

M [119868119899

119883119860⊤] = [

119868119899

119860119883⊤]119861119860

⊤

L [119868119899

119883119860⊤] = [

119868119899

119860119883⊤]

(30)

or if and only if its dual form is

[minus119860119883⊤

119868119899]M = [minus119883119860

⊤119868119899]

[minus119860119883⊤

119868119899]L = 119860119861

⊤[minus119883119860

⊤119868119899]

(31)

Armed with the property given in Theorem 5 andLemma 7 we can now tackle the problemof determining howthe deflating subspace is related to the solution of (1)

Theorem 9 Let 119860 119861 and 119862 isin R119899times119899 be given in (1) and let

M [1198801

1198811

] = [1198802

1198812

]1198791 (32a)

L [1198801

1198811

] = [1198802

1198812

]1198792 (32b)

where [ 119880119894119881119894

] is full rank 119894 = 1 2 Assume that the set of 120590(119861119860⊤)is ⊤-reciprocal free Then one has

Abstract and Applied Analysis 5

(1) 1198801= 1198802= 0 if 120590(119879

1minus 1205821198792) = 120590(119868

119899minus 120582119860119861

⊤)

(2) 1198801and 119880

2are nonsingular if 119879

1minus 1205821198792sim 119861119860⊤minus 120582119868119899

Moreover if 119860 is nonsingular then 119883 = 1198811119880minus1

1119860minus⊤

=

119880minus⊤

2119881⊤

2119860minus⊤ is the unique solution of (1)

Proof From (32a) and (32b) we get

119861119860⊤1198801= 11988021198791 (33a)

minus119862119860⊤1198801+ 1198811= 11988121198791 (33b)

1198801= 11988021198792 (33c)

119860119862⊤1198801+ 119860119861⊤1198811= 11988121198792 (33d)

(i) It follows from (33a) and (33c) that since 120590(119861119860⊤

minus

120582119868119899) cap 120590(119879

1minus 1205821198792) = 120601 we have 119880

1= 1198802

= 0 byLemma 7

(ii) It can be seen that there exist two nonsingularmatrices 119880 and 119881 such that

M [0

119880] = [

0

119881]1198792

L [0

119880] = [

0

119881]1198791

(34)

Hence together with (32a) and (32b) we have

M [0 1198801

119880 1198811

] = [0 1198802

119881 1198812

] [1198792

0

0 1198791

]

L [0 1198801

119880 1198811

] = [0 1198802

119881 1198812

] [1198791

0

0 1198792

]

(35)

Since 120590(Mminus120582L) = 120590(119861119860⊤minus120582119868119899)cup120590(119868

119899minus120582119860119861

⊤) and 120590(119861119860

⊤minus

120582119868119899) cap 120590(119868

119899minus 120582119860119861

⊤) = 120601 by Theorem 5 and Lemma 6 the

matrix [0 1198801

119880 1198811

] is nonsingular Together with (33c)1198801and119880

2

are nonsingularLet119883

119894= 119881119894119880minus1

119894 119894 = 1 2 then form (33b) and (33d)

119860119862⊤+ 119860119861⊤1198831= 11988121198792119880minus1

1= 1198832

minus119862119860⊤+ 1198831= 11988121198791119880minus1

1= 1198832119861119860⊤

(36)

or

119860119862⊤+ 119860119861⊤1198831= 1198832

119860119862⊤+ 119860119861⊤119883⊤

2= 119883⊤

1

(37)

Since the set of 120590(119860119861⊤) = 120590(119861119860

⊤) is ⊤-reciprocal free

together with

119883⊤

1minus 1198832minus 119860119861⊤(119883⊤

1minus 1198832)⊤

= 0 (38)

we get 1198831= 119883⊤

2 If 119860 is nonsingular it is easy to verify that

two matrices 1198831119860minus⊤ and 119883

⊤

2119860minus⊤ are both satisfying ⊤-Stein

equation (1) The proof of part (ii) is complete

Remark 10 (1) It is easily seen that [ 119868119899119883119860⊤ ] and [

1198801

1198812

] both spanthe unique deflating subspace of M minus 120582L corresponding tothe set of 120590(119861119860⊤) Otherwise in part (ii) we know that 119879

2

is nonsingular We then are able to transform the formulaedefined in (32a) and (32b) into the generalized eigenvalueproblem as follows

M [1198801

1198811

] = L [1198801

1198811

]119861119860⊤ (39)

That is some numerical methods for the computation ofthe eigenspace ofMminus120582L corresponding to the set of 120590(119861119860⊤)can be designed and solved (1)

(2) Since the transport of the unique solution 119883 of (1)is equal to the unique solution 119884 of the following matrixequation

119884 = 119861⊤119884119860⊤+ 119862⊤ (40)

analogous to the consequences of Theorem 9 the similarresults can be obtainedwith respect to (40) if119861 is nonsingularHowever we point out that (1) can be solved by computingdeflating subspaces of other matrix pencils For instance welet

M1minus 120582L

1= [

119860⊤119861 0

minus119862 minus 119860119862⊤119861 119868119899

] minus 120582 [119868119899

0

0 119860119861⊤] (41)

Assume that the set of 120590(119861119860⊤) is ⊤-reciprocal free it can beshown thatM

1[119868119899

119883] = L

1[119868119899

119883] 119860⊤119861 and it has similar results

to the conclusion ofTheorem 9The unique solution119883 of (1)can be found by computing deflating subspaces of the matrixpencil M

1minus 120582L

1without the assumption of the singularity

of 119860 and 119861

4 Computational Methods for Solving (1)Numerical methods for solving (1) have received great atten-tion in theory and in practice and can be found in [5 6] forthe Krylov subspace methods and in [15ndash17] for the Smith-type iterative methods In particular Smith-type iterativemethods are only workable in the case 120588(119860119861

⊤) lt 1 where

120588(119860119861⊤) denotes the spectral radius of 119860119861

⊤ In the recentyears a structure algorithm has been studied for (1) [4] viaPQZ decomposition which consists of transforming intoSchur form by a PQZ decomposition and then solving theresulting triangular system by way of back-substitution Inthis section we revisit these numericalmethods and point outthe advantages and drawbacks of all algorithms

41 Krylov Subspace Methods Since the ⊤-Stein equation isessentially a linear system (7) we certainly can use the Krylovsubspace methods to solve (7) See for example [5 6] andthe reference cited therein The general idea for applying theKrylov subspace methods is defining the ⋆-Stein operatorTas T 119883 rarr 119883 minus 119860119883

⊤119861 and its adjoint liner operator T as

Tlowast 119884 rarr 119884 minus 119861119884⊤119860 such that ⟨T(119883) 119884⟩ = ⟨119883Tlowast(119884)⟩

Here 119883 119884 isin R119898times119899 and the notion ⟨sdot sdot⟩ is denoted as theFrobenius inner product Then the iterative method basedon the Krylov subspaces for (1) is as follows

6 Abstract and Applied Analysis

(i) The conjugate gradient (CG) method [5]

119883119896+1

= 119883119896+

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2119875119896

119877119896+1

= 119862 minusT (119883119896+1

) = 119877119896minus

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2T (119875119896)

119863119896+1

= Tlowast(119877119896+1

) +

1003817100381710038171003817119877119896+11003817100381710038171003817

2

10038171003817100381710038171198771198961003817100381710038171003817

2119863119896

(42)

with an initial matrix 1198830and the corresponding initial

conditions

1198770= 119862 minusT (119883

0) 119863

0= Tlowast(1198770) (43)

Note that when the solvability conditions of Theorem 4are met the CG method is guaranteed to converge in a finitenumber of iterations for any initial matrix119883

0

42 The Bartels-Stewart-Like Algorithm [18] In this sectionwe focus on the discussion of the Bartels-Stewart algorithmwhich is known to be a numerical stable algorithm to solve⊤-Stein equationsThis method is to solve (1) by means of thePQZ decomposition [18] Its approach has been discussed in[4] and can be summarized as follows From Lemma 3 weknow that there exist two unitary matrices 119875 and 119876 (see [3]for the computation procedure) such that

119875119883119876 minus 119875119860119876 sdot 119876119867119883⊤119875⊤sdot 119875119861119876 = 119875119862119876 (44)

With 119860 = 119875119860119876 and 119861⊤

= 119876119867119861⊤119875119867 being uppertriangular

the transformed equation looks like

[11988311

11990912

11990921

11990922

] minus [11986011

11988612

0 11988622

] [119883⊤

11119909⊤

21

119909⊤

12119909⊤

22

] [11986111

0

21

22

]

= [11986211

11988812

11988821

11988822

]

(45)

with119883 = [1111990912

1199092111990922

] We then have

11990922

minus 11988622119909⊤

2222

= 11988822 (46)

11990921

minus 11988622119909⊤

1211986111

= 11988821

+ 11988622119909⊤

2221 (47)

11990912

minus 11986011119909⊤

2122

= 11988812

+ 11988612119909⊤

2222 (48)

11988311

minus 11986011119883⊤

1111986111

= 11986211

+ 11988612119909⊤

1211986111

+ 11986011119909⊤

2121

+ 11988612119909⊤

2221

(49)

Thus the Bartels-Stewart algorithm can easily be constructedby first solving 119909

22from (46) using 119909

22to obtain 119909

12and 119909

21

from (47) and (48) and then repeating the same discussionas (46)ndash(48) by taking advantage of the property of 119860

11and

11986111being lower triangular matrices from (49)

43 Smith-Type Iterative Methods Recently a class methodreferred to as the Smith accelerative iteration has been studiedfor solving the Sylvester matrix equation [15] The Smithaccelerative iteration has attracted much interests in manypapers (see [7 17] and the references therein) because of itsnice numerical behavior a quadratic convergence rate lowcomputational cost and high numerical reliability despite thelack of a rigorous error analysis Since the Sylvester matrixequation can be transformed into the Stein matrix equationwith a suitable transformation Zhou et al proposed Smith-type iterative methods (including the Smith accelerativeiteration Smith (ℓ) iteration and 119903-Smith iteration) forsolving the Stein matrix equation [17] and applying Smith-type iterative methods to (1) [7] In this section we try toexplain the Smith accelerative iteration based on the invariantsubspace method and summarize the recent results from [7]

Originally the Smith-type iterative methods are devel-oped to solve the standard Stein equation

119883 = A119883B +C ABC isin R119899times119899

(50)

As mentioned before the unknown119883 is highly related to thegeneralized eigenspace problems

[B 0

minusC 119868] [

119868

119883] = [

119868 0

0 A] [

119868

119883]B (51a)

or

A [119883 119868] [B 0

0 119868] = [119883 119868] [

119868 0

minusC A] (51b)

Premultiplying (51a) by the matrix [ B 0minusAC 119868 ] and postmul-

tiplying (51b) by the matrix [119868 0

minusCB A ] we get

[B2 0

minusC minusACB 119868119899

] [119868119899

119883] = [

119868119899

0

0 A2] [

119868119899

119883]B2

A2[119883 119868119899] [

B2 0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minusC minusACB A2]

(52)

Then for any positive integer 119896 gt 0 we obtain

[B2119896minus1

0

minus119862119896

119868119899

] [119868119899

119883] = [

119868119899

0

0 A2119896minus1] [

119868119899

119883]B2119896minus1

A2119896minus1

[119883 119868119899] [

B2119896minus1

0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minus119862119896

A2119896minus1]

(53)

where the sequence C119896 is defined by

119862119896= 119862119896minus1

+A2119896minus1

119862119896minus1

B2119896minus1

119896 ge 1 (54a)

1198620= C (54b)

The explicit expression of 119862119896is given as follows

119862119896=

2119896minus1

sum

119894=1

A119894CB119894 (55)

Abstract and Applied Analysis 7

Under the condition 120588(A)120588(B) lt 1 it is easy to see that 119862119896

is convergent and

lim sup119896rarrinfin

2119896

radic1003817100381710038171003817119883 minus 119862

119896

1003817100381710038171003817 le 120588 (A) 120588 (B) (56)

that is 119862119896converges quadratically to 119883 as 119896 rarr infin

This iterative method (54a) and (54b) is called the Smithaccelerative iteration [15] In recent years some modifiediterative methods are so-called Smith-type iteration whichare based on Smith iteration and improve its speed ofconvergence See for example [17] and the references citedtherein

Since the condition 120588(A)120588(B) lt 1 implies that theassumptions of Theorem 4 hold (1) is equivalent to (5) Wecan apply the Smith iteration to the (1) with the substitution(ABC) = (119860119861

⊤ 119860⊤119861 119862+119860119862

⊤119861) One possible drawback

of the Smith-type iterativemethods is that they cannot alwayshandle the case when there exist eigenvalues 120582 120583 isin 120590(119860

⊤119861)

such that 120582120583 = minus1 even if the unique solution119883 exists Basedon the solvable conditions given in this work it is possible todevelop a specific technique working on the particular caseand it is a subject currently under investigation

5 Error Analysis

Error analysis is a way for testing the stability of a numericalalgorithm and evaluating the accuracy of an approximatedsolution In the subsequent discussion we want to considerthe backward error and perturbation bounds for solving (1)

As indicated in (44) matrices 119860 and 119861⊤ are both upper-

triangularWe can then apply the error analysis for triangularlinear systems in [19 Section 31] and [20] to obtain

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 119888119898119899

u (1 +1003817100381710038171003817100381711986010038171003817100381710038171003817119865

1003817100381710038171003817100381711986110038171003817100381710038171003817119865

)1003817100381710038171003817100381711988310038171003817100381710038171003817119865

(57)

where 119888119898119899

is a content depending on the dimensions 119898 and119899 and u is the unit roundoff Since the PQZ decomposition isa stable process it is true that

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 1198881015840

119898119899u (1 + 119860

119865119861119865) 119883119865 (58)

with a modest multiple 1198881015840119898119899

Note that the inequality of the form (58) can serve

as a stopping criterion for terminating iterations generatedfrom the Krylov subspace methods [5 6] and Smith-typeiterative methods [15ndash17] In what follows we will derivethe error associated with numerical algorithms following thedevelopment in [20 21]

51 Backward Error Like the discussion of the ordinarySylvester equations [20] the normwise backward error of anapproximate solution 119884 of (1) is defined by

120578 (119884) equiv min 120598 119884 = (119862 + 120575119862) + (119860 + 120575119860)119884⊤

(119861 + 120575119861)

120575119860119865le 120598120572 120575119861

119865le 120598120573 120575119862

119865le 120598120574

(59)

where 120572 equiv 119860119865 120573 equiv 119861

119865 and 120574 equiv 119862

119865 Let R equiv 120575119862 +

120575119860119884⊤119861 + 119860119884

⊤120575119861 + 120575119860119884

⊤120575119861 which implies that R = 119884 minus

119860119884⊤119861 minus 119862 It can be seen that the residualR satisfies

R119865le 120578 (119884) (120574 + 119884

119865120572120573 (2 + 120578 (119884))) (60)

From (60) we know that a small backward error indeedimplies a small relative residual R Since the coefficientmatrices in (1) include nonlinearity it appears to be an openproblem to obtain the theoretical backward error with respectto the residual Again similar to the Sylvester equationdiscussed in [20 Section 162] the conditions under whicha ⊤-Stein equation has a well-conditioned solution remainunknown

52 Perturbation Bounds Consider the perturbed equation

119883 + 120575119883 = (119860 + 120575119860) (119883 + 120575119883)⊤

(119861 + 120575119861) + (119862 + 120575119862)

(61)

Let 119878(119883) = 119883minus119860119883⊤119861 be the corresponding⊤-Stein operator

We then have 119878(120575119883) = 120575119862+119860(119883+120575119883)⊤120575119861+120575119860(119883+120575119883)

⊤(119861+

120575119861) With the application of norm it follows that

120575119883119865le10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

120575119862119865+ 120575119878

119865(119883119865+ 120575119883

119865) (62)

where 120575119878119865

equiv 119860119865120575119861119865+ 120575119860

119865(119861119865+ 120575119861

119865) When

120575119878119865is small enough so that 1 ge 119878

minus1119865sdot 120575119878

119865 we can

rearrange the previous result to be

120575119883119865

119883119865

le

10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

1 minus1003817100381710038171003817119878minus11003817100381710038171003817119865

sdot 120575119878119865

(120575119862119865

119883119865

+ 120575119878119865) (63)

With 119862119865

= 119878(119883)119865

le 119878119865sdot 119883119865and the condition

number 120581(119878) equiv 119878119865

sdot 119878minus1119865 we arrive at the standard

perturbation result120575119883119865

119883119865

le120581 (119878)

1 minus 120581 (119878) sdot 120575119878119865119878119865

(120575119862119865

119862119865

+120575119878119865

119878119865

)

(64)

Thus the relative error in119883 is controlled by those in119860 119861 and119862 magnified by the condition number 120581(119878)

On the other hand we can also drop the high-order termsin the perturbation to obtain

120575119883 minus 119860120575119883⊤119861 = 119860119883

⊤120575119861 + 120575119860119883

⊤119861 + 120575119862 (65)

We then rewrite the system in terms of

Q vec (120575119883) = [(119883⊤119861)⊤

otimes 119868119898119868119899otimes (119860119883

⊤) 119868119898119899

] [

[

vec (120575119860)

vec (120575119861)vec (120575119862)

]

]

(66)

where Q = 119868119898119899

minus (119861⊤

otimes 119860)P Let 120577 = max120575119860119865119860119865

120575119861119865119861119865 120575119862

119865119862119865 It can be shown that

120575119883119865

119883119865

le radic3Ψ120577 (67)

where Ψ=Qminus1[120572(119883⊤119861)⊤otimes 119868119898

120573119868119899otimes (119860119883

⊤)120574119868119898119899

]2119883119865

8 Abstract and Applied Analysis

A possible disadvantage of the perturbation bound (67)which ignores the consideration of the underlying structureof the problem is that it overestimates the effect of theperturbation on the data But this ldquouniversalrdquo perturbationbound is accessible to any given matrices 119860 119861 and 119862 of (1)

Unlike the perturbation bound (67) it is desirable toobtain a posteriori error bound by assuming 120575119860 = 120575119861 = 0

and 120575119862 = 119883 minus 119860119883⊤119861 minus 119862 in (61) This assumption gives rise

to

120575119883119865

119883119865

le

10038171003817100381710038171003817119875minus1100381710038171003817100381710038172119877119865

119883119865

(68)

It is true that while doing numerical computation this boundgiven in (68) provides a simpler way for estimating the errorof the solution of (1)

6 Conclusion

In this note we propose a novel approach to the necessaryand sufficient conditions for the unique solvability of thesolution 119883 of the ⊤-Stein equation for square coefficientmatrices in terms of the analysis of the spectra 120590(119860

⊤119861)

Solvability conditions have been derived and algorithmshave been proposed in [4 8] by using PQZ decompositionOn the other hand one common procedure to solve theStein-type equations is by means of the invariant subspacemethod We believe that our discussion is the first whichimplements the techniques of the deflating subspace forsolving ⊤-Stein matrix equation and might also give riseto the possibility of developing an advanced and effectivesolver in the future Also we obtain the theoretical residualanalysis backward error analysis and perturbation boundsfor measuring accurately the error in the computed solutionof (1)

Acknowledgments

The author wishes to thank Proffessor Eric King-wah Chu(Monash University Australia) Proffessor Matthew M Lin(National Chung Cheng University Taiwan) and two anony-mous referees for many interesting and valuable suggestionson the paperThis research work is partially supported by theNational Science Council (NSC101-2115-M-150-002) and theNational Center for Theoretical Sciences in Taiwan

References

[1] H W Braden ldquoThe equations 119860⊤119883 plusmn 119883⊤119860 = 119861rdquo SIAM Journal

on Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021998

[2] X Ma J Han and F Pu ldquoA study of solution existence formatrix equation 119860119883 + 119860

⊤119862 = 119861rdquo Journal of Shenyang Institute

of Aeronautical Engineering vol 4 pp 64ndash66 2003[3] A W Bojanczyk G H Golub and P Van Dooren ldquoPeri-

odic schur decomposition algorithms and applicationsrdquo inAdvanced Signal Processing Algorithms Architectures andImplementations vol 1770 of Proceedings of SPIE pp 31ndash42 SanDiego Calif USA July 1992

[4] C-Y Chiang E K-W Chu andW-W Lin ldquoOn the lowast-Sylvesterequation 119860119883 plusmn 119883

lowast119861lowastrdquo Applied Mathematics and Computation

vol 218 no 17 pp 8393ndash8407 2012[5] Y Su and G Chen ldquoIterative methods for solving linear matrix

equation and linear matrix systemrdquo International Journal ofComputer Mathematics vol 87 no 4 pp 763ndash774 2010

[6] M Wang X Cheng and M Wei ldquoIterative algorithms forsolving the matrix equation 119860119883119861 + 119862119883

⊤119863 = 119864rdquo Applied

Mathematics andComputation vol 187 no 2 pp 622ndash629 2007[7] B Zhou J Lam and G-R Duan ldquoToward solution of matrix

equation 119860119891(119883)119861 + 119862rdquo Linear Algebra and its Applications vol435 no 6 pp 1370ndash1398 2011

[8] K D Ikramov and O Yu Vorontsov ldquoThe matrix equation119883+

119860119883⊤119861 = 119862 conditions for unique solvability and a numerical

algorithm for its solutionrdquo Doklady Mathematics vol 85 pp265ndash267 2012

[9] D S Bernstein Matrix Mathematics Theory Facts and For-mulas Princeton University Press Princeton NJ USA 2ndedition 2009

[10] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1994 Correctedreprint of the 1991 original

[11] F ZhangMatrix Theory Basic Results and Techniques Univer-sitext Springer 1999

[12] D A Bini B Iannazzo and B Meini Numerical Solution ofAlgebraic Riccati Equations Society for Industrial and AppliedMathematics (SIAM) Philadelphia Pa USA 2012

[13] I Gohberg P Lancaster and L Rodman Matrix PolynomialsComputer Science and Applied Mathematics Academic PressNew York NY USA 1982

[14] C-Y Chiang andMM Lin ldquoDeflating subspaces and Sylvesterequationsrdquo Tech Rep NCTS Taiwan 2013

[15] R A Smith ldquoMatrix equation119883119860+ 119861119883 = 119862rdquo SIAM Journal onApplied Mathematics vol 16 pp 198ndash201 1968

[16] T Penzl ldquoA cyclic low-rank Smith method for large sparseLyapunov equationsrdquo SIAM Journal on Scientific Computingvol 21 no 4 pp 1401ndash1418 1999

[17] B Zhou J Lam and G-R Duan ldquoOn Smith-type iterativealgorithms for the Stein matrix equationrdquo Applied MathematicsLetters vol 22 no 7 pp 1038ndash1044 2009

[18] RH Bartels andGW Stewart ldquoSolution of thematrix equation119860119883+119883119861 = 119862 algorithm 432rdquoCommunications of the ACM vol15 no 9 pp 820ndash826 1972

[19] G H Golub and C F Van Loan Matrix Computations JohnsHopkins Studies in the Mathematical Sciences Johns HopkinsUniversity Press Baltimore Md USA 3rd edition 1996

[20] N J Higham Accuracy and Stability of Numerical Algo-rithms Society for Industrial andAppliedMathematics (SIAM)Philadelphia Pa USA 2nd edition 2002

[21] A R Ghavimi and A J Laub ldquoBackward error sensitivityand refinement of computed solutions of algebraic Riccatiequationsrdquo Numerical Linear Algebra with Applications vol 2no 1 pp 29ndash49 1995

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Function Spaces

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Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

Note that for the ordinary eigenvalue problem if theeigenvalues are different then the eigenvectors are linearlyindependent This property is also true for every regularmatrix pencil and is demonstrated as follows For a detailedproof the reader is referred to [13 Theorem 73] and [14Theorem 42]

Theorem5 Given a pair of 119899 times 119899matrix119860 and119861 if thematrixpencil 119860 minus 120582119861 is regular then its Jordan chains correspondingto all finite and infinite eigenvalues carry the full spectralinformation about the matrix pencil and consist of 119899 linearlyindependent vectors

Lemma 6 Let 119860 minus 120582119861 isin C119899times119899 be a regular matrix pencilAssume that matrices 119883

119894 119884119894isin C119899times119899119894 119894 = 1 2 are full rank

and satisfy the following equations

119860119883119894= 119884119894119877119894 (22a)

119861119883119894= 119884119894119878119894 (22b)

where 119877119894and 119878

119894 119894 = 1 2 are square matrices of size 119899

119894times 119899119894

Then

(i) 119877119894minus 120582119878119894isin C119899119894times119899119894 are regular matrix pencils for 119894 = 1 2

(ii) if 120590(1198771minus 1205821198781) cap 120590(119877

2minus 1205821198782) = 120601 then the matrix

[1198831

1198832] isin R119899times(1198991+1198992) is full rank

We also need the following useful lemma

Lemma 7 Given two regular matrix pencils119860119894minus120582119861119894isin C119899119894times119899119894

1 le 119894 le 2 consider the following equations with respect to119880119881 isin C1198991times1198992

1198601119880 = 119881119860

2 (23a)

1198611119880 = 119881119861

2 (23b)

Then if 120590(1198601minus 1205821198611) cap 120590(119860

2minus 1205821198612) = 120601 (23a) has the unique

solution 119880 = 119881 = 0

Proof For 1198992= 1 we get

1198601119906 = 1198862V

1198611119906 = 1198872V

(24)

where 1198862 1198872

isin C 119906 V isin C1198991times1 We may without loss ofgenerality assume that 119887

2= 0 then119860

1119906 = (119886

21198872)1198611119906 and thus

119906 = V = 0 Now for any 1198992

gt 1 consider the generalizedSchur decomposition of 119860

2minus 1205821198612 We can assume that 119860

2=

[119886119894119895] and 119861

2= [119887119894119895] are upper triangular matrices (ie 119886

119894119895=

119887119894119895= 0 1 le 119895 lt 119894 le 119899

2) Denote that the 119894th columns of 119880 and

119881 are 119906119894and V119894 respectively Thus

1198601119906119894=

119894

sum

119896=1

119886119896119894V119896 (25a)

1198611119906119894=

119894

sum

119896=1

119887119896119894V119896 (25b)

for 119894 = 1 2 1198992

If 119894 = 1 we obtained 1199061

= V1

= 0 from the abovediscussion Given an integer 119894 such that 1 le 119894 lt 119899

2and assume

that 119906119896= V119896= 0 for 1 le 119896 le 119894 we claim 119906

119894+1= V119894+1

= 0indeed from (25a) and (25b) we have

1198601119906119894+1

= 119886119894+1119894+1

V119894+1

1198611119906119894+1

= 119887119894+1119894+1

V119894+1

(26)

Again the case 119896 = 119894 + 1 is following the special case 1198992= 1

By using mathematical induction we prove this lemma

Corollary 8 Given119860 isin C119899times119899 andΛ isin C119896times119896 if 120590(119860)cap120590(Λ) =

120601 then the equation with respect to 119880 isin C119899times119896

119860119880 = 119880Λ (27)

has the unique solution 119880 = 0

Now we have enough tools to analyze the solution of (1)associated with some deflating spaces We first establish animportant matrix pencil let the matrix pencil M minus 120582L bedefined as

M minus 120582L = [119861119860⊤

0

minus119862119860⊤

119868119899

] minus 120582 [119868119899

0

119860119862⊤

119860119861⊤] isin R

2119899times2119899

(28)

it is clear that

120590 (M minus 120582L) = 120590 (119861119860⊤) cup 120590 (119868

119899minus 120582119860119861

⊤) (29)

a direct calculation shows that119883 is a solution of the (1) if andonly if

M [119868119899

119883119860⊤] = [

119868119899

119860119883⊤]119861119860

⊤

L [119868119899

119883119860⊤] = [

119868119899

119860119883⊤]

(30)

or if and only if its dual form is

[minus119860119883⊤

119868119899]M = [minus119883119860

⊤119868119899]

[minus119860119883⊤

119868119899]L = 119860119861

⊤[minus119883119860

⊤119868119899]

(31)

Armed with the property given in Theorem 5 andLemma 7 we can now tackle the problemof determining howthe deflating subspace is related to the solution of (1)

Theorem 9 Let 119860 119861 and 119862 isin R119899times119899 be given in (1) and let

M [1198801

1198811

] = [1198802

1198812

]1198791 (32a)

L [1198801

1198811

] = [1198802

1198812

]1198792 (32b)

where [ 119880119894119881119894

] is full rank 119894 = 1 2 Assume that the set of 120590(119861119860⊤)is ⊤-reciprocal free Then one has

Abstract and Applied Analysis 5

(1) 1198801= 1198802= 0 if 120590(119879

1minus 1205821198792) = 120590(119868

119899minus 120582119860119861

⊤)

(2) 1198801and 119880

2are nonsingular if 119879

1minus 1205821198792sim 119861119860⊤minus 120582119868119899

Moreover if 119860 is nonsingular then 119883 = 1198811119880minus1

1119860minus⊤

=

119880minus⊤

2119881⊤

2119860minus⊤ is the unique solution of (1)

Proof From (32a) and (32b) we get

119861119860⊤1198801= 11988021198791 (33a)

minus119862119860⊤1198801+ 1198811= 11988121198791 (33b)

1198801= 11988021198792 (33c)

119860119862⊤1198801+ 119860119861⊤1198811= 11988121198792 (33d)

(i) It follows from (33a) and (33c) that since 120590(119861119860⊤

minus

120582119868119899) cap 120590(119879

1minus 1205821198792) = 120601 we have 119880

1= 1198802

= 0 byLemma 7

(ii) It can be seen that there exist two nonsingularmatrices 119880 and 119881 such that

M [0

119880] = [

0

119881]1198792

L [0

119880] = [

0

119881]1198791

(34)

Hence together with (32a) and (32b) we have

M [0 1198801

119880 1198811

] = [0 1198802

119881 1198812

] [1198792

0

0 1198791

]

L [0 1198801

119880 1198811

] = [0 1198802

119881 1198812

] [1198791

0

0 1198792

]

(35)

Since 120590(Mminus120582L) = 120590(119861119860⊤minus120582119868119899)cup120590(119868

119899minus120582119860119861

⊤) and 120590(119861119860

⊤minus

120582119868119899) cap 120590(119868

119899minus 120582119860119861

⊤) = 120601 by Theorem 5 and Lemma 6 the

matrix [0 1198801

119880 1198811

] is nonsingular Together with (33c)1198801and119880

2

are nonsingularLet119883

119894= 119881119894119880minus1

119894 119894 = 1 2 then form (33b) and (33d)

119860119862⊤+ 119860119861⊤1198831= 11988121198792119880minus1

1= 1198832

minus119862119860⊤+ 1198831= 11988121198791119880minus1

1= 1198832119861119860⊤

(36)

or

119860119862⊤+ 119860119861⊤1198831= 1198832

119860119862⊤+ 119860119861⊤119883⊤

2= 119883⊤

1

(37)

Since the set of 120590(119860119861⊤) = 120590(119861119860

⊤) is ⊤-reciprocal free

together with

119883⊤

1minus 1198832minus 119860119861⊤(119883⊤

1minus 1198832)⊤

= 0 (38)

we get 1198831= 119883⊤

2 If 119860 is nonsingular it is easy to verify that

two matrices 1198831119860minus⊤ and 119883

⊤

2119860minus⊤ are both satisfying ⊤-Stein

equation (1) The proof of part (ii) is complete

Remark 10 (1) It is easily seen that [ 119868119899119883119860⊤ ] and [

1198801

1198812

] both spanthe unique deflating subspace of M minus 120582L corresponding tothe set of 120590(119861119860⊤) Otherwise in part (ii) we know that 119879

2

is nonsingular We then are able to transform the formulaedefined in (32a) and (32b) into the generalized eigenvalueproblem as follows

M [1198801

1198811

] = L [1198801

1198811

]119861119860⊤ (39)

That is some numerical methods for the computation ofthe eigenspace ofMminus120582L corresponding to the set of 120590(119861119860⊤)can be designed and solved (1)

(2) Since the transport of the unique solution 119883 of (1)is equal to the unique solution 119884 of the following matrixequation

119884 = 119861⊤119884119860⊤+ 119862⊤ (40)

analogous to the consequences of Theorem 9 the similarresults can be obtainedwith respect to (40) if119861 is nonsingularHowever we point out that (1) can be solved by computingdeflating subspaces of other matrix pencils For instance welet

M1minus 120582L

1= [

119860⊤119861 0

minus119862 minus 119860119862⊤119861 119868119899

] minus 120582 [119868119899

0

0 119860119861⊤] (41)

Assume that the set of 120590(119861119860⊤) is ⊤-reciprocal free it can beshown thatM

1[119868119899

119883] = L

1[119868119899

119883] 119860⊤119861 and it has similar results

to the conclusion ofTheorem 9The unique solution119883 of (1)can be found by computing deflating subspaces of the matrixpencil M

1minus 120582L

1without the assumption of the singularity

of 119860 and 119861

4 Computational Methods for Solving (1)Numerical methods for solving (1) have received great atten-tion in theory and in practice and can be found in [5 6] forthe Krylov subspace methods and in [15ndash17] for the Smith-type iterative methods In particular Smith-type iterativemethods are only workable in the case 120588(119860119861

⊤) lt 1 where

120588(119860119861⊤) denotes the spectral radius of 119860119861

⊤ In the recentyears a structure algorithm has been studied for (1) [4] viaPQZ decomposition which consists of transforming intoSchur form by a PQZ decomposition and then solving theresulting triangular system by way of back-substitution Inthis section we revisit these numericalmethods and point outthe advantages and drawbacks of all algorithms

41 Krylov Subspace Methods Since the ⊤-Stein equation isessentially a linear system (7) we certainly can use the Krylovsubspace methods to solve (7) See for example [5 6] andthe reference cited therein The general idea for applying theKrylov subspace methods is defining the ⋆-Stein operatorTas T 119883 rarr 119883 minus 119860119883

⊤119861 and its adjoint liner operator T as

Tlowast 119884 rarr 119884 minus 119861119884⊤119860 such that ⟨T(119883) 119884⟩ = ⟨119883Tlowast(119884)⟩

Here 119883 119884 isin R119898times119899 and the notion ⟨sdot sdot⟩ is denoted as theFrobenius inner product Then the iterative method basedon the Krylov subspaces for (1) is as follows

6 Abstract and Applied Analysis

(i) The conjugate gradient (CG) method [5]

119883119896+1

= 119883119896+

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2119875119896

119877119896+1

= 119862 minusT (119883119896+1

) = 119877119896minus

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2T (119875119896)

119863119896+1

= Tlowast(119877119896+1

) +

1003817100381710038171003817119877119896+11003817100381710038171003817

2

10038171003817100381710038171198771198961003817100381710038171003817

2119863119896

(42)

with an initial matrix 1198830and the corresponding initial

conditions

1198770= 119862 minusT (119883

0) 119863

0= Tlowast(1198770) (43)

Note that when the solvability conditions of Theorem 4are met the CG method is guaranteed to converge in a finitenumber of iterations for any initial matrix119883

0

42 The Bartels-Stewart-Like Algorithm [18] In this sectionwe focus on the discussion of the Bartels-Stewart algorithmwhich is known to be a numerical stable algorithm to solve⊤-Stein equationsThis method is to solve (1) by means of thePQZ decomposition [18] Its approach has been discussed in[4] and can be summarized as follows From Lemma 3 weknow that there exist two unitary matrices 119875 and 119876 (see [3]for the computation procedure) such that

119875119883119876 minus 119875119860119876 sdot 119876119867119883⊤119875⊤sdot 119875119861119876 = 119875119862119876 (44)

With 119860 = 119875119860119876 and 119861⊤

= 119876119867119861⊤119875119867 being uppertriangular

the transformed equation looks like

[11988311

11990912

11990921

11990922

] minus [11986011

11988612

0 11988622

] [119883⊤

11119909⊤

21

119909⊤

12119909⊤

22

] [11986111

0

21

22

]

= [11986211

11988812

11988821

11988822

]

(45)

with119883 = [1111990912

1199092111990922

] We then have

11990922

minus 11988622119909⊤

2222

= 11988822 (46)

11990921

minus 11988622119909⊤

1211986111

= 11988821

+ 11988622119909⊤

2221 (47)

11990912

minus 11986011119909⊤

2122

= 11988812

+ 11988612119909⊤

2222 (48)

11988311

minus 11986011119883⊤

1111986111

= 11986211

+ 11988612119909⊤

1211986111

+ 11986011119909⊤

2121

+ 11988612119909⊤

2221

(49)

Thus the Bartels-Stewart algorithm can easily be constructedby first solving 119909

22from (46) using 119909

22to obtain 119909

12and 119909

21

from (47) and (48) and then repeating the same discussionas (46)ndash(48) by taking advantage of the property of 119860

11and

11986111being lower triangular matrices from (49)

43 Smith-Type Iterative Methods Recently a class methodreferred to as the Smith accelerative iteration has been studiedfor solving the Sylvester matrix equation [15] The Smithaccelerative iteration has attracted much interests in manypapers (see [7 17] and the references therein) because of itsnice numerical behavior a quadratic convergence rate lowcomputational cost and high numerical reliability despite thelack of a rigorous error analysis Since the Sylvester matrixequation can be transformed into the Stein matrix equationwith a suitable transformation Zhou et al proposed Smith-type iterative methods (including the Smith accelerativeiteration Smith (ℓ) iteration and 119903-Smith iteration) forsolving the Stein matrix equation [17] and applying Smith-type iterative methods to (1) [7] In this section we try toexplain the Smith accelerative iteration based on the invariantsubspace method and summarize the recent results from [7]

Originally the Smith-type iterative methods are devel-oped to solve the standard Stein equation

119883 = A119883B +C ABC isin R119899times119899

(50)

As mentioned before the unknown119883 is highly related to thegeneralized eigenspace problems

[B 0

minusC 119868] [

119868

119883] = [

119868 0

0 A] [

119868

119883]B (51a)

or

A [119883 119868] [B 0

0 119868] = [119883 119868] [

119868 0

minusC A] (51b)

Premultiplying (51a) by the matrix [ B 0minusAC 119868 ] and postmul-

tiplying (51b) by the matrix [119868 0

minusCB A ] we get

[B2 0

minusC minusACB 119868119899

] [119868119899

119883] = [

119868119899

0

0 A2] [

119868119899

119883]B2

A2[119883 119868119899] [

B2 0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minusC minusACB A2]

(52)

Then for any positive integer 119896 gt 0 we obtain

[B2119896minus1

0

minus119862119896

119868119899

] [119868119899

119883] = [

119868119899

0

0 A2119896minus1] [

119868119899

119883]B2119896minus1

A2119896minus1

[119883 119868119899] [

B2119896minus1

0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minus119862119896

A2119896minus1]

(53)

where the sequence C119896 is defined by

119862119896= 119862119896minus1

+A2119896minus1

119862119896minus1

B2119896minus1

119896 ge 1 (54a)

1198620= C (54b)

The explicit expression of 119862119896is given as follows

119862119896=

2119896minus1

sum

119894=1

A119894CB119894 (55)

Abstract and Applied Analysis 7

Under the condition 120588(A)120588(B) lt 1 it is easy to see that 119862119896

is convergent and

lim sup119896rarrinfin

2119896

radic1003817100381710038171003817119883 minus 119862

119896

1003817100381710038171003817 le 120588 (A) 120588 (B) (56)

that is 119862119896converges quadratically to 119883 as 119896 rarr infin

This iterative method (54a) and (54b) is called the Smithaccelerative iteration [15] In recent years some modifiediterative methods are so-called Smith-type iteration whichare based on Smith iteration and improve its speed ofconvergence See for example [17] and the references citedtherein

Since the condition 120588(A)120588(B) lt 1 implies that theassumptions of Theorem 4 hold (1) is equivalent to (5) Wecan apply the Smith iteration to the (1) with the substitution(ABC) = (119860119861

⊤ 119860⊤119861 119862+119860119862

⊤119861) One possible drawback

of the Smith-type iterativemethods is that they cannot alwayshandle the case when there exist eigenvalues 120582 120583 isin 120590(119860

⊤119861)

such that 120582120583 = minus1 even if the unique solution119883 exists Basedon the solvable conditions given in this work it is possible todevelop a specific technique working on the particular caseand it is a subject currently under investigation

5 Error Analysis

Error analysis is a way for testing the stability of a numericalalgorithm and evaluating the accuracy of an approximatedsolution In the subsequent discussion we want to considerthe backward error and perturbation bounds for solving (1)

As indicated in (44) matrices 119860 and 119861⊤ are both upper-

triangularWe can then apply the error analysis for triangularlinear systems in [19 Section 31] and [20] to obtain

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 119888119898119899

u (1 +1003817100381710038171003817100381711986010038171003817100381710038171003817119865

1003817100381710038171003817100381711986110038171003817100381710038171003817119865

)1003817100381710038171003817100381711988310038171003817100381710038171003817119865

(57)

where 119888119898119899

is a content depending on the dimensions 119898 and119899 and u is the unit roundoff Since the PQZ decomposition isa stable process it is true that

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 1198881015840

119898119899u (1 + 119860

119865119861119865) 119883119865 (58)

with a modest multiple 1198881015840119898119899

Note that the inequality of the form (58) can serve

as a stopping criterion for terminating iterations generatedfrom the Krylov subspace methods [5 6] and Smith-typeiterative methods [15ndash17] In what follows we will derivethe error associated with numerical algorithms following thedevelopment in [20 21]

51 Backward Error Like the discussion of the ordinarySylvester equations [20] the normwise backward error of anapproximate solution 119884 of (1) is defined by

120578 (119884) equiv min 120598 119884 = (119862 + 120575119862) + (119860 + 120575119860)119884⊤

(119861 + 120575119861)

120575119860119865le 120598120572 120575119861

119865le 120598120573 120575119862

119865le 120598120574

(59)

where 120572 equiv 119860119865 120573 equiv 119861

119865 and 120574 equiv 119862

119865 Let R equiv 120575119862 +

120575119860119884⊤119861 + 119860119884

⊤120575119861 + 120575119860119884

⊤120575119861 which implies that R = 119884 minus

119860119884⊤119861 minus 119862 It can be seen that the residualR satisfies

R119865le 120578 (119884) (120574 + 119884

119865120572120573 (2 + 120578 (119884))) (60)

From (60) we know that a small backward error indeedimplies a small relative residual R Since the coefficientmatrices in (1) include nonlinearity it appears to be an openproblem to obtain the theoretical backward error with respectto the residual Again similar to the Sylvester equationdiscussed in [20 Section 162] the conditions under whicha ⊤-Stein equation has a well-conditioned solution remainunknown

52 Perturbation Bounds Consider the perturbed equation

119883 + 120575119883 = (119860 + 120575119860) (119883 + 120575119883)⊤

(119861 + 120575119861) + (119862 + 120575119862)

(61)

Let 119878(119883) = 119883minus119860119883⊤119861 be the corresponding⊤-Stein operator

We then have 119878(120575119883) = 120575119862+119860(119883+120575119883)⊤120575119861+120575119860(119883+120575119883)

⊤(119861+

120575119861) With the application of norm it follows that

120575119883119865le10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

120575119862119865+ 120575119878

119865(119883119865+ 120575119883

119865) (62)

where 120575119878119865

equiv 119860119865120575119861119865+ 120575119860

119865(119861119865+ 120575119861

119865) When

120575119878119865is small enough so that 1 ge 119878

minus1119865sdot 120575119878

119865 we can

rearrange the previous result to be

120575119883119865

119883119865

le

10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

1 minus1003817100381710038171003817119878minus11003817100381710038171003817119865

sdot 120575119878119865

(120575119862119865

119883119865

+ 120575119878119865) (63)

With 119862119865

= 119878(119883)119865

le 119878119865sdot 119883119865and the condition

number 120581(119878) equiv 119878119865

sdot 119878minus1119865 we arrive at the standard

perturbation result120575119883119865

119883119865

le120581 (119878)

1 minus 120581 (119878) sdot 120575119878119865119878119865

(120575119862119865

119862119865

+120575119878119865

119878119865

)

(64)

Thus the relative error in119883 is controlled by those in119860 119861 and119862 magnified by the condition number 120581(119878)

On the other hand we can also drop the high-order termsin the perturbation to obtain

120575119883 minus 119860120575119883⊤119861 = 119860119883

⊤120575119861 + 120575119860119883

⊤119861 + 120575119862 (65)

We then rewrite the system in terms of

Q vec (120575119883) = [(119883⊤119861)⊤

otimes 119868119898119868119899otimes (119860119883

⊤) 119868119898119899

] [

[

vec (120575119860)

vec (120575119861)vec (120575119862)

]

]

(66)

where Q = 119868119898119899

minus (119861⊤

otimes 119860)P Let 120577 = max120575119860119865119860119865

120575119861119865119861119865 120575119862

119865119862119865 It can be shown that

120575119883119865

119883119865

le radic3Ψ120577 (67)

where Ψ=Qminus1[120572(119883⊤119861)⊤otimes 119868119898

120573119868119899otimes (119860119883

⊤)120574119868119898119899

]2119883119865

8 Abstract and Applied Analysis

A possible disadvantage of the perturbation bound (67)which ignores the consideration of the underlying structureof the problem is that it overestimates the effect of theperturbation on the data But this ldquouniversalrdquo perturbationbound is accessible to any given matrices 119860 119861 and 119862 of (1)

Unlike the perturbation bound (67) it is desirable toobtain a posteriori error bound by assuming 120575119860 = 120575119861 = 0

and 120575119862 = 119883 minus 119860119883⊤119861 minus 119862 in (61) This assumption gives rise

to

120575119883119865

119883119865

le

10038171003817100381710038171003817119875minus1100381710038171003817100381710038172119877119865

119883119865

(68)

It is true that while doing numerical computation this boundgiven in (68) provides a simpler way for estimating the errorof the solution of (1)

6 Conclusion

In this note we propose a novel approach to the necessaryand sufficient conditions for the unique solvability of thesolution 119883 of the ⊤-Stein equation for square coefficientmatrices in terms of the analysis of the spectra 120590(119860

⊤119861)

Solvability conditions have been derived and algorithmshave been proposed in [4 8] by using PQZ decompositionOn the other hand one common procedure to solve theStein-type equations is by means of the invariant subspacemethod We believe that our discussion is the first whichimplements the techniques of the deflating subspace forsolving ⊤-Stein matrix equation and might also give riseto the possibility of developing an advanced and effectivesolver in the future Also we obtain the theoretical residualanalysis backward error analysis and perturbation boundsfor measuring accurately the error in the computed solutionof (1)

Acknowledgments

The author wishes to thank Proffessor Eric King-wah Chu(Monash University Australia) Proffessor Matthew M Lin(National Chung Cheng University Taiwan) and two anony-mous referees for many interesting and valuable suggestionson the paperThis research work is partially supported by theNational Science Council (NSC101-2115-M-150-002) and theNational Center for Theoretical Sciences in Taiwan

References

[1] H W Braden ldquoThe equations 119860⊤119883 plusmn 119883⊤119860 = 119861rdquo SIAM Journal

on Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021998

[2] X Ma J Han and F Pu ldquoA study of solution existence formatrix equation 119860119883 + 119860

⊤119862 = 119861rdquo Journal of Shenyang Institute

of Aeronautical Engineering vol 4 pp 64ndash66 2003[3] A W Bojanczyk G H Golub and P Van Dooren ldquoPeri-

odic schur decomposition algorithms and applicationsrdquo inAdvanced Signal Processing Algorithms Architectures andImplementations vol 1770 of Proceedings of SPIE pp 31ndash42 SanDiego Calif USA July 1992

[4] C-Y Chiang E K-W Chu andW-W Lin ldquoOn the lowast-Sylvesterequation 119860119883 plusmn 119883

lowast119861lowastrdquo Applied Mathematics and Computation

vol 218 no 17 pp 8393ndash8407 2012[5] Y Su and G Chen ldquoIterative methods for solving linear matrix

equation and linear matrix systemrdquo International Journal ofComputer Mathematics vol 87 no 4 pp 763ndash774 2010

[6] M Wang X Cheng and M Wei ldquoIterative algorithms forsolving the matrix equation 119860119883119861 + 119862119883

⊤119863 = 119864rdquo Applied

Mathematics andComputation vol 187 no 2 pp 622ndash629 2007[7] B Zhou J Lam and G-R Duan ldquoToward solution of matrix

equation 119860119891(119883)119861 + 119862rdquo Linear Algebra and its Applications vol435 no 6 pp 1370ndash1398 2011

[8] K D Ikramov and O Yu Vorontsov ldquoThe matrix equation119883+

119860119883⊤119861 = 119862 conditions for unique solvability and a numerical

algorithm for its solutionrdquo Doklady Mathematics vol 85 pp265ndash267 2012

[9] D S Bernstein Matrix Mathematics Theory Facts and For-mulas Princeton University Press Princeton NJ USA 2ndedition 2009

[10] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1994 Correctedreprint of the 1991 original

[11] F ZhangMatrix Theory Basic Results and Techniques Univer-sitext Springer 1999

[12] D A Bini B Iannazzo and B Meini Numerical Solution ofAlgebraic Riccati Equations Society for Industrial and AppliedMathematics (SIAM) Philadelphia Pa USA 2012

[13] I Gohberg P Lancaster and L Rodman Matrix PolynomialsComputer Science and Applied Mathematics Academic PressNew York NY USA 1982

[14] C-Y Chiang andMM Lin ldquoDeflating subspaces and Sylvesterequationsrdquo Tech Rep NCTS Taiwan 2013

[15] R A Smith ldquoMatrix equation119883119860+ 119861119883 = 119862rdquo SIAM Journal onApplied Mathematics vol 16 pp 198ndash201 1968

[16] T Penzl ldquoA cyclic low-rank Smith method for large sparseLyapunov equationsrdquo SIAM Journal on Scientific Computingvol 21 no 4 pp 1401ndash1418 1999

[17] B Zhou J Lam and G-R Duan ldquoOn Smith-type iterativealgorithms for the Stein matrix equationrdquo Applied MathematicsLetters vol 22 no 7 pp 1038ndash1044 2009

[18] RH Bartels andGW Stewart ldquoSolution of thematrix equation119860119883+119883119861 = 119862 algorithm 432rdquoCommunications of the ACM vol15 no 9 pp 820ndash826 1972

[19] G H Golub and C F Van Loan Matrix Computations JohnsHopkins Studies in the Mathematical Sciences Johns HopkinsUniversity Press Baltimore Md USA 3rd edition 1996

[20] N J Higham Accuracy and Stability of Numerical Algo-rithms Society for Industrial andAppliedMathematics (SIAM)Philadelphia Pa USA 2nd edition 2002

[21] A R Ghavimi and A J Laub ldquoBackward error sensitivityand refinement of computed solutions of algebraic Riccatiequationsrdquo Numerical Linear Algebra with Applications vol 2no 1 pp 29ndash49 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

(1) 1198801= 1198802= 0 if 120590(119879

1minus 1205821198792) = 120590(119868

119899minus 120582119860119861

⊤)

(2) 1198801and 119880

2are nonsingular if 119879

1minus 1205821198792sim 119861119860⊤minus 120582119868119899

Moreover if 119860 is nonsingular then 119883 = 1198811119880minus1

1119860minus⊤

=

119880minus⊤

2119881⊤

2119860minus⊤ is the unique solution of (1)

Proof From (32a) and (32b) we get

119861119860⊤1198801= 11988021198791 (33a)

minus119862119860⊤1198801+ 1198811= 11988121198791 (33b)

1198801= 11988021198792 (33c)

119860119862⊤1198801+ 119860119861⊤1198811= 11988121198792 (33d)

(i) It follows from (33a) and (33c) that since 120590(119861119860⊤

minus

120582119868119899) cap 120590(119879

1minus 1205821198792) = 120601 we have 119880

1= 1198802

= 0 byLemma 7

(ii) It can be seen that there exist two nonsingularmatrices 119880 and 119881 such that

M [0

119880] = [

0

119881]1198792

L [0

119880] = [

0

119881]1198791

(34)

Hence together with (32a) and (32b) we have

M [0 1198801

119880 1198811

] = [0 1198802

119881 1198812

] [1198792

0

0 1198791

]

L [0 1198801

119880 1198811

] = [0 1198802

119881 1198812

] [1198791

0

0 1198792

]

(35)

Since 120590(Mminus120582L) = 120590(119861119860⊤minus120582119868119899)cup120590(119868

119899minus120582119860119861

⊤) and 120590(119861119860

⊤minus

120582119868119899) cap 120590(119868

119899minus 120582119860119861

⊤) = 120601 by Theorem 5 and Lemma 6 the

matrix [0 1198801

119880 1198811

] is nonsingular Together with (33c)1198801and119880

2

are nonsingularLet119883

119894= 119881119894119880minus1

119894 119894 = 1 2 then form (33b) and (33d)

119860119862⊤+ 119860119861⊤1198831= 11988121198792119880minus1

1= 1198832

minus119862119860⊤+ 1198831= 11988121198791119880minus1

1= 1198832119861119860⊤

(36)

or

119860119862⊤+ 119860119861⊤1198831= 1198832

119860119862⊤+ 119860119861⊤119883⊤

2= 119883⊤

1

(37)

Since the set of 120590(119860119861⊤) = 120590(119861119860

⊤) is ⊤-reciprocal free

together with

119883⊤

1minus 1198832minus 119860119861⊤(119883⊤

1minus 1198832)⊤

= 0 (38)

we get 1198831= 119883⊤

2 If 119860 is nonsingular it is easy to verify that

two matrices 1198831119860minus⊤ and 119883

⊤

2119860minus⊤ are both satisfying ⊤-Stein

equation (1) The proof of part (ii) is complete

Remark 10 (1) It is easily seen that [ 119868119899119883119860⊤ ] and [

1198801

1198812

] both spanthe unique deflating subspace of M minus 120582L corresponding tothe set of 120590(119861119860⊤) Otherwise in part (ii) we know that 119879

2

is nonsingular We then are able to transform the formulaedefined in (32a) and (32b) into the generalized eigenvalueproblem as follows

M [1198801

1198811

] = L [1198801

1198811

]119861119860⊤ (39)

That is some numerical methods for the computation ofthe eigenspace ofMminus120582L corresponding to the set of 120590(119861119860⊤)can be designed and solved (1)

(2) Since the transport of the unique solution 119883 of (1)is equal to the unique solution 119884 of the following matrixequation

119884 = 119861⊤119884119860⊤+ 119862⊤ (40)

analogous to the consequences of Theorem 9 the similarresults can be obtainedwith respect to (40) if119861 is nonsingularHowever we point out that (1) can be solved by computingdeflating subspaces of other matrix pencils For instance welet

M1minus 120582L

1= [

119860⊤119861 0

minus119862 minus 119860119862⊤119861 119868119899

] minus 120582 [119868119899

0

0 119860119861⊤] (41)

Assume that the set of 120590(119861119860⊤) is ⊤-reciprocal free it can beshown thatM

1[119868119899

119883] = L

1[119868119899

119883] 119860⊤119861 and it has similar results

to the conclusion ofTheorem 9The unique solution119883 of (1)can be found by computing deflating subspaces of the matrixpencil M

1minus 120582L

1without the assumption of the singularity

of 119860 and 119861

4 Computational Methods for Solving (1)Numerical methods for solving (1) have received great atten-tion in theory and in practice and can be found in [5 6] forthe Krylov subspace methods and in [15ndash17] for the Smith-type iterative methods In particular Smith-type iterativemethods are only workable in the case 120588(119860119861

⊤) lt 1 where

120588(119860119861⊤) denotes the spectral radius of 119860119861

⊤ In the recentyears a structure algorithm has been studied for (1) [4] viaPQZ decomposition which consists of transforming intoSchur form by a PQZ decomposition and then solving theresulting triangular system by way of back-substitution Inthis section we revisit these numericalmethods and point outthe advantages and drawbacks of all algorithms

41 Krylov Subspace Methods Since the ⊤-Stein equation isessentially a linear system (7) we certainly can use the Krylovsubspace methods to solve (7) See for example [5 6] andthe reference cited therein The general idea for applying theKrylov subspace methods is defining the ⋆-Stein operatorTas T 119883 rarr 119883 minus 119860119883

⊤119861 and its adjoint liner operator T as

Tlowast 119884 rarr 119884 minus 119861119884⊤119860 such that ⟨T(119883) 119884⟩ = ⟨119883Tlowast(119884)⟩

Here 119883 119884 isin R119898times119899 and the notion ⟨sdot sdot⟩ is denoted as theFrobenius inner product Then the iterative method basedon the Krylov subspaces for (1) is as follows

6 Abstract and Applied Analysis

(i) The conjugate gradient (CG) method [5]

119883119896+1

= 119883119896+

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2119875119896

119877119896+1

= 119862 minusT (119883119896+1

) = 119877119896minus

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2T (119875119896)

119863119896+1

= Tlowast(119877119896+1

) +

1003817100381710038171003817119877119896+11003817100381710038171003817

2

10038171003817100381710038171198771198961003817100381710038171003817

2119863119896

(42)

with an initial matrix 1198830and the corresponding initial

conditions

1198770= 119862 minusT (119883

0) 119863

0= Tlowast(1198770) (43)

Note that when the solvability conditions of Theorem 4are met the CG method is guaranteed to converge in a finitenumber of iterations for any initial matrix119883

0

42 The Bartels-Stewart-Like Algorithm [18] In this sectionwe focus on the discussion of the Bartels-Stewart algorithmwhich is known to be a numerical stable algorithm to solve⊤-Stein equationsThis method is to solve (1) by means of thePQZ decomposition [18] Its approach has been discussed in[4] and can be summarized as follows From Lemma 3 weknow that there exist two unitary matrices 119875 and 119876 (see [3]for the computation procedure) such that

119875119883119876 minus 119875119860119876 sdot 119876119867119883⊤119875⊤sdot 119875119861119876 = 119875119862119876 (44)

With 119860 = 119875119860119876 and 119861⊤

= 119876119867119861⊤119875119867 being uppertriangular

the transformed equation looks like

[11988311

11990912

11990921

11990922

] minus [11986011

11988612

0 11988622

] [119883⊤

11119909⊤

21

119909⊤

12119909⊤

22

] [11986111

0

21

22

]

= [11986211

11988812

11988821

11988822

]

(45)

with119883 = [1111990912

1199092111990922

] We then have

11990922

minus 11988622119909⊤

2222

= 11988822 (46)

11990921

minus 11988622119909⊤

1211986111

= 11988821

+ 11988622119909⊤

2221 (47)

11990912

minus 11986011119909⊤

2122

= 11988812

+ 11988612119909⊤

2222 (48)

11988311

minus 11986011119883⊤

1111986111

= 11986211

+ 11988612119909⊤

1211986111

+ 11986011119909⊤

2121

+ 11988612119909⊤

2221

(49)

Thus the Bartels-Stewart algorithm can easily be constructedby first solving 119909

22from (46) using 119909

22to obtain 119909

12and 119909

21

from (47) and (48) and then repeating the same discussionas (46)ndash(48) by taking advantage of the property of 119860

11and

11986111being lower triangular matrices from (49)

43 Smith-Type Iterative Methods Recently a class methodreferred to as the Smith accelerative iteration has been studiedfor solving the Sylvester matrix equation [15] The Smithaccelerative iteration has attracted much interests in manypapers (see [7 17] and the references therein) because of itsnice numerical behavior a quadratic convergence rate lowcomputational cost and high numerical reliability despite thelack of a rigorous error analysis Since the Sylvester matrixequation can be transformed into the Stein matrix equationwith a suitable transformation Zhou et al proposed Smith-type iterative methods (including the Smith accelerativeiteration Smith (ℓ) iteration and 119903-Smith iteration) forsolving the Stein matrix equation [17] and applying Smith-type iterative methods to (1) [7] In this section we try toexplain the Smith accelerative iteration based on the invariantsubspace method and summarize the recent results from [7]

Originally the Smith-type iterative methods are devel-oped to solve the standard Stein equation

119883 = A119883B +C ABC isin R119899times119899

(50)

As mentioned before the unknown119883 is highly related to thegeneralized eigenspace problems

[B 0

minusC 119868] [

119868

119883] = [

119868 0

0 A] [

119868

119883]B (51a)

or

A [119883 119868] [B 0

0 119868] = [119883 119868] [

119868 0

minusC A] (51b)

Premultiplying (51a) by the matrix [ B 0minusAC 119868 ] and postmul-

tiplying (51b) by the matrix [119868 0

minusCB A ] we get

[B2 0

minusC minusACB 119868119899

] [119868119899

119883] = [

119868119899

0

0 A2] [

119868119899

119883]B2

A2[119883 119868119899] [

B2 0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minusC minusACB A2]

(52)

Then for any positive integer 119896 gt 0 we obtain

[B2119896minus1

0

minus119862119896

119868119899

] [119868119899

119883] = [

119868119899

0

0 A2119896minus1] [

119868119899

119883]B2119896minus1

A2119896minus1

[119883 119868119899] [

B2119896minus1

0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minus119862119896

A2119896minus1]

(53)

where the sequence C119896 is defined by

119862119896= 119862119896minus1

+A2119896minus1

119862119896minus1

B2119896minus1

119896 ge 1 (54a)

1198620= C (54b)

The explicit expression of 119862119896is given as follows

119862119896=

2119896minus1

sum

119894=1

A119894CB119894 (55)

Abstract and Applied Analysis 7

Under the condition 120588(A)120588(B) lt 1 it is easy to see that 119862119896

is convergent and

lim sup119896rarrinfin

2119896

radic1003817100381710038171003817119883 minus 119862

119896

1003817100381710038171003817 le 120588 (A) 120588 (B) (56)

that is 119862119896converges quadratically to 119883 as 119896 rarr infin

This iterative method (54a) and (54b) is called the Smithaccelerative iteration [15] In recent years some modifiediterative methods are so-called Smith-type iteration whichare based on Smith iteration and improve its speed ofconvergence See for example [17] and the references citedtherein

Since the condition 120588(A)120588(B) lt 1 implies that theassumptions of Theorem 4 hold (1) is equivalent to (5) Wecan apply the Smith iteration to the (1) with the substitution(ABC) = (119860119861

⊤ 119860⊤119861 119862+119860119862

⊤119861) One possible drawback

of the Smith-type iterativemethods is that they cannot alwayshandle the case when there exist eigenvalues 120582 120583 isin 120590(119860

⊤119861)

such that 120582120583 = minus1 even if the unique solution119883 exists Basedon the solvable conditions given in this work it is possible todevelop a specific technique working on the particular caseand it is a subject currently under investigation

5 Error Analysis

Error analysis is a way for testing the stability of a numericalalgorithm and evaluating the accuracy of an approximatedsolution In the subsequent discussion we want to considerthe backward error and perturbation bounds for solving (1)

As indicated in (44) matrices 119860 and 119861⊤ are both upper-

triangularWe can then apply the error analysis for triangularlinear systems in [19 Section 31] and [20] to obtain

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 119888119898119899

u (1 +1003817100381710038171003817100381711986010038171003817100381710038171003817119865

1003817100381710038171003817100381711986110038171003817100381710038171003817119865

)1003817100381710038171003817100381711988310038171003817100381710038171003817119865

(57)

where 119888119898119899

is a content depending on the dimensions 119898 and119899 and u is the unit roundoff Since the PQZ decomposition isa stable process it is true that

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 1198881015840

119898119899u (1 + 119860

119865119861119865) 119883119865 (58)

with a modest multiple 1198881015840119898119899

Note that the inequality of the form (58) can serve

as a stopping criterion for terminating iterations generatedfrom the Krylov subspace methods [5 6] and Smith-typeiterative methods [15ndash17] In what follows we will derivethe error associated with numerical algorithms following thedevelopment in [20 21]

51 Backward Error Like the discussion of the ordinarySylvester equations [20] the normwise backward error of anapproximate solution 119884 of (1) is defined by

120578 (119884) equiv min 120598 119884 = (119862 + 120575119862) + (119860 + 120575119860)119884⊤

(119861 + 120575119861)

120575119860119865le 120598120572 120575119861

119865le 120598120573 120575119862

119865le 120598120574

(59)

where 120572 equiv 119860119865 120573 equiv 119861

119865 and 120574 equiv 119862

119865 Let R equiv 120575119862 +

120575119860119884⊤119861 + 119860119884

⊤120575119861 + 120575119860119884

⊤120575119861 which implies that R = 119884 minus

119860119884⊤119861 minus 119862 It can be seen that the residualR satisfies

R119865le 120578 (119884) (120574 + 119884

119865120572120573 (2 + 120578 (119884))) (60)

From (60) we know that a small backward error indeedimplies a small relative residual R Since the coefficientmatrices in (1) include nonlinearity it appears to be an openproblem to obtain the theoretical backward error with respectto the residual Again similar to the Sylvester equationdiscussed in [20 Section 162] the conditions under whicha ⊤-Stein equation has a well-conditioned solution remainunknown

52 Perturbation Bounds Consider the perturbed equation

119883 + 120575119883 = (119860 + 120575119860) (119883 + 120575119883)⊤

(119861 + 120575119861) + (119862 + 120575119862)

(61)

Let 119878(119883) = 119883minus119860119883⊤119861 be the corresponding⊤-Stein operator

We then have 119878(120575119883) = 120575119862+119860(119883+120575119883)⊤120575119861+120575119860(119883+120575119883)

⊤(119861+

120575119861) With the application of norm it follows that

120575119883119865le10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

120575119862119865+ 120575119878

119865(119883119865+ 120575119883

119865) (62)

where 120575119878119865

equiv 119860119865120575119861119865+ 120575119860

119865(119861119865+ 120575119861

119865) When

120575119878119865is small enough so that 1 ge 119878

minus1119865sdot 120575119878

119865 we can

rearrange the previous result to be

120575119883119865

119883119865

le

10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

1 minus1003817100381710038171003817119878minus11003817100381710038171003817119865

sdot 120575119878119865

(120575119862119865

119883119865

+ 120575119878119865) (63)

With 119862119865

= 119878(119883)119865

le 119878119865sdot 119883119865and the condition

number 120581(119878) equiv 119878119865

sdot 119878minus1119865 we arrive at the standard

perturbation result120575119883119865

119883119865

le120581 (119878)

1 minus 120581 (119878) sdot 120575119878119865119878119865

(120575119862119865

119862119865

+120575119878119865

119878119865

)

(64)

Thus the relative error in119883 is controlled by those in119860 119861 and119862 magnified by the condition number 120581(119878)

On the other hand we can also drop the high-order termsin the perturbation to obtain

120575119883 minus 119860120575119883⊤119861 = 119860119883

⊤120575119861 + 120575119860119883

⊤119861 + 120575119862 (65)

We then rewrite the system in terms of

Q vec (120575119883) = [(119883⊤119861)⊤

otimes 119868119898119868119899otimes (119860119883

⊤) 119868119898119899

] [

[

vec (120575119860)

vec (120575119861)vec (120575119862)

]

]

(66)

where Q = 119868119898119899

minus (119861⊤

otimes 119860)P Let 120577 = max120575119860119865119860119865

120575119861119865119861119865 120575119862

119865119862119865 It can be shown that

120575119883119865

119883119865

le radic3Ψ120577 (67)

where Ψ=Qminus1[120572(119883⊤119861)⊤otimes 119868119898

120573119868119899otimes (119860119883

⊤)120574119868119898119899

]2119883119865

8 Abstract and Applied Analysis

A possible disadvantage of the perturbation bound (67)which ignores the consideration of the underlying structureof the problem is that it overestimates the effect of theperturbation on the data But this ldquouniversalrdquo perturbationbound is accessible to any given matrices 119860 119861 and 119862 of (1)

Unlike the perturbation bound (67) it is desirable toobtain a posteriori error bound by assuming 120575119860 = 120575119861 = 0

and 120575119862 = 119883 minus 119860119883⊤119861 minus 119862 in (61) This assumption gives rise

to

120575119883119865

119883119865

le

10038171003817100381710038171003817119875minus1100381710038171003817100381710038172119877119865

119883119865

(68)

It is true that while doing numerical computation this boundgiven in (68) provides a simpler way for estimating the errorof the solution of (1)

6 Conclusion

In this note we propose a novel approach to the necessaryand sufficient conditions for the unique solvability of thesolution 119883 of the ⊤-Stein equation for square coefficientmatrices in terms of the analysis of the spectra 120590(119860

⊤119861)

Solvability conditions have been derived and algorithmshave been proposed in [4 8] by using PQZ decompositionOn the other hand one common procedure to solve theStein-type equations is by means of the invariant subspacemethod We believe that our discussion is the first whichimplements the techniques of the deflating subspace forsolving ⊤-Stein matrix equation and might also give riseto the possibility of developing an advanced and effectivesolver in the future Also we obtain the theoretical residualanalysis backward error analysis and perturbation boundsfor measuring accurately the error in the computed solutionof (1)

Acknowledgments

The author wishes to thank Proffessor Eric King-wah Chu(Monash University Australia) Proffessor Matthew M Lin(National Chung Cheng University Taiwan) and two anony-mous referees for many interesting and valuable suggestionson the paperThis research work is partially supported by theNational Science Council (NSC101-2115-M-150-002) and theNational Center for Theoretical Sciences in Taiwan

References

[1] H W Braden ldquoThe equations 119860⊤119883 plusmn 119883⊤119860 = 119861rdquo SIAM Journal

on Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021998

[2] X Ma J Han and F Pu ldquoA study of solution existence formatrix equation 119860119883 + 119860

⊤119862 = 119861rdquo Journal of Shenyang Institute

of Aeronautical Engineering vol 4 pp 64ndash66 2003[3] A W Bojanczyk G H Golub and P Van Dooren ldquoPeri-

odic schur decomposition algorithms and applicationsrdquo inAdvanced Signal Processing Algorithms Architectures andImplementations vol 1770 of Proceedings of SPIE pp 31ndash42 SanDiego Calif USA July 1992

[4] C-Y Chiang E K-W Chu andW-W Lin ldquoOn the lowast-Sylvesterequation 119860119883 plusmn 119883

lowast119861lowastrdquo Applied Mathematics and Computation

vol 218 no 17 pp 8393ndash8407 2012[5] Y Su and G Chen ldquoIterative methods for solving linear matrix

equation and linear matrix systemrdquo International Journal ofComputer Mathematics vol 87 no 4 pp 763ndash774 2010

[6] M Wang X Cheng and M Wei ldquoIterative algorithms forsolving the matrix equation 119860119883119861 + 119862119883

⊤119863 = 119864rdquo Applied

Mathematics andComputation vol 187 no 2 pp 622ndash629 2007[7] B Zhou J Lam and G-R Duan ldquoToward solution of matrix

equation 119860119891(119883)119861 + 119862rdquo Linear Algebra and its Applications vol435 no 6 pp 1370ndash1398 2011

[8] K D Ikramov and O Yu Vorontsov ldquoThe matrix equation119883+

119860119883⊤119861 = 119862 conditions for unique solvability and a numerical

algorithm for its solutionrdquo Doklady Mathematics vol 85 pp265ndash267 2012

[9] D S Bernstein Matrix Mathematics Theory Facts and For-mulas Princeton University Press Princeton NJ USA 2ndedition 2009

[10] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1994 Correctedreprint of the 1991 original

[11] F ZhangMatrix Theory Basic Results and Techniques Univer-sitext Springer 1999

[12] D A Bini B Iannazzo and B Meini Numerical Solution ofAlgebraic Riccati Equations Society for Industrial and AppliedMathematics (SIAM) Philadelphia Pa USA 2012

[13] I Gohberg P Lancaster and L Rodman Matrix PolynomialsComputer Science and Applied Mathematics Academic PressNew York NY USA 1982

[14] C-Y Chiang andMM Lin ldquoDeflating subspaces and Sylvesterequationsrdquo Tech Rep NCTS Taiwan 2013

[15] R A Smith ldquoMatrix equation119883119860+ 119861119883 = 119862rdquo SIAM Journal onApplied Mathematics vol 16 pp 198ndash201 1968

[16] T Penzl ldquoA cyclic low-rank Smith method for large sparseLyapunov equationsrdquo SIAM Journal on Scientific Computingvol 21 no 4 pp 1401ndash1418 1999

[17] B Zhou J Lam and G-R Duan ldquoOn Smith-type iterativealgorithms for the Stein matrix equationrdquo Applied MathematicsLetters vol 22 no 7 pp 1038ndash1044 2009

[18] RH Bartels andGW Stewart ldquoSolution of thematrix equation119860119883+119883119861 = 119862 algorithm 432rdquoCommunications of the ACM vol15 no 9 pp 820ndash826 1972

[19] G H Golub and C F Van Loan Matrix Computations JohnsHopkins Studies in the Mathematical Sciences Johns HopkinsUniversity Press Baltimore Md USA 3rd edition 1996

[20] N J Higham Accuracy and Stability of Numerical Algo-rithms Society for Industrial andAppliedMathematics (SIAM)Philadelphia Pa USA 2nd edition 2002

[21] A R Ghavimi and A J Laub ldquoBackward error sensitivityand refinement of computed solutions of algebraic Riccatiequationsrdquo Numerical Linear Algebra with Applications vol 2no 1 pp 29ndash49 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

(i) The conjugate gradient (CG) method [5]

119883119896+1

= 119883119896+

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2119875119896

119877119896+1

= 119862 minusT (119883119896+1

) = 119877119896minus

10038171003817100381710038171198771198961003817100381710038171003817

2

10038171003817100381710038171198751198961003817100381710038171003817

2T (119875119896)

119863119896+1

= Tlowast(119877119896+1

) +

1003817100381710038171003817119877119896+11003817100381710038171003817

2

10038171003817100381710038171198771198961003817100381710038171003817

2119863119896

(42)

with an initial matrix 1198830and the corresponding initial

conditions

1198770= 119862 minusT (119883

0) 119863

0= Tlowast(1198770) (43)

Note that when the solvability conditions of Theorem 4are met the CG method is guaranteed to converge in a finitenumber of iterations for any initial matrix119883

0

42 The Bartels-Stewart-Like Algorithm [18] In this sectionwe focus on the discussion of the Bartels-Stewart algorithmwhich is known to be a numerical stable algorithm to solve⊤-Stein equationsThis method is to solve (1) by means of thePQZ decomposition [18] Its approach has been discussed in[4] and can be summarized as follows From Lemma 3 weknow that there exist two unitary matrices 119875 and 119876 (see [3]for the computation procedure) such that

119875119883119876 minus 119875119860119876 sdot 119876119867119883⊤119875⊤sdot 119875119861119876 = 119875119862119876 (44)

With 119860 = 119875119860119876 and 119861⊤

= 119876119867119861⊤119875119867 being uppertriangular

the transformed equation looks like

[11988311

11990912

11990921

11990922

] minus [11986011

11988612

0 11988622

] [119883⊤

11119909⊤

21

119909⊤

12119909⊤

22

] [11986111

0

21

22

]

= [11986211

11988812

11988821

11988822

]

(45)

with119883 = [1111990912

1199092111990922

] We then have

11990922

minus 11988622119909⊤

2222

= 11988822 (46)

11990921

minus 11988622119909⊤

1211986111

= 11988821

+ 11988622119909⊤

2221 (47)

11990912

minus 11986011119909⊤

2122

= 11988812

+ 11988612119909⊤

2222 (48)

11988311

minus 11986011119883⊤

1111986111

= 11986211

+ 11988612119909⊤

1211986111

+ 11986011119909⊤

2121

+ 11988612119909⊤

2221

(49)

Thus the Bartels-Stewart algorithm can easily be constructedby first solving 119909

22from (46) using 119909

22to obtain 119909

12and 119909

21

from (47) and (48) and then repeating the same discussionas (46)ndash(48) by taking advantage of the property of 119860

11and

11986111being lower triangular matrices from (49)

43 Smith-Type Iterative Methods Recently a class methodreferred to as the Smith accelerative iteration has been studiedfor solving the Sylvester matrix equation [15] The Smithaccelerative iteration has attracted much interests in manypapers (see [7 17] and the references therein) because of itsnice numerical behavior a quadratic convergence rate lowcomputational cost and high numerical reliability despite thelack of a rigorous error analysis Since the Sylvester matrixequation can be transformed into the Stein matrix equationwith a suitable transformation Zhou et al proposed Smith-type iterative methods (including the Smith accelerativeiteration Smith (ℓ) iteration and 119903-Smith iteration) forsolving the Stein matrix equation [17] and applying Smith-type iterative methods to (1) [7] In this section we try toexplain the Smith accelerative iteration based on the invariantsubspace method and summarize the recent results from [7]

Originally the Smith-type iterative methods are devel-oped to solve the standard Stein equation

119883 = A119883B +C ABC isin R119899times119899

(50)

As mentioned before the unknown119883 is highly related to thegeneralized eigenspace problems

[B 0

minusC 119868] [

119868

119883] = [

119868 0

0 A] [

119868

119883]B (51a)

or

A [119883 119868] [B 0

0 119868] = [119883 119868] [

119868 0

minusC A] (51b)

Premultiplying (51a) by the matrix [ B 0minusAC 119868 ] and postmul-

tiplying (51b) by the matrix [119868 0

minusCB A ] we get

[B2 0

minusC minusACB 119868119899

] [119868119899

119883] = [

119868119899

0

0 A2] [

119868119899

119883]B2

A2[119883 119868119899] [

B2 0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minusC minusACB A2]

(52)

Then for any positive integer 119896 gt 0 we obtain

[B2119896minus1

0

minus119862119896

119868119899

] [119868119899

119883] = [

119868119899

0

0 A2119896minus1] [

119868119899

119883]B2119896minus1

A2119896minus1

[119883 119868119899] [

B2119896minus1

0

0 119868119899

] = [119883 119868119899] [

119868119899

0

minus119862119896

A2119896minus1]

(53)

where the sequence C119896 is defined by

119862119896= 119862119896minus1

+A2119896minus1

119862119896minus1

B2119896minus1

119896 ge 1 (54a)

1198620= C (54b)

The explicit expression of 119862119896is given as follows

119862119896=

2119896minus1

sum

119894=1

A119894CB119894 (55)

Abstract and Applied Analysis 7

Under the condition 120588(A)120588(B) lt 1 it is easy to see that 119862119896

is convergent and

lim sup119896rarrinfin

2119896

radic1003817100381710038171003817119883 minus 119862

119896

1003817100381710038171003817 le 120588 (A) 120588 (B) (56)

that is 119862119896converges quadratically to 119883 as 119896 rarr infin

This iterative method (54a) and (54b) is called the Smithaccelerative iteration [15] In recent years some modifiediterative methods are so-called Smith-type iteration whichare based on Smith iteration and improve its speed ofconvergence See for example [17] and the references citedtherein

Since the condition 120588(A)120588(B) lt 1 implies that theassumptions of Theorem 4 hold (1) is equivalent to (5) Wecan apply the Smith iteration to the (1) with the substitution(ABC) = (119860119861

⊤ 119860⊤119861 119862+119860119862

⊤119861) One possible drawback

of the Smith-type iterativemethods is that they cannot alwayshandle the case when there exist eigenvalues 120582 120583 isin 120590(119860

⊤119861)

such that 120582120583 = minus1 even if the unique solution119883 exists Basedon the solvable conditions given in this work it is possible todevelop a specific technique working on the particular caseand it is a subject currently under investigation

5 Error Analysis

Error analysis is a way for testing the stability of a numericalalgorithm and evaluating the accuracy of an approximatedsolution In the subsequent discussion we want to considerthe backward error and perturbation bounds for solving (1)

As indicated in (44) matrices 119860 and 119861⊤ are both upper-

triangularWe can then apply the error analysis for triangularlinear systems in [19 Section 31] and [20] to obtain

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 119888119898119899

u (1 +1003817100381710038171003817100381711986010038171003817100381710038171003817119865

1003817100381710038171003817100381711986110038171003817100381710038171003817119865

)1003817100381710038171003817100381711988310038171003817100381710038171003817119865

(57)

where 119888119898119899

is a content depending on the dimensions 119898 and119899 and u is the unit roundoff Since the PQZ decomposition isa stable process it is true that

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 1198881015840

119898119899u (1 + 119860

119865119861119865) 119883119865 (58)

with a modest multiple 1198881015840119898119899

Note that the inequality of the form (58) can serve

as a stopping criterion for terminating iterations generatedfrom the Krylov subspace methods [5 6] and Smith-typeiterative methods [15ndash17] In what follows we will derivethe error associated with numerical algorithms following thedevelopment in [20 21]

51 Backward Error Like the discussion of the ordinarySylvester equations [20] the normwise backward error of anapproximate solution 119884 of (1) is defined by

120578 (119884) equiv min 120598 119884 = (119862 + 120575119862) + (119860 + 120575119860)119884⊤

(119861 + 120575119861)

120575119860119865le 120598120572 120575119861

119865le 120598120573 120575119862

119865le 120598120574

(59)

where 120572 equiv 119860119865 120573 equiv 119861

119865 and 120574 equiv 119862

119865 Let R equiv 120575119862 +

120575119860119884⊤119861 + 119860119884

⊤120575119861 + 120575119860119884

⊤120575119861 which implies that R = 119884 minus

119860119884⊤119861 minus 119862 It can be seen that the residualR satisfies

R119865le 120578 (119884) (120574 + 119884

119865120572120573 (2 + 120578 (119884))) (60)

From (60) we know that a small backward error indeedimplies a small relative residual R Since the coefficientmatrices in (1) include nonlinearity it appears to be an openproblem to obtain the theoretical backward error with respectto the residual Again similar to the Sylvester equationdiscussed in [20 Section 162] the conditions under whicha ⊤-Stein equation has a well-conditioned solution remainunknown

52 Perturbation Bounds Consider the perturbed equation

119883 + 120575119883 = (119860 + 120575119860) (119883 + 120575119883)⊤

(119861 + 120575119861) + (119862 + 120575119862)

(61)

Let 119878(119883) = 119883minus119860119883⊤119861 be the corresponding⊤-Stein operator

We then have 119878(120575119883) = 120575119862+119860(119883+120575119883)⊤120575119861+120575119860(119883+120575119883)

⊤(119861+

120575119861) With the application of norm it follows that

120575119883119865le10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

120575119862119865+ 120575119878

119865(119883119865+ 120575119883

119865) (62)

where 120575119878119865

equiv 119860119865120575119861119865+ 120575119860

119865(119861119865+ 120575119861

119865) When

120575119878119865is small enough so that 1 ge 119878

minus1119865sdot 120575119878

119865 we can

rearrange the previous result to be

120575119883119865

119883119865

le

10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

1 minus1003817100381710038171003817119878minus11003817100381710038171003817119865

sdot 120575119878119865

(120575119862119865

119883119865

+ 120575119878119865) (63)

With 119862119865

= 119878(119883)119865

le 119878119865sdot 119883119865and the condition

number 120581(119878) equiv 119878119865

sdot 119878minus1119865 we arrive at the standard

perturbation result120575119883119865

119883119865

le120581 (119878)

1 minus 120581 (119878) sdot 120575119878119865119878119865

(120575119862119865

119862119865

+120575119878119865

119878119865

)

(64)

Thus the relative error in119883 is controlled by those in119860 119861 and119862 magnified by the condition number 120581(119878)

On the other hand we can also drop the high-order termsin the perturbation to obtain

120575119883 minus 119860120575119883⊤119861 = 119860119883

⊤120575119861 + 120575119860119883

⊤119861 + 120575119862 (65)

We then rewrite the system in terms of

Q vec (120575119883) = [(119883⊤119861)⊤

otimes 119868119898119868119899otimes (119860119883

⊤) 119868119898119899

] [

[

vec (120575119860)

vec (120575119861)vec (120575119862)

]

]

(66)

where Q = 119868119898119899

minus (119861⊤

otimes 119860)P Let 120577 = max120575119860119865119860119865

120575119861119865119861119865 120575119862

119865119862119865 It can be shown that

120575119883119865

119883119865

le radic3Ψ120577 (67)

where Ψ=Qminus1[120572(119883⊤119861)⊤otimes 119868119898

120573119868119899otimes (119860119883

⊤)120574119868119898119899

]2119883119865

8 Abstract and Applied Analysis

A possible disadvantage of the perturbation bound (67)which ignores the consideration of the underlying structureof the problem is that it overestimates the effect of theperturbation on the data But this ldquouniversalrdquo perturbationbound is accessible to any given matrices 119860 119861 and 119862 of (1)

Unlike the perturbation bound (67) it is desirable toobtain a posteriori error bound by assuming 120575119860 = 120575119861 = 0

and 120575119862 = 119883 minus 119860119883⊤119861 minus 119862 in (61) This assumption gives rise

to

120575119883119865

119883119865

le

10038171003817100381710038171003817119875minus1100381710038171003817100381710038172119877119865

119883119865

(68)

It is true that while doing numerical computation this boundgiven in (68) provides a simpler way for estimating the errorof the solution of (1)

6 Conclusion

In this note we propose a novel approach to the necessaryand sufficient conditions for the unique solvability of thesolution 119883 of the ⊤-Stein equation for square coefficientmatrices in terms of the analysis of the spectra 120590(119860

⊤119861)

Solvability conditions have been derived and algorithmshave been proposed in [4 8] by using PQZ decompositionOn the other hand one common procedure to solve theStein-type equations is by means of the invariant subspacemethod We believe that our discussion is the first whichimplements the techniques of the deflating subspace forsolving ⊤-Stein matrix equation and might also give riseto the possibility of developing an advanced and effectivesolver in the future Also we obtain the theoretical residualanalysis backward error analysis and perturbation boundsfor measuring accurately the error in the computed solutionof (1)

Acknowledgments

The author wishes to thank Proffessor Eric King-wah Chu(Monash University Australia) Proffessor Matthew M Lin(National Chung Cheng University Taiwan) and two anony-mous referees for many interesting and valuable suggestionson the paperThis research work is partially supported by theNational Science Council (NSC101-2115-M-150-002) and theNational Center for Theoretical Sciences in Taiwan

References

[1] H W Braden ldquoThe equations 119860⊤119883 plusmn 119883⊤119860 = 119861rdquo SIAM Journal

on Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021998

[2] X Ma J Han and F Pu ldquoA study of solution existence formatrix equation 119860119883 + 119860

⊤119862 = 119861rdquo Journal of Shenyang Institute

of Aeronautical Engineering vol 4 pp 64ndash66 2003[3] A W Bojanczyk G H Golub and P Van Dooren ldquoPeri-

odic schur decomposition algorithms and applicationsrdquo inAdvanced Signal Processing Algorithms Architectures andImplementations vol 1770 of Proceedings of SPIE pp 31ndash42 SanDiego Calif USA July 1992

[4] C-Y Chiang E K-W Chu andW-W Lin ldquoOn the lowast-Sylvesterequation 119860119883 plusmn 119883

lowast119861lowastrdquo Applied Mathematics and Computation

vol 218 no 17 pp 8393ndash8407 2012[5] Y Su and G Chen ldquoIterative methods for solving linear matrix

equation and linear matrix systemrdquo International Journal ofComputer Mathematics vol 87 no 4 pp 763ndash774 2010

[6] M Wang X Cheng and M Wei ldquoIterative algorithms forsolving the matrix equation 119860119883119861 + 119862119883

⊤119863 = 119864rdquo Applied

Mathematics andComputation vol 187 no 2 pp 622ndash629 2007[7] B Zhou J Lam and G-R Duan ldquoToward solution of matrix

equation 119860119891(119883)119861 + 119862rdquo Linear Algebra and its Applications vol435 no 6 pp 1370ndash1398 2011

[8] K D Ikramov and O Yu Vorontsov ldquoThe matrix equation119883+

119860119883⊤119861 = 119862 conditions for unique solvability and a numerical

algorithm for its solutionrdquo Doklady Mathematics vol 85 pp265ndash267 2012

[9] D S Bernstein Matrix Mathematics Theory Facts and For-mulas Princeton University Press Princeton NJ USA 2ndedition 2009

[10] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1994 Correctedreprint of the 1991 original

[11] F ZhangMatrix Theory Basic Results and Techniques Univer-sitext Springer 1999

[12] D A Bini B Iannazzo and B Meini Numerical Solution ofAlgebraic Riccati Equations Society for Industrial and AppliedMathematics (SIAM) Philadelphia Pa USA 2012

[13] I Gohberg P Lancaster and L Rodman Matrix PolynomialsComputer Science and Applied Mathematics Academic PressNew York NY USA 1982

[14] C-Y Chiang andMM Lin ldquoDeflating subspaces and Sylvesterequationsrdquo Tech Rep NCTS Taiwan 2013

[15] R A Smith ldquoMatrix equation119883119860+ 119861119883 = 119862rdquo SIAM Journal onApplied Mathematics vol 16 pp 198ndash201 1968

[16] T Penzl ldquoA cyclic low-rank Smith method for large sparseLyapunov equationsrdquo SIAM Journal on Scientific Computingvol 21 no 4 pp 1401ndash1418 1999

[17] B Zhou J Lam and G-R Duan ldquoOn Smith-type iterativealgorithms for the Stein matrix equationrdquo Applied MathematicsLetters vol 22 no 7 pp 1038ndash1044 2009

[18] RH Bartels andGW Stewart ldquoSolution of thematrix equation119860119883+119883119861 = 119862 algorithm 432rdquoCommunications of the ACM vol15 no 9 pp 820ndash826 1972

[19] G H Golub and C F Van Loan Matrix Computations JohnsHopkins Studies in the Mathematical Sciences Johns HopkinsUniversity Press Baltimore Md USA 3rd edition 1996

[20] N J Higham Accuracy and Stability of Numerical Algo-rithms Society for Industrial andAppliedMathematics (SIAM)Philadelphia Pa USA 2nd edition 2002

[21] A R Ghavimi and A J Laub ldquoBackward error sensitivityand refinement of computed solutions of algebraic Riccatiequationsrdquo Numerical Linear Algebra with Applications vol 2no 1 pp 29ndash49 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

Under the condition 120588(A)120588(B) lt 1 it is easy to see that 119862119896

is convergent and

lim sup119896rarrinfin

2119896

radic1003817100381710038171003817119883 minus 119862

119896

1003817100381710038171003817 le 120588 (A) 120588 (B) (56)

that is 119862119896converges quadratically to 119883 as 119896 rarr infin

This iterative method (54a) and (54b) is called the Smithaccelerative iteration [15] In recent years some modifiediterative methods are so-called Smith-type iteration whichare based on Smith iteration and improve its speed ofconvergence See for example [17] and the references citedtherein

Since the condition 120588(A)120588(B) lt 1 implies that theassumptions of Theorem 4 hold (1) is equivalent to (5) Wecan apply the Smith iteration to the (1) with the substitution(ABC) = (119860119861

⊤ 119860⊤119861 119862+119860119862

⊤119861) One possible drawback

of the Smith-type iterativemethods is that they cannot alwayshandle the case when there exist eigenvalues 120582 120583 isin 120590(119860

⊤119861)

such that 120582120583 = minus1 even if the unique solution119883 exists Basedon the solvable conditions given in this work it is possible todevelop a specific technique working on the particular caseand it is a subject currently under investigation

5 Error Analysis

Error analysis is a way for testing the stability of a numericalalgorithm and evaluating the accuracy of an approximatedsolution In the subsequent discussion we want to considerthe backward error and perturbation bounds for solving (1)

As indicated in (44) matrices 119860 and 119861⊤ are both upper-

triangularWe can then apply the error analysis for triangularlinear systems in [19 Section 31] and [20] to obtain

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 119888119898119899

u (1 +1003817100381710038171003817100381711986010038171003817100381710038171003817119865

1003817100381710038171003817100381711986110038171003817100381710038171003817119865

)1003817100381710038171003817100381711988310038171003817100381710038171003817119865

(57)

where 119888119898119899

is a content depending on the dimensions 119898 and119899 and u is the unit roundoff Since the PQZ decomposition isa stable process it is true that

10038171003817100381710038171003817119862 minus (119883 minus 119860119883

⊤119861)

10038171003817100381710038171003817119865le 1198881015840

119898119899u (1 + 119860

119865119861119865) 119883119865 (58)

with a modest multiple 1198881015840119898119899

Note that the inequality of the form (58) can serve

as a stopping criterion for terminating iterations generatedfrom the Krylov subspace methods [5 6] and Smith-typeiterative methods [15ndash17] In what follows we will derivethe error associated with numerical algorithms following thedevelopment in [20 21]

51 Backward Error Like the discussion of the ordinarySylvester equations [20] the normwise backward error of anapproximate solution 119884 of (1) is defined by

120578 (119884) equiv min 120598 119884 = (119862 + 120575119862) + (119860 + 120575119860)119884⊤

(119861 + 120575119861)

120575119860119865le 120598120572 120575119861

119865le 120598120573 120575119862

119865le 120598120574

(59)

where 120572 equiv 119860119865 120573 equiv 119861

119865 and 120574 equiv 119862

119865 Let R equiv 120575119862 +

120575119860119884⊤119861 + 119860119884

⊤120575119861 + 120575119860119884

⊤120575119861 which implies that R = 119884 minus

119860119884⊤119861 minus 119862 It can be seen that the residualR satisfies

R119865le 120578 (119884) (120574 + 119884

119865120572120573 (2 + 120578 (119884))) (60)

From (60) we know that a small backward error indeedimplies a small relative residual R Since the coefficientmatrices in (1) include nonlinearity it appears to be an openproblem to obtain the theoretical backward error with respectto the residual Again similar to the Sylvester equationdiscussed in [20 Section 162] the conditions under whicha ⊤-Stein equation has a well-conditioned solution remainunknown

52 Perturbation Bounds Consider the perturbed equation

119883 + 120575119883 = (119860 + 120575119860) (119883 + 120575119883)⊤

(119861 + 120575119861) + (119862 + 120575119862)

(61)

Let 119878(119883) = 119883minus119860119883⊤119861 be the corresponding⊤-Stein operator

We then have 119878(120575119883) = 120575119862+119860(119883+120575119883)⊤120575119861+120575119860(119883+120575119883)

⊤(119861+

120575119861) With the application of norm it follows that

120575119883119865le10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

120575119862119865+ 120575119878

119865(119883119865+ 120575119883

119865) (62)

where 120575119878119865

equiv 119860119865120575119861119865+ 120575119860

119865(119861119865+ 120575119861

119865) When

120575119878119865is small enough so that 1 ge 119878

minus1119865sdot 120575119878

119865 we can

rearrange the previous result to be

120575119883119865

119883119865

le

10038171003817100381710038171003817119878minus110038171003817100381710038171003817119865

1 minus1003817100381710038171003817119878minus11003817100381710038171003817119865

sdot 120575119878119865

(120575119862119865

119883119865

+ 120575119878119865) (63)

With 119862119865

= 119878(119883)119865

le 119878119865sdot 119883119865and the condition

number 120581(119878) equiv 119878119865

sdot 119878minus1119865 we arrive at the standard

perturbation result120575119883119865

119883119865

le120581 (119878)

1 minus 120581 (119878) sdot 120575119878119865119878119865

(120575119862119865

119862119865

+120575119878119865

119878119865

)

(64)

Thus the relative error in119883 is controlled by those in119860 119861 and119862 magnified by the condition number 120581(119878)

On the other hand we can also drop the high-order termsin the perturbation to obtain

120575119883 minus 119860120575119883⊤119861 = 119860119883

⊤120575119861 + 120575119860119883

⊤119861 + 120575119862 (65)

We then rewrite the system in terms of

Q vec (120575119883) = [(119883⊤119861)⊤

otimes 119868119898119868119899otimes (119860119883

⊤) 119868119898119899

] [

[

vec (120575119860)

vec (120575119861)vec (120575119862)

]

]

(66)

where Q = 119868119898119899

minus (119861⊤

otimes 119860)P Let 120577 = max120575119860119865119860119865

120575119861119865119861119865 120575119862

119865119862119865 It can be shown that

120575119883119865

119883119865

le radic3Ψ120577 (67)

where Ψ=Qminus1[120572(119883⊤119861)⊤otimes 119868119898

120573119868119899otimes (119860119883

⊤)120574119868119898119899

]2119883119865

8 Abstract and Applied Analysis

A possible disadvantage of the perturbation bound (67)which ignores the consideration of the underlying structureof the problem is that it overestimates the effect of theperturbation on the data But this ldquouniversalrdquo perturbationbound is accessible to any given matrices 119860 119861 and 119862 of (1)

Unlike the perturbation bound (67) it is desirable toobtain a posteriori error bound by assuming 120575119860 = 120575119861 = 0

and 120575119862 = 119883 minus 119860119883⊤119861 minus 119862 in (61) This assumption gives rise

to

120575119883119865

119883119865

le

10038171003817100381710038171003817119875minus1100381710038171003817100381710038172119877119865

119883119865

(68)

It is true that while doing numerical computation this boundgiven in (68) provides a simpler way for estimating the errorof the solution of (1)

6 Conclusion

In this note we propose a novel approach to the necessaryand sufficient conditions for the unique solvability of thesolution 119883 of the ⊤-Stein equation for square coefficientmatrices in terms of the analysis of the spectra 120590(119860

⊤119861)

Solvability conditions have been derived and algorithmshave been proposed in [4 8] by using PQZ decompositionOn the other hand one common procedure to solve theStein-type equations is by means of the invariant subspacemethod We believe that our discussion is the first whichimplements the techniques of the deflating subspace forsolving ⊤-Stein matrix equation and might also give riseto the possibility of developing an advanced and effectivesolver in the future Also we obtain the theoretical residualanalysis backward error analysis and perturbation boundsfor measuring accurately the error in the computed solutionof (1)

Acknowledgments

The author wishes to thank Proffessor Eric King-wah Chu(Monash University Australia) Proffessor Matthew M Lin(National Chung Cheng University Taiwan) and two anony-mous referees for many interesting and valuable suggestionson the paperThis research work is partially supported by theNational Science Council (NSC101-2115-M-150-002) and theNational Center for Theoretical Sciences in Taiwan

References

[1] H W Braden ldquoThe equations 119860⊤119883 plusmn 119883⊤119860 = 119861rdquo SIAM Journal

on Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021998

[2] X Ma J Han and F Pu ldquoA study of solution existence formatrix equation 119860119883 + 119860

⊤119862 = 119861rdquo Journal of Shenyang Institute

of Aeronautical Engineering vol 4 pp 64ndash66 2003[3] A W Bojanczyk G H Golub and P Van Dooren ldquoPeri-

odic schur decomposition algorithms and applicationsrdquo inAdvanced Signal Processing Algorithms Architectures andImplementations vol 1770 of Proceedings of SPIE pp 31ndash42 SanDiego Calif USA July 1992

[4] C-Y Chiang E K-W Chu andW-W Lin ldquoOn the lowast-Sylvesterequation 119860119883 plusmn 119883

lowast119861lowastrdquo Applied Mathematics and Computation

vol 218 no 17 pp 8393ndash8407 2012[5] Y Su and G Chen ldquoIterative methods for solving linear matrix

equation and linear matrix systemrdquo International Journal ofComputer Mathematics vol 87 no 4 pp 763ndash774 2010

[6] M Wang X Cheng and M Wei ldquoIterative algorithms forsolving the matrix equation 119860119883119861 + 119862119883

⊤119863 = 119864rdquo Applied

Mathematics andComputation vol 187 no 2 pp 622ndash629 2007[7] B Zhou J Lam and G-R Duan ldquoToward solution of matrix

equation 119860119891(119883)119861 + 119862rdquo Linear Algebra and its Applications vol435 no 6 pp 1370ndash1398 2011

[8] K D Ikramov and O Yu Vorontsov ldquoThe matrix equation119883+

119860119883⊤119861 = 119862 conditions for unique solvability and a numerical

algorithm for its solutionrdquo Doklady Mathematics vol 85 pp265ndash267 2012

[9] D S Bernstein Matrix Mathematics Theory Facts and For-mulas Princeton University Press Princeton NJ USA 2ndedition 2009

[10] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1994 Correctedreprint of the 1991 original

[11] F ZhangMatrix Theory Basic Results and Techniques Univer-sitext Springer 1999

[12] D A Bini B Iannazzo and B Meini Numerical Solution ofAlgebraic Riccati Equations Society for Industrial and AppliedMathematics (SIAM) Philadelphia Pa USA 2012

[13] I Gohberg P Lancaster and L Rodman Matrix PolynomialsComputer Science and Applied Mathematics Academic PressNew York NY USA 1982

[14] C-Y Chiang andMM Lin ldquoDeflating subspaces and Sylvesterequationsrdquo Tech Rep NCTS Taiwan 2013

[15] R A Smith ldquoMatrix equation119883119860+ 119861119883 = 119862rdquo SIAM Journal onApplied Mathematics vol 16 pp 198ndash201 1968

[16] T Penzl ldquoA cyclic low-rank Smith method for large sparseLyapunov equationsrdquo SIAM Journal on Scientific Computingvol 21 no 4 pp 1401ndash1418 1999

[17] B Zhou J Lam and G-R Duan ldquoOn Smith-type iterativealgorithms for the Stein matrix equationrdquo Applied MathematicsLetters vol 22 no 7 pp 1038ndash1044 2009

[18] RH Bartels andGW Stewart ldquoSolution of thematrix equation119860119883+119883119861 = 119862 algorithm 432rdquoCommunications of the ACM vol15 no 9 pp 820ndash826 1972

[19] G H Golub and C F Van Loan Matrix Computations JohnsHopkins Studies in the Mathematical Sciences Johns HopkinsUniversity Press Baltimore Md USA 3rd edition 1996

[20] N J Higham Accuracy and Stability of Numerical Algo-rithms Society for Industrial andAppliedMathematics (SIAM)Philadelphia Pa USA 2nd edition 2002

[21] A R Ghavimi and A J Laub ldquoBackward error sensitivityand refinement of computed solutions of algebraic Riccatiequationsrdquo Numerical Linear Algebra with Applications vol 2no 1 pp 29ndash49 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

A possible disadvantage of the perturbation bound (67)which ignores the consideration of the underlying structureof the problem is that it overestimates the effect of theperturbation on the data But this ldquouniversalrdquo perturbationbound is accessible to any given matrices 119860 119861 and 119862 of (1)

Unlike the perturbation bound (67) it is desirable toobtain a posteriori error bound by assuming 120575119860 = 120575119861 = 0

and 120575119862 = 119883 minus 119860119883⊤119861 minus 119862 in (61) This assumption gives rise

to

120575119883119865

119883119865

le

10038171003817100381710038171003817119875minus1100381710038171003817100381710038172119877119865

119883119865

(68)

It is true that while doing numerical computation this boundgiven in (68) provides a simpler way for estimating the errorof the solution of (1)

6 Conclusion

In this note we propose a novel approach to the necessaryand sufficient conditions for the unique solvability of thesolution 119883 of the ⊤-Stein equation for square coefficientmatrices in terms of the analysis of the spectra 120590(119860

⊤119861)

Solvability conditions have been derived and algorithmshave been proposed in [4 8] by using PQZ decompositionOn the other hand one common procedure to solve theStein-type equations is by means of the invariant subspacemethod We believe that our discussion is the first whichimplements the techniques of the deflating subspace forsolving ⊤-Stein matrix equation and might also give riseto the possibility of developing an advanced and effectivesolver in the future Also we obtain the theoretical residualanalysis backward error analysis and perturbation boundsfor measuring accurately the error in the computed solutionof (1)

Acknowledgments

The author wishes to thank Proffessor Eric King-wah Chu(Monash University Australia) Proffessor Matthew M Lin(National Chung Cheng University Taiwan) and two anony-mous referees for many interesting and valuable suggestionson the paperThis research work is partially supported by theNational Science Council (NSC101-2115-M-150-002) and theNational Center for Theoretical Sciences in Taiwan

References

[1] H W Braden ldquoThe equations 119860⊤119883 plusmn 119883⊤119860 = 119861rdquo SIAM Journal

on Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021998

[2] X Ma J Han and F Pu ldquoA study of solution existence formatrix equation 119860119883 + 119860

⊤119862 = 119861rdquo Journal of Shenyang Institute

of Aeronautical Engineering vol 4 pp 64ndash66 2003[3] A W Bojanczyk G H Golub and P Van Dooren ldquoPeri-

odic schur decomposition algorithms and applicationsrdquo inAdvanced Signal Processing Algorithms Architectures andImplementations vol 1770 of Proceedings of SPIE pp 31ndash42 SanDiego Calif USA July 1992

[4] C-Y Chiang E K-W Chu andW-W Lin ldquoOn the lowast-Sylvesterequation 119860119883 plusmn 119883

lowast119861lowastrdquo Applied Mathematics and Computation

vol 218 no 17 pp 8393ndash8407 2012[5] Y Su and G Chen ldquoIterative methods for solving linear matrix

equation and linear matrix systemrdquo International Journal ofComputer Mathematics vol 87 no 4 pp 763ndash774 2010

[6] M Wang X Cheng and M Wei ldquoIterative algorithms forsolving the matrix equation 119860119883119861 + 119862119883

⊤119863 = 119864rdquo Applied

Mathematics andComputation vol 187 no 2 pp 622ndash629 2007[7] B Zhou J Lam and G-R Duan ldquoToward solution of matrix

equation 119860119891(119883)119861 + 119862rdquo Linear Algebra and its Applications vol435 no 6 pp 1370ndash1398 2011

[8] K D Ikramov and O Yu Vorontsov ldquoThe matrix equation119883+

119860119883⊤119861 = 119862 conditions for unique solvability and a numerical

algorithm for its solutionrdquo Doklady Mathematics vol 85 pp265ndash267 2012

[9] D S Bernstein Matrix Mathematics Theory Facts and For-mulas Princeton University Press Princeton NJ USA 2ndedition 2009

[10] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1994 Correctedreprint of the 1991 original

[11] F ZhangMatrix Theory Basic Results and Techniques Univer-sitext Springer 1999

[12] D A Bini B Iannazzo and B Meini Numerical Solution ofAlgebraic Riccati Equations Society for Industrial and AppliedMathematics (SIAM) Philadelphia Pa USA 2012

[13] I Gohberg P Lancaster and L Rodman Matrix PolynomialsComputer Science and Applied Mathematics Academic PressNew York NY USA 1982

[14] C-Y Chiang andMM Lin ldquoDeflating subspaces and Sylvesterequationsrdquo Tech Rep NCTS Taiwan 2013

[15] R A Smith ldquoMatrix equation119883119860+ 119861119883 = 119862rdquo SIAM Journal onApplied Mathematics vol 16 pp 198ndash201 1968

[16] T Penzl ldquoA cyclic low-rank Smith method for large sparseLyapunov equationsrdquo SIAM Journal on Scientific Computingvol 21 no 4 pp 1401ndash1418 1999

[17] B Zhou J Lam and G-R Duan ldquoOn Smith-type iterativealgorithms for the Stein matrix equationrdquo Applied MathematicsLetters vol 22 no 7 pp 1038ndash1044 2009

[18] RH Bartels andGW Stewart ldquoSolution of thematrix equation119860119883+119883119861 = 119862 algorithm 432rdquoCommunications of the ACM vol15 no 9 pp 820ndash826 1972

[19] G H Golub and C F Van Loan Matrix Computations JohnsHopkins Studies in the Mathematical Sciences Johns HopkinsUniversity Press Baltimore Md USA 3rd edition 1996

[20] N J Higham Accuracy and Stability of Numerical Algo-rithms Society for Industrial andAppliedMathematics (SIAM)Philadelphia Pa USA 2nd edition 2002

[21] A R Ghavimi and A J Laub ldquoBackward error sensitivityand refinement of computed solutions of algebraic Riccatiequationsrdquo Numerical Linear Algebra with Applications vol 2no 1 pp 29ndash49 1995

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of