researcharticle a method of indefinite krylov subspace for...

Research ArticleA Method of Indefinite Krylov Subspace for Eigenvalue Problem

M Aliyari andM Ghasemi Kamalvand

Department of Mathematics Lorestan University Khorramabad Iran

Correspondence should be addressed to M Ghasemi Kamalvand ghasemimluacir

Received 7 November 2017 Revised 1 March 2018 Accepted 3 April 2018 Published 16 May 2018

Academic Editor Raffaele Solimene

Copyright copy 2018 M Aliyari and M Ghasemi Kamalvand This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

We describe an indefinite state of Arnoldirsquos method for solving the eigenvalues problems In the following we scrutinize theindefinite state of Lanczosrsquo method for solving the eigenvalue problems and we show that this method for the 119869-Hermitian matricesworks much better than Arnoldirsquos method

1 Introduction

Around the early 1950s the idea of Krylov subspace iterationwas established by Cornelius Lanczos and Walter ArnoldiLanczosrsquo method was based on two mutually orthogonalvector sequences and his motivation came from eigenvalueproblems In that context the most prominent feature ofthe method is that it reduces the original matrix to tridi-agonal form Lanczos later applied his method to solvelinear systems in particular symmetric ones Krylov subspaceiterations or Krylov subspace methods are iterative methodswhich are used as linear system solvers and also iterativesolvers of eigenvalue problems

On the other hand the indefinite inner product definedby 119869 = diag(1198951 119895119899) 119895119896 isin minus1 +1 arises frequentlyin applications It is used for example in the theory ofrelativity and in the research of the polarized light Moreon the applications of such products can be found in [1ndash6]These applications in other fields of science inspired us toresume Lanczos FOM and Arnoldirsquos methods in this men-tioned indefinite inner space Indefinite Arnoldirsquos methodis a process that constructs a 119869-orthonormal basis for thenondegenerated Krylov subspace the basis that we provedhave a particular commonproperty about the structure of theproduct of their vectors In the following iterative Arnoldirsquosmethod for eigenvalue problem has been renovated in thementioned indefinite inner product space and also a processis made which is useful for solving eigenvalue problem with119869-Hermitian coefficientmatrices Indefinite iterative Lanczosrsquo

method for eigenvalue problem is the same as Lanczosrsquomethod that has been restored in the indefinite inner productspace

This paper is organized as follows In Section 2 we recallthe standard inner product and indefinite inner product inC119899 In the following part of this section we show someexamples on indefinite inner product that will guide us to theobjectives of this article We express the indefinite Arnoldialgorithm for the computation of a 119869-orthonormal basis ofthe Krylov subspace and describe property of the basis inthe form of several propositions to review and in the end ofthis section we exhibit the indefinite version of the HermitianLanczos algorithm

In Section 3 we present a definition of the Ritz eigenpairof the a matrix and new algorithms from the indefiniteArnoldi and Lanczosrsquo algorithm that is named modifiedmethods by the Ritz approximation in the indefinite Arnoldiand Lanczosrsquo process This is our main goal and initiative inthis article

Since the initial vectors are very important at the startof each iteration to the above algorithms in this articlewe consider the Ritz eigenvector instead of the commoneigenvector in the start of the each repetition algorithm InSection 4 we present numerical examples and compare themodified method by the Ritz approximation in the Arnoldiand indefinite Lanczos algorithm Algorithms are writtenwith personal computer using MATLAB software and weillustrate examples by using these codes

HindawiMathematical Problems in EngineeringVolume 2018 Article ID 2919873 5 pageshttpsdoiorg10115520182919873

2 Mathematical Problems in Engineering

2 Indefinite Arnoldi and Lanczos Algorithms

Let C119899 be the 119899-dimensional space consisting of all columnvectors 119909 with complex coordinates 119909119894 119894 = 1 2 119899 Thestandard inner product in C119899 is denoted by (sdot sdot) Thus

(119909 119910) = 119899sum119894=1

119909119894119910119894 (1)

where 119909 = (1199091 119909119899)119879 and 119910 = (1199101 119910119899)119879 and the bardenotes the complex conjugation Remember that a function[sdot sdot] from C119899 times C119899 to C is called an indefinite inner productin C119899 if the following axioms are satisfied

(i) Linearity in the first argument[1205721199091 + 1205731199092 119910] = 120572 [1199091 119910] + 120573 [1199092 119910] (2)

for all 1199091 1199092 119910 isin C119899 and all complex numbers 120572 120573(ii) Antisymmetry[119909 119910] = [119910 119909] (3)

Example 1 Let 119909 = (1199091 119909119899)119879 119910 = (1199101 119910119899)119879 isin C119899 anddefine

[119909 119910] = 119903sum119894=1

119909120590(119894)119910120590(119894) minus 119899sum119894=119903+1

119909120590(119894)119910120590(119894) (4)

where 120590 is a permutation for which 120590(119894) = 119895119894 and 119895119894 isin1 119899 It is easy to see that [sdot sdot] is an indefinite innerproduct Indeed (i) (ii) properties are clearly confirmed Butfor nondegeneracy property if [119909 119910] = 0 for all 119910 isin C119899 thenin special case [119909 119910] = 0 for 119910 = (1199101 119910119899)119879 where

119910120590(119894) = 119909120590(119894) 119894 = 1 119903minus119909120590(119894) 119894 = 119903 + 1 119896 (5)

by choosing an appropriate permutation 120590Thus119903sum119894=1

119909120590(119894)119909120590(119894) minus (minus 119899sum119894=119903+1

119909120590(119894)119909120590(119894))= 119903sum119894=1

1003816100381610038161003816119909120590(119894)10038161003816100381610038162 + 119899sum119894=119903+1

1003816100381610038161003816119909120590(119894)10038161003816100381610038162 = 119899sum119894=1

1003816100381610038161003816119909120590(119894)10038161003816100381610038162 = 0 (6)

and this yields that 119909119894 = 0 for 119894 = 1 119899 So 119909 = 0By [3] we know that for every 119899times 119899 invertible Hermitian

matrix119867 the formula[119909 119910] = (119867119909 119910) 119909 119910 isin C119899 (7)

determines an indefinite inner product onC119899 and converselyfor every indefinite inner product [sdot sdot] on C119899 there exists an119899 times 119899 invertible and Hermitian matrix119867 such that (7) holdsand the established correspondence [sdot sdot] harr 119867 is a bijectionbetween the set of all indefinite inner products onC119899 and theset of all 119899 times 119899 invertible Hermitian matrices

Example 2 In Example (7) the corresponding nonsingularHermitian matrix to that indefinite inner product is writtenin the form 119869 = diag(plusmn1) wherein 119903 is the number of +1 and119899 minus 119903 is the number of minus1 This is because if[119909 119910] = (119869119909 119910) (8)

then[119909 119910] = (119869119909 119910) = 119910lowast119869119909 = (1199101 119910119899)119879 119869 (1199091 119909119899)= 119903sum119894=1

119909120590(119894)119910120590(119894) minus 119899sum119894=119903+1

119909120590(119894)119910120590(119894) (9)

and conversely if

[119909 119910] = 119903sum119894=1

119909120590(119894)119910120590(119894) minus 119899sum119894=119903+1

119909120590(119894)119910120590(119894) (10)

then it is clear that [119909 119910] = (119869119909 119910) for all 119909 119910 isin C119899Note For 119909 = (1199091 119909119899)119879 consider the relation [119909 119909] =119909lowast119869119909

That the result becomes zero is a very special case anddepends on how we choose the matrix 119869

Recently new algorithms have been presented in thisregard and have been compared in terms of the number ofreplays and the time required to run algorithms

Thesemethods could solve the linear systems of equationswith 119869-Hermitian coefficient matrices and also consideredArnoldi FOM and Lanczosrsquo methods for solving linearsystem of equations and resumed these methods in a spacethat is equipped with a special indefinite inner productand proposed new algorithms to run these methods Theindefinite Arnoldi algorithm for the computation of a 119869-orthonormal basis of the Krylov subspace

K119898 (119860 V0) = span V0 119860V0 1198602V0 119860119898minus1V0 (11)

where 119860 is 119899 times 119899 matrix and V119894rsquos are 119899 times 1 vectors is shownbelow

(1) Choose a vector 119909 such that [119909 119909] = 0(2) Define V1 = 119909radic|[119909 119909]|(3) For 119895 = 1 119898 Do(4) For 119894 = 1 119895 Do(5) Compute ℎ119894119895 fl [119860V119895 V119894] and 119905(V119894) = [V119894 V119894](6) Compute 119908119895 fl 119860V119895 minus sum119895119894=1 119905(V119894)ℎ119894119895V119894(7) ℎ119895+1119895 = radic|[119908119895 119908119895]|(8) if ℎ119895+1119895 = 0 then stop(9) V119895+1 = 119908119895ℎ119895+1119895(10) EndDo(11) EndDo

Here119867119898 is an upper Hessenberg matrixThe following proposition expressed indefinite Arnoldirsquos

algorithm (You can see [7 8])

Mathematical Problems in Engineering 3

Proposition 3 Assume that the indefinite Arnoldi algorithmdoes not stop before the 119898th step Then the vectors V1 V119898form a 119869-orthonormal basis of the Krylov subspaceK119898(119860 V)Proof Consider the following expression

ℎ119895+1119895V119895+1 = 119860V119895 minus 119895sum119894=1

119905 (V119894) ℎ119894119895V119894 119895 = 1 119898 (12)

Certainly that vectors V1 V119898 are 119869-orthonormal

Proposition 4 Define

(i) 119867119898 the 119898 times 119898 Hessenberg matrix whose nonzeroentries ℎ119894119895 are defined by indefinite Arnoldirsquos algorithm

(ii) 119881119898 the 119899 times 119898matrix with column vectors V1 V119898(iii) 119869 = diag(119905(V1) 119905(V119898))

Then the following relations are valid

119860119881119898 = 119881119898 119869119867119898 + ℎ119898+1119898V119898+1119890119879119898 (13)119881lowast119898119869119860119881119898 = 119867119898 (14)

In particular if119898 = 119899 then119881minus1119899 119860119881119899 = 119869119867119899 (15)

Proof By considering lines (4) (5) and (7) of indefiniteArnoldirsquos algorithm relation (13) is verified straightforwardlyIndeed in general (13) is the matrix representation of (12)

119860119881119898 = 119881119898 119869119867119898 + 119908119898119890119879119898 = 119881119898 119869119867119898 + ℎ119898+1119898V119898+1119890119879119898= (119881119898 119869 | V119898+1)( 1198671198980 0 ℎ119898+1119898) (16)

Now to see (14) left-multiply the relation (16) in 119881lowast119898119869 Weearn 119881lowast119898119869119860119881119898 = 119881lowast119898119869119881119898 119869119867119898 + ℎ119898+1119898119881lowast119898119869V119898+1119890119879119898 (17)

On the other hand given that the vectors V1 V119898 build a119869-orthonormal basis then we have the following

(1) According to the definition of V119898+1 V119898+1 = 0 orit is orthogonal to V1 V119898 that is [V119898+1 V119894] =Vlowast119894 119869V119898+1 = 0 for 119894 = 1 119898 Thus119881lowast119898119869V119898+1 = 0 (18)

(2) We have119881lowast119898119869119881119898 = ([V119894 V119895])119894119895 = diag ([V1 V1] [V119898 V119898])= 119869 (19)

In other words 119869119881lowast119898119869119881119898 = 119868

Therefore relation (17) can be summarized as follows119881lowast119898119869119860119881119898 = 119867119898 (20)

and by left-multiplying in 119869119869119881lowast119898119869119860119881119898 = 119869119867119898 (21)

Particularly if 119898 = 119899 relation (19) yields that 119869119881lowast119899 119869 = 119881minus1119899and thereby 119881minus1119899 119860119881119899 = 119869119867119899 (22)

Remark 5 It should be noted that a permutation matrix 119875 isavailable suchThat 119875119869119875 = 119869 (23)

We know that the indefinite Hermitian Lanczos algorithmis a simplified version of the indefinite Arnoldi algorithmthat is applied to 119869-Hermitian matrices The 119898th step of thealgorithm transforms matrix 119860 into a Hessenberg and 119869-Hermitian matrix in other words a tridiagonal matrix 119879119898when119898 is equal to the dimension of119860 119879119898 is similar to119860 Tosee more refer to [7ndash10] As can be seen in the following thismethod is expressed as a special case of the indefinite Arnoldimethod in the complex space for the special case whenmatrix119860 is 119869-Hermitian

The indefinite version of the Hermitian Lanczos algo-rithm can be formulated as follows

(1) Choose an initial vector V1 such thatradic|[V1 V1]| = plusmn1(2) Set 1205731 equiv 0 V0 = 0(3) For 119895 = 1 119898 Do(4) 120596119895 fl 119860V119895 minus 119905(V119895minus1)120573119895V119895minus1(5) 120572119895 fl [120596119895 V119895](6) 120596119895 fl 120596119895 minus 119905(V119895)120572119895V119895(7) ifradic|[120596119895 120596119895]| = 0 then stop

(8) V119895+1 = 119908119895radic|[120596119895 120596119895]|(9) 120573119895+1 fl 119905(V119895+1)radic|[120596119895 120596119895]|(10) EndDo

3 Indefinite Iterative Arnoldirsquos Method forEigenvalue Problem

Definition 6 Suppose that 119860 1198691015840 119867119898 and 119881119898 are the same asthose in Proposition (8) let (120582(119898)119894 119910(119898)119894 ) be an eigenpair of119867119898120582(119898)119894 is called Ritz value of 119860 and provides an approxima-tion for eigenvalue of 119860119906(119898)119894 = 119881119898119910(119898)119894 is called Ritz vector of 119860 and provides anapproximation for eigenvector of 119860


Proposition 7 Let 119910(119898)119894 be an eigenvector of 1198691015840119867119898 associatedwith the eigenvalue 120582(119898)119894 and 119906(119898)119894 the Ritz approximateeigenvector so that 119906(119898)119894 = 119881119898119910(119898)119894 Then

(119860 minus 120582(119898)119894 119868) 119906(119898)119894 = ℎ119898+1119898119890119879119898119910(119898)119894 V119898+1 (24)

Proof Let 119906(119898)119894 = 119881119898119910(119898)119894 and 1198691015840119898119867119898119910(119898)119894 = 120582(119898)119894 119910(119898)119894 We have

119860119881119898119910(119898)119894 = (1198811198981198691015840119898119867119898 + ℎ119898+1119898V119898+1119890119879119898) 119910(119898)119894119860119881119898119910(119898)119894 = 1198811198981198691015840119898119867119898119910(119898)119894 + ℎ119898+1119898V119898+1119890119879119898119910(119898)119894(119860119881119898 minus 120582(119898)119894 119881119898) 119910(119898)119894 = ℎ119898+1119898 (119890119879119898119910(119898)119894 ) V119898+1(119860 minus 120582(119898)119894 119868)119881119898119910(119898)119894 = ℎ119898+1119898 (119890119879119898119910(119898)119894 ) V119898+1(119860 minus 120582(119898)119894 119868) 119906(119898)119894 = ℎ119898+1119898 (119890119879119898119910(119898)119894 ) V119898+1(25)

One of way to circumvent the difficulty is to restart thealgorithm After a run with 119898 Arnoldi and Lanczosrsquo vectorswe compute the approximate eigenvector and use it as aninitial vector for the next run with Arnoldi and Lanczosrsquomethods This process which is the simplest of this kindis iterated to convergence The iterative indefinite Arnoldialgorithm can be formulated as follows

(1) Start choose an initial vector V1 and a dimension119898(2) Iterate perform 119898 steps of indefinite Arnoldirsquos algo-

rithm(3) Restart compute the approximate eigenvector 119906(119898)1

associated with the rightmost eigenvalue 120582(119898)1

If satisfied stop else set V1 equiv 119906(119898)1 and go to (2)Of course it should be noted that this algorithm is notbeneficial for us because in any case it is better to use thedefinite algorithm

In the following we have the same iterative indefiniteLanczos algorithm which can be formulated as follows

(1) Start choose an initial vector V1 and a dimension119898

Table 1

Method Iteration Time 119890Arnoldi 1 031 6 times 10minus14Indefinite Lanczos 1 028 1 times 10minus13Here 119890 = (119860 minus 120582(119898)119894 119868)119906

(119898)119894 2

(2) Iterate perform 119898 steps of indefinite Lanczosrsquo algo-rithm

(3) Restart compute the approximate eigenvector 119906(119898)1associated with the rightmost eigenvalue 120582(119898)1

If satisfied stop else set V1 equiv 119906(119898)1 and go to (2)

This algorithm is used for 119869-symmetric matrices since thenumber of flops in the performance of algorithm is certainlyless than that of the Arnoldi algorithm This is illustrated bythe examples in the next section

4 Numerical Example

In this section we present numerical examples and comparethe methods modified by the Ritz approximation in theArnoldi and indefinite Lanczos process

Algorithms are written with personal computer usingMATLAB software and the following examples are doneusing these codes

Example 1 Consider the matrices 119869 = ( 119868 00 minus119868 ) and 119860 =( 11986011 1198601211986021119860

22

) where11986011 11986022 are diagonal matrices with randomelements in (0 1) and also 11986012 11986021 are tridiagonal matriceswith random elements in (0 1) such that 11986012 = minus11986011987921 and Vis 119899 times 1 vector with random entries in (0 1)

Let V1 = Vradic|[V V]| in this case 119899 = 100 119898 = 50 Theresults are in Table 1

Example 2 Consider the matrices119869 = (11986811989910 minus11986811989910 11986811989910 minus11986811989910 11986811989910 minus11986811989910 11986811989910minus 11986811989910 11986811989910 minus11986811989910) (26)

where 119868 is identity matrix and 119860 isin 119872119899 in the form

119860 =(((((((((((((((((

11986011 11986012 11986013 11986014 11986015 11986016 11986017 11986018 11986019 11986011011986021 11986022 11986023 11986024 11986025 11986026 11986027 11986028 11986029 11986021011986031 11986032 11986033 11986034 11986035 11986036 11986037 11986038 11986039 11986031011986041 11986042 11986043 11986044 11986045 11986046 11986047 11986048 11986049 11986041011986051 11986052 11986053 11986054 11986055 11986056 11986057 11986058 11986059 11986051011986061 11986062 11986063 11986064 11986065 11986066 11986067 11986068 11986069 11986061011986071 11986072 11986073 11986074 11986075 11986076 11986077 11986078 11986079 11986071011986081 11986082 11986083 11986084 11986085 11986086 11986087 11986088 11986089 11986081011986091 11986092 11986093 11986094 11986095 11986096 11986097 11986098 11986099 119860910119860101 119860102 119860103 119860104 119860105 119860106 119860107 119860108 119860109 1198601010

)))))))))))))))))

(27)


Table 2

Method Iteration Time 119890Arnoldi 5 125 32 times 10minus8Indefinite Lanczos 3 064 43 times 10minus6

Table 3


(119898)119894 2

such that

minus119860119879119894119894+1 = 119860 119894+1119894 for 119894 = 1 9119860119879119894119894+2 = 119860 119894+2119894 for 119894 = 1 8minus119860119879119894119894+3 = 119860 119894+3119894 for 119894 = 1 7119860119879119894119894+4 = 119860 119894+1119894 for 119894 = 1 6minus119860119879119894119894+5 = 119860 119894+5119894 for 119894 = 1 5119860119879119894119894+6 = 119860 119894+6119894 for 119894 = 1 4minus119860119879119894119894+7 = 119860 119894+7119894 for 119894 = 1 3119860119879119894119894+8 = 119860 119894+8119894 for 119894 = 1 2minus119860119879119894119894+9 = 119860 119894+9119894 for 119894 = 1

(28)

wherein 119860 119894119895 119894 119895 = 1 10 are diagonal matrices with randomelements (entries) in (0 1)

In this case

(i) if 119899 = 50119898 = 30 the results are in Table 2

(ii) if 119899 = 200119898 = 50 the results are in Table 3

It should be noted that obviously the number of externaliterations depends on the number of internal iterationsselectedThe larger the119898 value is taken themore the externaliterations will decrease

5 Conclusion

We know that the matrices of 119869-Hermitian are importantand functional matrices We can find a way to find theeigenvalues of these types of matrices that are more efficientthan Arnoldirsquos method To do this we first constructedan indefinite Arnoldi method in order to be able to alsomake that indefinite Lanczos method We tried to study andcompare the Arnoldi and indefinite Lanczos methods It wasobserved that Arnoldirsquos method is not efficient in indefinitestatus but the indefinite Lanczos method for 119869-symmetric(Hermitian) matrices is much better than Arnoldirsquos method

Conflicts of Interest

The authors declare that they have no conflicts of interest

References

[1] K Appi Reddy and T Kurmayya ldquoMoore-Penrose inverses ofGram matrices leaving a cone invariant in an indefinite innerproduct spacerdquo Special Matrices vol 3 pp 155ndash162 2015

[2] N J Higham ldquoJ-orthogonal matrices properties and genera-tionrdquo SIAM Review vol 45 no 3 pp 504ndash519 2003

[3] A Kihcman and Z A Zhour ldquoThe representation and approx-imation for the weighted Minkowski inverse in Minkowskispacerdquo Mathematical and Computer Modelling vol 47 no 3-4pp 363ndash371 2008

[4] B C Levy ldquoA note on the hyperbolic singular value decompo-sitionrdquo Linear Algebra and its Applications vol 277 no 1-3 pp135ndash142 1998

[5] R Onn A O Steinhardt and A Bojanczyk ldquoThe hyperbolicsingular value decomposition and applicationsrdquo in proceedingsof the Applied Mathematics and Computing Trans 8th ArmyConf 108 93 pages NY USA 1991 ARO Rep 91-1

[6] X Sui and P Gondolo ldquoFactorizations into Normal Matrices inIndefinite Inner Product SpacesrdquoRings andAlgebras (mathRA)Numerical Analysis (mathNA) 31 Oct 2016

[7] Y Saad Iterative Methods for Sparse Linear Systems SIAM 2ndedition 2003

[8] Y Saad Numerical Methods for Large Eigenvalue Problems 2ndedition 2011

[9] A de laGarza ldquoAn iterativemethod for solving systems of linearequationsrdquo Union Carbide and Carbon Corb 1951 K-25 plantOak Ridge Tennessee Report K-731

[10] I Gohberg P Lancaster and L Rodman Indefinite linearalgebra and applications Birkhauser 2005

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2 Indefinite Arnoldi and Lanczos Algorithms

Let C119899 be the 119899-dimensional space consisting of all columnvectors 119909 with complex coordinates 119909119894 119894 = 1 2 119899 Thestandard inner product in C119899 is denoted by (sdot sdot) Thus

(119909 119910) = 119899sum119894=1

119909119894119910119894 (1)

where 119909 = (1199091 119909119899)119879 and 119910 = (1199101 119910119899)119879 and the bardenotes the complex conjugation Remember that a function[sdot sdot] from C119899 times C119899 to C is called an indefinite inner productin C119899 if the following axioms are satisfied

(i) Linearity in the first argument[1205721199091 + 1205731199092 119910] = 120572 [1199091 119910] + 120573 [1199092 119910] (2)

for all 1199091 1199092 119910 isin C119899 and all complex numbers 120572 120573(ii) Antisymmetry[119909 119910] = [119910 119909] (3)

Example 1 Let 119909 = (1199091 119909119899)119879 119910 = (1199101 119910119899)119879 isin C119899 anddefine

[119909 119910] = 119903sum119894=1

119909120590(119894)119910120590(119894) minus 119899sum119894=119903+1

119909120590(119894)119910120590(119894) (4)

where 120590 is a permutation for which 120590(119894) = 119895119894 and 119895119894 isin1 119899 It is easy to see that [sdot sdot] is an indefinite innerproduct Indeed (i) (ii) properties are clearly confirmed Butfor nondegeneracy property if [119909 119910] = 0 for all 119910 isin C119899 thenin special case [119909 119910] = 0 for 119910 = (1199101 119910119899)119879 where

119910120590(119894) = 119909120590(119894) 119894 = 1 119903minus119909120590(119894) 119894 = 119903 + 1 119896 (5)

by choosing an appropriate permutation 120590Thus119903sum119894=1

119909120590(119894)119909120590(119894) minus (minus 119899sum119894=119903+1

119909120590(119894)119909120590(119894))= 119903sum119894=1

1003816100381610038161003816119909120590(119894)10038161003816100381610038162 + 119899sum119894=119903+1

1003816100381610038161003816119909120590(119894)10038161003816100381610038162 = 119899sum119894=1

1003816100381610038161003816119909120590(119894)10038161003816100381610038162 = 0 (6)

and this yields that 119909119894 = 0 for 119894 = 1 119899 So 119909 = 0By [3] we know that for every 119899times 119899 invertible Hermitian

matrix119867 the formula[119909 119910] = (119867119909 119910) 119909 119910 isin C119899 (7)

determines an indefinite inner product onC119899 and converselyfor every indefinite inner product [sdot sdot] on C119899 there exists an119899 times 119899 invertible and Hermitian matrix119867 such that (7) holdsand the established correspondence [sdot sdot] harr 119867 is a bijectionbetween the set of all indefinite inner products onC119899 and theset of all 119899 times 119899 invertible Hermitian matrices

Example 2 In Example (7) the corresponding nonsingularHermitian matrix to that indefinite inner product is writtenin the form 119869 = diag(plusmn1) wherein 119903 is the number of +1 and119899 minus 119903 is the number of minus1 This is because if[119909 119910] = (119869119909 119910) (8)

then[119909 119910] = (119869119909 119910) = 119910lowast119869119909 = (1199101 119910119899)119879 119869 (1199091 119909119899)= 119903sum119894=1

119909120590(119894)119910120590(119894) minus 119899sum119894=119903+1

119909120590(119894)119910120590(119894) (9)

and conversely if

[119909 119910] = 119903sum119894=1

119909120590(119894)119910120590(119894) minus 119899sum119894=119903+1

119909120590(119894)119910120590(119894) (10)

then it is clear that [119909 119910] = (119869119909 119910) for all 119909 119910 isin C119899Note For 119909 = (1199091 119909119899)119879 consider the relation [119909 119909] =119909lowast119869119909

That the result becomes zero is a very special case anddepends on how we choose the matrix 119869

Recently new algorithms have been presented in thisregard and have been compared in terms of the number ofreplays and the time required to run algorithms

Thesemethods could solve the linear systems of equationswith 119869-Hermitian coefficient matrices and also consideredArnoldi FOM and Lanczosrsquo methods for solving linearsystem of equations and resumed these methods in a spacethat is equipped with a special indefinite inner productand proposed new algorithms to run these methods Theindefinite Arnoldi algorithm for the computation of a 119869-orthonormal basis of the Krylov subspace

K119898 (119860 V0) = span V0 119860V0 1198602V0 119860119898minus1V0 (11)

where 119860 is 119899 times 119899 matrix and V119894rsquos are 119899 times 1 vectors is shownbelow

(1) Choose a vector 119909 such that [119909 119909] = 0(2) Define V1 = 119909radic|[119909 119909]|(3) For 119895 = 1 119898 Do(4) For 119894 = 1 119895 Do(5) Compute ℎ119894119895 fl [119860V119895 V119894] and 119905(V119894) = [V119894 V119894](6) Compute 119908119895 fl 119860V119895 minus sum119895119894=1 119905(V119894)ℎ119894119895V119894(7) ℎ119895+1119895 = radic|[119908119895 119908119895]|(8) if ℎ119895+1119895 = 0 then stop(9) V119895+1 = 119908119895ℎ119895+1119895(10) EndDo(11) EndDo

Here119867119898 is an upper Hessenberg matrixThe following proposition expressed indefinite Arnoldirsquos

algorithm (You can see [7 8])



ℎ119895+1119895V119895+1 = 119860V119895 minus 119895sum119894=1

119905 (V119894) ℎ119894119895V119894 119895 = 1 119898 (12)






119860119881119898 = 119881119898 119869119867119898 + ℎ119898+1119898V119898+1119890119879119898 (13)119881lowast119898119869119860119881119898 = 119867119898 (14)



119860119881119898 = 119881119898 119869119867119898 + 119908119898119890119879119898 = 119881119898 119869119867119898 + ℎ119898+1119898V119898+1119890119879119898= (119881119898 119869 | V119898+1)( 1198671198980 0 ℎ119898+1119898) (16)


















(119860 minus 120582(119898)119894 119868) 119906(119898)119894 = ℎ119898+1119898119890119879119898119910(119898)119894 V119898+1 (24)










Table 1


(119898)119894 2





4 Numerical Example




22





119860 =(((((((((((((((((

11986011 11986012 11986013 11986014 11986015 11986016 11986017 11986018 11986019 11986011011986021 11986022 11986023 11986024 11986025 11986026 11986027 11986028 11986029 11986021011986031 11986032 11986033 11986034 11986035 11986036 11986037 11986038 11986039 11986031011986041 11986042 11986043 11986044 11986045 11986046 11986047 11986048 11986049 11986041011986051 11986052 11986053 11986054 11986055 11986056 11986057 11986058 11986059 11986051011986061 11986062 11986063 11986064 11986065 11986066 11986067 11986068 11986069 11986061011986071 11986072 11986073 11986074 11986075 11986076 11986077 11986078 11986079 11986071011986081 11986082 11986083 11986084 11986085 11986086 11986087 11986088 11986089 11986081011986091 11986092 11986093 11986094 11986095 11986096 11986097 11986098 11986099 119860910119860101 119860102 119860103 119860104 119860105 119860106 119860107 119860108 119860109 1198601010

)))))))))))))))))

(27)


Table 2


Table 3


(119898)119894 2

such that


(28)


In this case




5 Conclusion




References


















Journal of












Journal of







Volume 2018






Volume 2018










ℎ119895+1119895V119895+1 = 119860V119895 minus 119895sum119894=1

119905 (V119894) ℎ119894119895V119894 119895 = 1 119898 (12)






119860119881119898 = 119881119898 119869119867119898 + ℎ119898+1119898V119898+1119890119879119898 (13)119881lowast119898119869119860119881119898 = 119867119898 (14)



119860119881119898 = 119881119898 119869119867119898 + 119908119898119890119879119898 = 119881119898 119869119867119898 + ℎ119898+1119898V119898+1119890119879119898= (119881119898 119869 | V119898+1)( 1198671198980 0 ℎ119898+1119898) (16)


















(119860 minus 120582(119898)119894 119868) 119906(119898)119894 = ℎ119898+1119898119890119879119898119910(119898)119894 V119898+1 (24)










Table 1


(119898)119894 2





4 Numerical Example




22





119860 =(((((((((((((((((

11986011 11986012 11986013 11986014 11986015 11986016 11986017 11986018 11986019 11986011011986021 11986022 11986023 11986024 11986025 11986026 11986027 11986028 11986029 11986021011986031 11986032 11986033 11986034 11986035 11986036 11986037 11986038 11986039 11986031011986041 11986042 11986043 11986044 11986045 11986046 11986047 11986048 11986049 11986041011986051 11986052 11986053 11986054 11986055 11986056 11986057 11986058 11986059 11986051011986061 11986062 11986063 11986064 11986065 11986066 11986067 11986068 11986069 11986061011986071 11986072 11986073 11986074 11986075 11986076 11986077 11986078 11986079 11986071011986081 11986082 11986083 11986084 11986085 11986086 11986087 11986088 11986089 11986081011986091 11986092 11986093 11986094 11986095 11986096 11986097 11986098 11986099 119860910119860101 119860102 119860103 119860104 119860105 119860106 119860107 119860108 119860109 1198601010

)))))))))))))))))

(27)


Table 2


Table 3


(119898)119894 2

such that


(28)


In this case




5 Conclusion




References


















Journal of












Journal of







Volume 2018






Volume 2018










(119860 minus 120582(119898)119894 119868) 119906(119898)119894 = ℎ119898+1119898119890119879119898119910(119898)119894 V119898+1 (24)










Table 1


(119898)119894 2





4 Numerical Example




22





119860 =(((((((((((((((((

11986011 11986012 11986013 11986014 11986015 11986016 11986017 11986018 11986019 11986011011986021 11986022 11986023 11986024 11986025 11986026 11986027 11986028 11986029 11986021011986031 11986032 11986033 11986034 11986035 11986036 11986037 11986038 11986039 11986031011986041 11986042 11986043 11986044 11986045 11986046 11986047 11986048 11986049 11986041011986051 11986052 11986053 11986054 11986055 11986056 11986057 11986058 11986059 11986051011986061 11986062 11986063 11986064 11986065 11986066 11986067 11986068 11986069 11986061011986071 11986072 11986073 11986074 11986075 11986076 11986077 11986078 11986079 11986071011986081 11986082 11986083 11986084 11986085 11986086 11986087 11986088 11986089 11986081011986091 11986092 11986093 11986094 11986095 11986096 11986097 11986098 11986099 119860910119860101 119860102 119860103 119860104 119860105 119860106 119860107 119860108 119860109 1198601010

)))))))))))))))))

(27)


Table 2


Table 3


(119898)119894 2

such that


(28)


In this case




5 Conclusion




References


















Journal of












Journal of







Volume 2018






Volume 2018









Table 2


Table 3


(119898)119894 2

such that


(28)


In this case




5 Conclusion




References


















Journal of












Journal of







Volume 2018






Volume 2018















Journal of












Journal of







Volume 2018






Volume 2018








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