Nonexpansive Matrices with Applications to Solutions of Linear Systems by Fixed Point Iterations

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Volume 2010, Article ID 821928,13pages doi:10.1155/2010/821928

Research Article

Nonexpansive Matrices with Applications

to Solutions of Linear Systems by Fixed

Point Iterations

Teck-Cheong Lim

Department of Mathematical Sciences, George Mason University, 4400, University Drive, Fairfax, VA 22030, USA

Correspondence should be addressed to Teck-Cheong Lim,[email protected]

Received 28 August 2009; Accepted 19 October 2009

Academic Editor: Mohamed A. Khamsi

Copyrightq2010 Teck-Cheong Lim. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We characterizeimatrices which are nonexpansive with respect to some matrix norms, andii matrices whose average iterates approach zero or are bounded. Then we apply these results to iterative solutions of a system of linear equations.

Throughout this paper,Rwill denote the set of real numbers,Cthe set of complex numbers, andMnthe complex vector space of complexn×nmatrices. A function · :Mn → Ris a matrix normif for allA, B∈M_n, it satisfies the following five axioms:

1A ≥0;

2A0 if and only ifA0;

3cA|c|Afor all complex scalarsc;

4AB ≤ AB;

5AB ≤ A B.

Let| · |be a norm onCn. Define · onMnby

Amax

|x|1|Ax|. 1

This norm onMnis a matrix norm, called thematrix norm induced by| · |. A matrix norm on

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A∈Mncf.1, page 297. Indeed one can take · 2to be the matrix norm induced by the norm| · |onCndefined by

|x|Cx₁, 2

whereCxis the matrix inMnwhose columns are all equal tox. ForA∈Mn,ρAdenotes

the spectral radius ofA.

Let | · |be a norm in Cn. A matrix A ∈ Mn is acontraction relative to | · |if it is a

contraction as a transformation fromCnintoCn; that is, there exists 0≤λ <1 such that _Ax₋_Ay_≤_λ_x₋_y_{, x, y}_∈_Cn_. ₃

Evidently this means that for the matrix norm · induced by| · |,A < 1. The following theorem is well knowncf.1, Sections 5.6.9–5.6.12.

Theorem 1. For a matrixA∈M_n, the following are equivalent:

aAis a contraction relative to a norm inCn;

bA<1for some induced matrix norm · ;

cA<1for some matrix norm · ;

dlimk→ ∞Ak0;

eρA<1.

Thatbfollows fromcis a consequence of the previous remark about an induced matrix norm being less than a matrix norm. Since all norms onMn are equivalent, the limit ind

can be relative to any norm onMn, so thatdis equivalent to all the entries ofAkconverge

to zero ask → ∞, which in turn is equivalent to lim_k_{→ ∞}Akx0 for allx∈Cn.

In this paper, we first characterize matrices inMnthat arenonexpansiverelative to some

norm| · |onCn, that is,

_Ax₋_Ay_≤_x₋_y_{, x, y}_∈_Cn_. ₄

Then we characterize thoseA∈Mnsuch that

Ak 1_k

IAA2_{· · ·}_Ak−1 ₅

converges to zero ask → ∞, and those that{Ak:k0,1,2, . . .}is bounded.

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Theorem 2. For a matrixA∈Mn, the following are equivalent:

aAis nonexpansive relative to some norm onCn;

bA ≤1for some induced matrix norm · ;

cA ≤1for some matrix norm · ;

d{Ak:k0,1,2, . . .}is bounded;

eρA≤1, and for any eigenvalueλofAwith|λ|1, the geometric multiplicity is equal to the algebraic multiplicity.

Proof. As in the previous theorem,a,b, andcare equivalent. Assume thatbholds. Let the norm · be induced by a vector norm| · |ofCn. Then

Ak_x_≤_Ak_|_x_{| ≤}_Ak_|_x_{| ≤ |}_x_|_{, k}₀_,₁_,₂_{, . . . ,} ₆

proving thatAkxis bounded in norm| · |for everyx∈Cn. Takingxei, we see that the set

of all columns ofAk, k0,1,2, . . . ,is bounded. This proves thatAk, k0,1,2, . . . ,is bounded in maximum column sum matrix norm1, page 294, and hence in any norm inM_n. Note that the last part of the proof also follows from the Uniform Boundedness Principlesee, e.g.,

2, Corollary 21, page 66

Now we prove thatdimpliese. Suppose thatAhas an eigenvalueλwithλ > 1. Letxbe an eigenvector corresponding toλ. Then

Ak_x_|_λ_|k_x_{−→ ∞} ₇

ask → ∞, where · is any vector norm ofCn. This contradictsd. Hence|λ| ≤ 1. Now suppose thatλis an eigenvalue with|λ| 1 and the Jordan block corresponding toλis not diagonal. Then there exist nonzero vectorsv1, v2such thatAv1 λv1, Av2 v1λv2. Let

uv1v2. Then

Ak_u_λk−1_λ_k_v

1λkv2, 8

and Aku ≥ kv1 − v1 − v2. It follows that Aku, k 0,1,2, . . . , is unbounded, contradictingd. Hencedimpliese.

Lastly we prove thateimpliesc. Assume thateholds.Ais similar to its Jordan canonical form J whose nonzero oﬀ-diagonal entries can be made arbitrarily small by similarity 1, page 128. Since the Jordan block for each eigenvalue with modulus 1 is diagonal, we see that there is an invertible matrix S such that the l1-sum of each row of

SAS−1_{is less than or equal to 1, that is,}_SAS−1

∞≤1, where · ∞is the maximum row sum

matrix norm1, page 295. Define a matrix norm · byMSMS−1

∞. Then we have

A ≤1.

Letλbe an eigenvalue of a matrixA∈Mn. The index ofλ, denoted by indexλis the

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LetA∈Mn. Consider

Ak 1_k

IA· · ·Ak−1_. ₉

We callAk thek-average of A. As with Ak, we have Akx → 0 for every xif and only if

Ak → 0 inMn, and thatAkxis bounded for everyxif and only ifAkis bounded inMn. We

have the following theorem.

Theorem 3. LetA∈Mn. Then

aAk, k1,2, . . . ,converges to0if and only ifA ≤1for some matrix norm · and that

1is not an eigenvalue ofA,

bAk, k 1,2, . . . ,is bounded if and only ifρA≤ 1,indexλ≤2for every eigenvalueλ with|λ|1and that index1 1if1is an eigenvalue ofA.

Proof. First we prove the suﬃciency part ofa. Letxbe a vector inCn. Let

yk _k1

IA· · ·Ak−1_x_. ₁₀

ByTheorem 2for any eigenvaluesλofAeither|λ|<1 or|λ|1 and indexλ 1.

If Ais written in its Jordan canonical formA SJS−1_{, then the} _k_{-average of}_A _is

SJ_S−1_{, where}_J_{is the}_k_{-average of}_J_._J_{is in turn composed of the}_k_{-average of each of its}

Jordan blocks. For a Jordan block ofJof the form ⎛

⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝

λ 1

· · · 1

λ

⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠

, 11

|λ|must be less than 1. Itsk-average has constant diagonal and upper diagonals. LetD_jbe the constat value of itsjth upper diagonalD0being the diagonaland let Sj kDj. Then

Cm, n 0 forn > m

S0 1−λ

k

1−λ ,

SjCj, jCj1, jλ· · ·Ck−1, jλk−1−j, j 1,2, . . . , n−1.

12

Using the relationCm1, j−Cm, j Cm, j−1, we obtain

Sj−λSjSj−1−λk−jC

k, j. 13

Thus, we haveS0 → 1/1−λask → ∞. By induction, using13above and the fact that

λk−j_C_{k, j} _→ _{0 as}_k _{→ ∞}_{, we obtain}_S

j → 1/1−λj1ask → ∞. ThereforeDj Sj/k

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If the Jordan block is diagonal of constant valueλ, thenλ /1,|λ| ≤1 and thek-average of the block is diagonal of constant value1−λk/k1−λ O1/k.

We conclude thatAkO1/kand henceyk ≤ AkxO1/kask → ∞. Now we prove the necessity part of a. If 1 is an eigenvalue of A and x is a corresponding eigenvector, thenAkxx /0 for everykand of courseBkxfails to converge

to 0. Ifλis an eigenvalue ofAwith|λ|>1 andxis a corresponding eigenvector, then

Akx

λ

k₋₁

kλ−1

x ≥ |λ| k₋₁

k|λ−1| x. 14

which approaches to ∞ as k → ∞. If λ is an eigenvalue of A with |λ| 1, λ /1, and indexλ≥2, then there exist nonzero vectorsv1, v2such thatAv1 λv1, Av2 v1λv2. Then by using the identity

12λ3λ2· · · k−1λk−2 1−λk− 1

1−λ2 −k−1

λk−1

1−λ 15

we get

Akv2

1−λk−1

k1−λ2 −

1− 1

k

_λ_k₋₁ 1−λ

v1 1−λ

k

k1−λv2. 16

It follows that limk→ ∞Akv2does not exist. This completes the proof of parta.

Suppose that A satisfies the conditions in b and that A SJS−1 is the Jordan canonical form of A. Let λ be an eigenvalue of A and let v be a column vector of S corresponding to λ. If |λ| < 1, then the restriction B of A to the subspace spanned by

v, Av, A2_{v, . . .} _{is a contraction, and we have} _A

kv Bkv ≤ v. If |λ| 1, and λ /1,

then by conditions in b either Av λv, or there exist v1, v2 with v v2 such that

Av1 λv1, Av2 v1λv2. In the former case, we haveAk ≤ vand in the latter case, we see from16thatAkv Akv2is bounded. Finally ifλ1 then since index1 1, we haveAvvand henceA_kvv. In all cases, we proved thatA_kv, k0,1,2, . . . ,is bounded. Since column vectors ofSform a basis forCn, the suﬃciency part ofbfollows.

Now we prove the necessity part ofb. If Ahas an eigenvalue λ with|λ| > 1 and eigenvectorv, then as shown aboveA_kv → ∞ask → ∞. IfAhas 1 as an eigenvalue and index1 ≥ 2, then there exist nonzero vectorsv1, v2 such thatAv1 v1 andAv2 v1v2. ThenAkv2 k−1/2v2which is unbounded. Ifλis an eigenvalue ofAwith|λ|1, λ /1 and indexλ≥3, then there exist nonzero vectorsv1, v2andv3such thatAv1λv1, Av2

v1λv2andAv3 v2λv3. By expandingAjv3, j0,1,2, . . . , k−1 and using the identity

k−1

j2

Cj,2λj−2 1

1−λ2

1−λk−2

1−λ

1

2k−2λ

k−2_k₋₁_λ₋_k₁

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we obtain

Akv3

1

1−λ2

1−λk−2

k1−λ 1

2k−2λ

k−2 _k₋

1

k λ− k1

k

v1

1−λk−1

k1−λ2 −

1− 1

k

_λ_k₋₁ 1−λ

v2 1−λk

k1−λv3

18

which approaches to∞ask → ∞. This completes the proof.

We now consider applications of preceding theorems to approximation of solution of a linear systemAxb, whereA∈M_nandba given vector inCn. LetQbe a given invertible matrix inMn.xis a solution ofAx bif and only ifxis a fixed point of the mappingT

defined by

TxI−Q−1_A_x_Q−1_b. ₁₉

T is a contraction if and only if I −Q−1A is. In this case, by the well known Contraction Mapping Theorem, given any initial vector x0, the sequence of iterates xk Tkx0, k 0,1,2, . . . ,converges to the unique solution ofAx b. In practice, givenx0, each successive

xkis obtained fromxk−1by solving the equation

Qxk Q−Axk−1b. 20

The classical methods of Richardson, Jacobi, and Gauss-Seidelsee, e.g.,3haveQ I, D, and L respectively, where I is the identity matrix, D the diagonal matrix containing the diagonal of A, andL the lower triangular matrix containing the lower triangular portion ofA. Thus byTheorem 1we have the following known theorem.

Theorem 4. LetA, Q ∈ Mn, with Qinvertible. Letb, x0 ∈ Cn. IfρI −Q−1A < 1, thenA is

invertible and the sequencex_k, k1,2, . . . ,defined recursively by

Qxk Q−Axk−1b 21

converges to the unique solution ofAxb.

Theorem 4fails ifρI−Q−1_A _{1, For a simple 2} _× _{2 example, let}_Q_{I, b}₀_{, A}₂_I andx0any nonzero vector.

We need the following lemma in the proof of the next two theorems. For a matrix

A∈Mn, we will denoteRAandNAthe range and the null space ofArespectively.

Lemma 5. LetAbe a singular matrix inM_nsuch that the geometric multiplicity and the algebraic multiplicity of the eigenvalue 0 are equal, that is,index0 1. Then there is a unique projection

PAwhose range is the range ofAand whose null space is the null space ofA, or equivalently,Cn

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Proof. IfASJS−1_{is a Jordan canonical form of}_A_{where the eigenvalues 0 appear at the end} portion of the diagonal ofJ, then the matrix

PAS ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1

· ·

1

0

· ·

0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

S−1 ₂₂

is the required projection. ObviouslyAmapsRAintoRA. Ifz∈RAandAz 0, then

z∈NA∩RA {0}and soz0. This proves thatAis invertible onRA.

Remark 6. Under the assumptions ofLemma 5, we will call the component of a vectorcin

NAthe projection ofconNAalongRA. Note that by definition of index, the condition in the lemma is equivalent toNA2 NA.

Theorem 7. LetAbe a matrix inM_nandba vector inCn. LetQbe an invertible matrix inM_nand letB I−Q−1_A_{. Assume that}_ρ_B _≤ ₁_{and that}_index_λ ₁_{for every eigenvalue}_λ_of_B_with

modulus1, that is,Bis nonexpansive relative to a matrix norm. Starting with an initial vectorx0in

Cn_define_x

krecursively by

Qxk Q−Axk−1b 23

fork1,2, . . . .Let

yk x0x1_k· · ·xk−1. 24

IfAx bis consistent, that is, has a solution, thenyk, k 1,2, . . . ,converge to a solution vectorz with rate of convergenceyk−zO1/k. IfAxbis inconsistent, thenlimkxklimkyk ∞. More precisely,lim_kx_k/kcandlim_ky_k/kc/2, wherecQ−1_b_and_c_{is the projection of}_c

onNA NQ−1_A_along_R_Q−1_A_.

Proof. First we assume thatAis invertible so thatI−BQ−1_A_{is also invertible. Let}_T_{be the} mapping defined byTxBxc. ThenTkxBkxcBc· · ·Bk−1_c_{. Let}_s_c_Bc_{· · ·}_Bk−1_c_. Thens−Bsc−Bkcand hences I−B−1c−I−B−1Bkc I−B−1c−BkI−B−1c. Let

z I−B−1_c_A−1_b_._z_{is the unique solution of}_Ax_b_and

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Since the sequencexkin the theorem isTkx0, we have

yk 1_k

IB· · ·Bk−1_x

0−z zBkx0−z z. 26

SinceI−Bis invertible, 1 is not an eigenvalue ofB, and byTheorem 3 partayk−z Bkx0−z → 0 ask → ∞. Moreover, from the proof of the same theorem,yk−zO1/k.

Next we consider the case when Ais not invertible. Since Q is invertible, we have

RQ−1_A _Q−1_R_A_and_N_Q−1_A _N_A_{. The index of the eigenvalue 0 of}_Q−1_A_{is the} index of eigenvalue 1 ofBI−Q−1_A_{. Thus by}_{Lemma 5}_,_Cn_Q−1_R_A_⊕_N_A_{. For every} vectorv ∈ Cn, let vr andvn denote the component ofvin the subspaceQ−1RAand

NA, respectively.

Assume thatAxbis consistent, that is,b∈RA. Thenc∈RQ−1_A_{. By}_{Lemma 5}_, the restriction ofQ−1Aon its range is invertible, so there exists a uniquezinRQ−1Asuch thatQ−1_Az_c_{, or equivalently,}_I₋_B_z_c_{. For any vector}_x_{, we have}

Tk_x_Bk_x_c_Bc_{· · ·}_Bk−1_c

BK_xr_xn_I_B_{· · ·}_Bk−1_I₋_B_z

Bk_xr_xn_z₋_Bk_z

Bk_xr₋_z_xn_z_.

27

SinceBmapsRQ−1_A_into_R_Q−1_A_and_I₋_B_Q−1_A_{restricted to}_R_Q−1_A_{is invertible, we} can apply the preceding proof and conclude that the sequenceykas defined before converges

tozx₀nzandyk−zO1/k. NowAzAx0n Az Az Qcb, showing thatzis a solution ofAxb.

Assume now thatb /∈RA, that is,Ax bis inconsistent. Thenc /∈RQ−1_A_and_c

cr_cn_with_cn_/₀_._{As in the preceding case there exists a unique}_z_∈_R_Q−1_A_{such that}

I−Bz _cr_{. Note that for all}_y_∈_N_A_,_B_y_I₋_Q−1_A_y _y_{. Thus for any vector}_x and any positive integerj

xj Tjx

Bj_x_c_Bc_{· · ·}_Bj−1_c

Bj_xr_xn_I_B_{· · ·}_Bj−1_I₋_B_z_jcn

Bj_xr_xn_z₋_Bj_z_jcn

Bj_xr₋_z_xn_z_jcn_,

yk 1_k

xTx· · ·Tk−1_x

Bk

xr₋_z_xn_zk−1

2 c

n_,

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whereBk IB· · ·Bk−1. As in the preceding case,Bkxr−z, k0,1,2, . . .is bounded

andB_kxr−z, k1,2, . . . ,converges to 0. Thus lim_k_{→ ∞}x_k/k cnand lim_k_{→ ∞}y_k/k

cn_/_{2, and hence lim}_k_{→ ∞}_x_k_lim_k_{→ ∞}_y_k_∞_{. This completes the proof.}

Next we consider another kind of iteration in which the nonlinear case was considered in Ishikawa 4. Note that the type of mappings in this case is slightly weaker than nonexpansivitysee conditioncin the next lemma.

Lemma 8. LetBbe ann×nmatrix. The following are equivalent:

afor every0< μ <1, there exists a matrix norm · _μsuch thatμI 1−μB_μ≤1,

bfor every0< μ <1, there exists an induced matrix norm·μsuch thatμI1−μBμ≤1,

cρB≤1andindex1 1if1is an eigenvalue ofB.

Proof. As in the proof ofTheorem 2,aandbare equivalent. For 0 < μ < 1, denoteμI

1−μBbyBμ. Suppose now thataholds. Letλ be an eigenvalue ofB. Thenμ 1−

μλ is an eigenvalue of Bμ. By Theorem 2 |μ 1 −μλ| ≤ 1 for every 0 < μ < 1 and hence|λ| ≤1. If 1 is an eigenvalue ofB, then it is also an eigenvalue ofBμ. ByTheorem 2, the index of 1, as an eigenvalue of Bμ, is 1. Since obviously B and Bμ have the same eigenvectors corresponding to the eigenvalue 1, the index of 1, as an eigenvalue ofB, is also 1. This provesc.

Now assumecholds. Since|μ 1−μλ| < 1 for|λ| 1, λ /1, every eigenvalue of

Bμ, except possibly for 1, has modulus less than 1. Reasoning as above, if 1 is an eigenvalue ofBμ, then its index is 1. Therefore byTheorem 2,aholds. This completes the proof.

Theorem 9. LetA ∈ Mnandb ∈ Cn. LetQbe an invertible matrix inMn, andB I−Q−1A. SupposeρB≤1and thatindex1 1if1is an eigenvalue ofB. Let0 < μ <1be fixed. Starting with an initial vectorx0, definexk, yk, k0,1,2, . . . ,recursively by

y0x0,

Qxk Q−Ayk−1

b,

ykμyk−1

1−μx_k.

29

IfAxbis consistent, thenyk, k 0,1,2, . . . ,converges to a solution vectorzofAxbwith rate of convergence given by

_y_k₋_z_o_ζk_, ₃₀

whereζis any number satisfying

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IfAxbis inconsistent, thenlimk→ ∞yk∞; more precisely,

lim

k→ ∞ yk

k

1−μcn, 32

wherecnis the projection ofconNAalong RQ−1_A_.

Proof. Letc Q−1_{b, B}

1 μI 1−μB I−1−μQ−1A, andTx B1x 1−μc. Then

ykTkx0.

First we assume thatAis invertible. ThenI−B1 1−μQ−1Ais invertible and 1 is not an eigenvalue ofB1; thusρB1<1. Letz 1−μI−B1−1cA−1b. We have

ykTkx0

Bk

1x0

1−μcB1c· · ·Bk₁−1c

Bk

1x0

1−μ 1 1−μ

IB1· · ·B₁k−1

I−B1z

Bk

1x0−z z.

33

By a well known theoremsee, e.g.1,yk−zoζkfor everyζ > ρB1.

Assume now thatAis not invertible andb ∈ RA. Thenc is in the range ofQ−1_A_. Since B I − Q−1_A _{satisfies the condition in} _{Lemma 8}_, _Q−1_A _{satisfies the condition in} Lemma 5. Thus the restriction ofQ−1_A_{on its range is invertible and there exists}_z_in_R_Q−1_A such that Q−1_Az _c_{, or equivalently,} _I ₋_B

1z 1−μc. For any vector x x0, we have

ykTkx

Bk

1x

1−μcB1c· · ·Bk1−1c

Bk

1

xr_xn_I_B

1· · ·B₁k−1

I−B1z

Bk

1

xr_xn_z₋_Bk

1

z

Bk

1

xr₋_z_xn_z_.

34

SinceB1mapsRQ−1AintoRQ−1AandI−BQ−1Arestricted toRQ−1Ais invertible, we can apply the preceding proof and conclude that the sequenceyk, k0,1,2, . . .converges

toz xnzandyk−z oζk.zsolvesAx bsinceAz Axn Az Az

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Assume lastly thatb /∈RA, that is,Ax b is inconsistent. Thenc /∈RQ−1_A_and

ccr_cn_with_cn_/_{0. As before there exists}_z_∈_R_Q−1_A_{such that}_I₋_b

1z 1−μcr. Note thatB1p pforp∈NA. Then

ykTkx

Bk

1x

1−μcB1c· · ·Bk₁−1c

Bk

1

xr_xn_I_B

1· · ·Bk₁−1

I−B1zk

1−μcn

Bk

1

xr₋_z_xn_z_k₁₋_μ_cn_.

35

SinceBk₁xr−z, k0,1,2, . . . ,converges to 0, we have

lim

k→ ∞ yk

k

1−μcn, 36

and hence lim_k_{→ ∞}y_k∞. This completes the proof.

By takingQIand considering only nonexpansive matrices in Theorems7and9, we obtain the following corollary.

Corollary 10. LetAbe ann×nmatrix such thatI−A ≤1for some matrix norm · . Letbbe a vector inCn. Then:

astarting with an initial vectorx0inCndefinexkrecursively as follows:

xk I−Axk−1 b 37

fork1,2, . . . .Let

yk x0x1_k· · ·xk−1 38

fork 1,2, . . . .IfAxbis consistent, thenyk, k 1,2, . . . ,converges to a solution vectorzwith rate of convergence given by

_y_k₋_z_O1

k

. 39

IfAxbis inconsistent, thenlimk→ ∞xklimk→ ∞yk∞.

blet0< μ <1be a fixed number. Starting with an initial vectorx0, let

y0x0,

xk I−Ayk−1

b, ykμyk−1

1−μxk.

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_y_k₋_z_o_ζk ₄₁

maxμ1−μλ:λan eigenvalue ofB, λ /1< ζ <1. 42

IfAxbis inconsistent, thenlimk→ ∞yk∞.

Remark 11. If in the previous corollary,I−A<1, andμ0 in partb, the sequenceykxk

converges to a solution. This is the Richardson method, see for example,3. Even in this case, our method in partbmay yield a better approximation. For example if

A

1 0.9

−0.9 1

, 43

b 0, andx0 e1, then the nth iterate in the Richardson method is 0.9n away from the solution 0, while thenth iterate using the method in the corollary partbwithμ1/2 is less than0.5n/2.

Ann×nmatrixA a_ijis calleddiagonally dominantif

|aii| ≥ n

j1,j /i

_a_ij ₄₄

for alli1, . . . , n. IfAis diagonally dominant withaii/0 for everyiand ifQDorL, where

D is the diagonal matrix containing the diagonal ofA, and Lthe lower triangular matrix containing the lower triangular entries of A, then it is easy to prove thatI−Q−1_A

∞ ≤

1 where · ∞ denotes the maximum row sum matrix norm; see, for example,1,3. The

following follows from Theorems7and9.

Corollary 12. LetAbe a diagonally dominantn×nmatrix witha_ii_/0for alli1, . . . , n. LetQD orL, whereDis the diagonal matrix containing the diagonal ofA, andLthe lower triangular matrix containing the lower triangular entries ofA. Letbbe a vector inCn. Then:

astarting with an initial vectorx0inCndefinexkrecursively as follows:

Qxk Q−Axk−1 b 45

fork1,2, . . . .Let

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fork 1,2, . . . .IfAx bis consistent, thenyk, k 1,2, . . .converges to a solution vectorzwith rate of convergence given by

_y_k₋_z_O1

k

. 47

IfAxbis inconsistent, thenlimk→ ∞xklimk→ ∞yk∞.

bLet0< μ <1be a fixed number. Starting with an initial vectorx0, let

y0x0,

Qxk Q−Ayk−1

b, ykμyk−1

1−μx_k.

48

_y_k₋_z_o_ζk_, ₄₉

maxμ1−μλ:λan eigenvalue ofB, λ /1< ζ <1. 50

IfAxbis inconsistent, thenlimk→ ∞yk∞.

References

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2 N. Dunford and J. T. Schwartz,Linear Operators, Part I, Interscience Publishers, New York, NY, USA, 1957.

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