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We characterizei matrices which are nonexpansive with respect to some matrix norms, and ii matrices whose average iterates approach zero or are bounded.. A matrix norm on M n is called a

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Volume 2010, Article ID 821928, 13 pages

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,tlim@gmu.edu

Received 28 August 2009; Accepted 19 October 2009

Academic Editor: Mohamed A Khamsi

Copyrightq 2010 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 characterizei matrices which are nonexpansive with respect to some matrix norms, and ii 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,R will denote the set of real numbers, C the set of complex numbers, andM nthe complex vector space of complexn × n matrices A function · : M n → R is a

matrix norm if for all A, B ∈ M n, it satisfies the following five axioms:

1 A ≥ 0;

2 A 0 if and only if A 0;

3 cA |c|A for all complex scalars c;

4 A B ≤ A B;

5 AB ≤ A B.

Let| · | be a norm on Cn Define · on M nby

A max

This norm onM n is a matrix norm, called the matrix norm induced by| · | A matrix norm on

M n is called an induced matrix norm if it is induced by some norm onCn If · 1is a matrix norm onM n, there exists an induced matrix norm · 2onM nsuch thatA2 ≤ A1for all

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A ∈ M ncf 1, page 297 Indeed one can take · 2to be the matrix norm induced by the norm| · | on Cndefined by

whereCx is the matrix in M nwhose columns are all equal tox For A ∈ M n,ρA denotes

the spectral radius ofA.

Let | · | be a norm in Cn A matrix A ∈ M n is a contraction relative to | · | if it is a contraction as a transformation fromCnintoCn; that is, there exists 0≤ λ < 1 such that

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 matrix A ∈ M n , the following are equivalent:

a A is a contraction relative to a norm in C n ;

b A < 1 for some induced matrix norm · ;

c A < 1 for some matrix norm · ;

d limk → ∞ A k 0;

e ρA < 1.

Thatb follows from c is a consequence of the previous remark about an induced matrix norm being less than a matrix norm Since all norms onM n are equivalent, the limit ind can be relative to any norm onM n, so thatd is equivalent to all the entries of A kconverge

to zero ask → ∞, which in turn is equivalent to lim k → ∞ A k x 0 for all x ∈ C n

In this paper, we first characterize matrices inM n that are nonexpansive relative to some

norm| · | on Cn, that is,

Then we characterize thoseA ∈ M nsuch that

A k 1

k

I A A2 · · · A k−1

5

converges to zero ask → ∞, and those that {A k:k 0, 1, 2, } is bounded.

Finally we apply our theory to approximation of solution ofAx b using iterative

methodsfixed point iteration methods

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Theorem 2 For a matrix A ∈ M n , the following are equivalent:

a A is nonexpansive relative to some norm on C n ;

b A ≤ 1 for some induced matrix norm · ;

c A ≤ 1 for some matrix norm · ;

d {A k:k 0, 1, 2, } is bounded;

e ρA ≤ 1, and for any eigenvalue λ of A with |λ| 1, the geometric multiplicity is equal to

the algebraic multiplicity.

Proof As in the previous theorem,a, b, and c are equivalent Assume that b holds Let the norm · be induced by a vector norm | · | of Cn Then

A k x ≤A k |x| ≤ A k |x| ≤ |x|, k 0, 1, 2, , 6

proving thatA k x is bounded in norm | · | for every x ∈ C n Takingx e i, we see that the set

of all columns ofA k , k 0, 1, 2, , is bounded This proves that A k , k 0, 1, 2, , is bounded

in maximum column sum matrix norm1, page 294, and hence in any norm in Mn 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 thatd implies e Suppose that A has an eigenvalue λ with λ > 1.

Letx be an eigenvector corresponding to λ Then

ask → ∞, where · is any vector norm of C n This contradictsd Hence |λ| ≤ 1 Now

suppose thatλ is an eigenvalue with |λ| 1 and the Jordan block corresponding to λ is not

diagonal Then there exist nonzero vectorsv1, v2such thatAv1 λv1, Av2 v1 λv2 Let

u v1 v2 Then

A k u λ k−1 λ kv1 λ k v2, 8

and A k u ≥ kv1 − v1 − v2 It follows that A k u, k 0, 1, 2, , is unbounded,

contradictingd Hence d implies e

Lastly we prove thate implies c Assume that e holds A is 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−1is 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 · by M SMS−1∞ Then we have

A ≤ 1.

Letλ be an eigenvalue of a matrix A ∈ M n The index ofλ, denoted by indexλ is the

smallest value ofk for which rankA − λI k rankA − λI k11, pages 148 and 131 Thus conditione above can be restated as ρA ≤ 1, and for any eigenvalue λ of A with |λ| 1,

indexλ 1

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LetA ∈ M n Consider

A k 1kI A · · · A k−1

We callA k thek-average of A As with A k, we have A k x → 0 for every x if and only if

A k → 0 in M n, and thatA k x is bounded for every x if and only if A kis bounded inM n We have the following theorem

Theorem 3 Let A ∈ M n Then

a A k , k 1, 2, , converges to 0 if and only if A ≤ 1 for some matrix norm · and that

1 is not an eigenvalue of A,

b A k , k 1, 2, , is bounded if and only if ρA ≤ 1, indexλ ≤ 2 for every eigenvalue λ with |λ| 1 and that index1 1 if 1 is an eigenvalue of A.

Proof First we prove the su ﬃciency part of a Let x be a vector in C n Let

y k 1

k

I A · · · A k−1

ByTheorem 2for any eigenvaluesλ of A either |λ| < 1 or |λ| 1 and indexλ 1.

If A is written in its Jordan canonical form A SJS−1, then the k-average of A is

SJS−1, whereJis thek-average of J Jis in turn composed of thek-average of each of its

Jordan blocks For a Jordan block ofJ of the form

⎛

⎜

⎝

λ 1

· ·

· 1

λ

⎞

⎟

⎠

|λ| must be less than 1 Its k-average has constant diagonal and upper diagonals Let D jbe the constat value of itsjth upper diagonal D0being the diagonal and let Sj kD j Then

Cm, n 0 for n > m

S0 1− λ k

1− λ ,

S j Cj, j Cj 1, j λ · · · Ck − 1, j λ k−1−j , j 1, 2, , n − 1.

12

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

S j − λS j S j−1 − λ k−j Ck, j 13

Thus, we haveS0 → 1/1 − λ as k → ∞ By induction, using 13 above and the fact that

λ k−j Ck, j → 0 as k → ∞, we obtain S j → 1/1 − λ j1ask → ∞ Therefore D j S j /k O1/k as k → ∞.

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If the Jordan block is diagonal of constant valueλ, then λ / 1, |λ| ≤ 1 and the k-average

of the block is diagonal of constant value1 − λ k /k1 − λ O1/k.

We conclude thatA k O1/k and hence y k ≤ A k x O1/k as k → ∞.

Now we prove the necessity part of a If 1 is an eigenvalue of A and x is a

corresponding eigenvector, thenA k x x / 0 for every k and of course B k x fails to converge

to 0 Ifλ is an eigenvalue of A with |λ| > 1 and x is a corresponding eigenvector, then

A k x

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 vectors v1, v2such thatAv1 λv1, Av2 v1 λv2 Then by using the identity

1 2λ 3λ2 · · · k − 1λ k−2 1− λ k−1

1 − λ2 − k − 1 λ k−1

we get

A k v2

1− λ k−1

k1 − λ2 −

1−k1

λ k−1

1− λ

v1 1− λ k

k1 − λ v2. 16

It follows that limk → ∞ A k v2 does not exist This completes the proof of part a

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, A2v, is a contraction, and we have A k v B k v ≤ 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 v1 λv2 In the former case, we haveA k ≤ v and in the latter case,

we see from16 that A k v A k v2 is bounded Finally if λ 1 then since index1 1, we

haveAv v and hence A k v v In all cases, we proved that A k v, k 0, 1, 2, , is bounded.

Since column vectors ofS form a basis for C n, the suﬃciency part of b follows

Now we prove the necessity part ofb If A has an eigenvalue λ with |λ| > 1 and

eigenvectorv, then as shown above A k v → ∞ as k → ∞ If A has 1 as an eigenvalue and

index1 ≥ 2, then there exist nonzero vectors v1, v2 such thatAv1 v1 andAv2 v1 v2 ThenA k v2 k−1/2v2which is unbounded Ifλ is an eigenvalue of A with |λ| 1, λ / 1

and indexλ ≥ 3, then there exist nonzero vectors v1, v2andv3such thatAv1 λv1, Av2

v1 λv2andAv3 v2 λv3 By expandingA j v3, j 0, 1, 2, , k − 1 and using the identity

k−1

j2

Cj, 2 λ j−2 1

1 − λ2

1− λ k−2

1− λ

1

2k − 2λ k−2 k − 1λ − k 1

, 17

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we obtain

A k v3 1

1 − λ2

1− λ k−2

k1 − λ

1

2k − 2λ k−2k − 1

k λ −

k 1 k

v1

1− λ k−1

k1 − λ2 −

1− 1

k

λ k−1

1− λ

v2 1− λ k

k1 − λ v3

18

which approaches to∞ as k → ∞ This completes the proof.

We now consider applications of preceding theorems to approximation of solution of

a linear systemAx b, where A ∈ M nandb a given vector in C n LetQ be a given invertible

matrix inM n.x is a solution of Ax b if and only if x is a fixed point of the mapping T

defined by

Tx I − Q−1Ax Q−1b. 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 x k T k x0, k

0, 1, 2, , converges to the unique solution of Ax b In practice, given x0, each successive

x kis obtained fromx k−1by solving the equation

The classical methods of Richardson, Jacobi, and Gauss-Seidelsee, e.g., 3 have Q I, D,

and L respectively, where I is the identity matrix, D the diagonal matrix containing the

diagonal of A, and L the lower triangular matrix containing the lower triangular portion

ofA Thus byTheorem 1we have the following known theorem

Theorem 4 Let A, Q ∈ M n , with Q invertible Let b, x0 ∈ Cn If ρI − Q−1A < 1, then A is invertible and the sequence x k , k 1, 2, , defined recursively by

converges to the unique solution of Ax b.

Theorem 4fails ifρI − Q−1A 1, For a simple 2 × 2 example, let Q I, b 0, A 2I

andx0any nonzero vector

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

A ∈ M n, we will denoteRA and NA the range and the null space of A respectively.

Lemma 5 Let A be a singular matrix in M n such that the geometric multiplicity and the algebraic multiplicity of the eigenvalue 0 are equal, that is, index 0 1 Then there is a unique projection

P A whose range is the range of A and whose null space is the null space of A, or equivalently, C n

RA ⊕ NA Moreover, A restricted to RA is an invertible transformation from RA onto RA.

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Proof If A SJS−1is a Jordan canonical form ofA where the eigenvalues 0 appear at the end

portion of the diagonal ofJ, then the matrix

P A S

⎡

⎢

1

·

· 1 0

·

· 0

⎤

⎥

is the required projection ObviouslyA maps RA into RA If z ∈ RA and Az 0, then

z ∈ NA ∩ RA {0} and so z 0 This proves that A is invertible on RA.

Remark 6 Under the assumptions ofLemma 5, we will call the component of a vectorc in NA the projection of c on NA along RA Note that by definition of index, the condition

in the lemma is equivalent toNA2 NA.

Theorem 7 Let A be a matrix in M n and b a vector in C n Let Q be an invertible matrix in M n and let B I − Q−1A Assume that ρB ≤ 1 and that indexλ 1 for every eigenvalue λ of B with modulus 1, that is, B is nonexpansive relative to a matrix norm Starting with an initial vector x0 in

Cn define x k recursively by

for k 1, 2, Let

y k x0 x1 · · · x k−1

If Ax b is consistent, that is, has a solution, then y k , k 1, 2, , converge to a solution vector z with rate of convergence y k − z O1/k If Ax b is inconsistent, then lim k x k limk y k

∞ More precisely, lim k x k /k cand lim k y k /k c/2, where c Q−1b and cis the projection of c

on NA NQ−1A along RQ−1A.

Proof First we assume that A is invertible so that I − B Q−1A is also invertible Let T be the

mapping defined byTx Bxc Then T k x B k xcBc· · ·B k−1 c Let s cBc· · ·B k−1 c.

Thens − Bs c − B k c and hence s I − B−1c − I − B−1B k c I − B−1c − B k I − B−1c Let

z I − B−1c A−1b z is the unique solution of Ax b and

T k x B k x z − B k z B k x − z z. 25

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Since the sequencex kin the theorem isT k x0, we have

y k 1kI B · · · B k−1

x0− z z B k x0− z z. 26

SinceI − B is invertible, 1 is not an eigenvalue of B, and byTheorem 3 parta y k − z

B k x0−z → 0 as k → ∞ Moreover, from the proof of the same theorem, y k −z O1/k.

Next we consider the case when A is not invertible Since Q is invertible, we have RQ−1A Q−1RA and NQ−1A NA The index of the eigenvalue 0 of Q−1A is the

index of eigenvalue 1 ofB I − Q−1A Thus byLemma 5,Cn Q−1RA ⊕ NA For every

vectorv ∈ C n, let v r andv ndenote the component ofv in the subspace Q−1RA and

NA, respectively.

Assume thatAx b is consistent, that is, b ∈ RA Then c ∈ RQ−1A ByLemma 5, the restriction ofQ−1A on its range is invertible, so there exists a unique zinRQ−1A such

thatQ−1Az c, or equivalently, I − Bz c For any vector x, we have

T k x B k x c Bc · · · B k−1 c

B K

x r x n

I B · · · B k−1

I − Bz

B k

x r

 x n z− B k

z

B k

x r − z

 x n z.

27

SinceB maps RQ−1A into RQ−1A and I −B Q−1A restricted to RQ−1A is invertible, we

can apply the preceding proof and conclude that the sequencey kas defined before converges

toz x0n zandy k − z O1/k Now Az Ax n0  Az Az Qc b, showing

thatz is a solution of Ax b.

Assume now thatb /∈ RA, that is, Ax b is inconsistent Then c /∈ RQ−1A and c

c r c nwithc n / 0 As in the preceding case there exists a unique z∈ RQ−1A such that

I − Bz c r Note that for ally ∈ NA, By I − Q−1Ay y Thus for any vector x

and any positive integerj

x j T j x

B j x c Bc · · · B j−1 c

B j

x r x n

I B · · · B j−1

I − Bz jc n

B j

x r

 x n z− B j

z  jc n

B j

xr − z

 x n z jc n ,

y k 1kx Tx · · · T k−1 x

B kx r − z

 x n zk − 1

2 c n ,

28

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whereB k I B · · · B k−1 As in the preceding case, B k x r − z, k 0, 1, 2, is bounded

andB k x r −z, k 1, 2, , converges to 0 Thus lim k → ∞ x k /k cn and lim k → ∞ y k /k

c n /2, and hence lim k → ∞ x k limk → ∞ 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 condition c in the next lemma

Lemma 8 Let B be an n × n matrix The following are equivalent:

a for every 0 < μ < 1, there exists a matrix norm · μ such that μI 1 − μB μ ≤ 1,

b for every 0 < μ < 1, there exists an induced matrix norm · μ such that μI1−μB μ ≤ 1,

c ρB ≤ 1 and index1 1 if 1 is an eigenvalue of B.

Proof As in the proof ofTheorem 2,a and b are equivalent For 0 < μ < 1, denote μI 

1 − μB by Bμ Suppose now that a holds Let λ be an eigenvalue of B Then μ 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 of B, then it is also an eigenvalue of Bμ 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 assumec holds 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,a holds This completes the proof

Theorem 9 Let A ∈ M n and b ∈ C n Let Q be an invertible matrix in M n , and B I − Q−1A Suppose ρB ≤ 1 and that index1 1 if 1 is an eigenvalue of B Let 0 < μ < 1 be fixed Starting with an initial vector x0 , define x k , y k , k 0, 1, 2, , recursively by

y0 x0,

Qx k Q − Ay k−1 b,

y k μy k−11− μ x k

29

If Ax b is consistent, then y k , k 0, 1, 2, , converges to a solution vector z of Ax b with rate

of convergence given by

y k − z oζ k

where ζ is any number satisfying

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If Ax b is inconsistent, then lim k → ∞ y k ∞; more precisely,

lim

k → ∞

y k

k

where c n is the projection of c on NA along RQ−1A.

Proof Let c Q−1b, B1 μI 1 − μB I − 1 − μQ−1A, and Tx B1x 1 − μc Then

y k T k x0

First we assume thatA is invertible Then I − B1 1 − μQ−1A is invertible and 1 is

not an eigenvalue ofB1; thusρB1 < 1 Let z 1 − μI − B1−1c A−1b We have

y k T k x0

B k

1x01− μ c B1c · · · B k−1

1 c

B k

1x01− μ

1− μ

I B1 · · · B k−1

1

I − B1z

B k

1x0− z z.

33

By a well known theoremsee, e.g 1, y k − z oζ k for every ζ > ρB1

Assume now thatA is not invertible and b ∈ RA Then c is in the range of Q−1A.

Since B I − Q−1A satisfies the condition in Lemma 8, Q−1A satisfies the condition in

Lemma 5 Thus the restriction ofQ−1A on its range is invertible and there exists zinRQ−1A

such that Q−1Az c, or equivalently, I − B1z 1 − μc For any vector x x0, we have

y k T k x

B k

1x 1− μ c B1c · · · B k−1

1 c

B k

1

x r x n

I B1 · · · B k−1

1

I − B1z

B k

1

x r

 x n z− B k

1

z

B k

1

x r − z

 x n z.

34

SinceB1mapsRQ−1A into RQ−1A and I − B Q−1A restricted to RQ−1A is invertible,

we can apply the preceding proof and conclude that the sequencey k , k 0, 1, 2, converges

toz x n zandy k − z oζ k z solves Ax b since Az Ax n Az Az

Qc b.

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Proof If A ...

which approaches to< i>∞ as k → ∞ This completes the proof.

We now consider applications of preceding theorems to approximation of solution of

a linear systemAx b,... class="text_page_counter">Trang 5

If the Jordan block is diagonal of constant valueλ, then λ / 1, |λ| ≤ and the k-average

of the

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