ARnn

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xRn

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（4-如果對任意x(0)，

k時x(k)x*x(k)(x(k),x(k),,x(k))T

limx(k1)k

Blimx(k)k

x*x*Bx*三種基本的迭代一雅可比(Jacobi)迭代法。a11x1a12x2a1nxna21x1a22x2a2nxnan1x1an2x2annxnA(aij)nn非奇異，且aii0(i12,n)x1

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n二、高斯-賽德爾（Gauss-Seidel)迭代

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x(k1)Bx(k) (4-

fDL三、迭代法的收斂性Ax

x(k1)Bx(k) 設(shè)

是(1)的解x(k

k若有l(wèi)imx(k kk注意.limx(k)x*limx(k)x*.k kx(k)

為任意向量范

x*Bx*fx(k1

x*Bx(k)BxB(x(k)x*) Bk1x(0)x*令ε(k)x(k) ，ε(k1)Bε(k)B2ε(k1)Bklimε(k

0

(k

limBk1ε(0)

k

k

k ε(0) x(0)

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引

BRnn limBk

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存在非奇異矩陣P使得B=P-1JP,Bk=P-1JkJ J

. . J

i

1Jir Jir s 迭代法收斂性的判定條定理4.1(充要條件 迭代x(k1)Bx(k)

x(0) .

(B)1證明

若||B||

||x(k)x*

||B||

||x(k

x(k1)

(4-||x(k

x*

||B||k1||B.

||

x(0)

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||x*x(m)

B(x*

x(m1) x*

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x*x(0)||x*x(0)||為常數(shù)，||B||<1,||x*-x(m)||→0,

m

B(m) Bx(k)x(k B

x(k1)x(k p

x(k

x*(x(k)x*)p

x(k

x(kp

p

)x(kp

p例 x12x22x3

2x

A 121 2 2x2x321 2 2 D1(LU)

10 0 detIBJ

3 (B)0 detI得

detI(DL)1Udet(DL)1D-LU

3424(2)2 進一步得(BG)2

A(aij

n|aii||aijnjj

(i

等式成立，稱矩陣A為嚴(yán)格對角占優(yōu)陣 A為非奇異矩陣，det(A) 2

0

1

1 6 6

04 04 定理4.3（充分性條件）Ax中的A為嚴(yán)格對角占優(yōu)陣則Jacobi證（1）Jacobi0 0a

a1na

nn a2n D1(LU)

a22

|a|

(i1,2,,|aii

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(i1,2,,

所 1|a (i1,2,,|aii 即||

||

j

(D 的特征值 detI(DL)1Udet(DL)1det[(DL)U] det(DL)1 C(DL)U

anndet(C)det(DL)U （4-現(xiàn)在證明|| .用反證法，假設(shè)||1，又由|||aii||||aij|j j