迭代矩阵谱半径

则称集合 {1 , 2 , , n }
为A的谱. 记为 ch A
特征值取模最大
矩阵A的谱半径
( A)
max |
1 k n
k
|
注1: 当A是对称矩阵时, ||A||2 = (A)
注2: 对 Rn×n 中的范数|| ·||,有
(A) ≤ || A ||
5/15
定理4.1 迭代法 x(k+1) = B x(k) + f 收敛
1 2 2
A1 1 1
1
2 2 1
2 1 1
A2 1 1
1
1 1 2
试分析Jacobi 迭代法和 Seidel 迭代法的敛散性
0 2
(1)
BJ
1
0
2 2
(BJ ) 1
A1=[1,2,-2;1,1,1;2,2,1]
2
1 0
D=diag(diag(A1)); B1=D\(D-A1); max(abs(eig(B1)))
( k = 1, 2, 3, ······ )
|| (k)|| ≤ || B||k || (0)||
|| B|| < 1 所以
lim || (k) || lim || B ||k || (0) || 0
k
k
lim (k) 0
k
4/15
矩阵A的谱
设n阶方阵A的n个特征值为: 1 , 2 , , n
BJ
1
0
1
1 / 2 1 / 2 0
D=diag(diag(A2)) B2=D\(D-A2) max(abs(eig(Bj)))
(BJ ) 1.1180 Ans= 1.1180
9/15
0 BS 0
0
1/ 2 1/ 2
0
1/ 2 1 / 2 1 / 2
(BS ) 1/ 2
DL=tril(A2) B2=DL\(DL-A2) max(abs(eig(B2)))
Ans= 1.2604e-005
8/15
0 2 2 BS 0 2 3
0 0 2
DL=tril(A1) B1=DL\(DL-A1) max(abs(eig(B1)))
(BS ) 2 Ans= 2
(2) A2=[2, -1, 1; 1, 1, 1; 1, 1, -2]
0 1/ 2 1/ 2
12/15
所以 || x(k) x* || 1 || x(k1) x(k) ||
1 || B ||
x(k+1)–x(k) =(Bx(k)–f ) – (Bx(k-1)–f ) =B(x(k) – x(k-1) )
||x(k+1)–x(k)|| ≤ ||B || || x(k) – x(k-1) ||
Ans= 1/2
两种迭代法之间没有直接联系
对矩阵A1,求A1 x = b 的Jacobi迭代法收敛, 而Gauss-Seidel迭代法发散;
Hale Waihona Puke Baidu
对矩阵A2,求A2 x = b 的Jacobi迭代法发散, 而Gauss-Seidel迭代法收敛.
10/15
误差估计定理
定理4.2 :设x*为方程组 Ax=b 的解 若||B||<1,则对迭代格式 x(k+1) = B x(k) + f 有
(1) || x(k) x* || || B || || x(k) x(k1) ||
1 || B ||
(2)
|| x(k) x* || || B ||k || x(1) x(0) || 1 || B ||
11/15
证 由||B||<1,有
lim x(k) x *
k
x(k+1)–x* =(Bx(k)+f ) – (Bx*+f ) =B(x(k) – x* )
迭代法
x(k+1) = B x(k) + f
lim J k 0
k
lim
k
i
k
0
收敛 <=> lim Bk 0 k
(i = 1, 2,···, r)
| i | 1
max
1 i r
|
i
|
1
(i = 1, 2,···, r) 谱半径 (B) < 1
7/15
例 线性方程组 A x = b, 分别取系数矩阵为
|| x(k+1) – x* || ≤ ||B|| || x(k) – x* || 所以
||x(k+1) – x(k) ||= ||(x*– x(k)) – (x* – x(k+1))||
≥||(x*– x(k)) || – ||(x* – x(k+1))|| ≥ ||(x*– x(k))|| –||B|| ||(x* – x(k))|| = ( 1 - || B ||) ||(x* – x(k))||
误差估计:
|| x(k) x* ||
|| B ||
|| x(k ) x(k1) ||
1 || B ||
|| x(k) x* || || B ||k || x(1) x(0) || 1 || B ||
《数值分析》10
迭代法的收敛性
Convergence of iterative method
迭代矩阵谱半径
Spectral radius
对角占优矩阵
diagonally dominant matrix
原始方程: A x = b 迭代格式: x(k+1) = B x(k) + f
设方程组的精确解为 x*,则有
<=> 谱半径ρ(B) < 1
证: 对任何 n 阶矩阵B都存在非奇矩阵P使
B = P –1 J P
其中, J 为B的 Jordan 标准型
J1
J
J2
J r nn
其中, Ji 为Jordan块
i 1
Ji
1
i ni ni
6/15
其中,λi 是矩阵B的特征值, 由 B = P –1 J P B k = (P –1 J P) (P –1 J P) ···(P –1 J P)= P –1 J k P
(1) lim (k) 0
k
(2)
lim Bk 0
k
lim [x(k) x*] 0
k
lim x(k) x*
k
迭代格式 x(k+1) = B x(k) + f 收敛 ！！
3/15
命题 若||B||<1,则迭代法 x(k+1) =B x(k) +f 收敛
证: 由(k) = B (k-1),得 || (k)|| ≤ || B|| || (k-1)||
x* = B x* + f
x(k+1) – x*= B(x(k) – x*)
记 (k) = x(k) – x* ( k = 0, 1, 2, 3, ······ )
则有
(k+1) = B (k) (k) = B (k-1) ( k = 1, 2, 3, ······ )
2/15
(k) = B (k-1)=B2 (k-2)=···=Bk (0)