给出线性方程组 Hn*x = b，其中系数矩阵Hn为希尔伯特矩阵：     

假设 x ∗ =(1, 1, . . . , 1)T，b = Hnx ∗。若取 n = 6,8, 10，分别用 Jacobi

迭代法及 SOR迭代（ω = 1, 1:25,1:5）求解，比较计算结果。

MATLAB源码如下，运行Demo_Jacobi_SOR.m文件，附件包含另外两个函数文件，分别为：Jacobi.m与SOR.m。

Demo_Jacobi_SOR.m

clear all

clc

n=[6 8 10];

for i=1:length(n)

H{i}=hilb(n(i));

size_H{i}=size(H{i},1);

x_true{i}=ones(1,size_H{i});

b{i}=x_true{i}*H{i};

x_ini{i}=zeros(1,size_H{i});

%accuracy=5*(10^-6);

accuracy=0.01;

end

%Jacobi

disp('Jacobi');

for i=1:length(n)

    fprintf('The dimension is %d\n',n(i))

    [x{i},k{i}]=Jacobi(H{i},b{i},x_ini{i},accuracy,x_true{i});

end

%SOR

disp('SOR');

w=1;

disp('SOR w=1');

for i=1:length(n)

    fprintf('The dimension is %d\n',n(i))

    [q{i},e{i}]=SOR(H{i},b{i},x_ini{i},accuracy,x_true{i},w);

end

w=1.25;

disp('SOR w=1.25');

for i=1:length(n)

    fprintf('The dimension is %d\n',n(i))

    [z{i},x{i}]=SOR(H{i},b{i},x_ini{i},accuracy,x_true{i},w);

end

w=1.5;

disp('SOR w=1.5');

for i=1:length(n)

    fprintf('The dimension is %d\n',n(i))

    [v{i},b{i}]=SOR(H{i},b{i},x_ini{i},accuracy,x_true{i},w);

end

%[x,k]=Jacobi(A,b,x_ini,accuracy,x_true);

Jacobi.m

function [x,k]=Jacobi(A,b,x_ini,accuracy,x_true)

% input:  A—Left Matrix

%         b—Right Matrix

%         x_ini—initial value

%         accuracy—calculation precision between solution and true value

%         x_true—true valuer

% output: x—solution

%         k—iteration-1

n=size(A,1);

uint16 k;

k=1;

x{1}=x_ini;

    while(norm(x_true-x{k},Inf)>=accuracy)

        k=k+1;

        %disp(k-1);

        for i=1:1:n

            temp1=0;

            for j=1:1:i-1

              temp1=temp1+A(i,j)*x{k-1}(j);

            end

            temp2=0;

            for j=i+1:1:n

                temp2=temp2+A(i,j)*x{k-1}(j);

            end

            x{k}(i)=(b(i)-temp1-temp2)/A(i,i);

        end

        if (k==200000)

           break;

        end

    end

  fprintf('The iteration k=%d\n',k-1);

  disp('The solution is');

  disp(x{k});

  end

SOR.m

function [x,k]=SOR(A,b,x_ini,accuracy,x_true,w)

% input:  A—Left Matrix

%         b—Right Matrix

%         x_ini—initial value

%         accuracy—calculation precision between solution and true value

%         x_true—true value

%         w—relaxation factor

% output: x—solution

%         k—iteration-1

n=size(A,1);

uint16 k;

k=1;

x{1}=x_ini;

    while(norm(x_true-x{k},Inf)>accuracy)

        k=k+1;   

        %disp(k-1);

        for i=1:1:n

            temp1=0;

            if(i>1)

                for j=1:1:i-1

                    temp1=temp1+A(i,j)*x{k}(j);

                end

            end

            temp2=0;

            for j=i:1:n

                temp2=temp2+A(i,j)*x{k-1}(j);

            end

            x{k}(i)=x{k-1}(i)+w*(b(i)-temp1-temp2)/A(i,i);

        end

        if (k==200000)

           break;

        end    

    end

    

  fprintf('The iteration k=%d\n',k-1);

  disp('The solution is');

  disp(x{k});

  end

MATLAB运行结果为：

Jacobi

The dimension is 6

The iteration k=199999

The solution is

  Inf   Inf   Inf  Inf   Inf   Inf

The dimension is 8

The iteration k=199999

The solution is

  Inf   Inf   Inf  Inf   Inf   Inf  Inf   Inf

The dimension is 10

The iteration k=199999

The solution is

  Inf   Inf   Inf  Inf   Inf   Inf  Inf   Inf   Inf  Inf

SOR

SOR w=1

The dimension is 6

The iteration k=14508

The solution is

   0.9999    1.0016    0.9957   0.9994    1.0100    0.9933

The dimension is 8

The iteration k=42694

The solution is

   1.0001    0.9984    1.0076   0.9900    0.9971    1.0073   1.0068    0.9926

The dimension is 10

The iteration k=25508

The solution is

   1.0001    0.9980    1.0065   0.9985    0.9912    0.9970   1.0057    1.0091    1.0039   0.9900

SOR w=1.25

The dimension is 6

The iteration k=82840

The solution is

   1.0000    0.9995    1.0036   0.9908    1.0100    0.9961

The dimension is 8

The iteration k=50752

The solution is

   1.0001    0.9984    1.0073   0.9900    0.9983    1.0064   1.0060    0.9934

The dimension is 10

The iteration k=26267

The solution is

   1.0001    0.9983    1.0048   1.0012    0.9900    0.9971   1.0053    1.0087    1.0038   0.9906

SOR w=1.5

The dimension is 6

The iteration k=174587

The solution is

   1.0000    0.9994    1.0036   0.9908    1.0100    0.9961

The dimension is 8

The iteration k=52211

The solution is

   1.0001    0.9985    1.0071   0.9900    0.9996    1.0049   1.0059    0.9939

The dimension is 10

The iteration k=199999

The solution is

   1.0000    1.0007    0.9954   1.0113    0.9909    0.9986   0.9993    1.0049    1.0026   0.9962

运行结果分析：

          用雅克比迭代法进行线性方程系数矩阵为希尔伯特矩阵的求解，解是发散的，最终趋于无穷大。

         用SOR迭代法进行线性方程系数矩阵为希尔伯特矩阵的求解，解是收敛的，且收敛速度与矩阵的大小有关，但不是单调性的正相关或者负相关关系。