[c++] 幂法求特征向量

幂法的原理可参考此篇论文：http://d.wanfangdata.com.cn/Periodical/hnnydxxb2001Z1023

本文求解的是 3 阶矩阵最大特征值及其特征向量

下面是其 C++ 实现代码：

#include <iostream>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<iomanip>
using namespace std;

double A[3][3];
double Y[3]={1,1,1};
double X[3]={0,0,0};
int row=0; int col=0;
double max1=0;

void open_file()
{
    FILE *fp;
    fp = fopen("array.txt", "r");      //3*3矩阵由外部读入
    if(fp==NULL)
        cout<<"File opened failed!"<<endl;

    for(row=0;row<3;row++)
    {
        for(col= 0; col < 3; col ++)
            fscanf(fp, "%lf,",&A[row][col]);
        if(feof(fp)) break;
    }
    fclose(fp);
}

void mult()
{
    X[0]=0;X[1]=0;X[2]=0;
    for(row=0;row<3;row++)
    {
        for(col=0;col<3;col++)
            X[row] +=A[row][col]*Y[col];
    }
}

void to1()
{
    long double tmp=fabs(X[0]);
    for(int i=1;i<3;i++)
    {
        if(fabs(X[i])>tmp)
            tmp=fabs(X[i]);
    }
    for(int i=0;i<3;i++)
    {
        Y[i]=X[i]/tmp;
    }
    max1=tmp;
}

int main()
{
    cout <<setiosflags(ios::fixed);
    open_file();
    double ago=max1+100.0;
    double feature_vector[3];
    int k=1;
    while(fabs(max1-ago)>0.000001)
    {
          ago=max1;
          for(int j=0;j<3;j++)
          {
              feature_vector[j]=Y[j];
          }
          mult();
          to1();
          cout<<"k= "<<k<<"  ";
          for(int i=0;i<3;i++)
              cout<<X[i]<<" ";
          cout<<endl;
          k++;
    }
    cout<<endl<<"totally run "<<k-1<<" times"<<endl;
    cout<<endl<<"the matrix eigenvalue is "<<max1<<endl;
    cout<<endl<<"the feature vector is "<<"["<<feature_vector[0]<<" , "<<feature_vector[1]<<" , "<<feature_vector[2]<<"]"<<endl;

}

　　部分参数可修改用于扩展