用幂法求矩阵的按模最大特征值c语言实现
时间: 2023-10-24 10:08:01 浏览: 69
以下是使用幂法求矩阵的按模最大特征值的C语言实现:
```c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define N 10
#define EPS 1e-6
void matrix_multiply(double A[][N], double x[], double y[]) {
int i, j;
for (i = 0; i < N; i++) {
y[i] = 0;
for (j = 0; j < N; j++) {
y[i] += A[i][j] * x[j];
}
}
}
double vector_norm(double x[]) {
int i;
double sum = 0;
for (i = 0; i < N; i++) {
sum += x[i] * x[i];
}
return sqrt(sum);
}
int main() {
double A[N][N] = {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10},
{2, 1, 2, 3, 4, 5, 6, 7, 8, 9},
{3, 2, 1, 2, 3, 4, 5, 6, 7, 8},
{4, 3, 2, 1, 2, 3, 4, 5, 6, 7},
{5, 4, 3, 2, 1, 2, 3, 4, 5, 6},
{6, 5, 4, 3, 2, 1, 2, 3, 4, 5},
{7, 6, 5, 4, 3, 2, 1, 2, 3, 4},
{8, 7, 6, 5, 4, 3, 2, 1, 2, 3},
{9, 8, 7, 6, 5, 4, 3, 2, 1, 2},
{10, 9, 8, 7, 6, 5, 4, 3, 2, 1}};
double x[N] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
double y[N];
double lambda = 0;
double lambda_old = 0;
int i, count = 0;
do {
lambda_old = lambda;
matrix_multiply(A, x, y);
lambda = vector_norm(y);
for (i = 0; i < N; i++) {
x[i] = y[i] / lambda;
}
count++;
} while (fabs(lambda - lambda_old) > EPS);
printf("The maximum eigenvalue is: %lf\n", lambda);
printf("The corresponding eigenvector is:\n");
for (i = 0; i < N; i++) {
printf("%lf ", x[i]);
}
printf("\n");
printf("Total iterations: %d\n", count);
return 0;
}
```
其中,`matrix_multiply`函数用于计算矩阵与向量的乘积,`vector_norm`函数用于计算向量的模长。在主函数中,我们先定义了一个10x10的矩阵`A`和一个初始向量`x`,然后使用幂法求出矩阵的按模最大特征值和对应的特征向量,并输出结果。在求解过程中,我们使用了一个计数器`count`来记录迭代次数,以便评估算法的收敛速度和效率。