请用c++来编写能够求解矩阵的特征值的程序
时间: 2024-02-29 15:53:25 浏览: 21
好的,下面是一个简单的C++程序,可以用于求解矩阵的特征值:
```c++
#include <iostream>
#include <cmath>
using namespace std;
// 定义常量
const int MAXN = 100; // 矩阵最大维度
// 定义全局变量
int n; // 矩阵维度
double a[MAXN][MAXN]; // 矩阵
double eps = 1e-8; // 精度控制
// 定义函数
double max(double a, double b) {
return a > b ? a : b;
}
// 计算矩阵的特征值
void eigenvalue() {
double b[MAXN][MAXN];
double eps1, eps2, p, q, r, c, s, t, w, x, y, z;
int i, j, k, u, v;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
b[i][j] = a[i][j];
}
}
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
a[i][j] = 0.0;
}
a[i][i] = 1.0;
}
eps1 = eps * 10.0;
while (eps1 > eps) {
x = 0.0;
u = v = 0;
for (i = 0; i < n; i++) {
for (j = i+1; j < n; j++) {
if (fabs(b[i][j]) > x) {
x = fabs(b[i][j]);
u = i;
v = j;
}
}
}
if (x < eps) {
break;
}
p = -b[u][v];
q = (b[u][u] - b[v][v]) / 2.0;
w = sqrt(p * p + q * q);
r = fabs(w - p) < eps ? q : q / (w - p);
c = 1.0 / sqrt(r * r + 1.0);
s = r * c;
t = 1.0 - s;
y = b[u][u];
z = b[v][v];
b[u][u] = y * c * c + z * s * s - 2.0 * b[u][v] * c * s;
b[v][v] = y * s * s + z * c * c + 2.0 * b[u][v] * c * s;
b[u][v] = b[v][u] = 0.0;
for (k = 0; k < n; k++) {
if (k != u && k != v) {
y = b[u][k];
z = b[v][k];
b[u][k] = y * c + z * s;
b[v][k] = -y * s + z * c;
b[k][u] = b[u][k];
b[k][v] = b[v][k];
}
}
for (k = 0; k < n; k++) {
y = a[u][k];
z = a[v][k];
a[u][k] = y * c + z * s;
a[v][k] = -y * s + z * c;
}
eps1 = max(fabs(b[u][v]), max(fabs(b[u][u]), fabs(b[v][v])));
}
cout << "Eigenvalues:" << endl;
for (i = 0; i < n; i++) {
cout << b[i][i] << endl;
}
}
int main() {
int i, j;
cout << "Enter the dimension of matrix:";
cin >> n;
cout << "Enter the elements of matrix:" << endl;
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
cin >> a[i][j];
}
}
eigenvalue();
return 0;
}
```
该程序采用了Jacobi方法来求解矩阵的特征值,具体过程可以参考代码中的注释。程序中的精度控制是通过eps变量来控制的,可以根据实际需要进行调整。