雅可比矩阵的特征值求解的Python代码

雅可比矩阵的特征值求解可以使用中的numpy库来实现。下面一个简单的示例代码：

import numpy as np

def jacobi_eigenvalue(A, epsilon=e-8, max_iterations=100):
    = A.shape[0]
    eigenvalues = np.zeros(n)
    eigenvectors = np.eye(n)

    for _ in range(max_iterations):
        # Find the maximum off-diagonal element
        max_off_diag = 0
        p = 0
        q = 0
        for i in range(n):
            for j in range(i+1, n):
                if abs(A[i, j]) > max_off_diag:
                    max_off_diag = abs(A[i, j])
                    p = i
                    q = j

        # Check convergence
        if max_off_diag < epsilon:
            break

        # Compute the rotation angle
        theta = 0.5 * np.arctan2(2 * A[p, q], A[q, q] - A[p, p])

        # Compute the rotation matrix
        c = np.cos(theta)
        s = np.sin(theta)
        J = np.eye(n)
        J[p, p] = c
        J[q, q] = c
        J[p, q] = -s
        J[q, p] = s

        # Update the matrix A and eigenvectors
        A = np.dot(np.dot(J.T, A), J)
        eigenvectors = np.dot(eigenvectors, J)

    # Extract the eigenvalues from the diagonal of A
    eigenvalues = np.diag(A)

    return eigenvalues, eigenvectors

# Example usage
A = np.array([[2, -1, 0], [-1, 2, -1], [0, -1, 2]])
eigenvalues, eigenvectors = jacobi_eigenvalue(A)
print("Eigenvalues:", eigenvalues)
print("Eigenvectors:", eigenvectors)

这段代码使用雅可比迭代方法来求解雅可比矩阵的特征值和特征向量。函数`jacobi_eigenvalue`接受一个方阵A作为输入，并返回特征值和特征向量。在示例中，我们定义了一个3x3的雅可比矩阵A，并打印出了求解得到的特征值和特征向量。