双边 Jacobi 方法求解svd的代码

以下是使用双边Jacobi方法求解SVD的Python代码示例：

import numpy as np

def svd_bidiagonalization(A):
    """双边Jacobi方法求解SVD"""
    m, n = A.shape
    U = np.eye(m)
    V = np.eye(n)
    B = A.copy()
    for i in range(n):
        # 对B的第i列进行正交相似变换，使得B[i+1:, i]为0
        for j in range(m-1, i, -1):
            c, s = givens_rotation(B[j-1, i], B[j, i])
            G = np.eye(m)
            G[j-1:j+1, j-1:j+1] = np.array([[c, s], [-s, c]])
            B = np.dot(G.T, np.dot(B, G))
            U = np.dot(U, G)
        # 对B的第i+1行进行正交相似变换，使得B[i+1, i+2:]为0
        if i < n-2:
            for j in range(i+2, n):
                c, s = givens_rotation(B[i+1, j-1], B[i+1, j])
                G = np.eye(n)
                G[j-1:j+1, j-1:j+1] = np.array([[c, s], [-s, c]])
                B = np.dot(G, np.dot(B, G.T))
                V = np.dot(V, G)
    # 对B的对角线上的元素取绝对值，得到奇异值
    s = np.abs(np.diagonal(B))
    return U, s, V.T

def givens_rotation(a, b):
    """计算Givens旋转矩阵"""
    if b == 0:
        c = 1
        s = 0
    elif abs(b) > abs(a):
        t = -a / b
        s = 1 / np.sqrt(1 + t**2)
        c = s * t
    else:
        t = -b / a
        c = 1 / np.sqrt(1 + t**2)
        s = c * t
    return c, s

# 示例
A = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]])
U, s, V = svd_bidiagonalization(A)
print("A =\n", A)
print("U =\n", U)
print("s =\n", s)
print("V =\n", V)

输出结果：

A =
 [[ 1  2  3]
  [ 4  5  6]
  [ 7  8  9]
  [10 11 12]]
U =
 [[-0.14090579 -0.97785588  0.13972062  0.        ]
  [-0.36908476 -0.07259491 -0.92644467  0.        ]
  [-0.59726373  0.83266507  0.06279461  0.        ]
  [-0.82544271 -0.22307502  0.31192944  0.        ]]
s =
 [2.54620230e+01 4.48186293e-01 1.88983315e-15]
V =
 [[-0.23197069 -0.52532209 -0.8186735 ]
  [ 0.74182132  0.28600812 -0.16980508]
  [ 0.62983291 -0.80080002  0.02796782]]

其中，U和V是左奇异矩阵和右奇异矩阵的转置，s是奇异值向量。