帮我用Python写一个svd算法，不用numpy库

好的，我可以帮你编写一个Python的svd算法。以下是实现代码：

def svd(A):
    import math
    
    # 使用Jacobi方法进行奇异值分解
    
    # 定义Jacobi旋转操作
    def jacobi(A, p, q):
        if A[p][q] == 0:
            c = s = 1
        else:
            t = (A[q][q] - A[p][p]) / (2.0 * A[p][q])
            if t >= 0:
                t = 1 / (t + math.sqrt(1 + t * t))
            else:
                t = -1 / (-t + math.sqrt(1 + t * t))
            c = 1 / math.sqrt(1 + t * t)
            s = t * c
        
        B = A[:]
        for i in range(len(A)):
            if i != p and i != q:
                B[i][p] = c * A[i][p] - s * A[i][q]
                B[i][q] = s * A[i][p] + c * A[i][q]
                B[p][i] = B[i][p]
                B[q][i] = B[i][q]
        B[p][p] = c * c * A[p][p] - 2 * s * c * A[p][q] + s * s * A[q][q]
        B[q][q] = s * s * A[p][p] + 2 * s * c * A[p][q] + c * c * A[q][q]
        B[p][q] = B[q][p] = 0.0
        
        return B
    
    # 判断是否已收敛
    def converged(A, eps=1e-8):
        for i in range(len(A)):
            for j in range(len(A[0])):
                if i != j and abs(A[i][j]) > eps:
                    return False
        return True
    
    # 初始化A、U、V
    n = len(A)
    m = len(A[0])
    U = [[0.0] * n for i in range(n)]
    for i in range(n):
        U[i][i] = 1.0
    V = [[0.0] * m for i in range(m)]
    for i in range(m):
        V[i][i] = 1.0
        
    # 迭代Jacobi方法
    while not converged(A):
        for p in range(n):
            for q in range(p+1, n):
                B = jacobi(A, p, q)
                U = jacobi(U, p, q)
                V = jacobi(V, p, q)
                A = B
    
    # 计算奇异值和左右奇异向量
    S = [abs(A[i][i]) for i in range(min(n, m))]
    V = [[V[i][j] for i in range(m)] for j in range(m)]
    U = [[U[i][j] / S[j] for j in range(min(n, m))] for i in range(n)]
    
    # 返回结果
    return (U, S, V)
    
# 测试
A = [[1,2,3],[4,5,6],[7,8,9]]
U, S, V = svd(A)
print('U =', U)
print('S =', S)
print('V =', V)

这是一个基于Jacobi方法的SVD算法实现。由于您要求不使用numpy库，所以代码可能相对比较长，但应该是可行的。请尝试直接将代码复制到Python解释器中运行，看看是否正确。