请用Python实现svd算法，不用numpy库

好的，我会为您实现svd算法。

以下是代码示例：

def svd(A):
    # 首先计算A^T * A
    ATA = [[0]*len(A) for i in range(len(A))]
    for i in range(len(A)):
        for j in range(len(A)):
            for k in range(len(A[0])):
                ATA[i][j] += A[k][i] * A[k][j]
    
    # 计算ATA的特征值和特征向量
    e_values, e_vectors = power_iteration(ATA)
    
    # 构造特征值和特征向量矩阵
    e_values_matrix = [[0]*len(A) for i in range(len(A))]
    for i in range(len(A)):
        e_values_matrix[i][i] = e_values[i]
    e_vectors_matrix = [[0]*len(A) for i in range(len(A))]
    for i in range(len(A)):
        for j in range(len(A)):
            e_vectors_matrix[j][i] = e_vectors[i][j]
    
    # 计算A * 特征向量矩阵
    U = [[0]*len(A) for i in range(len(A[0]))]
    for i in range(len(A)):
        for j in range(len(A[0])):
            for k in range(len(A)):
                U[j][i] += A[k][i] * e_vectors_matrix[k][j] / math.sqrt(e_values[i])
    
    # 计算A^T * U
    V = [[0]*len(A) for i in range(len(A))]
    for i in range(len(A)):
        for j in range(len(A)):
            for k in range(len(A[0])):
                V[i][j] += A[k][i] * U[k][j] / math.sqrt(e_values[j])
    
    return U, e_values_matrix, V

def power_iteration(A):
    n = len(A)
    v = [1/math.sqrt(n)] * n
    for i in range(100):
        w = [0] * n
        for j in range(n):
            for k in range(n):
                w[j] += A[j][k] * v[k]
        v = w / math.sqrt(sum([x**2 for x in w]))
    Av = [0] * n
    for i in range(n):
        for j in range(n):
            Av[i] += A[i][j] * v[j]
    lambda_ = math.sqrt(sum([x**2 for x in Av]))
    return lambda_, [x/lambda_ for x in Av]

这是一个比较简单的svd实现，采用幂法来求矩阵的特征值与特征向量。需要注意的是，这个实现没有使用numpy库。