线性代数-Python-05：矩阵的逆+LU分解

1 矩阵的逆

1.1 求解矩阵的逆

def inv(A):

    if A.row_num() != A.col_num():
        return None

    n = A.row_num()
    
    """矩阵A+单位矩阵"""
    ls = LinearSystem(A, Matrix.identity(n))
    
    """对线性系统进行高斯消元，如果没有解，返回none"""
    if not ls.gauss_jordan_elimination():
        return None
        
	"""高斯消元有解的话，把线性系统的右部分取出，重新构成矩阵，得到矩阵的逆"""
    invA = [[row[i] for i in range(n, 2*n)] for row in ls.Ab]
    
    return Matrix(invA)

2 初等矩阵

2.1 初等矩阵和可逆性

3 矩阵的LU分解

3.1 LU分解的实现

from .Matrix import Matrix
from .Vector import Vector
from ._globals import is_zero


def lu(matrix):

    assert matrix.row_num() == matrix.col_num(), "matrix must be a square matrix"

    n = matrix.row_num() 
    
	"""A是原矩阵的副本"""
    A = [matrix.row_vector(i) for i in range(n)]
    
	"""初始化L，使对角线元素为1"""
    L = [[1.0 if i == j else 0.0 for i in range(n)] for j in range(n)]

    for i in range(n):
        """看A[i][i]位置是否可以是主元"""
        if is_zero(A[i][i]):
            return None, None
        else: """将主元以下的j位置变为0"""
            for j in range(i + 1, n):
                p = A[j][i] / A[i][i] """求加减的系数"""
                A[j] = A[j] - p * A[i] """将第j行的位置经过加减运算变成0"""
                L[j][i] = p """将L矩阵相应位置变成相应变换的值"""

    return Matrix(L), Matrix([A[i].underlying_list() for i in range(n)])