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问题描述

scipy/numpy/...的宇宙中是否存在用于矩阵高斯消除的标准方法?

有人通过Google找到了许多代码片段,但我希望尽可能使用受信任的"模块.

我终于发现,可以使用 LU分解完成此操作.这里的 U 矩阵表示线性系统的简化形式.

from numpy import array
from scipy.linalg import lu

a = array([[2.,4.,4.,4.],[1.,2.,3.,3.],[1.,2.,2.,2.],[1.,4.,3.,4.]])

pl, u = lu(a, permute_l=True)

然后u读取

array([[ 2.,  4.,  4.,  4.],
       [ 0.,  2.,  1.,  2.],
       [ 0.,  0.,  1.,  1.],
       [ 0.,  0.,  0.,  0.]])

取决于系统的可溶性,该基质具有上部三角形或梯形结构.在上述情况下,由于矩阵仅具有等级3,所以会出现零线.

Is there somewhere in the cosmos of scipy/numpy/... a standard method for Gauss-elimination of a matrix?

One finds many snippets via google, but I would prefer to use "trusted" modules if possible.

解决方案

I finally found, that it can be done using LU decomposition. Here the U matrix represents the reduced form of the linear system.

from numpy import array
from scipy.linalg import lu

a = array([[2.,4.,4.,4.],[1.,2.,3.,3.],[1.,2.,2.,2.],[1.,4.,3.,4.]])

pl, u = lu(a, permute_l=True)

Then u reads

array([[ 2.,  4.,  4.,  4.],
       [ 0.,  2.,  1.,  2.],
       [ 0.,  0.,  1.,  1.],
       [ 0.,  0.,  0.,  0.]])

Depending on the solvability of the system this matrix has an upper triangular or trapezoidal structure. In the above case a line of zeros arises, as the matrix has only rank 3.

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10-18 17:39