如何理解 scipy.linalg.lu_factor 的主元矩阵? [英] How to understand the pivot matrix of scipy.linalg.lu_factor?

问题描述

如何手动重建由 lu_factor?(A = PLU)

How can I manually reconstruct a matrix A that was factorized by lu_factor? (A = PLU)

由于矩阵P的设置，我目前的尝试都失败了.这是我目前所拥有的:

My current attempts all failed due to the setup of matrix P. Here is what I have so far:

A = np.random.rand(3,3)
lu, piv = lu_factor(A)

U = np.triu(lu)
L = np.tril(lu, -1)
L[np.diag_indices_from(L)] = 1.0

我正在寻找使这条线打印True的矩阵P:

I am looking for the matrix P that makes this line print True:

print np.allclose(A, np.dot(P, np.dot(L, U)))

感谢任何提示/链接/建议！

Any hint/link/suggestion is appreciated!

推荐答案

置换向量需要依次解释.如果 piv=[1,2,2] 则需要按顺序执行以下操作(使用从零开始的索引):

The permutation vector needs to be interpreted in sequence. If piv=[1,2,2] then the following needs to be done in sequence (with zero-based indexing):

1. 第 0 行随第 1 行变化
2. 新的第 1 行随着第 2 行的变化而变化
3. 新的第 2 行保持不变.

在代码中，这可以解决问题:

In code this would do the trick:

P = np.eye(3)
for i, p in enumerate(piv):
    Q = np.eye(3,3)
    q = Q[i,:].copy()
    Q[i,:] = Q[p,:]
    Q[p,:] = q
    P = np.dot(P, Q)

对于 piv=[1,2,2] P 是

[[ 0.  0.  1.]
 [ 1.  0.  0.]
 [ 0.  1.  0.]]

这可能不是一种计算 P 的非常快的方法，但它可以解决问题并回答问题.

This is probably not a very fast way of computing P but it does the trick and answers the question.

这篇关于如何理解 scipy.linalg.lu_factor 的主元矩阵?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！