那个scipy.linalg.lu（）为什么不返回与scipy.sparse.linalg.splu（）相同的L矩阵？

Question

I have the following piece of code where I compute the L matrix of a given square matrix using the command scipy.linalg.lu() and then I do the same thing again except then applied to the sparse form of the given matrix using scipy.sparse.linalg.splu(). This is the code:

import numpy as np
from scipy.sparse.linalg import splu
from scipy.sparse import csc_matrix
import scipy.linalg
A1 = csc_matrix([[1., 0, 0.], [5., 0, 2], [0, -1., 0]])
A2 = np.array([[1., 0, 0.], [5., 0, 2], [0, -1., 0]])
B = splu(A1)
P,L,U = scipy.linalg.lu(A2)
print(L);print(csr_matrix.todense(B.L))

Which returns the following:

[[ 1.   0.   0. ]
 [ 0.   1.   0. ]
 [ 0.2 -0.   1. ]]
[[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]

As we can see, these matrices are not the same. Am I misunderstanding what both commands do or is something else going wrong? Any help is appreciated. Thanks!

Answer 1

I think the answer here is that the "SuperLU" decomposition of a sparse matrix requires permutations of both the rows and columns (see the docs ):

Pr * A * Pc = L * U

These are provided by the mapping of indices in the perm_r and perm_c attributes. So,

Pr = csc_matrix((3,3))
Pr[B.perm_r, np.arange(3)] = 1
Pc = csc_matrix((3,3))
Pc[np.arange(3), B.perm_c] = 1

(Pr.T @ B.U @ B.L @ Pc.T).A

gives, as required:

array([[ 1.,  0.,  0.],
       [ 5.,  0.,  2.],
       [ 0., -1.,  0.]])

The same as the non-sparse result which requires only permutation of the L-matrix, P @ L @ U .