Python matrix LU decomposition

scipy.linalgProvides a series of matrix decomposition functions, the most basic of which must be LU decomposition.

L and U

LU decomposition, that is, the matrix $A$ decomposed into$LU$ , where$L$ is the lower triangular matrix,$U$ is an upper triangular matrix. For these two matrices,scipy.linalgprovided inkkthtril, triucan be$All\; the\; elements\; below\; or\; above\; the\; k$ diagonals are set to zero, so that the L matrix or U matrix can be obtained.

import numpy as np
import scipy.linalg as sl
x = np.random.rand(4,4)
print(sl.tril(x,-1))	# 返回见[1]

$$\left[\begin{array}{cccc}0.& 0.& 0.& 0.\\ 0.62594216& 0.& 0.& 0.\\ 0.16043717& 0.5820587& 0.& 0.\\ 0.24560828& 0.76599572& 0.1922379& 0.\end{array}\right]$$

print(sl.triu(x,2))		#

$$\left[\begin{array}{cccc}0.& 0.& 0.91943758& 0.2531733\\ 0.& 0.& 0.& 0.76514452\\ 0.& 0.& 0.& 0.\\ 0.& 0.& 0.& 0.\end{array}\right]$$

lu decomposition

LUDecomposition is the first matrix decomposition method to appear in almost any book on matrix algorithms. In scipy.linalg, functions such as are provided lu, lu_factor, lu_solvefor LU decomposition, respectively, and for solving Ax=bsimilar problems by LU decomposition.

LUThe decomposition example is as follows

import numpy as np
from scipy.linalg import lu
A = np.random.rand(4,4)
p, l, u = lu(A)
np.allclose(A - p @ l @ u, np.zeros((4, 4)))
# True

luIn addition to the parameters aused for finiteness checks and aoverridability, the function also has a permute_ldefault value of False, when it Trueis , two parameters will be returned pl, u. Among them pl==p@l.

lu_factor(a)Return the result of the decomposition in another form LU. There are two return values, namely , lu, pivL + U − I L+UILU$L+U-I$ ,$I$ is the identity matrix;pivrepresents the matrixPPThe position of the non-zero element in$P.$

lu_solve

lu_solveis according to $LU$ decomposition solves$PLUx-The\; problem\; of\; b$ , whose input is(lu, piv)andb, is tested as follows

>>> b = np.array([1,2,3,4])
>>> lu, pix = sl.lu_factor(A)
>>> x = lu_solve((lu, pix), b)
>>> A@x-b
array([ 2.22044605e-16, -4.44089210e-16,  0.00000000e+00,  0.00000000e+00])

Among them Ais the random matrix generated above, the value is

$$\left[\begin{array}{cccc}0.2495995& 0.59179571& 0.34236803& 0.78559552\\ 0.39427709& 0.36015762& 0.23789732& 0.09223244\\ 0.79701282& 0.40291763& 0.93215531& 0.10486747\\ 0.46812908& 0.58426202& 0.16560106& 0.96889267\end{array}\right]$$