LU Factorization

Copyright (C) 2021 Andreas Kloeckner

In part based on material by Edgar Solomonik

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In [54]:
import numpy as np

import numpy.linalg as la



np.set_printoptions(linewidth=150, suppress=True, precision=3)
In [55]:
A = np.random.randn(4, 4)

A
Out[55]:
array([[ 1.675, -1.63 ,  1.236, -0.329],
       [-0.018,  0.006, -1.171, -0.222],
       [-0.607, -0.098,  0.194, -1.705],
       [ 0.218,  0.441, -0.713, -0.787]])

Initialize L and U:

In [56]:
L = np.eye(len(A))

U = np.zeros_like(A)

Recall the "recipe" for LU factorization:

$$\let\B=\boldsymbol \begin{array}{cc} & \left[\begin{array}{cc} u_{00} & \B{u}_{01}^T\\ & U_{11} \end{array}\right]\\ \left[\begin{array}{cc} 1 & \\ \B{l}_{10} & L_{11} \end{array}\right] & \left[\begin{array}{cc} a_{00} & \B{a}_{01}\\ \B{a}_{10} & A_{11} \end{array}\right] \end{array}$$

Find $u_{00}$ and $u_{01}$. Check A - L@U.

In [57]:
#clear



U[0] = A[0]

A - L@U
Out[57]:
array([[ 0.   ,  0.   ,  0.   ,  0.   ],
       [-0.018,  0.006, -1.171, -0.222],
       [-0.607, -0.098,  0.194, -1.705],
       [ 0.218,  0.441, -0.713, -0.787]])

Find $l_{10}$. Check A - L@U.

In [58]:
#clear



L[1:,0] = A[1:,0]/U[0,0]

A - L@U
Out[58]:
array([[ 0.   ,  0.   ,  0.   ,  0.   ],
       [ 0.   , -0.011, -1.158, -0.226],
       [ 0.   , -0.689,  0.642, -1.824],
       [ 0.   ,  0.653, -0.874, -0.745]])

Recall $A_{22} =\B{l}_{21} \B{u}_{12}^T + L_{22} U_{22}$. Write the next step generic in terms of i.

After the step, print A-L@U and remaining.

In [59]:
i = 1

remaining = A - L@U
In [62]:
#clear

U[i, i:] = remaining[i, i:]

L[i+1:,i] = remaining[i+1:,i]/U[i,i]

remaining[i+1:, i+1:] -= np.outer(L[i+1:,i], U[i, i+1:])



i = i + 1



print(remaining)

print(A-L@U)

[[  0.      0.      0.      0.   ]
[  0.     -0.011  -1.158  -0.226]
[  0.     -0.689  72.62   12.212]
[  0.      0.653 -69.135  -2.43 ]]
[[ 0.  0.  0.  0.]
[ 0.  0.  0.  0.]
[ 0. -0.  0. -0.]
[ 0.  0.  0. -0.]]
In [ ]: