LU Factorization with Partial Pivoting

Copyright (C) 2021 Andreas Kloeckner

MIT License

Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated documentation files (the "Software"), to deal in the Software without restriction, including without limitation the rights to use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom the Software is furnished to do so, subject to the following conditions: The above copyright notice and this permission notice shall be included in all copies or substantial portions of the Software. THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.

import numpy as np
import numpy.linalg as la

np.set_printoptions(linewidth=150, suppress=True, precision=3)

n = 5

np.random.seed(235)
A = np.random.randn(n, n)
A[0,0] = 0
A

array([[ 0.   ,  0.769,  1.436, -1.441, -0.165],
       [-0.407, -1.935, -0.037,  0.178,  0.371],
       [ 0.271, -0.743, -1.049, -1.63 , -1.277],
       [-0.476, -1.349,  1.129, -1.424, -1.897],
       [-0.01 , -0.978, -1.146,  2.017,  1.023]])

Permutation Matrices

This function returns a matrix that swas rows i and j:

def row_swap_mat(i, j):
    P = np.eye(n)
    P[i] = 0
    P[j] = 0
    P[i, j] = 1
    P[j, i] = 1
    return P

Pivoted LU: Initialization

We're trying to obtain $PA=LU$. Initialize:

i = 0
I = np.eye(n)
P = I.copy()
Lsub = np.zeros_like(A)
U = np.zeros_like(A)
remaining = A.copy()

remaining

array([[ 0.   ,  0.769,  1.436, -1.441, -0.165],
       [-0.407, -1.935, -0.037,  0.178,  0.371],
       [ 0.271, -0.743, -1.049, -1.63 , -1.277],
       [-0.476, -1.349,  1.129, -1.424, -1.897],
       [-0.01 , -0.978, -1.146,  2.017,  1.023]])

First column

First, find the pivot as ipiv:

ipiv = i + np.argmax(np.abs(remaining[i:, i]))
ipiv

3

Swap the rows in remaining, record in P:

swap_mat = row_swap_mat(i, ipiv)
P = swap_mat @ P
remaining = swap_mat @ remaining
remaining

array([[-0.476, -1.349,  1.129, -1.424, -1.897],
       [-0.407, -1.935, -0.037,  0.178,  0.371],
       [ 0.271, -0.743, -1.049, -1.63 , -1.277],
       [ 0.   ,  0.769,  1.436, -1.441, -0.165],
       [-0.01 , -0.978, -1.146,  2.017,  1.023]])

Now carry out a step of LU, as above:

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

i = i + 1

print(P@A-(Lsub+I)@U)

[[ 0.     0.     0.     0.     0.   ]
 [ 0.    -0.78  -1.003  1.397  1.995]
 [ 0.    -1.513 -0.405 -2.443 -2.359]
 [ 0.     0.769  1.436 -1.441 -0.165]
 [ 0.    -0.949 -1.17   2.048  1.064]]

Subsequent columns

Find the pivot and perform the swaps so that you still have a valid $PA=LU$ factorization:

ipiv = i + np.argmax(np.abs(remaining[i:, i]))
swap_mat = row_swap_mat(i, ipiv)

P = swap_mat @ P
Lsub = swap_mat @ Lsub
remaining = swap_mat @ remaining

Here are some checks to make sure you're on the right track:

print("Should maintain the same 'zero fringe' as the previous step:")
print(P @ A - (Lsub+I) @ U)

print("Should be zero:")
print(remaining[i:, i:] - (P @ A - (Lsub+I) @ U)[i:, i:])

Should maintain the same 'zero fringe' as the previous step:
[[ 0.     0.     0.     0.     0.   ]
 [ 0.    -0.     0.     0.     0.   ]
 [ 0.     0.     0.    -0.     0.   ]
 [ 0.    -0.    -0.     0.     0.   ]
 [ 0.     0.     0.     0.     1.439]]
Should be zero:
[[0.]]

Carry out a step of LU, as always:

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

i = i + 1

print(P@A-(Lsub+I)@U)

[[ 0.  0.  0.  0.  0.]
 [ 0. -0.  0.  0.  0.]
 [ 0.  0.  0. -0.  0.]
 [ 0. -0. -0.  0.  0.]
 [ 0.  0.  0.  0.  0.]]

Inspect the result

P

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

I+Lsub

array([[ 1.   ,  0.   ,  0.   ,  0.   ,  0.   ],
       [-0.571,  1.   ,  0.   ,  0.   ,  0.   ],
       [ 0.   , -0.508,  1.   ,  0.   ,  0.   ],
       [ 0.021,  0.627, -0.745,  1.   ,  0.   ],
       [ 0.856,  0.515, -0.646,  0.583,  1.   ]])

U

array([[-0.476, -1.349,  1.129, -1.424, -1.897],
       [ 0.   , -1.513, -0.405, -2.443, -2.359],
       [ 0.   ,  0.   ,  1.23 , -2.682, -1.363],
       [ 0.   ,  0.   ,  0.   ,  1.583,  1.529],
       [ 0.   ,  0.   ,  0.   ,  0.   ,  1.439]])

Questions

• Why do we switch to maintaining Lsub instead of all of L?