LU Factorization with Partial Pivoting

Copyright (C) 2021 Andreas Kloeckner

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In [1]:
import numpy as np

import numpy.linalg as la



np.set_printoptions(linewidth=150, suppress=True, precision=3)
In [99]:
n = 5



np.random.seed(235)

A = np.random.randn(n, n)

A[0,0] = 0

A
Out[99]:
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:

In [100]:
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:

In [101]:
i = 0

I = np.eye(n)

P = I.copy()

Lsub = np.zeros_like(A)

U = np.zeros_like(A)

remaining = A.copy()



remaining
Out[101]:
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:

In [102]:
#clear

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

ipiv
Out[102]:
3

Swap the rows in remaining, record in P:

In [103]:
#clear

swap_mat = row_swap_mat(i, ipiv)

P = swap_mat @ P

remaining = swap_mat @ remaining

remaining
Out[103]:
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:

In [104]:
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:

In [114]:
#clear

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:

In [115]:
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:

In [116]:
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

In [117]:
P
Out[117]:
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.]])
In [118]:
I+Lsub
Out[118]:
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.   ]])
In [119]:
U
Out[119]:
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?
In [ ]: