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 swaps 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]:
ipiv = i + np.argmax(np.abs(remaining[i:, i]))
ipiv
Out[102]:
3

Swap the rows in remaining, record in P:

In [103]:
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]:
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?

Realize that P is a permutation of the "trailing" rows. Applying to Lsub avoids disturbing the diagonal.

Why is it mathematically correct to only apply the permutations to Lsub and not all of L?

Realize that P is a permutation of the "trailing" rows. If we applied to Lsub + I instead, then the pivots would need to go into the row to which the 1 from the diagonal was permuted. Instead, it goes into U[i,i], and leaving the diagonal of L undisturbed reflects that.
In [ ]: