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I have a rather naive question on LU factorization which probably should be easy to answer. Say I have a matrix with entries only from $\{0,1\}$. When can we expect to get an LU factorization of such a matrix(whenever it exists) with entries $(a)$ from integers? $(b)$ from $\{-1,0,+1\}$?

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  • $\begingroup$ In the current form, I don't think the question is on-topic, i.e. about theoretical computer science. $\endgroup$ – Kristoffer Arnsfelt Hansen Apr 8 '11 at 16:11
  • $\begingroup$ Actually it is related to a problem for getting capacity of 1-d constrained channels! $\endgroup$ – Turbo Apr 8 '11 at 18:45
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Here's a 5-by-5 matrix with a non-integer LU-decomposition.

A

A =

 1     1     0     1     0
 1     1     1     0     0
 1     0     1     1     0
 1     0     0     1     1
 0     1     0     1     1

[L U] = lu(A)

L =

1.0000         0         0         0         0
1.0000         0    1.0000         0         0
1.0000    1.0000         0         0         0
1.0000    1.0000   -1.0000   -0.5000    1.0000
     0   -1.0000    1.0000    1.0000         0

U =

1.0000    1.0000         0    1.0000         0
     0   -1.0000    1.0000         0         0
     0         0    1.0000   -1.0000         0
     0         0         0    2.0000    1.0000
     0         0         0         0    1.5000

Your questions seems related to unimodular and totally unimodular matrices.

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No. Consider the 4-by-4 matrix below.

 0     1     1     1
 1     0     1     1
 0     0     0     1
 1     1     0     1
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  • $\begingroup$ Indeed, as can be verified here. But is there only one LU decomposition of this matrix (apologies if this is a silly question)? $\endgroup$ – Anthony Labarre Apr 7 '11 at 14:30
  • $\begingroup$ comment: Hi this actually produces integral LU matrices. LU Decomposition: L: 1.000 0.000 0.000 0.000 0.000 1.000 0.000 0.000 1.000 1.000 1.000 0.000 0.000 0.000 0.000 1.000 U: 1.000 0.000 1.000 1.000 0.000 1.000 1.000 1.000 0.000 0.000 -2.000 -1.000 0.000 0.000 0.000 1.000 Permutation matrix: 0.000 1.000 0.000 0.000 1.000 0.000 0.000 0.000 0.000 0.000 0.000 1.000 0.000 0.000 1.000 0.000 $\endgroup$ – Turbo Apr 7 '11 at 16:49

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