Talk:Unimodular matrix

[Untitled]
Examples of totally unimodular matrices?

Charles Matthews 10:14, 13 Apr 2004 (UTC)

The definition of unimodular matrix is really with +1 or -1, and so I have deleted the special linear group link.

Simon Lacoste-Julien 01:02, 25 Apr 2004 (PST)

A number of definitions circulates on the web of totally unimodular matrices. The important difference is whether or not the matrix should be square. Some authors even claim that a totally unimodular matrix need to be unimodular itself (which would exclude the zero matrix). I think that now the correct definition is used.

Basten 11:50, 19 February 2007 (UTC)

kronecker
Does anybody know if the Kronecker product of two totally unimodular matrices is again totally unimodular?

integer matrices
I strongly think that you need an additional restriction of unimodular matrices being integer matrices.

Some sources ("Loop Parallelization", Utpal Banerjee) require the inverted matrix to be an integer matrix, even though I think this is unecessary.

--M1ck1 12:10, 1 December 2006 (UTC)

Ok, I also found 2 standard books on the subject that also agree on requiring integer matrices:
 * "Integer and Combinatorial Optimization", Geroge L. Nemhauser, Laurence A. Wolsey, 1988, page 189
 * "Theory of Linear and Integer Programming", Alexander Schrijver, 1989, page 49

--M1ck1 14:52, 1 December 2006 (UTC)

After some very casual browsing of sources, I don't see a consensus on whether the matrices have to be integer or not. Is there a definitive source, or is including "integer" in the definition conventional for some and not for others? More importantly, is the lack of consensus worth noting in the actual article rather than calling them integer and leaving it at that?

--Error9312 (talk) 20:19, 3 March 2009 (UTC)

Other concepts of "unimodular matrices"
I find a variety of definitions for “Unimodular” on the web as of 2007 Mar 15.

This Wikipedia article presumes a matrix over Z, the integers with determinate = ±1.

Planet Math requires a square matrix over some field and determinant = 1.

Wolfram requires real elements with det = 1 but generalizes by allowing elements from some polynomial domain if the matrix has an inverse in the same domain. (This suggests ring elements, requiring inverse matrix over same ring.)

Analysis, Manifolds and Physics (page 173 in my edition) specifies real entries and det=1.

All of these are groups but are generalizations in different directions. In physics contexts it seems to be a real matrix with det=1. Perhaps there needs to be a category split between the discrete and continuous cases. I have no idea what they should be called. Wikipedia has a good disambiguation pattern for cases where one word applies to different subjects. —The preceding unsigned comment was added by NormHardy (talk • contribs) 05:19, 16 March 2007 (UTC).

Mx=b where b is an integer matrix
o.O shouldnt that be M? —The preceding unsigned comment was added by 141.3.160.82 (talk) 11:03, 10 May 2007 (UTC).

Unitary Matrices
Naive question: A unitary matrix U has the property |det(U)|=1. Why is there no mention of unitary matrices in the unimodular matrix article and vice versa? I'm curious to know how they're related and under what circumstances matrices can be both unitary and unimodular.

--Error9312 (talk) 20:25, 3 March 2009 (UTC)

Yes since |det(U)|=1 for any unitary matrix U then any unitary matrix is also unimodular. Added to examples section

Jhmadden (talk) 03:59, 16 October 2014 (UTC)

I just took out mention of unitary and symplectic matrices from the examples, since the page says unimodular matrices are integer matrices, and unitary and symplectic matrices are not necessarily integer. But I see that it isn't clear whether they should be required to be integer. Maybe it would be better if the page said that some authors don't include the integer restriction in the definition? Then unitary and symplectic could be added back to the examples, with remarks that these are not necessarily integer. Clay Spence (talk) 19:47, 13 January 2015 (UTC)

determinant = zero
I learned that a unimodular matrix may have determinant -1 or 1 or 0. Unfortunately I do not have another source of this than my university script. -- Phil1881 (talk) 14:29, 23 August 2010 (UTC)

degenerate case? matrix, having only singular square submatrices is totally unimodular?
Under the definition given,

A totally unimodular matrix (TU matrix) is a matrix for which every square non-singular submatrix is unimodular.

I would believe that a matrix which has only singular square sub-matrices is also totally unimodular. Is this correct?

Or should the definition read

A totally unimodular matrix is a matrix for which every square submatrix is non-singular and unimodular?

Ngvrnd (talk) 15:44, 9 May 2011 (UTC)


 * Indeed such a matrix is totally unimodular. But if you request that every submatrix is singular, then also the 1x1 matrices are, that is, the matrix is a zero matrix. But it is totally unimodular. Your 2nd definition does not make sense, as the 1x1 matrices must all be non-singular and hence not 0. MatthiasWalter (talk) 16:11, 11 November 2011 (UTC)

link to software for testing TU?
Is it common / okay to add the following link as it is the (to the best of my knowledge) only implementation of a decomposition-based implementation for testing TU. It is also quite fast and outperforms all the trivial algorithms.


 * Software for testing total unimodularity by M. Walter and K. Truemper

MatthiasWalter (talk) 16:18, 11 November 2011 (UTC)

Square or non-square?
Article correctly asserts that a unimodular matrix must be square. Further down the page, at the network flow example, it shows a non-square matrix as an example of a unimodular matrix. Please fix. 101.170.213.70 (talk) 23:37, 25 September 2012 (UTC)

Abstract linear algebra
What is the meaning of the last paragraph? I understand: "If the ring is a field, then the matrix is invertible over the ring iff it is invertible over the field", which seems to be trivial. — Preceding unsigned comment added by Maformatiker (talk • contribs) 11:30, 26 July 2016 (UTC)

The unimodular matrices of order n form a group, which is denoted GL_n(Z).
That should be SL_n(Z). Volker1308 (talk) 17:10, 20 December 2019 (UTC)


 * I don't pretend to be a number theorist, yet I think that GL_n(Z) is correct. Note that the determinant of a unimodular matrix can be -1, which doesn't appear to be the case for SL_n(Z). Sanpitch (talk) 05:26, 16 April 2020 (UTC)