Talk:Hermite normal form

Redirect

This page should not redirect to reduced echelon form unless that article is modified to include information on the Hermite normal form. These are not the same. 66.76.11.9 05:08, 12 March 2007 (UTC)

The definitions look wrong

[The definitions seems to have been patched up as of July 14th 2013] — Preceding unsigned comment added by Bandanna (talk • contribs) 21:01, 14 July 2013 (UTC)

I don't think the definitions or the discussion of uniqueness are right. The upper-triangular HNF of $M$ is obtained by doing invertible integer row operations and is equal to $H=UM$ for some invertible integer matrix $U$ (the discussion of uniqueness gets this the wrong way round). The definition of upper-triangular HNF must therefore require each diagonal entry to be the unique maximum in its column not in its row, since it would be impossible to achieve the latter using row operations. For example, with the correct definitions ${\begin{pmatrix}1&2\\0&3\end{pmatrix}}$ is already in upper-triangular HNF, while it cannot be transformed by invertible integer row operations into a form satisfying the definition given.

Unfortunately, I have no reference for anything like the definition proposed for non-square matrices (which also looks wrong), so I am not in a position to attempt to fix the page myself. Bandanna (talk) 16:25, 21 June 2013 (UTC) — Preceding unsigned comment added by Bandanna (talk • contribs) 16:22, 21 June 2013 (UTC)

The definition is still wrong. The third point of the definition

for $j<i$, $m_{ii}>m_{ji}\geq 0$, i.e. in a given column, the entries above the diagonal are non-negative and less than the diagonal

must be

for $j>i$, $m_{ii}>m_{ij}\geq 0$, i.e. in a given row, the entries right of the diagonal are non-negative and less than the diagonal

and in case one defines the lower-left triangular version it must be

for $j<i$, $m_{ii}>m_{ij}\geq 0$, i.e. in a given row, the entries left of the diagonal are non-negative and less than the diagonal

Anyway, one has to look at the rows not at the columns. See at Wolfram or the second example on this Wiki page. In the second example the diagonal is completely zero but there are still non-zero elements in the upper triangle. This is a clear contradiction to the definition.153.96.12.26 (talk) 16:12, 12 March 2014 (UTC)Matthias

@Matthias: I think everything written until now (9 Jan 2015) is consistent. As far as I can tell your definition would also work if we demanded the multiplication to be from the other side ie $H=AU$ (instead of the current $H=UA$), but both options look fine to me. Bilingsley (talk) 14:41, 9 January 2015 (UTC)

Big update 25 June 2016

I made a large edit that changed the following things

• cleaned up the two definitions and how definitions for HNF are not well standardized. Also explained how these definitions are different (...how the U matrix acts)
• added applications section
• added links to code
• added refs for algorithms.
• added the U matrix for each example, added a non-square example.

DrWikiWikiShuttle (talk) 05:10, 26 June 2016 (UTC)