1
$\begingroup$

After describing the Hermite Normal Form (HNF), MathWorld explains:

The Hermite normal form for integer matrices is implemented in Mathematica as HermiteDecomposition[A], which however uses the "rows" convention (thus making $\mathbf H$ upper triangular) and replaces (the condition $h_{i\,j}\leq 0$ and $|h_{i\,j}| < h_{i\,i}$ for $j < i$) with balanced remainders $\pmod {h_{ii}}$.

Is there a way, using Mathematica and without much coding, to compute HNF in "columns" convention?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

Not too hard; all that's needed is a simple application of matrix identities:

ColumnHermiteDecomposition[mat_ /; MatrixQ[mat, IntegerQ]] :=
                           Transpose /@ HermiteDecomposition[Transpose[mat]]

Test:

mat = {{1, 2, 3, 2, 2}, {1, 2, 3, 4, 0}, {0, 5, 4, 2, 1}, {3, 2, 4, 0, 2}};
{u, t} = ColumnHermiteDecomposition[mat];

u
{{8, 24, 22, 7, 28}, {7, 21, 20, 6, 25}, {-11, -32, -30, -9, -38},
 {3, 8, 7, 2, 9}, {3, 7, 7, 2, 9}}

t
{{1, 0, 0, 0, 0}, {1, 2, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}}

mat.u == t
True
Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ I see that HermiteDecomposition[Transpose[mat]] works, but I don't quite see what lead you to use Transpose[mat]. Would you care to explain? (+1 because it works, and I couldn't get it to do so) $\endgroup$
    – rcollyer
    Commented Jun 5, 2013 at 18:51
  • 1
    $\begingroup$ @rcollyer, the row-oriented Hermite decomposition of $\mathbf A$ is $\mathbf U\mathbf A=\mathbf T$, with $\mathbf U$ unimodular and $\mathbf T$ upper triangular. The column-oriented form is $\mathbf A\mathbf V=\mathbf L$, with $\mathbf V$ unimodular and $\mathbf L$ lower triangular. You should now see how transposition affects the arrangement. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jun 5, 2013 at 18:55
  • $\begingroup$ I see the transposition, but my mind isn't making the leap to the relationship between $\DeclareMathOperator{\H}{H}\H_R(\mathbf{A}^T)$ and $\H_C(\mathbf{A})$. I can see that it is a reasonable guess that $\H_R(\mathbf{A}^T)=\{\mathbf{U}^\prime,\mathbf{T}^\prime\}$ should be related to $\{\mathbf{V},\mathbf{L}\}$, but I don't necessarily see why $\mathbf{A}(\mathbf{U}^\prime)^T$ should be $\mathbf{L}$. That is where my difficulty lies. I see that it works, but I don't see why. $\endgroup$
    – rcollyer
    Commented Jun 5, 2013 at 19:20
  • $\begingroup$ @rcollyer, transpose both sides of $\mathbf U\mathbf A^\top=\mathbf T$. What form does the transpose of an upper-triangular matrix take? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jun 5, 2013 at 19:30
  • $\begingroup$ I knew it was something simple; I just wasn't seeing it. Thanks. $\endgroup$
    – rcollyer
    Commented Jun 5, 2013 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.