### 5.48.12 Hermite normal form : ihermite

ihermite takes as argument a matrix A with coefficients
in ℤ.

ihermite returns two matrices U and B such that
B=U*A, U is invertible in ℤ (det(U) = ± 1)
and B is upper-triangular. Moreover,
the absolute value of the coefficients above the diagonal of B are
smaller than the pivot of the column divided by 2.

The answer is obtained by a Gauss-like reduction algorithm
using only operations of rows with integer coefficients
and invertible in ℤ.

Input :

A:=[[9,-36,30],[-36,192,-180],[30,-180,180]]; U,B:=ihermite(A)

Output :

[[9,-36,30],[-36,192,-180],[30,-180,180]], [[13,9,7],[6,4,3],[20,15,12]],[[3,0,30],[0,12,0],[0,0,60]]

Application: Compute a ℤ-basis of the kernel of a
matrix having integer coefficients

Let M be a matrix with integer coefficients.

Input :

(U,A):=ihermite(transpose(M)).

This returns U and A such that A=U*transpose(M) hence

transpose(A)=M*transpose(U).

The columns of transpose(A) which are identically 0 (at the right,
coming from the rows of A which are identically 0 at the bottom)
correspond to columns of transpose(U) which form a basis
of Ker(M). In other words, the rows of A
which are identically 0 correspond to rows of U
which form a basis of Ker(M).

Example

Let M:=[[1,4,7],[2,5,8],[3,6,9]]. Input

U,A:=ihermite(tran(M))

Output

U:=[[-3,1,0],[4,-1,0],[-1,2,-1]] and A:=[[1,-1,-3],[0,3,6],[0,0,0]]

Since A[2]=[0,0,0], a ℤ-basis of Ker(M) is
U[2]=[-1,2,-1].

Verification M*U[2]=[0,0,0].