Hermite normal form : ihermite

ihermite takes as argument a matrix A with coefficients in $\mathbb {Z}$.
ihermite returns two matrices U and B such that B=U*A, U is invertible in $\mathbb {Z}$ (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 $\mathbb {Z}$.
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 $\mathbb {Z}$-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 $\mathbb {Z}$-basis of Ker(M) is U[2]=[-1,2,-1].
Verification M*U[2]=[0,0,0].