Hermite normal form : ihermite

### 2.46.12 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]`.