Is there any way to find $x, y, z \in \mathbb{Z}$ such that, given $a, b, c \in \mathbb{Z},\ a, b, c > 0,\ (a, b, c) = 1$ matrix $$H = \begin{bmatrix} x & y \\ 0 & z \\ 0 & 0 \end{bmatrix}$$ is the Hermite normal form of $$A = \begin{bmatrix}c & 0 \\ 0 & c \\ a & b \end{bmatrix} ? $$ Note that I'm not looking for an algorithm, but rather for a generic answer, e.g. that heavily uses Bézout's lemma. It will be good if there is some beautiful matrix $U$ such that $U \times A = H$.
Side questions:
- Is it always true that $z = c$?
- Smith normal form $D = U \times A \times V$ will be good too.
Pre-multiplication of an $m\times n$ integer matrix $A$ by a unimodular integer matrix $U$ does not change the ideal generates by the $r\times r$ minors of $A$, that is leaves their highest common factor unchanged.
In your example
(i) we have that the HCF of the $1\times 1$ minors of $A$ is $\gcd(a,b,c)=1$, and the highest common factor of the $2\times 2$ minors is $\gcd(c^2,ac,bc)=c$.
(ii) we have that the HCF of the $1\times 1$ minors of $H$ is $\gcd(x,y,z)$, and the highest common factor of the $2\times 2$ minors is $\gcd(xz)=xz$.
I don't think more can be said.
With $c=6,a=2,b=3$ in $A$ we get $x=2,y=0,z=3$ in $H$: so we don't get $1,c$ on the "diagonal".