Let $A=(k_1\cdots k_m)\in\text{Mat}(n\times m,\mathbb Z)$, so $k_i\in\mathbb Z^n$. Then we have $A\mathbb Z^m=\langle k_1,\dots, k_m\rangle\subset\mathbb Z^n$.
With a given Matrix $A$ one can easily compute $A\mathbb Z^m$ by gaussian algorithm transformations until $A$ becomes diagonal. My question is: Do we have $A\mathbb Z^m=A^T\mathbb Z^m$ in the case $m=n$?
In the few cases I tried it resulted in the same subgroup. But I'm not sure if it was just a coincidence. Obviously we have $A\sim D$ for a certain diagonal Matrix $D$ and $D\sim D^T$. But is it correct to say $A^T\sim D^T$ so we can conclude $A\sim A^T$? ($\sim$ meaning that the matrices are equivalent in a sense that one can be transformed into the other).
Example: For $A=\begin{pmatrix}0&-1&1\\2&0&0\\4&1&3\end{pmatrix}$ we have $A\mathbb Z^3\cong2\mathbb Z\times \mathbb Z\times 4\mathbb Z\cong A^T\mathbb Z^3$.