Question regarding matrices with same image Let $A$ and $B$ be $m\times n$ matrices over $\mathbb{Z}$ such that image($A$) = image($B$), where $A$ and $B$ are considered as maps $\mathbb{Z}^n \rightarrow \mathbb{Z}^m$. Does there exist an invertible $n\times n$ matrix $P$ such that $A=BP$? what happens if we consider the matrices to be over $\mathbb{Q}$ instead of $\mathbb{Z}$? 
I would prefer some hints instead of full answer.
Thanks in advance. 
 A: Suppose that $A$ and $B$ are $m \times n$ integer matrices with the same image.  So, for any integer vector $x$, there exists an integer vector $y$ such that 
$$
Ax = By
$$
Now, take $x \in \mathcal B = \{e_1,\dots,e_n\}$ (the standard basis), and let $y_j$ denote the vector $y$ corresponding to $e_j$ as above.  Let $P$ be the matrix whose columns are $y_1,\dots,y_j$.
We note that for every $x \in \mathcal B$, we have
$$
Ax = BPx
$$
Since $\mathcal B$ is a generating set, we may conclude that $Ax = BPx$ for all $x \in \Bbb Z^n$.  That is, $A = BP$ as desired.
Note, however, that $P$ as chosen here need not be invertible.  Perhaps you can improve this construction (i.e. choose $y$ more carefully) in order to guarantee $P$ is invertible as a matrix over $\Bbb Q$.  Then (since $B = AP'$ for a matrix $P'$), you can show that $P$ must also be invertible as a matrix over $\Bbb Z$.

For the case of $\Bbb Q$ matrices: it is sufficient to note that if $A$ and $B$ have the same image, they can be "column reduced" (as opposed to row-reduced) to the same form.
