3
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ I perfectly understood the case for $\mathbb{Q}$, but I'm still confused about the $\mathbb{Z}$ case. How can you choose those $y$'s more carefully is not clear to me. $\endgroup$ May 12, 2016 at 5:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .