Do equivalent matrices represent the same module homomorphism Say if we have $R$-module homomorphisms $\theta, \phi :R^m \rightarrow R^n$ which are represented by matrices $A$ and $B$ respectively. Can we say that if $A$ and $B$ can be obtained from each other by elementary row and column operations, that then they represent the same $R$-module homomorphism in the same we can say that linear maps between vector spaces are the same if the matrices representing them are equivalent?
 A: (Sorry for my bad English.)
If $A$ and $B$ are translated by elementary operation, there is regular matrices $P,Q$ s.t. $A=PBQ.$ ($P\in GL(n), Q\in GL(m)$).
Now we denote $f: R^n\to R^n$ induced by $P$, $g: R^m\to R^m$ induced by $Q$, because of $P,Q$ is regular $f,g$ is isomorphic.
$A=PBQ$ implies $\theta=f\circ\phi\circ g $.
Your sense "the same" is equivalent to $\theta=f\circ\phi\circ g $.
A: In order to represent a homomorphism of free modules $R^n \to R^m$ with a matrix you need to make a explicit choice of basis for $R^n$ and $R^m$. Two matrices $A$ and $B$ are equivalent if there are invertible matrices $P$ and $Q$ with $A = Q^{-1}BP$. In this case, you can think of the matrices $A$ and $B$ as representing the same module homomorphism under different choices of bases for $R^n$ and $R^m$, where $P$ and $Q$ are the respective change of basis matrices.
A: I would say no for the following reason.
Think about the case when $R$ is a field and $m=n$.
We know every invertible $n\times n$ matrix is transformable to the identity matrix using elementary row and column operations.  Indeed, this is one common strategy for computing the inverse matrix.
Would you want to believe that every invertible transformation $F^n\to F^n$ "is the same homomorphism" as the identity homomorphism?
I think it would be a rather unusual version of sameness... I'm not sure what it would capture beyond the rank of the transformation.
