The case of a graded module is handled analogously. Therefore, we restrict ourself to the local settings.

Let two modules *M* and *M*' be given by their representation matrices
*A* and *A*'. Assume the modules to be isomorphic. There are two
quadratic matrices *U* and *V* such that *UAV*=*A*'. By the above
properties of the representations both matrices have to be invertible,
and
.

Now, let's look from the other side: Starting with two representation
matrices *A* and *A*' a necessary condition for an isomorphism of
the represented modules *M* and *M*' is an identical size of both
matrices. Denote this size by
.
The sufficient
condition for an isomorphism is again the existence of matrices
and
with *UAV*=*A*'. That means, we have to
resolve the equation

and to look for a pair (

First, we determine the module of all possible transformations
.
Let us consider the matrix *A* as map

just by multiplying with

to be the map induced by the multiplication with

given by is a well defined module homomorphism and its kernel is the module