I'm trying to characterize equivalence classes of matrices over a vector space.


Specifically, let $V$ be a vector space over a field $K$, let $M \in M\_n(V)$ be an $n \times n$ matrix with entries in $V$ and let $A,B \in M\_n(K)$ be $n \times n$ matrices with entries in $K$. I want to characterize the equivalence class formed by all matrices formed as


$$A M B$$


i.e, equivalent matrices, but over a vector field instead of a ring, and the invertible "side" matrices are picked from the base field, not the vector field itself.


The problem is related to constructing the Jacobian variety of a curve.


So far, I've been experimenting with the $2\times2$ case:


$$\left[\matrix{v\_1 & v\_3 \\ v\_2 & v\_4}\right]$$


Let's multiply the equation out:


$$AMB = \left[\matrix{a & c \\ b & d}\right] \left[\matrix{v\_1 & v\_3 \\ v\_2 & v\_4}\right] \left[\matrix{e & g \\ f & h}\right]$$


$$=\left[\matrix{eav\_1 + ebv\_2 + gav\_3 + gbv\_4 && fav\_1+fbv\_2+hav\_3+hbv\_4 \\ ecv\_1+edv\_2 +gcv\_3+gdv\_4 && fcv\_1+fdv\_2+hcv\_3+hdv\_4}\right]$$


Obviously, all of the vectors have to be in the linear space spanned by the original four vectors. So, is picking a four-dimensional subspace of $V$ enough to specify our equivalence class?


No. Let the transformed matrix have the form:


$$\left[\matrix{
A\_1v\_1 + A\_2v\_2 + A\_3v\_3 + A\_4v\_4 &&
C\_1v\_1 + C\_2v\_2 + C\_3v\_3 + C\_4v\_4 \\
B\_1v\_1 + B\_2v\_2 + B\_3v\_3 + B\_4v\_4 &&
D\_1v\_1 + D\_2v\_2 + D\_3v\_3 + D\_4v\_4
}\right]$$


We see that the obvious equality $ea\cdot gb = eb \cdot ga$ implies that $A\_1A\_4 = A\_2A\_3$. Likewise, $B\_1B\_4=B\_2B\_3$, $C\_1C\_4=C\_2C\_3$ and $D\_1D\_4=D\_2D\_3$.


There are also similar cross-relationships between the four vectors, but the restriction I just derived is enough to show that a four-dimensional subspace alone isn't enough to specify an equivalence class. The vectors have to picked from a particular three-dimensional subvariety, and it's a subvariety, not a subspace, because the relationships $A\_1A\_4 = A\_2A\_3$, etc., are not linear.


So, specifying a equivalence class requires specifying a four-dimensional subspace, then picking a certain three-dimensional subvariety from that space. I've left out considerations like the vectors being linearly dependent, but I think this is enough to give the sense of what I'm trying to accomplish.


Has anybody seen anything like this? Any idea how to characterize these equivalence classes?



Equivalence of matrices over a vector space
If $\dim V = d$, then this is equivalent to the problem of simultaneous similarity of $d$-tuples of $n \times n$ matrices over $K$. I'll prove the equivalence below, but let me first discuss the consequences for your question. The latter is a famous "wild" problem (in the formal sense of representation-wildness, see, e.g. Chapter 19 of Simson & Skowronski Vol. 3). This means that classifying the equivalence classes here is universal for classifying the representations of arbitrary finite-dimensional algebras over $K$, and hence is sometimes said to be "hopeless." Nonetheless, there is some hope and quite a lot is known. For example, there is a polynomial-time algorithm to decide when two such matrices of vectors lie in the same orbit [[Sergeichuk](http://www.ams.org/mathscinet-getitem?mr=1782204); [Brooksbank-Luks](http://www.ams.org/mathscinet-getitem?mr=2464805); [Chistov-Ivanyos-Karpinski](http://www.ams.org/mathscinet-getitem?mr=1809971)].


Sergeichuk also gives a "normal form," although the normal form is (necessarily) quite a bit more complicated than that for single matrices, and is recursive in nature. 


Proof of equivalence: choose a basis $v\_1, \dotsc, v\_d$ of $V$. Write the $(i,j)$ entry of your matrix as $\sum\_{k=1}^d T\_{ijk} v\_k$. Then your transformation of $A M B$ transforms the tensor $T$ into another tensor $T'\_{ijk} = \sum\_{i'=1}^n \sum\_{j'=1}^n A\_{ii'} B\_{j'j} T\_{i'j'k}$. If you look at what this action is, it's the same as if we instead considered the $d$-tuple of matrices $(M\_1, \dotsc, M\_d)$ where $(M\_k)\_{ij} = T\_{ijk}$ under the action $A(M\_1, \dotsc, M\_d)B = (AM\_1 B, \dotsc, A M\_d B)$. This problem is wild for any $d \geq 3$. For $d=1$ the normal form is obvious and classical. For $d=2$ it is the Kronecker normal form for pencils, which I guess is also classical, though perhaps less well-known (e.g. not often taught in first courses on linear algebra).


