I have posted this [elsewhere](http://www.mathlinks.ro/viewtopic.php?t=198806) and got only a partial reply. I don't know whether this qualifies the question for an open-problem tag; if it does, please anyone insert it.



> 
> Let $L$ be a field, and $K$ a subfield of $L$. Let $n$ and $m$ be two nonnegative integers.
> 
> 
> For any $n\times n$ matrix $A \in \operatorname{M}\_n\left(K\right)$ and any $m\times m$-matrix $B \in \operatorname{M}\_m\left(K\right)$, and any field $S$ containing $K$, we define
> \begin{align}
> \rho\_{S}\left( A,B\right)
> := \max\left\{\operatorname{Rank} Q
> \mid Q\in\operatorname{M}\_{n,m}\left( S\right) ;
> \ AQ = QB\right\}.
> \end{align}
> We can call this number $\rho\_{S}\left( A,B\right)$ the "conjugacy rank" of the matrices $A$ and $B$ over the field $S$.
> 
> 
> 


(Note that if $n = m$, then this conjugacy rank is directly connected with conjugacy -- i.e., similarity -- of matrices: Namely, in this case, we have $\rho\_{S}\left( A,B\right) = n$ if and only if the matrices $A$ and $B$ are conjugate to each other in the ring $\operatorname{M}\_{n}\left( S\right)$.)



> 
> My question is: Do we have $\rho\_{K}\left( A,B\right) = \rho\_{L}\left( A,B\right)$ for any two matrices $A \in \operatorname{M}\_n\left(K\right)$ and $B \in \operatorname{M}\_m\left(K\right)$ ?
> 
> 
> 


This can be shown in the case of $n = m \leq \left\vert K\right\vert$ by a "polynomials which vanish everywhere must be identically $0$" argument. Besides, in the case of $\rho\_{L}\left( A,B\right) = n = m$, it can be shown using the rational canonical form. I am interested in the most general case of the problem -- neither restricting $\left\vert K\right\vert$, nor $\rho\_{L}\left( A,B\right)$, nor requiring $n = m$.


What also might be of help: For any field $S$ containing $K$, the space
$$R\_{S}\left( A,B\right) = \left\{ Q\in\operatorname{M}\_{n,m}\left( S\right) \mid AQ = QB\right\}$$
is a subspace of the vector space $\operatorname{M}\_{n,m}\left(S\right)$.
Besides, every basis of the space $R\_{K}\left( A,B\right)$ is also a basis of the space $R\_{S}\left( A,B\right)$ for every field $S$ containing $K$. However, *this alone is not enough*; you can easily construct a subspace of $\operatorname{M}\_{n}\left(\mathbb{F}\_p\right)$ that consists of singular matrices only but loses this property when extended into $\operatorname{M}\_{n}\left(\mathbb{F}\_{p^2}\right)$.



"Conjugacy rank" of two matrices over field extension
I think this is true, and can be proved by brute force: write an explicit formula for conjugacy rank. I'll prefer to restate the problem in terms of modules.


To an $n\times n$ matrix $A$ over a field $K$, associate the $K[x]$-module $M$ that is $K^n$
as a vector space, while $x$ acts as $A$. Everywhere below, all $K[x]$-modules are finite-dimensional as $K$-vector spaces. Then your definition becomes as follows:


Let $M$ and $N$ be two $K[x]$-modules. Define their conjugacy rank $\rho(M,N)$ to be
the maximal dimension (over $K$) of a $K[x]$-module that is simultaneously isomorphic to a submodule of $M$ and a quotient-module of $N$. We aim to prove that $\rho(M,N)$ is stable under field extensions of $K$.


By structure theorem for modules over PID, we can write $M\simeq\bigoplus K[x]/f\_i$,
where invariant factors $f\_i=f\_i(M)\in K[x]$ satisfy $f\_{i+1}|f\_i$. (We set $f\_i = 1$ when $i$ is larger than the number of invariant factors.) It is easy to check the following claim:


Lemma: $M'$ is isomorphic to a quotient of $M$ if and only $f\_i(M')|f\_i(M)$. The same criterion holds for $M'$ being isomorphic to a submodule of $M$.


Corollary: There is unique up to isomorphism maximal-dimensional module $M'$ that is simultaneously isomorphic to a submodule of $M$ and a quotient-module of $N$; its invariant factors are given by $f\_i(M')=gcd(f\_i(M),f\_i(N))$.


Since the formula for $M'$ is stable under field extensions of $K$, the claim follows.


