$\newcommand{\rank}{\operatorname{rank}}\newcommand{\im}{\operatorname{im}}$ Given $A,B\in M_{n\times n}(k)$, define $\rank(A,B):=\dim(\im A+\im B)$. I'm looking for results regarding relationships between $\rank(A,B)$ and $\rank(A^t,B^t)$ under certain constraints on $A$ and $B$.
Here is a motivating example: Suppose we restrict out attention to matrices $A$ and $B$ which have all of their non-zero entries in the first three rows. Then certainly $\rank(A,B)\le 3$, and for $n>3$, equality will hold generically. Regarding the transposes, $A^t$ and $B^t$ will annihilate all but the first three coordinate vectors. Generically, the first three columns of each of $A^t$ and $B^t$ will be linearly independent, and so generically $\rank(A^t,B^t)>3$.
The intuition I glean from this example is that "short and wide" matrices yield smaller ranks than "tall and thin" matrices. One way to quantify this is to pick a subspace of $W\subseteq M_{n\times n}(\mathbb F_q)$ and sum $q^{-\rank(A,B)}$ over all $A,B\in W$. We define \[f(W)=\sum_{A,B\in W}q^{-\rank(A,B)}.\] The motivating example above would then be translated as saying that, in the case where $W$ is the subspace of all matrices with zeros outside the first three rows, we have $f(W)\ge f(W^t)$.
It would be nice to have a complete characterization of those subspaces $W$ spanned by $e_{i,j}\in M_{n\times n}(\mathbb F_q)$ for which $f(W)\ge f(W^t)$.
Are results of this form known? How might one go about proving a result like this? For simple subspaces $W$, one can write down the number of matrices of each rank explicitly, but as $W$ becomes more complicated, this quickly becomes too tedious. An important special case of this problem seems to be the case when $W=\{X\in M_{n\times n}(\mathbb F_q)\mid X_{n,1}=\cdots=X_{n,k}=0\}$.
I should add that my particular interest is when $A$ and $B$ are nilpotent commuting matrices over a finite field, though I'm not sure if this information is relevant to the problem.
