$A$ is a matrix of size $n\times r$. $B$ is a matrix of size $r\times n$. The rank of $A$ and $B$ are both equal to $r$. Assuming $r < n$.
My question is: $\def\rank{\operatorname{rank}}\rank(AB) = \rank(BA)$?
PS: It's easy to prove that $\rank(AB)=r$.
Choose subsets $I,J$ of size $r$ among $\{1,2,\ldots,n\}$, let $B$ be obtained by selecting rows with index in $I$ from the identity matrix $I_n$, and $A$ by taking from $I_n$ columns with index in $J$. Then $AB$ has rank $r$ but $BA$ has rank $|I\cap J|$, which is usually less than $r$.
To answer the question to characterise the cases where $BA$ does have rank $r$: the linear map for $A$ is injective with image (in $\Bbb R^n$) of dimension $r$, and that of $B$ is surjective with a kernel of dimension $n-r$; the rank of $BA$ will be less than $r$ by the dimension of the intersection of these two subspaces, so it will be $r$ if and only if the intersection is $\{0\}$ (they are complementary subspaces).