$\DeclareMathOperator\Im{Im}\DeclareMathOperator\Ker{Ker}$Let $A$ be a discrete valuation ring with uniformizer $p$. Let $X, Y\in M_n(A)$ be square matrices such that $XY=YX$, and let $X^T$, $Y^T$ be their transposes. Assume that $$p^dA^n\subset \Im X+\Im Y.$$
Does this imply that $$p^{f(d)}A^n \subset \Im X^T+\Im Y^T,$$ where $f : \mathbb{N}\to \mathbb{N}$ is a function that does not depend on the matrices $X$ and $Y$, and not even on their size $n$? (It might depend on the dvr $A$, if needed.)
Remark:
To answer the question of Luc Guyot in the comments, note the following fact. This implies that, under the above assumption, at least the matrix $[X^T, Y^T]$ has rank $n$.
Fact: If $X, Y\in M_n(K)$ are commuting matrices, then $\Im X+\Im Y=K^n$ if and only if $\Ker X\cap \Ker Y=0$.
Proof of the fact: We may assume that $K$ is algebraically closed. Then, splitting the vector space as the direct sum of the characteristic spaces of $X$ (which are stable by $Y$), we may assume that $X$ has a single eigenvalue. If this eigenvalue is nonzero, then $\Ker X=0$ and we are done. Otherwise $X$ is nilpotent. Arguing the same way with $Y$, we may assume that both $X$ and $Y$ are nilpotent. But then, since they commute, the assumption that $\Im X+\Im Y=K^n$ implies that $n=0$.
