Almost every $m\times n$ real matrix is Dirichlet approximable Let $\| \cdot \|$ denote the maximum norm in Euclidean spaces.
Consider the set $D_{m,n}$ of $m \times n $ real matrices satisfying that the system of inequalities
$$\|Aq-p\|^m < \frac{1}{T}, \|q\|^n < T$$
have a nontrivial integer solutions for all large enough $T$. $p\in \mathbb Z^m, q \in \mathbb Z^n$
By Dirichlet approximation, $D_{1,1}=\mathbb R$. I wonder what happens if $(m,n) \ne (1,1)$. Does $D_{m,n}$ necessarily have full measure? I am not sure if there is an elementary proof or ergodic theory/dynamics (things like Dani's correspondence) is needed.
Source of question: the last paragraph of
https://arxiv.org/pdf/1709.04082.pdf#page=2#
 A: I think that $D_{m,n}$ is just the set of all $m \times n$ matrices $A$ for all $m,n$. The proof is basically the same as that of Dirichlet approximation, i.e., Pigeonhole Principle.
For any $m, n$, any $A$, and any $k$, define $T = (2k)^{mn}$. Then, take the set $S$ of vectors $q \in \mathbb{Z}^n$ with all entries in $[-2^{m-1} k^m, 2^{m-1} k^m)$. Then $|S| = (2^m k^m)^n = (2k)^{mn}$.
Now, consider all vectors $Aq$, $q \in S$, as elements of $[0,1)^m$ (by taking coordinates mod 1). If we partition $[0,1)^m$ into half-open subcubes of side length $(2k)^{-n}$, then the number of such cubes is $(2k)^{mn} = |S|$. We have two cases; either some $Aq'$ and $Aq''$ lie in the same such subcube for $q' \neq q'' \in S$, or every such subcube contains exactly one $Aq$ for $q \in S$.
In the first case, there exists $p \in \mathbb{Z}^m$ so that $A(q' - q'') - p$ has all entries with absolute value less than $(2k)^{-n}$. In the second case, there exists $p \in \mathbb{Z}^m$ so that $Aq - p$ has all entries with absolute value less than $(2k)^{-n}$. If we're in the first case, we redefine $q = q' - q''$, and note that in either case, $q$ has all entries with absolute value less than $(2k)^m$  (either because $q \in S$ or because $q$ is the difference of vectors in $S$.)
Now, $\|q\|^n < (2k)^{mn} = T$, and $\|Aq - p\|^m < (2k)^{-mn} = T^{-1}$, and since $k$ was arbitrary, we get such a solution for infinitely many $T$.
