Here's a geometrical reformulation of the problem that yields an $O(n \log n)$ solution for $m=2$ and suggests a context that may yield good answers for arbitrary fixed $m$.
Denote the $i$-th coordinate of $u_k$ and $v_k$ by $u_k^{(i)}$ and $v_k^{(i)}$ respectively. Consider the $n$ vectors of length $m$: $$ x_i = (u_1^{(i)}, u_2^{(i)}, u_3^{(i)}, \ldots u_n^{(i)}) \in {\bf R}^m =: V $$ ($i=1,2,\ldots,n$), and the $n$ dual vectors $$ y_j = (v_1^{(j)}, v_2^{(j)}, v_3^{(j)}, \ldots v_n^{(j)}) \in V^*. $$ ($j=1,2,\ldots,n$). Then the problem asks for the maximum of $y_j(x_i)$ as $i,j$ range independently over $\lbrace 1, 2, \dots, n \rbrace$.
Note that given $A$ there are many choices of $u_k$ and $v_k$, but the choice is tantamount to a choice of basis on $V$ and of dual basis on $V^*$, so geometrically our $y_j(x_i)$ problem depends only on $A$.
Now it's clear that the maximizing $x_i$ and $y_j$ must be vertices of the convex hull of $\lbrace x_i \rbrace$ and $\lbrace y_j \rbrace$ respectively. This recovers the known solution for $m=1$, when any bounded convex subset of $V$ has (at most) two vertices, which can be found in $O(n)$ comparisons.
For $m=2$, it is still known how to find the vertices of the convex hull (in order) in $O(n \log n)$ steps [see for instance the "Convex hull algorithms" Wikipage for references]. Once we know the vertices of the convex hull of the $x_i$, we can for each $j$ find the maximal $y_j(x_i)$ in $O(\log n)$ steps, making the overall computational cost still $O(n \log n)$.
Finding the convex hull and its structure for $m=3$, and for larger fixed $m$, is harder, but at least there's some literature on this problem, and experts who can suggest good ways to proceed.
This all assumes that we don't run into difficulties like $m=1$, $u_1 = (-1,2.54)$, $u_2 = (-1,.3937)$ where there are two or more very close candidates for the maximum. To deal with that, we might assume that we can do exact arithmetic (perhaps the coordinates are quantized with fixed denominator), or tolerate an error that can be brought below $\epsilon$ in $O(\log(1/\epsilon))$ steps.