TITLE: Finding minimum (or maximum) element of a low rank matrix. 
QUESTION [14 upvotes]: Let $A\in\mathbb{R}^{n\times n}$ and suppose that $A$ is of rank $m\leq n$. Moreover suppose we know $u_1,\ldots, u_m \in\mathbb{R}^{n\times 1}$ and $v_1,\ldots, v_m \in\mathbb{R}^{n\times 1}$ such that 
$
A = \sum_{k=1}^m u_kv_k^T
$
Is there a faster way than $\mathcal{O}(n^2)$ for finding the minimum (or maximum) element of $A$?  Here I'm thinking of $m$ being small in comparison to $n$.  
For instance, if $m=1$ it is easy to find the minimum (or maximum) element just by finding the minimum of $u_1$ and then of $v_1$, taking $\mathcal{O}(n)$.
For instance, if $m=1$ then to compute the maximum you can do, 
$
\max\{\min u_1\cdot \min v_1,\min u_1\cdot\max v_1, \max u_1\cdot\min v_1,\max u_1\cdot\max v_1\}
$
which is $\mathcal{O}(n)$.
Thanks.

REPLY [2 votes]: We can reformulate the problem as follows: given vectors $u_1,\ldots,u_n,v_1,\ldots,v_n \in \mathbb{R}^m$, find $\max | \langle u_i,v_j \rangle |$ over all $i,j \in [n]$. 
Now, if an approximate answer is good enough, I think that for small $m$ an approximate answer with multiplicative error of $1+\epsilon$ can be found in time $\approx n m + (1/\epsilon)^m$. Take an $\epsilon$-net of the $m$-dimensional sphere (that is, a set of points on the sphere such that each other point on the sphere is $\epsilon$-close to one of them). Such a net has size $ (1/\epsilon)^{O(m)}$. Now, hash the vectors $u_1,\ldots,u_n$ to the sphere. Let $u'_i=u_i / \|u_i\|$ be the image of $u_i$. Find the point on the net closest to $u'_i$ (trivially this can be done in time linear in the size of the net, but probably there are some geometric algorithms to do this is constant or near constant time). For each point in the net store the vector close to it with maximal norm. Do the same for $v_1,\ldots,v_n$. To finish, go over all pairs of points in the net and compute the inner products of their associated vectors.