Let $A = xy^T$ be a rank-$1$ matrix, and suppose every entry of $A$ is in $[0,1]$. We can create a binary matrix $A_{\rm rounded}$ by setting 
$$ [A_{\rm rounded}]_{ij} = \begin{cases} 1 & \mbox{ with probability } A_{ij} \\ 0 & \mbox{ with probability } 1-A_{ij} \end{cases} $$

Is it possible to recover $A$ by looking at the principal eigenvector of $A_{\rm rounded}$? 

I wrote a quick MATLAB simulation that suggests the answer is yes. Here is my code:


> n=20000; %matrix size
> 
> %generate matrix; 
> 
> x = rand(n,1);
> 
> y = rand(n,1);
> 
> A = x*y';
> 
> 
> 
> %generate rounded matrix
> 
> B = zeros(n,n);
> 
> for i=1:n
> 
>     for j =1:n
> 
>         if rand>A(i,j)
> 
>             B(i,j)=0;
> 
>         else
> 
>             B(i,j) =1;
> 
>         end
> 
>     end
> 
> end
> 
> %eigendecomposition of B
> 
> [v,d]=eig(B);
> 
> %figure out the index of the principal eigenvector;
> 
> m = max(abs(d));
> 
> mm = max(m);
> 
> i = find(m==mm);
> 
> %compare principal eigenvector to true answer
> 
> y = v(:,i)./x; 
> 
> %ideally, y is a multiple of the all-ones vector. 
> 
> %check how far this is from being the case
> 
> J = eye(n,n)-(1/n)*ones(n,n);
> 
> norm(J*y)

Running this code with matrix size of 20,000 took me a few hours and returned the following result: if $v$ is the principal eigenvector of $A_{\rm rounded}$, and $v/x$ is the elementwise ratio of $v$ and $x$, then $ ||(I - (1/n) {\bf 1} {\bf 1}^T) v/x||_2 \approx 0.06$. Given that $v$ and $x$ are in $\mathbb{R}^{20,000}$, this strongly suggests the answer is positive. 

Since the answer is likely yes, is this something that is present in the literature? And is there a simple argument to see that recovery is possible, in this or a related model?