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ofer zeitouni
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The diagonal elements just shift the spectrum (and the top eigenvalue) by $1$. So we may assume they are $0$.

You are essentially dealing with a symmetric matrix whose entries above the diagonal are iid, sum of a Bernoulli $\{0,1\}$ of parameter (=mean) $q=2k/n^2$ and of a uniform random variable on $[0,1]$. The mean is therefore $p=q+(1-q)/2$ and the variance is $\sigma^2=q+(1-q)/3-p^2$. Since, for any value of q, the mean is bounded below and so is the variance, you are dealing with the perturbation by $\sqrt{N}$ times a Wigner matrix of a matrix of rank $1$ and norm $pN$. Thus, your top eigenvalue will concentrate around $pn$ ($\pm O(\sqrt{N}))$ by estimates on the top eigenvalue of a Wigner matrix and Weyl's inequalities.

I earlier referred to https://arxiv.org/pdf/math/0605624v1.pdf, but this is a very different scaling, so I scraped that answer.

ofer zeitouni
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