Let $X$ be an $n\times n$ symmetric random matrix whose diagonal is fixed as $1$, and every element in the upper triangle (excluding the diagonal) is drawn from Bernoulli($p$). The elements in the lower triangle is then set to ensure symmetry.
First ask a simple question, for such a random matrix, let $\lambda_1(X)$ denotes the largest eigenvalue of $X$, then
Do $\lambda_1(\frac{X}{\sqrt n})$ and $\frac{\lambda_1(X)}{\sqrt n}$ have the same distribution, as $n\to \infty$?
I think they are probably not, otherwise further discussion is meaningless. It looks like $\lambda_1(\frac{X}{\sqrt n})$ obeys semi-circle law, but if $\frac{\lambda_1(X)}{\sqrt n}$ obeys the same distribution, $\frac{\lambda_1(X)}{n}$ will simply concentrate at $0$, which is not possible.
Now let $\sigma^2$ be the variance of the distribution. In the answer of Bound for largest eigenvalue of symmetric matrices of uniform random variables over $[0,1]$ and fixed $1$s along diagonal and scattered $1$s, it states
It is a standard fact that there is a transition: if $p\leq \sigma$, then $\lambda_1(\frac{X}{\sqrt n})$ concentrates at $2\sigma$; if $p>\sigma$, then $\lambda_1(\frac{X}{\sqrt n})$ concentrates at $p+\sigma^2/p$.
The answer also indicates
The above claim holds if we replace the Bernoulli distribution by a uniform distribution, and let $p$ be its mean, and $\sigma^2$ be its variance.
I know in random matrix theory, it is standard to study the distribution of eigenvalues of $\frac{X}{\sqrt n}$, and it obeys semi-circle law. My question is, rather than the largest/smallest eigenvalue of $X/\sqrt n$,
Is there similar result for $\frac{\lambda_1(X)}{n}$; if so please help provide some explanation or reference. (sorry, there was a typo)
A simulation shows $\frac{\lambda_1(X)}{n}$ looks like Gaussian (10000 samples of $1000\times 1000$ matrices, Bernoulli distribution parameter by 0.7).
