Let $X$ be an $n\times n$ symmetric random matrix whose diagonal is fixed as $1$, and every element above the diagonal is drawn from Bernoulli($p$). The elements in the lower triangle is then set to ensure symmetry.
My first question is, for such a random matrix, let $\lambda(X)$ denote the spectrum of $X$, and let $\lambda_1(X)$ denotes the largest eigenvalue of $X$, then
Do $\lim_{n\to \infty}\lambda(\frac{X}{\sqrt n})$ and $\lim_{n\to \infty}\frac{\lambda(X)}{\sqrt n}$ have the same distribution?
I think they are probably not. Such matrix is a Wigner matrix, so as $n \to \infty$, $\lambda(\frac{X}{\sqrt n})$ obeys semi-circle law over a finite range. If $\frac{\lambda(X)}{\sqrt n}$ also obeys the semi-circle distribution, $\frac{\lambda(X)}{n}$ will simply concentrate at $0$, which is not possible.
I am not in the random matrix area, please correct if I am wrong.
I know in random matrix theory, it is standard to study the distribution of eigenvalues of $\frac{X}{\sqrt n}$ as $n \to \infty$. My second question is,
Instead of $\lim_{n \to \infty}\lambda(\frac{X}{\sqrt n})$ or $\lim_{n \to \infty}\lambda_1(\frac{X}{\sqrt n})$, is there similar result for $\frac{\lambda(X)}{n}$ or $\frac{\lambda_1(X)}{n}$; if so please help provide some explanation or reference.
A simulation shows $\frac{\lambda_1(X)}{n}$ looks like Gaussian (10000 samples of $1000\times 1000$ matrices, Bernoulli distribution parameter by 0.7).
