TITLE: Intuition for scale of the largest eigenvalue of symmetric Gaussian matrix
QUESTION [8 upvotes]: Let $X$ be $n \times n$ matrix whose matrix elements are independent identically distributed normal variables with zero mean and variance of $\frac{1}{2}$. Then
$$
     A = \frac{1}{2} \left(X + X^\top\right)
$$
is a random matrix from GOE ensemble with weight $\exp(-\operatorname{Tr}(A^2))$. Let $\lambda_\max(n)$ denote its largest eigenvalue. The soft edge limit asserts convergence of $\left(\lambda_\max(n)-\sqrt{n}\right) n^{1/6}$ in distribution as $n$ increases. 

Q: I am seeking to get an intuition (or better yet, a simple argument) for why the largest eigenvalue scales like $\sqrt{n}$.

REPLY [4 votes]: The scaling follows from the Wigner semicircle law. Proof of the Wigner semicircle law is outlined in section 2.5 of the review "Orthogonal polynomials ensembles in probability theory" by W. König, Probability Surveys, vol. 2 (2005), pp. 385-447.