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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 $\sqrt{n}$.

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up vote 2 down vote accepted

Consider a Frobenius norm squared of $A$: $$ \|A\|^2 = \sum_{i,j} a_{i,j}^2 = 2 \sum_{i<j}a_{i,j}^2 + \sum_{i} a_{ii}^2 = \sum_{i=1}^n \lambda_i^2 $$ The expected value of $\|A\|^2$ is easy to find: $$ \mathbb{E}\left( \|A\|^2 \right) = 2 \sum_{i<j} \mathbb{E}\left(a_{i,j}^2\right) + \sum_{i} \mathbb{E}\left( a_{ii}^2 \right) = 2 \sum_{i<j} \frac{1}{4} + \sum_{i} \frac{1}{2} = \frac{n(n+1)}{4} < n \mathbb{E}\left(\lambda_\max(n)^2\right) $$ this establishes $\lambda_\max(n)$ scales at least as $\sqrt{n}$

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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.

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