# Intuition for scale of the largest eigenvalue of symmetric Gaussian matrix

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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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 $\sqrt{n}$