probability - 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 $\sqrt{n}$.

Answer

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