I am analyzing the the following problem:
\begin{equation} \max \limits_{x \in D} x^t A x \end{equation}
where $D \subseteq \mathbb{R}^n$ is a set and $A$ is random matrix with all independent entries of the following form. $(A_{ii})_{1 \leq i \leq n}$ are all iid Gaussian random variables with mean zero and variance two and $(A_{ij})_{1 \leq i < j \leq n}$ are all iid Gaussian random variables with mean zero and variance one. The rest of the entries are defined trough the symmetry constraint $A_{ij} = A_{ji}$ for $1 \leq i \leq j \leq n$.
Now we know that (see this lecture Lemma 2) the distribution of $A$ is invariant under orthogonal conjugation. Since $A$ is symmetric we can hence assume that $A$ has the same distribution as the diagonal matrix $\Lambda$ having the eigenvalues $\lambda_1(A), \ldots \lambda_n(A)$ of $A$ as diagonal entries (just compute some eigendecomposition of $A$).
Question: Is then also the above maximization problem equivalent to
\begin{equation} \max \limits_{x \in D} x^t \Lambda x \quad ? \end{equation}
This would simplify the original problem a lot since the quadratic form is a function of the eigenvalues only. If it is not equivalent could you shortly argue why that is. This might be a trivial question but I am new to this topic. Thank you in advance.
If $A = U^t\Lambda U$ for orthogonal matrix $U$, then $x^t A x = (U x)^t \Lambda (U x)$. So maximizing $x^t A x$ for $x \in D$ is equivalent to maximizing $y^t \Lambda y$ for $y \in U(D)$. If $D$ happens to be invariant under orthogonal transformations (e.g. the unit ball), your two problems are the same. Otherwise they are not.
EDIT: Here's an amusing example. Take $n=2$ and $D = \{e_1, e_2, v\}$ where $e_1$ and $e_2$ are the standard unit vectors and $v = (e_1+e_2)/\sqrt{2}$. If $\Lambda$ is a diagonal matrix, $v^t \Lambda v = \frac{1}{2} e_1^t \Lambda e_1 + \frac{1}{2} e_2^t \Lambda e_2$, so $\max_{x \in D} x^t \Lambda x = \max_{x \in \{e_1, e_2\}} x^t \Lambda x$. On the other hand, if $A_{12}$ is large enough compared to $A_{11}$ and $A_{22}$, $v^t A v > \max(e_1^t A e_1, e_2^t A e_2)$. Thus $\max_{x\in D} x^t A x$ and $\max_{x \in \{e_1, e_2\}} x^t A x$ have different distributions.