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Let $\lambda_1(A_n),...,\lambda_n(A_n)$ be the random eigenvalues of a random $(n \times n)$ matrix $A_n$. We can define the empirical spectral measure $\mu_n^{A_n}$ on $(\mathbb{C},\mathcal{B}(\mathbb{C}))$ by $\mu_n^{A_n} = \frac{1}{n} \sum\limits_{j=1}^n \delta_{\lambda_j(A_n)}$. For many interesting random matrix ensembles $(A_n)_{n \in \mathbb{N}}$ there are results of the form $$ n\big( \mu^{A_n}_n(f) - \mu(f) \big) \xrightarrow{n \rightarrow \infty}_{\mathcal{D}} X_f $$ for certain (sufficiently smooth) functions $f : \mathbb{C} \rightarrow \mathbb{R}$, a determinsitic measure $\mu$ and a Gaussian process $X_f$.

My question is if anyone knows if such a result is known for iid matrices, that is when the entries of $A_n$ are all independent identically distributed with mean zero and variance $\frac{1}{n}$. By the Circular Caw we know, that in this case $\mu$ would need to be the uniform distribution on the unit sphere $B_1^{\mathbb{C}}(0)$.

An example for a random matrix ensemble, where such a result is known, is the Wigner ensemble. See Theorem 9.2 of Bai and Silversteins 'Spectral Analysis of Large Dimensional Random Matrices'.

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  • $\begingroup$ you want the spectral density of the real Ginibre ensemble? the eigenvalues fall into two sets, of order $\sqrt n$ of them condense on the real axis and the remaining uniformly fill a disc of radius $\simeq\sqrt n$. $\endgroup$ Commented Nov 23, 2021 at 12:15
  • $\begingroup$ The spectral density would contain all the necessary information to prove such a result, but proving it is far from easy. The methods used for the Wigner ensemble break down, when the measure $\mu$ has support with non-empty interior. It is very well possible, that such a result is not yet provable. $\endgroup$ Commented Nov 23, 2021 at 13:08
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    $\begingroup$ If I understand your question correctly, perhaps this paper of Rider and Silverstein (and its followup work) answer your question? mathscinet.ams.org/mathscinet-getitem?mr=2294978 $\endgroup$
    – Terry Tao
    Commented Nov 23, 2021 at 19:41
  • $\begingroup$ Thank you! This is exactly what I was looking for. $\endgroup$ Commented Nov 23, 2021 at 19:57

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