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The crucial point is that the eigenvectors are not unique, as David Handelman commented. There is a gauge freedom since you can change their phase at will. If you want the map $U\to(D,V)$ to be biject …

The joint probability distribution of the eigenvalues of $S$ is proportional to $$ \rho(S)=e^{-{\rm Tr}(S)}\det(S)^{a/2}|\Delta(S)|,$$ where $a=m-d-1$ and $\Delta(S)$ is the Vandermonde. The average v …

Complementing the answer by Carlo, if you take all $k$'s equal you have $$f_{\rm GUE(d)}(k,...,k)\propto \int dX e^{ik{\rm Tr}(X)}e^{-\frac{d}{2}{\rm Tr}(X^2)}.$$ Taking $x$ to be any real diagonal el …

The spectrum of random matrices has some universality properties, in the sense that they do not depend on details of the distribution of the matrix elements. For instance, the spectral gap, i.e. the s …

I do not know what is exactly the KS philosophy, or much number theory for that matter, but maybe I can tell you a few things. Take the Riemann zeta function, for instance. It was discovered by Montgo …

I think this text by Eynard, Kimura and Ribault may interest you. There are some of those diagrams in chapter 2, but there are also nice connections to algebraic geometry, loop equations and integrabl …

If $A$ is a Gaussian random matrix as you describe, then the ensemble of matrices given by $A^TA$ is known as the Wishart ensemble, or the Laguerre ensemble. It has been extensively studied, and you c …

The diagonal elements of $P=\frac{1}{N}MM^T$, like $$P_{11}=\frac{1}{N}\sum_{i=1}^NX_i^2,$$ satisfy $ \langle P_{11}\rangle=1$ and $ \langle P_{11}^2\rangle=1+2/N$ (variance decreases like $N^{-1}$). …

I think the following papers are closely related to what you are looking for: How many eigenvalues of a Gaussian random matrix are positive? (2011) Index Distribution of Gaussian Random Matrices (20 …

The Oxford handbook of random matrix theory (Oxford University Press, 2011), edited by G. Akemann, J. Baik, P. Di Francesco, is an excellent reference, which covers a wide variety of properties and ap …

I am writing more than a year after the question was posted, only to spell out some more details regarding the calculation implicit in the solution presented by Suvrit, and to clarify the dependence o …

On the one hand, you can compute the average of the trace of $X^2$ by using the eigenvalues, $\langle {\rm Tr}(X^2)\rangle=\sum_{i}\langle \lambda_{i}\rangle$. If the eigenvalues must be $O(1)$, this …

Your first expression for the potential $$ \Phi(\omega)=\frac{1}{N}\log E_J \int d^2z...$$ is equivalent to $$ e^{N\Phi(\omega)}=E_J \int d^2z...$$ Ok? Writing the expectation in $J$ according to it …

I think the essence is the central limit theorem. If you compute the traces of powers of your random matrix, they will be the sum of many independent random variables and will be Gaussian distributed …

I'll post an answer to spell out all the details. You have $$G(ω)=\frac{1}{N}E\left[{\rm Tr}\frac{1}{Iω−J}\right]=\frac{1}{N}E\left[\sum_\lambda\frac{1}{ω−\lambda}\right]$$ This can be written as $$ …