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Statistics of spectral properties of matrix-valued random variables.
1
vote
Normalizing Entries In Defining Random Matrices (Wigner Matrix)
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 …
1
vote
Accepted
Alternative formula of a Green's function for average density of eigenvalues of random matrix
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 $$ …
3
votes
Accepted
A question from Zeitouni's Introduction to Random Matrices
I think the essence is the central limit theorem. If you compute the traces of powers of your random matrix, they will be the sum of many independent random variables and will be Gaussian distributed …
7
votes
Accepted
Are random circulant matrices almost orthonormal?
The diagonal elements of $P=\frac{1}{N}MM^T$, like
$$P_{11}=\frac{1}{N}\sum_{i=1}^NX_i^2,$$
satisfy $ \langle P_{11}\rangle=1$ and $ \langle P_{11}^2\rangle=1+2/N$ (variance decreases like $N^{-1}$).
…
5
votes
Distribution of dot product of two unit random vectors
If $u$ is uniformly distributed over the sphere, we can write it as $u=Uv$, where $U$ is a unitary transformation uniformly distributed over the unitary group. Then the quantity $|u\cdot v|^2$ is just …
1
vote
Spectral gap of $AA^{T}$ for Bernoulli random matrix A
The spectrum of random matrices has some universality properties, in the sense that they do not depend on details of the distribution of the matrix elements. For instance, the spectral gap, i.e. the s …
3
votes
1
answer
158
views
Integrability of complex gaussian random matrix model
It is known that the partition function
$$ \mathcal{Z}_1=\int dH e^{-N{\rm Tr}(H^2)}e^{-NV(H)},$$ where the integral is over $N\times N$ hermitian matrices $H$, with the potential $$ V(H)=\sum_{j\ge 1 …
2
votes
Calculate correlation values of an ensemble of $N\times N$ real asymmetric random matrix fro...
Just expanding a bit on the comment by Beenakker. If you write the Gaussian measure in terms of the matrix elements,
$$ \prod_{ij}\exp[−\frac{N}{2(1−τ^2)}(J_{ij}^2-\tau J_{ij}J_{ji})]$$
You can see …
7
votes
Moments of the trace of orthogonal matrices
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 …
5
votes
1
answer
519
views
Riemann-Hilbert approach to Selberg integral
I am interested in matrix integrals, and I have seen many mentions to a certain Riemann-Hilbert approach that indicate that this is a very powerful tool to can be used in this area, when coupled with …
1
vote
1
answer
128
views
Distribution of top left block from unitary symmetric matrices
If $U$ is a $N\times N$ random unitary matrix uniformly distributed with respect to Haar measure, a $M\times M$ block $A$ from it has distribution given by
$$ \det(1-AA^\dagger)^{N-2M}.$$
If $O$ is a …
2
votes
Non combinatorial random matrix theory
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 …
1
vote
A Gaussian integral over complex variables by a defined Green's function for a Gaussian ense...
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 …
4
votes
1
answer
317
views
Average of product of matrix elements in the special orthogonal group
Given two lists $i$ and $j$ of $2n$ positive integers less than $N$, Collins and Sniady have computed, in Integration with respect to the Haar measure on unitary, orthogonal and symplectic group (see …
1
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Eigenvectors of random unitary matrices
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 …