This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 18.338J/16.394J: The Mathematics of Infinite Random Matrices Numerical Methods in Random Matrices PerOlof Persson Handout #7, Tuesday, September 28, 2004 1 Largest Eigenvalue Distributions In this section, the distributions of the largest eigenvalue of matrices in the ensembles are studied. His tograms are created first by simulation, then by solving the Painlev e II nonlinear differential equation. 1.1 Simulation The Gaussian Unitary Ensemble (GUE) is defined as the Hermitian n n matrices A , where the diagonal elements x jj and the upper triangular elements x jk = u jk + iv jk are independent Gaussians with zeromean, and braceleftBigg Var( x jj ) = 1 , 1 j n, Var( u jk ) = Var( v jk ) = 1 2 , 1 j < k n. (1) Since a sum of Gaussians is a new Gaussian, an instance of these matrices can be created conveniently in MATLAB: A=randn(n)+i*randn(n); A=(A+A)/2; The largest eigenvalue of this matrix is about 2 n . To get a distribution that converges as n , the shifted and scaled largest eigenvalue max is calculated as max = n 1 6 ( max 2 n ) . (2) It is now straightforward to compute the distribution for max by simulation: for ii=1:trials A=randn(n)+i*randn(n); A=(A+A)/2; lmax=max(eig(A)); lmaxscaled=n^(1/6)*(lmax2*sqrt(n)); % Store lmax end % Create and plot histogram The problem with this technique is that the computational requirements and the memory requirements grow fast with n , which should be as large as possible in order to be a good approximation of infinity. Just storing the matrix A requires n 2 doubleprecision numbers, so on most computers today n has to be less than 10 4 . Furthermore, computing all the eigenvalues of a full Hermitian matrix requires a computing time proportional to n 3 . This means that it will take many days to create a smooth histogram by simulation, even for relatively small values of n . To improve upon this situation, another matrix can be studied that has the same eigenvalue distribution as A above. In [1], it was shown that this is true for the following symmetric matrix when = 2: H 1 2 N (0 , 2) ( n 1) ( n 1) N (0 , 2) ( n 2) . . . . . . . . . 2 N (0 , 2) N (0 , 2) . (3) Here, N (0 , 2) is a zeromean Gaussian with variance 2, and d is the squareroot of a 2 distributed number with d degrees of freedom. Note that the matrix is symmetric, so the subdiagonal and the superdiagonal are always equal. This matrix has a tridiagonal sparsity structure, and only 2 n doubleprecision numbers are required to store an instance of it. The time for computing the largest eigenvalue is proportional to n , either using Krylov subspace based methods or the method of bisection [2]....
View
Full
Document
 Fall '02
 DimitriBertsekas

Click to edit the document details