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handout7 - 18.338J/16.394J: The Mathematics of Infinite...

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Unformatted text preview: 18.338J/16.394J: The Mathematics of Infinite Random Matrices Numerical Methods in Random Matrices Per-Olof 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 zero-mean, 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 straight-forward 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)*(lmax-2*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 double-precision 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 zero-mean Gaussian with variance 2, and d is the square-root 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 double-precision 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]....
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handout7 - 18.338J/16.394J: The Mathematics of Infinite...

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