Extreme eigenvalues of real symmetric matrix with main diagonal variance twice non-diagonal - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T08:28:57Z http://mathoverflow.net/feeds/question/100354 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100354/extreme-eigenvalues-of-real-symmetric-matrix-with-main-diagonal-variance-twice-no Extreme eigenvalues of real symmetric matrix with main diagonal variance twice non-diagonal JD Vlok 2012-06-22T13:34:51Z 2012-06-22T13:34:51Z <p><strong>Main question</strong></p> <p>Suppose there exists a random real symmetric $N \times N$ matrix <strong>$A$</strong> with the main diagonal elements distributed according to $\mathcal N(\mu = 0, \sigma^2 = 4N)$, while all non-diagonal elements are distributed according to $\mathcal N(0, 2N)$. What are the limits of the eigenvalue spectrum (i.e. the lower limit of the smallest eigenvalue and the upper limit of the largest eigenvalue)?</p> <p><strong>More information</strong></p> <p>$A = X^T W + W^TX$ where each element of $X$ equals $\pm 1$ and each element of $W$ is independently drawn from $\mathcal N(0,1)$. Each element of $X^TW$ (and $W^TX$) is the sum of $N$ i.i.d. random variables such that each element is distributed according to $\mathcal N(0,N)$. Summing $X^T W$ and $W^T X$ will therefore result in the matrix as described in the main question above.</p> <p><strong>Possible solutions</strong></p> <ol> <li><p>Wigner's semicircle law might provide a solution, though the elements of $A$ are not i.i.d. (esp. the difference in variance on the main diagonal)</p></li> <li><p>Xingzhi Zhan's "Extremal eigenvalues of real symmetric matrices with entries in an interval" (SIAM J. Matrix Anal. Appl. Vol. 27, No. 3, pp. 851-860) provides a solution to some extent. The only issue is limiting the normal distribution, which does not provide an <em>exact</em> solution.</p></li> </ol>