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Given a $n \times n$ symmetric random matrix whose diagonal elements are $1$, and the elements of $k\times k, k<n$ leading principal sub-matrix are $1$. All other values are i.i.d. uniformly randomly drawn from $[0,1]$.

For example, it could look like $\left( {\begin{array}{*{20}{c}} 1&1&*&* \\ 1&1&*&* \\ *&*&1&* \\ *&*&*&1 \end{array}} \right)$, where "$*$" is a decimal uniformly drawn from [0,1].

Anyone knows what this type of random matrix this, and its eigenvalue distribution, and the probability that its largest eigenvalue are bigger than $\frac{n}{2}$? Anyone can provide any reference to solve this problem? Thanks!


PS: I have been looking into random matrix materials, and they seem to assume Gaussian random matrix rather than this drawn from uniform. In Gaussian case the eigenvalue distribution is semi-circle.

PS2: I find this simulation. If every element is uniform from [0,1], a simulation shows the eigenvalue is possibly still semi-circle. https://blogs.sas.com/content/iml/2012/05/16/the-curious-case-of-random-eigenvalues.html. I am still searching proof for this simpler case. Our case is more complicated with the top-left principal sub-matrix has all its elements being 1.

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Let $A$ be your matrix. If $e$ is the vector of all $1$'s, the greatest eigenvalue of $A$ is at least $\dfrac{e^T A e}{e^T e} =\dfrac{e^T A e}{n}$. Now $e^TAe$ is the sum of the entries of $A$, which $\ge n^2/2$ with probability $> 1/2$ (since the sum of the entries that are not fixed at $1$ has a symmetric distribution about its mean).

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  • $\begingroup$ In fact, with probability $1/2$, this quantity is at least $(n^{2}+k^{2})/2$. $\endgroup$ Jan 12, 2018 at 3:40

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