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arXiv 1503.07926math.PRhep-thmath-phmath.MPnlin.SI
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一个大的随机矩阵没有实特征值的概率是多少？

What is the probability that a large random matrix has no real eigenvalues?

Eugene Kanzieper, Mihail Poplavskyi, Carsten Timm, Roger Tribe, Oleg Zaboronski

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23 pages, 1 figure
发表信息：
The Annals of Applied Probability 2016, Vol. 26, No. 5, 2733-2753
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摘要：

研究了从实Ginibre系综中抽样的随机$2n\times 2n$矩阵具有$2k$实特征值的概率$p_{2n,2k}$的大$n$极限.我们证明了，$$\lim_{n\rightarrow \infty}\frac {1}{\sqrt{2n}} \log p_{2n,2k}=\lim_{n\rightarrow \infty}\frac {1}{\sqrt{2n}} \log p_{2n,0}= -\frac{1}{\sqrt{2π}}ζ\left(\frac{3}{2}\right),$$其中$ζ$是黎曼ζ函数。此外，对于任何非负整数序列$(k_n)_{n\geq 1}$，$$\lim_{n\rightarrow \infty}\frac {1}{\sqrt{2n}} \log p_{2n,2k_n}=-\frac{1}{\sqrt{2π}}ζ\left(\frac{3}{2}\right),$$提供了$\lim_{n\rightarrow \infty} \left(n^{-1/2}\log(n)\right) k_{n}=0$ .

英文摘要：

We study the large-$n$ limit of the probability $p_{2n,2k}$ that a random $2n\times 2n$ matrix sampled from the real Ginibre ensemble has $2k$ real eigenvalues. We prove that, $$\lim_{n\rightarrow \infty}\frac {1}{\sqrt{2n}} \log p_{2n,2k}=\lim_{n\rightarrow \infty}\frac {1}{\sqrt{2n}} \log p_{2n,0}= -\frac{1}{\sqrt{2π}}ζ\left(\frac{3}{2}\right),$$ where $ζ$ is the Riemann zeta-function. Moreover, for any sequence of non-negative integers $(k_n)_{n\geq 1}$, $$\lim_{n\rightarrow \infty}\frac {1}{\sqrt{2n}} \log p_{2n,2k_n}=-\frac{1}{\sqrt{2π}}ζ\left(\frac{3}{2}\right),$$ provided $\lim_{n\rightarrow \infty} \left(n^{-1/2}\log(n)\right) k_{n}=0$.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

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