Let $G_\mathbb{R}\in\mathbb{R}^{n\times n}$ and $G_\mathbb{C}\in\mathbb{C}^{n\times n}$ denote the real and complex Ginibre random matrices, i.e. random matrices with independent real/complex Gaussian entries of zero mean and variance $\mathbb{E}\lvert(G_K)_{ij}\rvert^2=n^{-1}$ for $K=\mathbb{R},\mathbb{C}$. Denote the singular values of the matrix $X$ by $\sigma_1(X)\ge\dots\ge \sigma_n(X)$. It follows from the work of Edelman that \begin{equation}\mathbb{P}\Big( \sigma_n(G_K)\le\frac{x}{n}\Big) = \begin{cases} 1-e^{-x^2/2-x}+o(1),&K=\mathbb{R}\\ 1-e^{-x^2},&K=\mathbb{C}. \end{cases}\tag{*}\label{sing val}\end{equation} In particular, the smallest singular value is of order $n^{-1}$.

Now I am wondering about additive perturbations to $G_K$, say $G_K+\lambda I$, where $\lambda$ is some real (or even complex) parameter and $I$ is the identity matrix. In general the singular values of $G_K$ give little information about the singular values of $G_K+\lambda I$. The most one can hope for is bounds like $$\sigma_n(G_K+\lambda I)\ge \lvert\lambda\rvert-\sigma_1(G_K).$$ It is known that $\sigma_1(G_K)$ is Tracy-Widom distributed around $2$, so in particular, $\sigma_n(G_k+\lambda I)$ is bounded away from $0$ as long as $\lvert \lambda\rvert>2$.

Question: Is the analogue of $\eqref{sing val}$, i.e. the distribution of $\sigma_n(G_K+\lambda I)$ known for $G_K+\lambda I$? If exact formulae are not available, I would be interested in the average scaling of $\sigma_n(G_K+\lambda I)$. I guess there should be some phase transition of the type $$\mathbb E \sigma_n(G_K+\lambda I)\sim\begin{cases}n^{-1},&\lvert\lambda\rvert<c\\ n^{-?}, &\lvert\lambda\rvert=c,\\ 1, &\lvert\lambda\rvert>c.\end{cases}$$ I think $c=1$, but am unsure about the critical exponent.

  • $\begingroup$ the scaling $1/n$ is always a lower bound, no matter what $\lambda$ is, e.g. by results of Sankar, Spielman and Teng cs.yale.edu/homes/spielman/Research/nopivotdas.pdf (Theorem 3.3). I am not sure about complementary upper bounds, especially in the critical regime $\endgroup$ – ofer zeitouni Apr 12 at 18:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.