$M$ is an iid random matrix with $M_{ij} \sim \mathcal{N}(0,\frac{g^2}N)$ except that the diagonal entries are $-1$.

I am to compute, in the limit $N\to\infty$,

  1. the eigenvalue/singular value spectrum/distribution. I.e. we order the eigenvalues according to their relevant values such that $\lambda_1 \ge \lambda_2 \ge \dots \ge \lambda_N$, and then we hope to know the mean of each $\lambda_i$) of a random matrix $M$: $$\bar \lambda_1, \bar\lambda_2, \dots, \bar \lambda_N$$
  2. the mean ($\ln$) thereof: $$\langle \ln \lambda \rangle = \frac{\ln\lambda_1 + \ln\lambda_2 + \dots + \ln\lambda_N}N.$$

It seems we can calculate $\langle \ln \lambda \rangle$ using $\det(M)$. Then how can we compute the determinant (also a statistics) of such a random matrix?

We possibly need to use the Wigner semicircular law.
Possible methods in random matrices or more general probability include

reductions, the Fourier method, the moment method, the Lindeberg swapping trick, individual swapping, Stein's method, Predecessor comparison.


I hope to know any books or papers about solving the above (or other) statistics of this (or simpler similar) random matrix, or of other 'symmetric' (w.r.t. probability distribution of each entry) matrices, or of other kinds of (large) square random matrices. The literature of random matrices is vast so I hope to narrow down the scope of my research. I can read them to find clues. Exact solutions are welcome but not necessarily expected right now.

(This section is unnecessary for answering the question:) An example of calculation is given as (note that $\mu_i$ below is actually eigenvalues denoted as $\lambda_i$ above. I hope to know what is more general than the example):


This seems to be a seminal paper in the topic: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.60.1895

A closely related post: Spectrum of large random asymmetric matrices with correlation

  • $\begingroup$ When we know the probability distribution of $\lambda$ or $\lambda_i$, the other relational statistics seems to simply follow from the standard probability calculation, e.g. $\det(M)=\int (\prod_i \lambda_i) \mu(\lambda_i, \dots, \lambda_N)$, where $\mu$ is the probability measure/density, though the details are still unclear. $\endgroup$ Jun 9, 2022 at 11:10

1 Answer 1


If I slightly modify your prescription, taking i.i.d. Gaussians (mean zero, variance $g^2/N$) for all matrix elements, including the diagonal elements, then your matrix is a member of the (real) Ginibre ensemble; for large $N$ this modification only subtracts 1 from each eigenvalue.

The density of the eigenvalues in the Ginibre ensemble is given by Girko's circular law, and the density of the singular values is given by the Marchenko-Pastur quarter circular law, see for example Around the circular law. These densities refer to the bulk of the spectrum, in addition there are of order $\sqrt N$ real eigenvalues, distributed uniformly in the interval $(-1,1)$. This line of real eigenvalues is clearly visible in the numerical data shown below (for $N=100$); notice that the line is surrounded by a narrow depletion region, since the complex eigenvalues are repelled from the real axis. See the plot below.

Eigenvalues in the complex plane of 200 real matrices of size 100 × 100 in the Ginibre ensemble, from https://arxiv.org/abs/1305.2924

  • $\begingroup$ 1. Why 'for large 𝑁 this modification only subtracts 1 from each eigenvalue'? You mean eigenvalues of $\Lambda$ are those of the Ginibre ensemble minus $1$?// 2. Is it true that for the complex Ginibre ensemble, the eigenvalues is homogeneously distributed within a disk $O_d$ of the complex plane (plot in en.wikipedia.org/wiki/Circular_law + Sommers's paper), and therefore for the real Ginibre ensemble, the probability density of eigenvalues will be projection of $O_d$ onto the real line, and become (actually twice, but the constant will not alter the prob distribution) semi-circular? $\endgroup$ Jun 6, 2022 at 10:56
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    $\begingroup$ 1) the difference between your matrix and the Ginibre matrix is the unit matrix plus a random Gaussian of variance $1/N$ on each diagonal element; the unit matrix simply subtracts 1 from each eigenvalue and the Gaussians become irrelevant for large $N$; 2) no, the real Ginibre ensemble still has complex eigenvalues, because the matrix is not symmetric; the eigenvalue density consists of the same unit disc as in the complex Ginibre ensemble, with the addition of $\sqrt N$ eigenvalues pinned to the real axis (surrounded by a narrow depletion strip) $\endgroup$ Jun 6, 2022 at 11:10
  • $\begingroup$ 1. I see, $\lambda_M v = M v = (G-I) v = G v-I v = \lambda_G v- v = (\lambda_G -1) v$, where $G$ is the real Ginibre ensemble. A minor difference is that $M$ has -1 in diagonal, while $G-I$ has gaussians$-1$. This might cause further complexity when $g$ is large but for small $g$ the approximation seems good enough. For very large $g$ perhaps $M$ would behave like $G$, while for $g$ near 1 $M$'s behavior might be unpredictable. $\endgroup$ Jun 6, 2022 at 11:21
  • $\begingroup$ Why '...still has complex eigenvalues, because the matrix is not symmetric'? $\endgroup$ Jun 6, 2022 at 12:55
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    $\begingroup$ a real matrix will have real eigenvalues if $M_{ij}=M_{ji}$; in your case that symmetry does not apply; its eigenvalues may be complex, and in fact what happens is that of order $\sqrt N$ of its eigenvalues are real, the remaining having a nonzero imaginary part. $\endgroup$ Jun 6, 2022 at 13:37

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