I have asked this in MSE [here](https://math.stackexchange.com/questions/2854782/random-complex-eigenvalues-and-averages-of-traces), but got no interesting answers.


Suppose I have a random matrix $M$ of dimension $N$ which is real, but not symmetric. Suppose I know that, for large $N$, the marginal distribution of its eigenvalues is uniform over the unit disk in the complex plane. What does that tell me about the average values of traces, $\langle {\rm Tr}(M^n)\rangle$?


Of course $\langle {\rm Tr}(M^n)\rangle \to 0$ as $N\to\infty$, but can I know the asymptotics?


Does it follow from the "disk law" for the eigenvalues that the singular values have a quarter-circle distribution?



Random complex eigenvalues and averages of traces
Summary: I don't think the average $\langle {\rm Tr}(M^n)\rangle \to 0$ for $N\rightarrow\infty$, at least for $n$ even I will argue this average is $\propto\sqrt N$ because of the contributions from eigenvalues on the real axis.




---


• Consider first the eigenvalues $\lambda\_p=r\_p e^{i\phi\_p}$ of $M$ off the real axis. Because $M$ is real these must come in complex conjugate pairs, $\lambda\_p=\lambda^\ast\_{N-p}$, so 
$${\rm Tr}\, M^n=2\sum\_{p=1}^{N/2} r\_p^n \cos n\phi\_p,$$
and the average 
$$\langle{\rm Tr}\,M^n\rangle\_{\rm complex}=N\int\_0^\infty dr \int\_0^{2\pi}d\phi\, P(r,\phi)\, r^{n+1} \cos n\phi.$$
Under the assumption that the eigenvalue density $P(r,\phi)$ in the complex plane is rotationally invariant in the unit disc this average is zero.


• Now turn to the eigenvalues on the real axis. There are $N\_{\rm real}=c\sqrt N$ real eigenvalues for $N\gg 1$, where the coefficient $c$ depends on the distribution of $M$. (For a Gaussian $M$ one has $c=\sqrt{2/\pi}$.) The real eigenvalues are uniformly distributed in the interval $(-1,1)$. So their contribution to the average is
$$\langle {\rm Tr}\,M^n\rangle\_{\rm real}=c\sqrt{N}\int\_{-1}^{1} r^n dr=\begin{cases}
\frac{2c}{n+1}\sqrt{N}&\text{if n is even},\\
0&\text{if n is odd}.
\end{cases}$$


It is likely that the deviations from uniformity in the distribution of complex eigenvalues also give a nonzero contribution of the same order $\sqrt N$ as those from the real axis. (Basically because the eigenvalues that are pinned to the real axis repel nearby eigenvalues with nonzero imaginary part, see plot below.)




---


![](https://ilorentz.org/beenakker/MO/Ginibre_etal.png)
![](https://ilorentz.org/beenakker/MO/squarerootlaw.png)


Illustration of the eigenvalue distribution in the complex plane for three different ensembles of real random matrices (of size $N=100$). You can see the eigenvalues pinned to the real axis (they appear as a horizontal black line of dense points), surrounded by a depletion zone (white region of few points). The ensemble of Hamiltonian matrices has an additional symmetry that also pins eigenvalues to the imaginary axis. In each case the number $N\_X$ of real eigenvalues scales as $\sqrt N$, with a different numerical prefactor (red line for Hamiltonian ensemble, blue for skew-GOE, Ginibre has $\sqrt{2N/\pi}$).  

Figures from [arxiv:1305.2924.](https://arxiv.org/abs/1305.2924)


