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I am reading a paper using random matrix theory to calculate Lyapunov spectrum.

What particularly confuses me is

Why is the Lyapunov spectrum simply the inverse of $\chi(x)$ (the probabilistic cummulative distribution of the eigenvalues of the Jacobian $D_{ij}(t_s)$; i.e. the integration of (the eigenvalues' real parts') semicircular density function)?

Moreover, shouldn't it be the Oseledets $\Lambda$ instead of $D_{ij}(t_s)$? Because when $N\to \infty$, Lyapunov spectrum approximates the probability distribution of eigenvalues of the Jacobian/Oseledets matrix?

Related: The eigenvalue/singular values of (large) square random matrices

UPDATE:

$\chi(\lambda_i) = P(\lambda \le \lambda_i) = \frac{N+1-i}N$ for $\lambda_i$ is decreasing. Then $$\chi^{-1}\left(\frac{N+1-i}N\right) = \lambda_i.$$

Why the Oseledets $\Lambda$ instead of $D_{ij}(t_s)$? Perhaps because in this case $D_{ij}$ is constant ($J-I$). So $\Lambda=(T_t^T T_t)^{\frac 1 {2t}}=((D^t)^T (D^t))^{\frac 1 {2t}}=(D^T D)^{\frac12}$.

Why the real parts of eigenvalues of $D$? Possibly the singular values of a matrix are always real parts of the eigenvalues? (NOT sure.) For example, let $A=\left[ \begin{matrix} 1 & i \\ -i & i \end{matrix}\right].$

As said here, $D$'s eigenvalues follow the circular law.

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  • $\begingroup$ It is perhaps simply because $\frac{\#\lambda\in[i, i+n] }N = P(\lambda\in[i,i+n])$. Image we do a Monte Carlo experiment of 'dropping' a $\lambda$ into its range $[\min\lambda, \max\lambda]$ ('a line segment'), then # $\lambda$s falling into $[\lambda_i, \lambda_j]$ is proportional to the prob of $[\lambda_i, \lambda_j]$, = $\chi(\lambda_j)-\chi(\lambda_i)$. I.e. difference between 2 values of cumulative prob $\chi$ = prob of the random var being in the interval, $\chi(b)-\chi(a)=P([a,b])$.//equ10 is $\chi(\lambda_i) = P(\lambda \le \lambda_i) = \frac{N+1-i}N$ for $\lambda_i$ is decreasing. $\endgroup$
    – Charlie Chang
    Commented Jun 9, 2022 at 10:25
  • $\begingroup$ .. i.e. the value of a random var = $\chi^{-1}$(the corresponding cumulative prob). More formally, using probability *measure* (here as the density of $\lambda$), $P(\lambda\in[\lambda_i, \lambda_j]) = \int_{\lambda_i}^{\lambda_j} \mu(\lambda) d\lambda=\chi(\lambda_j)-\chi(\lambda_i)$. So, no magic here. $\endgroup$
    – Charlie Chang
    Commented Jun 9, 2022 at 10:37

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