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Let $A_n$ be the matrix product of $n$ i.i.d. N-by-N random complex matrices. The matrix distribution is not fixed and can be tuned to suit specific solution if needed, as long as it's not too "special". One possible choice is with i.i.d. uniform or Gaussian entries with 0 mean. What I want to know is the asymptotic bahavior of the ratio $r_n$ between the second greatest singular value and the greatest of $A_n$ with respect to $n$. (Dimension of the matrices $N$ is fixed and may not be large)

This question comes from a physics problem and we actually want to prove that this ratio will decay exponentially with respect to $n$: $r_{n\rightarrow\infty}\propto exp({-\alpha n}) $, with probability 1, which is supported by some numerical simulation.

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So this is correct. The theorem that you need is the multiplicative ergodic theorem. Expressing it in your language, it states that $\frac 1n\log s_i(A_n)\to\lambda_i$, where $s_i$ is the $i$th singular value of the matrix product; and $\lambda_i$ is the $i$th Lyapunov exponent of the system. In order to have what you want, you require that $\lambda_1$, the largest Lyapunov exponent of the system, is strictly larger than $\lambda_2$, the second largest Lyapunov exponent. This is guaranteed by a theorem of Guivarc'h and Raugi ("Products of Random Matrices: Convergence Theorems") that applies in any of the models for the randomness that you are assuming. That paper gives criteria that guarantee that the top Lyapunov exponent is simple (which is exactly what you want). Later papers due to Gol'dsheid and Margulis gave criteria for simplicity of the entire Lyapunov spectrum.

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  • $\begingroup$ Thanks for such a quick answer! Just one more question: I looked up the multiplicative ergodic theorem but I can only find its description in terms of real matrices instead of complex ones that we need. Is the extension to complex domain straightforward? $\endgroup$ – Z. Zhang Apr 7 '17 at 14:37
  • $\begingroup$ Raghunathan's proof in the Israel Journal of Mathematics applies to matrices with entries in $\mathbb C$. I'm less sure of where you would find the simplicity results, although this is almost certainly a fairly straightforward generalization of the real case. $\endgroup$ – Anthony Quas Apr 7 '17 at 15:00
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An addendum to Anthony's answer: Very explicit conditions on the simplicity of the top Lyapunov exponent (and much other cool stuff) can be found in the paper by Breuillard and Gelander:

Breuillard, E.; Gelander, T., On dense free subgroups of Lie groups, J. Algebra 261, No.2, 448-467 (2003). ZBL1014.22007.

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