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My question is highlighted in bold at the end.

$\mathrm{\underline{Background}}$

Consider a product of i.i.d. $d\times d$ random matrices $A_{i}$ (with $\mathbb{E}\log\left\Vert A_{i}\right\Vert <\infty$) acting on a non-zero vecor $X$, i.e. $$ A_{n}\cdots A_{1}X. $$ The Lyapunov exponents are used to describe the exponential growth properties of $$ \left\Vert A_{n}\cdots A_{1}X\right\Vert . $$ Thus we define the the Lyapunov exponent as $$ \lambda\left(X\right):=\lim_{n\to\infty}\frac{1}{n}\log\left\Vert A_{n}\cdots A_{1}X\right\Vert \;\;(1). $$ According to the Muliplicative Ergodic Theorem (or Furstenberg-Kesten Theorem) the number of distinct values $p$ that ($1$) can take is at most $d$, i.e. we have $$ \lambda_{d}\leq\cdots\leq\lambda_{1}. $$ Now, let $\sigma_{i,n}$ be the $i$th singular value of the matrix product $A_{n}\cdots A_{1}$ such that $$ \sigma_{d,n}\leq\cdots\leq\sigma_{1,n}. $$ Furstenberg-Kesten Theorem states that $$ \lim_{n\to\infty}\frac{1}{n}\log\sigma_{i,n}=\lambda_{i}.\;\;(2) $$ The crux of my question (to follow) revolves around ($2$).

$\underline{\mathrm{Question\;Setup: Independent\;nonidentical\;matrices}}$

Consider random scalars $a_{i},b_{i},c_{i}$ where $b_{i}s$ are i.i.d. with $\mathbb{E}\log\left|b_{1}\right|<\infty$, $c_{i}$s are i.i.d. $\mathbb{E}\log\left|c_{1}\right|<\infty$ and $a_{i}s$ are independent but for any $i$ there exists a finite non-zero not-necessarily unitary constant $\alpha_{i}$ such that $a_{1}\overset{d}{=}\alpha_{i}a_{i}$ ($\overset{d}{=}$ denotes equality in distribution) with $\mathbb{E}\log\left|a_{1}\right|<\infty$ . It is easy to show that with $B_{i}$ defined as $$ B_{i}:=\left(\begin{array}{cc} a_{i} & b_{i}\\ 0 & c_{i} \end{array}\right) $$ $$ \mathbb{E}\log\left\Vert B_{i}\right\Vert <\infty\Longleftrightarrow\mathbb{E}\log\left|a_{1}\right|<\infty,\mathbb{E}\log\left|b_{1}\right|<\infty,\mathbb{E}\log\left|c_{1}\right|<\infty. $$ The Lyapunov index of $a_{i}$ is given by $$ \lim_{n\to\infty}\frac{1}{n}\log\left|a_{n}\cdots a_{1}\right|=\lim_{n\to\infty}\frac{1}{n}\log\left|\alpha_{n}\cdots\alpha_{1}\right|+\mathbb{E}\log\left|a_{1}\right|, $$ that of $b_{i}$ is given by $$ \mathbb{E}\log\left|b_{1}\right| $$ and that of $c_{i}$ is given by $$ \mathbb{E}\log\left|c_{1}\right|. $$

With $\sigma_{i,n}$ $i=1,2$ defined as the singular values of $A_{n}\cdots A_{1}$, it can be shown that $$ \frac{1}{n}\log\sigma_{1,n}\to\max\left\{ \lim_{n\to\infty}\frac{1}{n}\log\left|\alpha_{n}\cdots\alpha_{1}\right|+\mathbb{E}\log\left|a_{1}\right|,\mathbb{E}\log\left|c_{1}\right|\right\} \;\; (3) $$ and $$ \frac{1}{n}\log\sigma_{2,n}\to\min\left\{ \lim_{n\to\infty}\frac{1}{n}\log\left|\alpha_{n}\cdots\alpha_{1}\right|+\mathbb{E}\log\left|a_{1}\right|,\mathbb{E}\log\left|c_{1}\right|\right\} \;\; (4). $$ Here is my question: As stated previously, according to Furstenberg for i.i.d. matrices $A_{i}$, the limiting behavior of the singular values coincide with the Lyapunov exponents of $A_{i}$, see ($2$). Is this true if one considers my non-i.i.d. setup? In other words, is it right to say that ($3$) and ($4$) are the Lyapunov exponents of $B_i$?

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Yes, it is true. However, in your question you mix up a number of things (to begin with, it is Kesten, not Keston). The Multiplicative Ergodic Theorem is not the same as the Furstenberg-Kesten theorem. You can find answers to all your questions in this article.

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