**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 $B\_i$? 



Lyapunov Exponents for independent-nonidentically distributed matrices?
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](http://www.ams.org/mathscinet-getitem?mr=947327).


