Consider the the state of a system at time $n$, $X\_n$, as the action of a product of i.i.d. $d\times d$ random matrices acting on a $d$ dimensional vector $X\_0$, so we have


$$X\_n = A\_n \cdots A\_1X\_0.$$


Suppose $\mathbb E\log \Vert A\_i \Vert <\infty$ and $A\_i$ takes the form


$$A\_i =
\left[\begin{array}{cc}
H\_i & B\_i\\
\underline{0} & 1
\end{array}\right]
$$
where $H\_i$ is a $(d-1) \times (d-1)$ random matrix, $B\_i$ is a $(d-1)\times 1$ random vector and $\underline{0}$ is the zeros row vector. From [[1](http://projecteuclid.org/euclid.aop/1176991593), Thrm. $5$], it can be seen that the Lyapunov exponents 
$$\lambda\_i = \lim\_{n\to\infty} \frac{1}{n}\log \sigma\_i(A\_n\cdots A\_1)$$
($\sigma\_i(A)$ denotes $i$th ordered singular value of $A$) of the above random dynamical system (RDS) are given by those of $H\_i$ and an extra equal to $0$. Note, the Lyapunov exponents describe the exponential growth / decay characteristics of the system. Intuitively, this makes sense because if we consider initial states given by
$$
X\_{0}=\left[\begin{array}{c}
x\_{0}\\
\vdots\\
x\_{d-1}\\
0
\end{array}\right]
$$
it is easy to see that the growth / decay of the system is entirely governed by $H\_i$. Similarly, if all elements are $0$ except the final, the system state will display no growth (as suggested by an exponent of $0$).


Now, consider the dual / adjoint / transpose system given by
$$X\_n = A\_1^\dagger \cdots A\_n^\dagger X\_0$$
($A^\dagger$ denotes conjugate transpose of $A$). Because 
$$\sigma\_i(A\_n\cdots A\_1) = \sigma\_i(A\_1^\dagger\cdots A\_n^\dagger),$$ the Lyapunov exponents of the systems should be equal, **right?** However, if we inspect the behaviour of the transpose system by considering 
$$X\_{0}=\left[\begin{array}{c}
\hat{X}\_{0}\\
x\_{d}
\end{array}\right]$$ ($\hat X\_0$ a $d-1$ dimensional vector) we see that, e.g.,
$$X\_2 = \left[\begin{array}{c}
H\_{1}^{\dagger}H\_{2}^{\dagger}\hat{X}\_{0}\\
B\_{1}^{\dagger}H\_{1}^{\dagger}\hat{X}\_{0}+B\_{2}^{\dagger}\hat{X}\_{0}+x\_d
\end{array}\right].$$ And in general, the $n$th state will always contain terms in its $d$th dimension displaying no growth or decay. Thus, in particular, $\Vert X\_n\Vert \not\to 0$. Consequently, the trasnpose system can never have exponential decay behaviour (i.e., negative Lyapunov exponents).


**My Question**: If $H\_i$ has negative Lyapunov exponents will the transpose system have negative Lyapunov exponents? Theory suggests that it will, but intuition suggests that no initial states can decay exponentially (i.e., have magnitudes that decay exponentially). Can anyone help me with my confusion?



Lyapunov exponents of dual / adjoint / transpose random dynamical system (RDS)
The theory is right! If $H\_i$ has negative Lyapunov exponents, then images of any vector under the transpose system will be exponentially close to a single vector (everything is being compressed in the initial components). There is then a co-dimension 1 subspace where the components of this vector cancel.


The difficulty is that to decide if you're in this subspace or not, you need to follow the entire orbit - whereas the structure of the matrices before meant that you could "see" which part belonged to which block. 


