Lyapunov exponents of dual / adjoint / transpose random dynamical system (RDS)

Question

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, 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?

Answer 1

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.

Answer 2

No. The "theory" which you refer to is actually wrong. The mistake is that you confuse several conditions on a sequence of invertible matrices $B_n$. In the order of strengthening these conditions are

(1) Existence of exponential growth rates of respective singular values of $B_n$;

(2) Existence of Lyapunov exponents $\chi(x)=\lim_n \frac1n \log\|B_n x\|$ for any non-zero vector $x$;

(3) Lyapunov regularity, i.e., additionally to (2) existence of $\chi_{\det}=\lim_n \frac1n \log |\det B_n|$ and its coincidence with the sum of Lyapunov exponents (taken with their multiplicity).

It is precisely the third condition which was originally introduced by Lyapunov, and the one which is used in numerous applications. The reason why it is often confused with the other ones is that they become equivalent under the additional assumption that the "increments" $A_n=B_n B_{n-1}^{-1}$ (so that $B_n=A_n \dots A_1$) are moderate in the sense that $\log\|A_n\|$ and $\log|\det A_n |$ are both sublinear (in particular, this assumption is always satisfied when $A_i$ are stationary with $\mathbf E \log\|A_i\|, \mathbf E \log\|A_i^{-1}\|<\infty$). However, without this assumption condition (1) does not imply even existence of the limits from condition (2), let alone full Lyapunov regularity! In particular, Lyapunov regularity of a sequence $B_n$ does not imply Lyapunov regularity of the transposed sequence $B_n^\dagger$ in spite of precisely the same behavior of singular values for both sequences. In your setup, in fact, the sequence $B_n^\dagger$ is a.e. not Lyapunov regular for any reasonably non-degenerate distribution of $A_i$.