Lyapunov Exponents for $n$-Dimensional Matrix $A$

Question

I am wondering whether my solve is correct. I know how to solve the 2, or 3 dimension of the state matrix. But what if the state matrix goes to n-dim? Here is what I tried:

To find the Lyapunov exponents for the matrix $A$, where $ A = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{bmatrix} $ and $a_{ij}$ are real random variables with $\mathbb{E}[\log |a_{ij}|] < \infty$, you can use Oseledet's multiplicative ergodic theorem.

The Lyapunov exponents $\lambda_k$ are defined as: $ \lambda_k = \lim_{{m \to \infty}} \frac{1}{m} \log \left\| \prod_{{i=1}}^m A_i \right\|, $ where $A_i$ are the random matrices in the product. In your case, $A_i = A$ for all $i$.

To calculate $\prod_{i=1}^m A_i$, you can compute $A^m$ for any positive integer $m$ using standard matrix multiplication. The expression for $A^m$ will involve $a_{ij}$ raised to the power $m$ in the matrix entries.

After finding $A^m$, you can compute its operator norm, denoted as $\|A^m\|$, which is the maximum singular value of the matrix $A^m$. Finally, you can calculate the Lyapunov exponents using the formula: $ \lambda_k = \lim_{{m \to \infty}} \frac{1}{m} \log \|A^m\|. $ Finding a general closed-form expression for the Lyapunov exponents for $n$-dimensional matrices with random entries like $A$ can be quite challenging and may depend on the specific distributions of the $a_{ij}$ entries. The approach involves computing the matrix powers and singular values, which can become increasingly complicated as $n$ grows. Clarify the question when $m\to n$ of the dimension of the state matrix, how to find an exponent of the matrix above.

You might need to work with specific distributions or assumptions to simplify the calculations.

Answer 1

I think your question would be what is the difference between $A$ is 2 or 3 dimensions and $A$ is an n-dim state matrix when finding the Lyapunov exponent?

If so, the most regular method is Oseledet's multiplicative ergodic. Let me first use a 2-dim state matrix to show how to adopt this theorem for finding Exponent:

To find the Lyapunov exponents for the given matrix $A$, you can use Oseledet's multiplicative ergodic theorem. First, let's define the matrix $A$ as:

$$ A = \begin{bmatrix} 0 & a \\ 1 & b \end{bmatrix} $$

where a and b are real random variables with finite expectations, i.e., $\mathbb{E}[\log |a|] < \infty$ and $\mathbb{E}[\log |b|] < \infty$.

The Lyapunov exponents are defined as:

$$ \lambda_k = \lim_{n \to \infty} \frac{1}{n} \log \left\| \prod_{i=1}^n A_i \right\| $$

where $A_i$ are the random matrices in the product. In your case, $A_i = A$ for all $i$, as given. To find the Lyapunov exponents, we need to calculate the product of these matrices and then compute the limit.

Let's calculate $\prod_{i=1}^n A_i$:

$$ \prod_{i=1}^n A_i = A^n = \begin{bmatrix} 0 & a \\ 1 & b \end{bmatrix}^n $$

And

$ \lambda_k = \lim_{n \to \infty} \frac{1}{n} \log \|A^n\| $.

Similarity, for $n$-dim matrix, one can use Oseledet's multiplicative ergodic theorem to find the Lyapunov exponents for the matrix $A$, where $$ A = \begin{bmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \ldots & a_{nn} \end{bmatrix} $$ and $a_{ij}$ are real random variables with $$\mathbb{E}[\log |a_{ij}|] < \infty$$

The Lyapunov exponents $\lambda_k$ are defined as: $$ \lambda_k = \lim_{{m \to \infty}} \frac{1}{m} \log \left\| \prod_{{i=1}}^m A_i \right\|, $$ where $A_i$ are the random matrices in the product. In your case, $A_i = A$ for all $i$.

Remark that when calculating $\prod_{i=1}^m A_i$, the most regular method is to compute $A^m$ for any positive integer $m$ using standard matrix multiplication. The expression for $A^m$ will involve $a_{ij}$ raised to the power $m$ in the matrix entries.

After finding $A^m$, you can compute its operator norm, denoted as $\|A^m\|$, which is the maximum singular value of the matrix $A^m$. Finally, you can calculate the Lyapunov exponents using the formula: $$ \lambda_k = \lim_{{m \to \infty}} \frac{1}{m} \log \|A^m\|. $$

Finding a general closed-form expression for the Lyapunov exponents for $n$-dimensional matrices with random entries like $A$ can be quite challenging and may depend on the specific distributions of the $a_{ij}$ entries.