We prove that for an irreducible subshift of finite type $(\Sigma_A, \sigma)$ with transition matrix $A$, the topological entropy is $h_{top}(\sigma) = \log \lambda(A)$ where $\lambda(A)$ is the spectral radius.

Step 1: Counting periodic orbits

A period-$n$ orbit corresponds to a closed loop in the transition graph. The number of such loops equals $\text{tr}(A^n)$—the trace of $A^n$ counts walks of length $n$ from each state to itself.

By the spectral theorem, for large $n$:

$\text{tr}(A^n) = \sum_{i=1}^k \lambda_i^n \approx \lambda(A)^n$

since $\lambda(A) > |\lambda_i|$ for all other eigenvalues $\lambda_i$ (by Perron-Frobenius).

Step 2: Entropy via orbit growth

Topological entropy measures exponential growth of distinguishable orbits. One definition is:

$h_{top}(\sigma) = \lim_{n \to \infty} \frac{1}{n} \log N_n$

where $N_n$ is the number of distinct orbit segments of length $n$. For SFTs, $N_n = \text{tr}(A^n)$ (each allowed word of length $n$ corresponds to an $n$-step path).

Step 3: Taking the limit

$h_{top}(\sigma) = \lim_{n \to \infty} \frac{1}{n} \log \text{tr}(A^n) = \lim_{n \to \infty} \frac{1}{n} \log(\lambda(A)^n + \text{lower order terms})$

Since $\lambda(A)^n$ dominates for large $n$:

$h_{top}(\sigma) = \lim_{n \to \infty} \frac{1}{n} \cdot n \log \lambda(A) = \log \lambda(A)$

Step 4: Rigor via Perron-Frobenius

The Perron-Frobenius theorem guarantees that for irreducible $A$:

• $\lambda(A)$ is positive and simple
• All other eigenvalues satisfy $|\lambda_i| < \lambda(A)$

This ensures the asymptotic $\text{tr}(A^n) \sim c \lambda(A)^n$ for some constant $c > 0$, justifying the limit calculation.

Conclusion: The topological entropy of an SFT equals the logarithm of the spectral radius of its transition matrix.