Parry measure

Remarks

The matrix $P$ is stochastic and $p^T P = p^T$, so these values are consistent and $\mu$ is a $\sigma$-invariant measure. Its entropy is $\log \lambda$, which is the topological entropy of $\sigma$ on $\Sigma_A$, and it is the only invariant probability measure with this entropy. Example: the golden mean shift has $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ and $\lambda = \varphi = (1 + \sqrt5)/2$. Here $u = v = (\varphi, 1)$ up to scaling, since $\varphi + 1 = \varphi^2$. After normalisation, $p_1 = \varphi^2/(\varphi^2 + 1)$ and $p_2 = 1/(\varphi^2 + 1)$, and $P_{11} = 1/\varphi$, $P_{12} = 1/\varphi^2$, $P_{21} = 1$, $P_{22} = 0$. Each row of $P$ sums to $1$, because $1/\varphi + 1/\varphi^2 = 1$. Non-example: on the same shift, the Markov measure with $P_{11} = P_{12} = 1/2$ is invariant for its stationary vector but is not the Parry measure, and its entropy is smaller than $\log \varphi$.