Let $B$ be an integral matrix and suppose that there is a primitive (irreducible and aperiodic) matrix $A$ such that $\lambda_A = \lambda_B$ (i.e., their spectral radius are the same). Moreover, suppose that there are integral matrices R and S with adequate orders and some $l \geq 0$such that $AR=RB$ and $SR = B^l$. I need to show that $B$ is spectrally Perron, i.e., there is a simple eigenvalue $\lambda \geq 1$ of $B$ such that $\lambda > |\mu|$ for all other eigenvalues $\mu$ of $B$.
Since $A$ is primitive, we already know that $\lambda_A \geq 1$, so that we get for free that $\lambda_B \geq 1$. Also, from Perron Theorem, we also know that $\lambda_A$ is simple (with respect to the matrix A, of course) and $\lambda > |\mu|$ for all other eigenvalues $\mu$ of $A$. We need to show that those properties are still true for $B$ and I was trying to use the existence of $R$ and $S$ to conclude this. One thing I know is that $\lambda_B^l$ is an eigenvalue for $SR$, but I am not sure how to use this information.
Any help is appreciated.