Is the Perron-Frobenius dimension of a G-Set given by its cardinality? Given a ring $R$ with finite additive basis $\{e_i\}_{i=1}^{n}$, such that $e_i e_j=\sum c_{ijk}e_k$ with $c_{ijk}\in \mathbb{N}$, we define the Perron-Frobenius dimension $FPDim(e_i)$ of a basis element $e_i$ to be the maximal positive real eigenvalue of matrix $M_{e_i}$, multiplication by $e_i$. This exists by the Perron-Frobenius theorem, and we extend by linearity to all of $R$.
Now take $B(G)$ to be the burnside ring of a finite group $G$, with basis given by isomorphism classes of transitive actions of $G$. One can directly check for $G\cong C_p$ that $FPDim(X)=|X|$. Does this hold in general for finite $G$-sets?
Note that our ring $B(G)$ is not necessarily transitive in the sense of Etingof's Tensor Categories (Definition 3.3.1), so this doesn't seem to follow immediately from the results in there.
 A: For $H \subset G$, write $e_H$ for the basis element of $B(G)$ corresponding to the $G$-set $G/H$. Recall that for $K \subset G$, 
$$
e_H\cdot e_K = \sum\limits_{H g K \in H \backslash G / K} e_{H \cap g K g^{-1}}.
$$
Say that $K$ is subconjugate to $H$ if $K$ is conjugate to a subgroup of $H$. This gives rise to a partial order on the set of conjugacy classes of subgroups of $G$, and hence on the basis elements $e_H$. With respect to this partial ordering, $M_{e_H}$ is lower-triangular for each $H \subset G$. The eigenvalues are then the diagonal elements. The diagonal element corresponding to $e_{\{\textrm{id}_G\}}$ is $|G/H|$. The diagonal element corresponding to $e_K$ for $K \subset G$ is bounded above by the number of $(H, K)$-double cosets in $G$, which is bounded above by $|G/H|$.
A: Lets fix an orbit $X = G/H$. It suffices to determine the asymptotic growth of the trace of multiplication by $X^n$,  since the maximal positive eigenvalue(s) dominates  the sum $\sum_i \lambda_i^n$. 
This trace is the sum  $$\sum_{K \subset G}  \langle G/K, X^n \times G/K \rangle,$$ where the sum is over conjugacy classes of subgroups,  and the bracket counts the number of disjoint orbits of type $G/K$.  
Notice that the term $\langle G, X^n \times G \rangle$ equals $|X|^n$ (one free orbit for each $x^n \times e$).  Conversely, $\langle G/K, X^n \times G/K \rangle$ must have size $\leq |X|^n |G/K|/|G/K| = |X|^n$. 
Thus the exponent of the order growth is $|X|$,  and so $|X|$ is the maximal eigenvalue.  The class $[G]$ is an eigenvector corresponding to this eigenvalue.
Edit:  As Darij points out, in this case Perron--Frobenius only guarantees that $$ max_{\lambda \in {\rm eig}(A)} |\lambda|$$ can be achieved by some  positive real $\lambda$.  There may be other complex eigenvalues of the same absolute value. In this case more argument is required to ensure that there is not cancellation.
Edit2: Here is a strategy to resolve the issue. Let $z_1, \dots, z_k$ be the eigenvalues achieving the maximum.  Assume we can show that there exists an eplison such that  $$S = \{n \in \mathbb N ~|~ Re(z_i^n) \geq -\lambda + \epsilon ~ \forall i \}$$  is infinite.  Then there would be an infinite subset $S \subset \mathbb N$ and a $C \in \mathbb R_{> 1}$ such that $C \lambda^s \geq Tr(A^s) \geq 1/C \lambda^s$  for all $s \in S$.  
In our case,  we have   $$|X|^n \leq {\rm Tr}(A^n) \leq \#\{\text{ subgroups }\}|X|^n$$ for all $n$. Restricting to the infinite subset $S$ we would see that $\lambda = |X|$.
