I have a question about the proof of proposition $3.3.6(3)$ in "Tensor Categories" by Etingof et al.. 


This part states that for $A$, transitive unital $\mathbb Z\_+$-ring, there is a unique character taking non-negative values on the basis elements.


The proof uses the fact that if $\chi$ is a character, and $f$ is a vector with entries $\chi(Y)$ for basis elements $Y$, then $f$ is an eigenvector of the matrix of left multiplication by $\sum\_{X\in I}X$, where $I$ is a set of basis elements.


But I don't see why this holds. As I understand we want an equation:
$$
\sum\_{X,Y\in I} \chi(Y) XY = \lambda \sum\_{Z\in I} \chi(Z) Z ,
$$
to hold, so $\sum\_{X,Y\in I} \chi(Y) c\_{XY}^Z = \lambda \chi(Z)$. But I do not understand what to do next.


In bibliographical notes there is a link to the paper <https://arxiv.org/pdf/math/0203060.pdf> (Lemma 8.3). Which uses more or less the same fact with $X$ instead of $\sum\_{X\in I} X$ as a matrix, which is even more mysterious to me.


Can someone please clarify this for me?



Uniqueness of character for Z_+-rings
Hm, I probably figured it out myself, but I won't delete the question since I think the formulation in the proof is slightly misguiding. 


Use $b\_i$ to denote basis elements ($i \in I$), a take element $y = \sum y\_i b\_i$ and denote $\chi\_i = \chi(b\_i)$. We have $\chi(y) = \sum y\_i \chi\_i$. But if the multiplication by $x$ -- any element of the ring is given by matrix $A\_{ij}$ we have $xy = \sum b\_iA\_{ij}y\_j$. Now take character of both sides to obtain:
$$
\chi(x) \sum \chi\_j y\_j = \sum \chi\_i A\_{ij} y\_j \ ,
$$
since this holds we have $\chi(x) \chi\_i = \sum \chi\_i A\_{ij}$, so the row vector $\chi\_i$ is indeed the eigenvector of $A\_{ij}$. Now the rest of the proof works the same since the can use the Frobenius-Perron theorem.


