Why is arbitrary linear representation of $S_3$ spanned by action of $\tau \in A_3$? Quoted this question.
I am reading  the book on representation theory by Fulton and Harris in GTM.  I came across this paragraph.

[..] we will start our analysis of an arbitrary representation $W$ of $S_3$ by looking just at the action of the abelian subgroup $\mathfrak A_3 = \mathbb Z/3 \subset \mathfrak S_3$ on $W$. This yields a very simple decomposition: if we take $\tau$ to be any generator of $\mathfrak A_3$ (that is, any three-cycle), the space $W$ is spanned by eigenvectors $v_i$ for the action of $\tau$, whose eivenvalues are of course all powers of a cube root of unity $\omega = e^{2\pi i/3}.$

Why is $W$ spanned by eigenvectors of $\tau$?
What I know is that $A_3$ is abelian, so it must be so. But how do I derive it from the construction of $\tau$?
Thank you in advance!
 A: Just echoing Tobias's remarks:
This part of the analysis does not use anything about $S_3$. The representation $\rho:S_3 \to \operatorname{GL}(W)$ is only used to define the single linear transformation $\rho(\tau)=\rho((123))$. This linear transformation has minimal polynomial dividing $x^3-1$, which has no multiple roots, so this linear transformation has a basis of eigenvectors. Said differently, the vector space $W$ has a basis of eigenvectors of $\rho(\tau)$. It does not claim anything about the representation $\rho$ (even when they call the representation $W$ to confuse you).
The nice thing about doing this is that we get the distinction between the 1-dimensional irreducible representations (with eigenvalue 1, but there are two non-isomorphic such representations of $S_3$) and the 2-dimensional irreducible representation (with eigenvalue $\omega$ or $\omega^2$). This is only so-so nice pedagogically because 1 eigenvalue gives two different representations, and 2 eigenvalues give one representation.
In the second half of the book, you'll look at better abelian subgroups whose eigenvalues actually uniquely specify the irreducible representations. Unfortunately $S_3=\operatorname{GL}(2,2)$ is very exceptional and doesn't have such a subgroup (the diagonal matrices form a subgroup of order 1).
