On the eigenvalues of a linear transformation $\tau$ such that $\tau^3 = \mathrm{id}$ 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 $\omega = e^{2\pi i/3}.$

I know the eigenvalues of $\tau$ are some cube root of unity, but I don't know why all of the cube roots of unity are eigenvalues of $\tau$.
I tried to calculate the minimal polynomial $f$ of $\tau$.  It is easily seen that $f$ divides $X^3 - 1$, but I cannot go further in this regard, since I do not know anything about $W$ except the fact that it is finite-dimensional.
I would be most grateful if you could help me understand the paragraph.
 A: "All eigenvalues are powers of $\omega$" is not the same as "All powers of $\omega$ are eigenvalues".
A: The last sentence is remarkably ambiguous: I can stare at it and watch my perception of its meaning flicker back and forth between "Every eigenvalue of $\tau$ is a power of $\omega = e^{2 \pi i/3}$ and "The eigenvalues of $\tau$ give all of the powers of $\omega$."
Mathematically it is the former which is true -- as the authors very well know -- so that is the intended meaning.  It is enough to observe that a complex representation of the cyclic group $C_3 = \langle \tau \ | \ \tau^3 = 1 \rangle$ need not have $\omega$ as an eigenvalue.  For instance, there is always the trivial representation, whose only eigenvalue is $1$.  More generally $C_3$ has precisely $3$ irreducible complex representations: they are all one-dimensional and consist of scaling by $1$, $\omega$ or $\omega^2$.  By taking direct sums of these -- or, in simpler terms, taking a diagonal matrix with whatever combination of $1,\omega,\omega^2$ you want on the diagonal -- you can get whatever eigenvalues you want (among cube roots of $1$!) with whatever multiplicities.
That's over $\mathbb{C}$.  If we are working over a subfield $K$ of $\mathbb{C}$ which doesn't contain a primitive cube root of unity -- e.g. $\mathbb{R}$ or any of its subfields, e.g. $\mathbb{Q}$ -- then in any representation in which $\omega$ appears as an eigenvalue, so also must $\omega^2$.  There are lots of ways to see this, but perhaps the most purely representation-theoretic is to notice that instead of three one-dimensional irreducible representations of $C_3$, in this case there is the trivial (one-dimensional) representation and an irreducible two-dimensional representation with irreducible minimal polynomial $t^2+t+1$.  And this comes, for instance, from decomposing the group ring $K[C_3]$ into a product of simple rings:
$K[C_3] \cong K[t]/(t^3-1) = K[t]/((t-1)(t^2+t+1))$
$ \cong K[t]/(t-1) \oplus K[t]/(t^2+t+1) \cong K \oplus K[\omega]$.   
