In §1.3 of Fulton & Harris the authors guide the reader through a classification of the linear representations of $S_3$. Given a representation $\rho:S_3\to \mathrm{GL}(W)$ they first restrict to the abelian subgroup $A_3$ and use the fact it's a finite abelian group to conclude the image of its generator $\tau$ (rotation by a cube root of unity) decomposes $W$ into its eigenspaces. If I understand correctly, this is because $\tau$ is of finite order whence its minimal polynomial divides a separable polynomial and is thereby itself separable, which over $\mathbb C$ is equivalent to diagonalizability.
The authors moreover write the all eigenspaces are actually 1-dimensional.
Later on, to understand $\rho$ we are left with understanding how the second generator $\sigma$ (reflection) interacts with the eigenspaces of $\tau$. A calculation using the relations of $S_3$ (viewed as $D_3)$ proved that if $v$ is a $\tau$-eigenvector with eigenvalue $\lambda$, then $\sigma v$ is a $\tau$-eigenvector with eigenvalue $\lambda ^2$. From there we are left to check cases by possible $\tau$-eigenvalue $\lambda$.
In the case $\lambda=1$ there's a separation of cases I don't understand. It is claimed that given a $\tau$-eigenvector $v$ of eigenvalue $1$, the vectors $v,\sigma v$ may or not be independent. How can this be if the eigenspaces of $\tau $ are 1-dimensional (as suggested by the penultimate centered equation on p9)? The vectors $v,\sigma v$ are both $\tau$-eigenvectors of eigenvalue 1.
The $\alpha_i$ on page 9 don't have to be distinct, so the eigenspaces don't have to be one-dimensional, a priori (and, in fact, since $W$ is an arbitrary representation there, they will not be one-dimensional in general).