Let $V$ be a representation of the Weyl group. For any reflection $\sigma_{\alpha}$ (where $\alpha$ is a root), we know that $V$ has two eigenspaces with eigenvalues $1$ and $-1$. The $-1$-eigenspace is one dimensional.
I came across a statement like this in unapologetic mathematician but was a little confused. To me it seems like the $1$-eigenspace has to be one dimensional. So, for example, if $\sigma_{\alpha}$ is a reflection across a certain line, then everything outside the line perpendicular to $\sigma_{\alpha}$ would belong to the $-1$-eigenspace (I'm not sure how a single element can generate this space). But only the elements on the reflection line would belong to the $1$-eigenspace.
In a reflection representation, $\sigma_\alpha$ acts as the reflection at a hyperplane, which means that the hyperplane is fixed and the vectors orthogonal to it get multiplied by $-1$. Hence the eigenspace of $-1$ is indeed $1$-dimensional.
However, in an arbitrary representation of the Weyl group, abstract reflections do not necessarily act as reflections at hyperplanes, so one should probably add this assumption in your context.