The example is given below:
But I do not understand the example given after the word in particular, why $\operatorname{spec}(\alpha) = \varnothing$? where $\operatorname{spec}(\alpha)$ is the set of all eigenvalues of $\alpha$ and $\sigma_{c}: v \rightarrow cv$ for $c \in F.$
In the picture that you post is the explanation. Let me make it clearer. Suppose $\operatorname{spec}(\alpha)\neq\varnothing$, then there is some $c$ in $\operatorname{spec}(\alpha)$. Let's say $\alpha(v)=cv$ for some non-zero vector $v$. Now, $$\alpha^2(v)=\alpha(\alpha(v))=\alpha(cv)=c\alpha(v)=c^2v$$ and since $\alpha^2=-\sigma_1$ it follows that $-1v=c^2v$. Rearranging this gives us $(c^2+1)v=0_V$. But $v\neq 0_V$, so $c^2+1=0$, which is impossible.