In the book of linear algrebra by Werner Greub, at page $120$, it is given that
To show that every map doesn't have an eigenvector, it says
Define $\phi$ in $E$ such that $\phi x_1 = x_2 $ and $\phi x_2 =
> -x_1$, where $x_1, x_2$ forms a basis for $E$. Assume $a = \delta_1 x_1 + \delta_2 x_2$ is an eigenvector.Then $\phi a = \lambda a$, and
hence $$\delta_1^2 + \delta_2^2 = 0 \Rightarrow \delta_1 = 0 =
> \delta_2$$
But the fact that $$\delta_1^2 + \delta_2^2 = 0$$ doesn't imply that $\delta_1 = 0 = \delta_2$ since
$$\delta_1^2 + \delta_2^2 = 0 \Rightarrow \delta_1^2 = – \delta_2^2$$
$$\Rightarrow \delta_1 * \delta_2^{-1} = – \delta_1^{-1} * \delta_2 $$
$$\Rightarrow – (\delta_1 * \delta_2^{-1}) = (\delta_1*\delta_2^{-1})^{-1} $$
i.e $k^{-1} = -k$
and this can happen in a field.
For example, in $\mathbb{Z}_5$
$$2 * 3 = 1$$
$$2 + 3 = 0$$
and $$3 = -2$$
So where is the problem ?
Best Answer
The problem is that he is (implicitly) working over $\Bbb R$. The point is not to define a linear map which doesn't have any eigenvalue over any field (what does that even mean?), but to show that there are fields out there with linear maps that don't have an eigenvalue.