If $T$ is an operator on a vector space $V$ over $\mathbb{R}$, can the eigenvalues be complex? So if we have an operator $T \in \mathcal{L}(V)$, where $V$ is a vector space over a field $F$, and $F$ = $\mathbb{R}$ or $\mathbb{C}$. Then, the eigenvalues $\lambda$ could be real- or complex-valued. Correct?
Now, if $T \in \mathcal{L}(V)$ but $F$ = $\mathbb{R}$, then can $T$ necessarily have complex-valued eigenvalues? If so, when (what criteria must be met)?
Thanks. I'm genuinely curious.

Suppose $T$ is an operator on a real vector space. The minimal polynomial of $T$ may have complex roots, but only the real roots are actually eigenvalues.
Example: Let $T$ be the transformation on $\mathbb{R}^2$ whose matrix with respect to the standard basis is $$ A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}_.$$
The minimal polynomial of $T$ is $x^2+1$. The roots of which are $i, -i$. But there are no non-zero vectors $v \in \mathbb{R}^2$ such that $Av = iv$ or $Av = -iv$.
