The following question may be somewhat ill-posed due to a lack of experience in dealing with fields.
Let $T$ be a linear operator over a finite dimensional complex vector space. If the only (complex) eigenvalues of $T$ are zero, then does this mean that the only eigenvalues of $T$ in any subfield of $\mathbb{C}$ are zero too? The converse of this is definitely not true.
More generally, for any field $K$ whose additive identity is the same "zero" as that of $\mathbb{C}$, are the only eigenvalues of $T$ in such a field 0?
I am asking this question for I was posed a question about linear operators over a vector space whose field over which was not specified.
$\textbf{Edit :}$ Perhaps I should provide context to what I am asking. I would like to show that if an operator on a vector space over some field $K$ is nilpotent, then the operator in some basis of the vector space is upper triangular with all zeros on the diagonal.
If the vector space is complex this is trivial, and over a general field I know how to prove this without talking about eigenvalues and characteristic polynomials (by noting that one has the ascending chain of nullspaces of powers of $T$ that eventually equals the whole space).
I have added the homework tag too because the problem I am asking now is related to a homework problem. Namely, the one on nilpotent operators I have mentioned above.