Suppose I have an $n\times n$ nilpotent matrix $A$. If the entries are from any field, then I can show that all eigenvalues are zero and the trace is zero. Indeed, if we consider the algebraic closure of the field then the Jordan normal form $J$ of $A$ must be nilpotent, so all its eigenvalues are zero, which is of course in the original field itself. Also $\operatorname{tr}(J)$ is zero, and using the fact that $\operatorname{tr}(AB)=\operatorname{tr}(BA)$, it follows that $A$ (which is similar to $J$) must also have zero trace.
But now I want to consider the case where the entries of $A$ are from a finite dimensional associative division algebra $D$ over a field $K$. If $K$ is algebraically closed then we are back in the case above since the only finite dimensional division algebra over an algebraically closed field is the field itself. But I'm having some difficulty with the case where $K$ is not necessarily algebraically closed - are the above still true?
For simplicity let's assume that $K$ has characteristic $0$ but is not necessarily algebraically closed. The proof above (for a field) does not seem applicable in this case - at least I can't convince myself of it. I can't use the idea of algebraic closure, so I do not know if there exists any eigenvalues in $D$. Also, since commutativity does not in general hold in $D$, I do not know if the trace is invariant under a change of basis.
The difficulty seems to be that I don't know what results continue to hold for a division algebra. Any ideas of a good way to think about this?