Nilpotent matrices over field of characteristic zero

Solution 1:

The existence of a Jordan normal form of a (finite dimensional) vector space endomorphism $\phi$ does not depend on the fact that ground the field is algebraically closed, just that the minimal polynomial of $\phi$ splits over the ground field. Since an endomorphism is nilpotent if and only if its minimal polynomial is of the form $X^k$ for some $k\in\mathbf N$, nilpotent matrices always have a Jordan normal form.

This gives you point (b). Point (a) is a consequence of the fact that in characterestic $0$ the coefficients of the characteristic polynomial of $\phi$ can be expressed in terms of the values $\operatorname{tr}(\phi^k)$ for $0<k\leq n)$ via Newton's identities; if all of those traces are $0$ then the characteristic polynomial is $X^n$ and $\phi$ nilpotent. Apart from its characteristic, this does not depend on the ground field.