# Nilpotent matrices over field of characteristic zero

Let $$K$$ be a field of characteristic zero and let $$A$$ be an $$n \times n$$ matrix over $$K$$. Prove the following:

(a) $$N$$ is nilpotent iff $$\mathrm{tr}(N^m)=0$$ for all $$0 iff $$\mathrm{tr}(N^m)=0$$ for all $$m \in \mathbb{N_{+}}$$.

(b) If $$N$$ is nilpotent, then $$N$$ is similar to a strictly upper triangular matrix.

Additional Query: Is above true if $$K$$ is not assumed to be algebraically closed? (Then one can't apply Jordan normal form anymore.)

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.
• In fact, looking at the proof of the Newton identities, we see that b) holds whenever $\mathrm{char}(K)>n$ Sep 6 '12 at 12:26