Direct proof that nilpotent matrix has zero trace Does anyone know a proof from the first principles that a nilpotent matrix has zero trace. No eigenvalues, no characteristic polynomials, just definition and basic facts about bases and matrices.
 A: I assume you want the trace of a matrix $A\in M_n(F)$ to be defined as the sum of the diagonal elements and that you take the coefficients in a (commutative) field $F$. Here is an approach using only basic facts about bases and matrices.
1) Recall the trace is commutative $\mathrm{tr}(AB)=\mathrm{tr}(BA)$, as shown by the usual computations. In particular the trace is invariant under similarity (change of basis): $\mathrm{tr}(PAP^{-1})=\mathrm{tr} A$ for every $P$ invertible in $M_n(F)$.
2) If $A$ is nilpotent with degree $k$ (i.e. $A^k=0$ but $A^{k-1}\neq 0$), we have the following flag
$$
\{0\}\subseteq \ker A\subseteq \ker A^2\subseteq \ldots\subseteq \ker A^k=F^n
$$
where dimensions are strictly increasing.
Starting with a basis of $\ker A$, we can complete it into a basis of $\ker A^2$ and so on until we get a basis of $F^n$. If $P$ denotes the corresponding change of basis matrix, then $PAP^{-1}$ is strictly upper-triangular as $A(\ker A^j)\subseteq \ker A^{j-1}$. In particular, the diagonal of $PAP^{-1}$ is zero whence $\mathrm{tr}(PAP^{-1})=0$.
A: If you're working over a field of characteristic $p$, you can view this as a consequence of the identity $Tr(X^p) = Tr(X)^p$, which has a short combinatorial proof by expanding out $Tr(X)^p$. Even though this doesn't apply in generality (i.e. characteristic 0), I still think it is worth mentioning.
