If $A$ is nilpotent matrix then $tr(A)=0$ I want to prove if $A$ is nilpotent matrix then $tr(A)=0$. I have seen some answers on MSE about this question but none of them were clear enough to be understood for me. I appreciate if someone explains the proof in mathematical notation rather than a general explanation about it.

If $A\in\mathbb{K}^{n\times n}$ is nilpotent, then its characteristic polynomial is $\chi_A (\lambda)=(-1)^n\lambda^n$ and in general por any $M\in\mathbb{K}^{n\times n}$, $\chi_M (\lambda)=(-1)^n\lambda^n+(-1)^{n-1}(\text{trace }M)\lambda^{n-1}+\cdots+\det M$ so, $\text{trace }A=0$.
