Given a matrix $A \in R^{n \times n}$ which is normal ($AA^H=A^HA$ where $A^H$ is hermitian of $A$) and nilpotent ($A^k=0$ for some $k$). Now we need to show that $A=0$.
I tried to show in the following way,
we know that, $AA^H=A^HA$
pre-multiply by $A^{k-1}\implies A^kA^H=A^{k-1}A^HA$
Now, we have $0 = A^{k-1}A^HA$, since $A$ nilpotent.
I am not sure how to proceed from here to show $A=0$. Can someone help me in this problem?