TITLE: $A$ is normal and nilpotent, show $A=0$
QUESTION [17 upvotes]: 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 $A=0$.
(This is essentially exercise 5(b) in sec. 80 on p.162 of Paul R. Halmos' Finite-Dimensional Vector Spaces.)
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?

REPLY [25 votes]: Here is a proof without using eigenvalues or diagonalization. In the below we prove the statement that "if $A^k=0$ and $k>1$ then $A^{k-1}=0$". The result then follows immediately.

Let $B=A^{k-1}$. Then $B$ is normal and $B^2=0$ (because $k>1$).
For all $x$, we have $\|B^\ast Bx\|^2 = (B^\ast Bx)^\ast (B^\ast Bx) = x^\ast B^\ast BB^\ast Bx=x^\ast B^\ast B^\ast BBx=0.$
Therefore $B^\ast Bx=0$ and in turn $\|Bx\|^2 = x^\ast B^\ast Bx=0$ for all $x$.
So $Bx=0$ for all $x$. That is, $B=A^{k-1}=0$.