$A$ is normal and nilpotent, show $A=0$

Question

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?

Answer 1

All the eigenvalues of a nilpotent matrix must be zero (this can be seen by taking powers of the Jordan canonical form). A normal matrix is diagonalizable. So $A=U \Lambda U^H$ where $\Lambda$ is the diagonal matrix containing the eigenvalues on the diagonal. But $\Lambda$ must be zero because $A$ is nilpotent. So $A=U 0 U^H=0$.

Answer 2

If you can use the spectral theorem then you know that $A$ is similar to a diagonal matrix $D$. Since $A$ is nilpotent, so is $D$. But then $D$ needs to be $0$.

Answer 3

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.

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