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

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?

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, $B=A^{k-1}=0$.
• Very nice. And it ties nicely with attempting to do the simple case $k=2$ for $A$ first.
– lhf
Sep 30, 2011 at 16:54

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 $A=U 0 U^H=0$.

• You don't need the Jordan canonical form. Suppose $A^k = 0$, and let $\lambda$ be an eigenvalue of $A$ with nonzero eigenvector $x$. Then $0 = A^k x = \lambda^k x$, so $\lambda = 0$. Sep 30, 2011 at 13:44
• @manos: is it a standard result that all normal matrix are diagonalisable? can u point me to the reference Sep 30, 2011 at 14:07
• @Learner, see en.wikipedia.org/wiki/Spectral_theorem#Normal_matrices. Normal matrices are exactly the ones for which the spectral theorem holds: en.wikipedia.org/wiki/Normal_matrix#Consequences
– lhf
Sep 30, 2011 at 14:10
• @Learner, Axler's Linear Algebra Done Right explains this in chapter 7, which is freely available.
– lhf
Sep 30, 2011 at 14:15
• Normal matrices are diagonal over $\mathbb{C}$ but OP is asking about real matrices. So this answer doesn't really work in general? Dec 5, 2019 at 15:40

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 $0$.