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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 $A^{k-1}$ => $A^kA^H=A^{k-1}A^HA$
Now,we have $0 = A^{k-1}A^HA$ Since $A$ nilpotent.

Iam not sure how to proceed from here to show $A=0$. Can someone help me in this problem?

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up vote 12 down vote accepted

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

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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 $\lambda = 0$. – Nate Eldredge Sep 30 '11 at 13:44
Thanks, good point. – Manos Sep 30 '11 at 13:53
@manos: is it a standard result that all normal matrix are diagonalisable? can u point me to the reference – Learner Sep 30 '11 at 14:07
@Learner, see Normal matrices are exactly the ones for which the spectral theorem holds: – lhf Sep 30 '11 at 14:10
@Learner, Axler's Linear Algebra Done Right explains this in chapter 7, which is freely available. – lhf Sep 30 '11 at 14:15

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, $B=A^{k-1}=0$.
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Very nice. And it ties nicely with attempting to do the simple case $k=2$ for $A$ first. – lhf Sep 30 '11 at 16:54

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 $0$.

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