17
$\begingroup$

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?

$\endgroup$

3 Answers 3

26
$\begingroup$

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, $B=A^{k-1}=0$.
$\endgroup$
1
  • $\begingroup$ Very nice. And it ties nicely with attempting to do the simple case $k=2$ for $A$ first. $\endgroup$
    – lhf
    Sep 30, 2011 at 16:54
22
$\begingroup$

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

$\endgroup$
5
  • 2
    $\begingroup$ 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 $\lambda = 0$. $\endgroup$ Sep 30, 2011 at 13:44
  • $\begingroup$ @manos: is it a standard result that all normal matrix are diagonalisable? can u point me to the reference $\endgroup$
    – Learner
    Sep 30, 2011 at 14:07
  • $\begingroup$ @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 $\endgroup$
    – lhf
    Sep 30, 2011 at 14:10
  • 2
    $\begingroup$ @Learner, Axler's Linear Algebra Done Right explains this in chapter 7, which is freely available. $\endgroup$
    – lhf
    Sep 30, 2011 at 14:15
  • $\begingroup$ Normal matrices are diagonal over $\mathbb{C}$ but OP is asking about real matrices. So this answer doesn't really work in general? $\endgroup$
    – Macrophage
    Dec 5, 2019 at 15:40
7
$\begingroup$

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 $0$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .