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Is it true that if $A$ is nilpotent, then $A^{*}A$ is also nilpotent? ($A$ here is an $n$ by $n$ complex matrix)

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If $A$ and $A^\ast$ commute yes...but an answer in full generality, I'll have to think. –  user21436 Apr 20 '12 at 19:16
    
and also when $A^{*}=A$ I mean when self adjoint –  Bunuelian Trick Apr 20 '12 at 19:21
    
Have you tried any examples at all? –  Chris Eagle Apr 20 '12 at 19:23
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The only way $A^* A$ can be nilpotent is when it is actually $0$. –  Robert Israel Apr 20 '12 at 19:29
    
@M.Krov It is not at all hard. Suppose $A$ is nilpotent of index $k$, that is $A^{k-1} \neq 0$, then, look at $(A^\ast A)^k=A^\ast A\underbrace{ \cdots} A^\ast A$ and see what it means for $A$ and $A^\ast$ to commute. –  user21436 Apr 20 '12 at 19:34

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up vote 2 down vote accepted

Not in general: take $A:=\pmatrix{0&1\\ 0&0}$, then $A$ is nilpotent since $A^2=0$, but $$A^*A=\pmatrix{0&0\\1&0}\cdot\pmatrix{0&1\\ 0&0} =\pmatrix{0&0\\ 0&1}$$ which is not nilpotent.

In fact if $A^*A$ is nilpotent, since it's a Hermitian matrix it's diagonalizable, so we can write $A^*A=P^*DP$ where $D$ is diagonal. Then fact that $A^*A$ is nilpotent gives us that $D=0$ hence $A^*A=0$.

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What if $A$ and $A^*$ commute? according to one of the comments above, the product $A^*A$ should be nilpotent. Can you show how to prove it? –  M.Krov Apr 20 '12 at 19:31
    
The product of two commuting nilpotent matrices $A$ and $B$ which commute is nilpotent (if $A^m=0$ and $B^p=0$ then $(AB)^{m+p}=A^{m+p}B^{m+p}=0$). –  Davide Giraudo Apr 20 '12 at 19:37
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Isn't it enough to have one of them nilpotent? @DavideGiraudo (That is relax that $B$ is nilpotent..., we may prove the same result with less assumptions.) –  user21436 Apr 20 '12 at 19:42

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