Show that $AB$ is nilpotent

Question

Let $A,B$ be two $n\times n$ matrices. If $A^2B-2ABA+BA^2=0$ and $A$ is nilpotent, that is, there exists a positive integer $k$ such that $A^k=0$. Show that $AB$ is nilpotent.

If $k=2$, it is OK. Since $(AB)^2=ABAB=\frac{1}{2}(A^2B+BA^2)B=0$. But if $k\geq 3$, what

The given it's $[A,[A,B]]=0.$ – Michael Rozenberg Apr 15 '19 at 08:37 — Michael Rozenberg, Apr 15 '19 at 08:37

Tojra · Answer 1 · 2019-04-15T08:25:00.170

0

Given $A$ is nilpotent, means there exists a real k such that $A^m=0$ for all $m\ge k$ and $A^p\ne0$ for all $p\lt k$. So you have initial equation $A^2B-2ABA+BA^2=0$. Now left multiply the equation both sides by $A^{k-2}$. You get : $0+2A^{k-1}BA+A^{k-2}BA^2=0$. Now,take $A^{k-2}$ common. Since it is non zero, the other part has to be zero. So $2ABA+BA^2=0$. Now you right Multiply both sides by $A^{k-2}$. Now you get $2ABA^{k-1}=0$. Following similar argument,you conclude $AB=0$. So obviously,$AB$ is nilpotent as $(AB)^k=0$ for any $k\ge1$

edited Apr 15 '19 at 08:25

answered Apr 15 '19 at 08:03

Tojra

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The determinant of $A^{k-2}$ is zero, $A$ being nilpotent. – Helmut Apr 15 '19 at 08:14
Sorry, a mistake – Tojra Apr 15 '19 at 08:26
@Tojrah "since it is non zero, the other part has to be zero" - why is this true? – abhi01nat Apr 15 '19 at 08:48
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The matrix product $XY$ can be zero with both matrices being nonzero. – egreg Apr 17 '19 at 08:50
May be there is some kind of assumption for the question – Tojra Apr 17 '19 at 10:43

Show that $AB$ is nilpotent

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