# Show that $AB$ is nilpotent

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 $[A,[A,B]]=0.$ Commented Apr 15, 2019 at 8:37
• It is a duplicate Commented Apr 17, 2019 at 16:57

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 $$k\ge1$$
• The determinant of $A^{k-2}$ is zero, $A$ being nilpotent. Commented Apr 15, 2019 at 8:14
• The matrix product $XY$ can be zero with both matrices being nonzero. Commented Apr 17, 2019 at 8:50