How prove $A^2=0$,if $AB-BA=A$ 
let $A_{2\times 2}$ matrix, and  The matrix $B$ is  order  square,such
  $$AB-BA=A$$ show that
  $$A^2=0$$

My idea: since $$Tr(AB)=Tr(BA)$$ so
$$Tr(A)=Tr(AB-BA)=Tr(AB)-Tr(BA)=0$$
Question:2 
if $A_{n\times n}$ matrix,and the matrix $B$ is order square,such
$$AB-BA=A$$
then we also have
$$A^2=0?$$
and then I can't Continue .Thank you

An alternative geometric approach:


*

*We have $A\in \mathfrak{sl}_2$ and it can be assumed that $B\in \mathfrak{sl}_2$ as well. Hence $A$, $B$ can be seen as vectors $\vec{a},\vec{b}\in \mathbb{C}^3$. In this picture, $[A,B]\sim \vec{a}\wedge \vec{b}$ and $\operatorname{Tr}AB\sim \vec{a}\cdot\vec{b}.$

*Now since $\vec{a}\wedge \vec{b}\sim \vec{a}$,
taking the scalar product with $\vec{a}$, we get $\operatorname{Tr}A^2\sim\vec{a}\cdot \vec{a}=0 $. As for $A\in\mathfrak{sl}_2$ one has $A^2=\frac{\operatorname{Tr}A^2}{2} I$, the result follows.
