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
You say that $tr(A) = tr(AB)-tr(BA)=0$. Therefore Cayley Hamilton equation tells us that $A^2 = \det(A) I_2$. 

On the other hand we have $A^2= A^2B-ABA=ABA-BA^2$. Therefore $2A^2 = A^2B-BA^2=0$ since $A^2$ is a multiple of the identity.
$AB - BA = A \Rightarrow A^2B - ABA = A^2 \Rightarrow$ $$-ABA = A^2(I-B)$$
$AB - BA = A \Rightarrow ABA - BA^2 = A^2 \Rightarrow$ $$ABA = (I+B)A^2$$
Adding this identities: $0 = A^2(I-B) + (I+B)A^2$