Question: Let the matrix $A=(a\_{ij})=Mat(2, R)$, satisfies $A^{2011}=0$

Prove that: $A^2=0$

Thanks in advance!

Answer: 
The purpose of this exercise, of course, is to illustrate that if $A$ is an $n\times n$ nilpotent matrix, then $A^n=0$. However, since $n=2$ here, we have a more elementary proof.

As $A^{2011}=0$, $A$ is singular. So, its two rows are linearly dependent and hence $A=uv^\top$ for some vectors $u$ and $v$. It follows that for $m\ge2$, $A^m=(uv^\top)^m=uv^\top uv^\top\cdots uv^\top=(v^\top u)^{m-1}A$. So, ...
$A^n=O\_2=> det(A^n)=0=>det(A)=0=>A^2=Tr(A).A=>A^n=[Tr(A)]^{n-1}.A=>[Tr(A)]^{n-1}.A=O\_2.$

I. $Tr(A)=0=>A^2=Tr(A).A=O\_2;$

II. $A=O\_2=>A^2=O\_2.$