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

Prove that: $A^2=0$

Thanks in advance!

Answer: 
Consider the minimal polynomial $p(x)$ of $A$. By hypothesis, $p(x) \mid x^{2011}$, whence $p(x) = x^k$ for some $k\in \mathbb{N}$. Also $deg(p(x)) \leq 2$, and so $p(x) = x$ or $p(x) = x^2$.

In either case, $A^2 = 0$.
$A^n=O\_2=> det(A^n)=0=>det(A)=0=>A^2=Tr(A).A=>A^n=[Tr(A)]^{n-1}.A=>[Tr(A)]^{n-1}.A=O\_2.$

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

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