# assume $A\in M_n(\mathbb C)$,$A^2=0$ how prove $\exists C,B\in M_n(\mathbb C)$ such that A=BC and CB=0?

assume $A\in M_n(\mathbb C)$,$A^2=0$ how prove $\exists C,B\in M_n(\mathbb C)$ such that $A=BC$ and $CB=0$.
Hint: Let $C$ be projection onto a complement of the kernel of $A$ and let $B = A$. Then you should be able to show that $BC = A$ but $CB = 0$.
• @MaisamHedyelloo: I should have called it $C$ instead of $B$ to make my answer clearer, I have edited to reflect that. – Jim Feb 22 '13 at 20:24
WLOG, you may assume that $A$ is in Jordan form. Then $A^2=0$ means that every Jordan block of $A$ is either the zero scalar or a $2\times 2$ nilpotent Jordan block. Now, Jim's answer to another question that was posted 3 hours ago is useful too.