1
$\begingroup$

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 $CB=0$.
Thanks in advance

$\endgroup$
1
$\begingroup$

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 $CB = 0$.

$\endgroup$
  • $\begingroup$ @MaisamHedyelloo: I should have called it $C$ instead of $B$ to make my answer clearer, I have edited to reflect that. $\endgroup$ – Jim Feb 22 '13 at 20:24
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.