# Nilpotent

• Nov 29th 2011, 03:41 PM
dwsmith
Nilpotent
$\displaystyle A^k=0$ for some $\displaystyle k>0$

If A and B are nilpotent with AB=BA, then A+B is nilpotent.

Suppose $\displaystyle B^n=0$.

Then $\displaystyle (A+B)^{k+n}=\sum_{i=0}^{k+n}\binom{k+n}{i}A^iB^{k+ n-i}$

if $\displaystyle 0\leq i<k$, then $\displaystyle B^{k+n-i}=0$ since $\displaystyle k+n-i>n$

if $\displaystyle k\leq i\leq k+n$, then $\displaystyle A^i=0$.

Does this prove it?
• Nov 29th 2011, 04:41 PM
alexmahone
Re: Nilpotent
Quote:

Originally Posted by dwsmith
$\displaystyle A^k=0$ for some $\displaystyle k>0$

If A and B are nilpotent with AB=BA, then A+B is nilpotent.

Suppose $\displaystyle B^n=0$.

Then $\displaystyle (A+B)^{k+n}=\sum_{i=0}^{k+n}\binom{k+n}{i}A^iB^{k+ n-i}$

if $\displaystyle 0\leq i<k$, then $\displaystyle B^{k+n-i}=0$ since $\displaystyle k+n-i>n$

if $\displaystyle k\leq i\leq k+n$, then $\displaystyle A^i=0$.

Does this prove it?

Yes.

Note that $\displaystyle (A+B)^{k+n}=\sum_{i=0}^{k+n}\binom{k+n}{i}A^iB^{k+ n-i}$ because $\displaystyle AB=BA$.