# Math Help - Nilpotent

1. ## Nilpotent

$A^k=0$ for some $k>0$

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

Suppose $B^n=0$.

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

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

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

Does this prove it?

2. ## Re: Nilpotent

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

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

Suppose $B^n=0$.

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

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

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

Does this prove it?
Yes.

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