Prove: the sum of two nilpotent and exchangable matrices is nipotent.

Question

If $A$ and $B$ are two $n\times n$ nilpotent matrices, and they are exchangable: $AB = BA$, it is said that the sum $A+B$ is also nilpotent. Could you pls give me some hint how to prove that?

Answer 1

Suppose that $A^m=B^q=0$, with $m\geq 1$ and $q\geq 1$. As $AB=BA$, we can expand $(A+B)^{m+q}$ with the binomial theorem:

$$(A+B)^{m+q}=\sum_{k=0}^{m+q}\binom{m+q}{k}A^kB^{m+q-k}$$ In the sum above, if $k\geq m$, then $A^{k}=A^{m}A^{k-m}=0$; if $k\leq m$, then $m+q-k\geq q$, and in the same manner $B^{m+q-k}=0$. Hence $(A+B)^{m+q}=0$, and $A+B$ is nilpotent.