TITLE: Prove that $(A+B)^n = 0$
QUESTION [7 upvotes]: Let $A$ and $B$ be $n \times n$ real matrices such that $A^n = B^n = 0$ and $AB = BA$. Prove that $(A+B)^n = 0$.

We have $$(A+B)^n = A^n+A^{n-1}B\binom{n}{1}+\cdots+AB^{n-1}\binom{n}{n-1}+B^n.$$ I tried proving this for just the case $n = 2$. We have $(A+B)^2 = A^2+2AB+B^2 = 2AB$ since $A^2 = B^2 = 0$. Then I didn't see how to show that $AB = 0$.

REPLY [6 votes]: If you expand $(A+B)^{2n}$ with the binomial theorem, you'll see that $A^n=B^n=0$ implies each term is $0$ because each term has $A^j B^k$ where either $j\geq n$ or $k\geq n$.  Hence $A+B$ is a nilpotent matrix.  If $C$ is an $n$-by-$n$ nilpotent matrix, then $C^n=0$.