Problem: Show that if A and B are n×n nilpotent matrices that commute, then AB is also nilpotent
I ended up proving that if either A or B is nilpotent, then AB is nilpotent, so you don't need both matrices to be nilpotent. Is my proof correct? My proof seems kind of trivial, so I'm not sure if it's correct.
Let m >= 1. If A is nilpotent then $A^m = 0$. Hence $A^m*B^m=0$ Hence $(AB)^m=0$ (Shown from previous question.)
You need to be careful. You're setting $m$ to be an arbitrary positive integer then claiming that $A^m=0$. What if you had taken $m=1$? There are certainly matrices $A$ for with $A^2=0$ but $A^1\neq0$.
Here's how you should word it: $A$ is nilpotent, so for some $m\ge1$ we have $A^m=0$. Then [the rest of your proof, which is fine].