Let $F$ be a field, and $a$ a nilpotent element of $M_n(F)$ (the $n \times n$ matrix algebra over $F$). Prove that $a^n = 0$.

Question

Let $F$ be a field, and $a$ a nilpotent element of $M_n(F)$ (the $n \times n$ matrix algebra over $F$). Prove that $a^n = 0$.

I know that all I need to prove is that if $a$ has order $m$, then $m \leq n$. I started by assuming, by contradiction, that $m > n$. Then $m = nq + r$, $0 \leq r < n$. Hence $ 0 = a^m = (a^n)^q \cdot a^r$. This is where I get stuck. I don't know that $M_n(F)$ has no zero divisors, so I can't proceed with this type of argument (I think). I would appreciate some hints on how to prove this. It's a homework problem, so I'd rather not have a complete solution.

Answer 1

Hint If $a$ is nilpotent then $a^k=0$ for some $k$.

Now look at the minimal polynomial of $a$. It must divide $X^k$ and the characteristic polynomial of $a$....