> 
> Let $A$ be a $n \times n$ real matrix. Show that if $A^m = 0$, then
> $\mbox{rank}(A) \leq \frac{m-1}{m}{n}$
> 
> 
> 

My attempt:  

If $m=1$, then $A=0$ so $\mbox{rank}(A)=0$.  

If $m=2$, we have $\mbox{im}(A) \subset \ker(A)$ so $2\operatorname{rank}(A) \leq \dim \mbox{im}(A) + \dim \ker(A)=n$  

For arbitrary $m$, I want to use induction. $B=A|\_{\mbox{im}(A)}$ satisfies $B^{m-1}=0$ so $\mbox{rank}(B) \leq \frac{m-2}{m-1}\mbox{rank}(A)$. Thus $\dim \ker B \geq \mbox{rank}(A)- \frac{m-2}{m-1}\mbox{rank}(A) =\frac{1}{m-1}\mbox{rank}(A)$ by rank-nullity theorem. Thus $n=\dim \ker A + \mbox{rank}(A) \geq \dim \ker B +\mbox{rank}(A)\geq \frac m {m-1}\mbox{rank}(A)$

Is this ok?