Prove there exists $v$ such that $f^3(v) = f(f(f(v))) \neq 0$ 
Let $V$ be a vector space of dimension $n$ and $f: V \to V$. If $\dim Im f \geq 2n/3$, then prove there exists $v$ such that $f^3(v) = f(f(f(v))) \neq 0$

The only thing I can deduce is that $\dim \ker f \leq n/3$.

$dimImf =dim(Kerf_{\mid Im f}+dim(Im f^2)$ since $dim(Ker f)<n/3$ and $dim(Imf)\geq {{2n}\over 3}$, we deduce that $dim(Imf^2)>n/3$,
$dimImf^2=dim(Ker f_{\mid Imf^2}+dim(Imf^3)$ since $dim(Ker f)<n/3$ and $dim(Imf^2)>n/3$, we deduce that $dimImf^3>0$
