If $T$ and $T^2$ have same image, then $N(T) \cap Im(T)={0}$ Let V be a finite dimensional vectorial space and $T:V \rightarrow V$ a linear operator. Show that if  $T$ and $T^2$ have same image, then $N(T) \cap Im(T)={0}$.

My first attempt to solve this problem was think in the matrix associated to the linear transformation. I tried to use the fact that the span of the column vectors of the matrix T was equal to the span of vectors of the matrix T^2, but as I don't know the dimension of the matrix T, this problem got very complex.
Consider the endomorphism $T$ restricted to its image. This gives an endomorphism $T'$ of $T(V)$. The assumption that $T^2$ and $T$ have the same image implies that $T'$ is surjective. Hence, since $T(V)$ is finite-dimensional, $T'$ is also injective. This implies that $\mathrm{Ker} \ T \cap T(V)=0$.
The proof is indeed rather sketchy: If $w$ belongs to $N(T) \cap \mbox{Im} T$, then $w=T(v)=0$, for some $v$ in $V$. Using the hypothesis $\mbox{Im} T=\mbox{Im} T^2$, we get
the sequence of identities 
$$w=T(w)=T(0)=0$$
so that $w\in \{0\},$ as desired.