Does Im$(T)\cap$Ker$(T)={0}$ implies $T^2=T$, where $T\in\mathcal{L}(V)$, $V$ is finite dimensional vector space?

Question

I know that in a finite dimensional vector space $V$ if $T^2=T$, then $\operatorname{Im}(T)\cap\operatorname{Ker}(T)={0}$, where $T:V\to V$ is linear map. But my question is- Does the converse true?
My intuition says that the answer is NO. Even I can prove if $\operatorname{Im}(T)\cap\operatorname{Ker}(T)={0}$ then $\operatorname{Rank}(T^2)=\operatorname{Rank}(T)$.
If the answer is NO, can anybody give an example where $\operatorname{Im}(T)\cap\operatorname{Ker}(T)={0}$ but $T^2\ne T$?
Thanks for assistance in advance.

Answer 1

Take $T(x)=2x$, $kerT=0$, $imT=V$, where $dimV\geq 1$.