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

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 $$T^2\ne T$$?

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

• Do you mean $\dim V$? – Chickenmancer Oct 31 '18 at 17:24