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.


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

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

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