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I ran into this exercise whilst going through the lecture notes and I'm not sure how to proceed on the second part, where I need to show that $ImT \cap KerT = {0}$. Any guidance and hints are much, much appreciated.

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Thank you very much for your time.

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    Hint: Let $v$ in $\mathrm{Im}(T)\cap\ker(T)$ then $v$ is in $\mathrm{Im}(T)$ hence ... and $v$ is in $\ker(T)$ hence ... but $T^2=T$ hence ... – 2017-01-06
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    @Did Ah, I see. Just to check I've gotten this finally: $v$ in $ImT$ then $v = T(u)$ for some u. Since $v$ is in $KerT$ also $Tv = TTu = 0$, since $T^2 = T$ $u$ is in KerT. Finally since $u$ must be in $KerT$ $v = Tu = 0$. So all $v$ in $ImT \cap KerT$ is 0. – 2017-01-06
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    Indeed. $ $ $ $ – 2017-01-06

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First note that for any $v\in ImT$, we have $Tv=v$, because $v=Tu$ for some $u$ and $Tv=T^2u=Tu=v$. Now with that fact if $v \in kerT\cap ImT$, then $0=Tv=v$