Let $\mathbb K$-vector space be and let $T \in L(V, V)$. Show that the following statements are equivalent.
a) $Im(T) \cap Ker(T) = ${$0$};
b) If $T(T(v)) = 0$, then $T(v) = 0$.
$(i) \to (ii)$ : $(i)$ says that nothing in the $im(T)$ gets mapped to zero except for 0. So, if x is in
the $im(T)$ then $T(x) = 0 \to x = 0$. But $T(v)$ is in $im(T)$, thus $T(T(v)) = 0 \to T(v) = 0$.
$(ii) \to (i)$: Let x is in both the $Im(T)$ and $Ker(T)$. Since x is in $Im(T)$, $x = T(v)$ for some $v$. But then x in the $Ker(T) \to T(x) = 0$ which implies T(T(v)) = 0. So $(ii)$ implies $T(v) = 0$ or equivalently $x = 0$. Therefore $Im(T) \cap Ker(T) = ${$0$}.
That was my attempt, is it acceptable?
Thanks.
Best Answer
Yes, your proof is correct, well done!
(There's not much more to add, I'm afraid...)