If $T\colon V\to V$ is linear then$\text{ Im}(T) = \ker(T)$ implies $T^2 = 0$ I'm trying to show that if $V$ is finite dimensional and $T\colon V\to V$ is linear then$\text{ Im}(T) = \ker(T)$ implies $T^2 = 0$.
I've tried taking a $v$ in the kernel and then since it's in the kernel we know its in the image so there is a $w$ such that $T(w) = v$, so then $TT(w) = 0$, but thats only for a specific $w$?
Thanks

Hint: Note that $T^2$ should be read as $T\circ T$. You wish to show that $(\forall v\in V)((T\circ T)(v)=0)$. I think you can do this.
