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: $T^2=0$ iff $\operatorname{Im} T\subseteq \operatorname{ker} T$. And you don't even need $T$ to be linear.
