If $T^2$ is the zero transformation, then $T^{\dagger} \circ T^{\dagger}$ is the zero transformation

Question

I have no idea how to approach this proof. I understand that this implies that the $\operatorname{im}(T) \subseteq \ker(T)$, so $\ker(T^{\dagger}) \subseteq \operatorname{im}(T^{\dagger})$, but I can't come up with anything past this point. Can someone give me a hint, or am I on the right track?

Answer 1

If $T^2 = 0$ then $\operatorname{im}(T) \subseteq \ker(T)$ which implies that $\ker(T)^{\perp} \subseteq \operatorname{im}(T)^{\perp}$.

To show that $T^{\dagger} \circ T^{\dagger} = 0$, it is enough to show that $T^{\dagger} \circ T^{\dagger}$ sends $\operatorname{im}(T)$ to zero (since $\operatorname{im}(T)^{\perp}$ is already sent to zero by $T^{\dagger}$). But $T^{\dagger}$ always sends $\operatorname{im}(T)$ to $\ker(T)^{\perp} \subseteq \operatorname{im}(T)^{\perp}$ and so $T^{\dagger} \circ T^{\dagger}$ sends $\operatorname{im}(T)$ to zero.