Solution
= Solution
Let $w=Tv$. If $T^2v=0$, then $Tw=0$, and part (i) gives $\|T^*w\|=\|Tw\|=0$. Thus $T^*w=0$ and
$$
\|w\|^2=\langle Tv,w\rangle=\langle v,T^*w\rangle=0.
$$
The <positive definiteness> of the <inner product> gives $w=0$. Since the reverse <kernel> inclusion always holds, we obtain
$$
\boxed{T^2v=0\Longrightarrow Tv=0,\qquad \ker T^2=\ker T}.
$$
This is <kernel stabilization for a normal operator>.