Kernel stabilization for a normal operator ($\ker T^k=\ker T(k\ge 1)$)

For a normal operator on a real or complex inner product space, equality of $\Vert Tv\Vert$ and $\Vert T^{\ast }v\Vert$ gives $\ker T=\ker T^{\ast }$. If $T^{2}v=0$, put $w=Tv$. Then $Tw=0$ forces $T^{\ast }w=0$, and the adjoint operator identity givesThus $Tv=0$ and $\ker T^2=\ker T$. Applying this implication successively to $T^{k-2}v,T^{k-3}v,\dots ,v$ provesIn particular a nilpotent normal operator is zero. The argument uses no spectral theorem.