= Kernel stabilization for a normal operator
{title2=$\ker T^k=\ker T\quad(k\geq1)$}
For a <normal operator> on a real or complex <inner product space>, equality of $\|Tv\|$ and $\|T^*v\|$ gives $\ker T=\ker T^*$. If $T^2v=0$, put $w=Tv$. Then $Tw=0$ forces $T^*w=0$, and the <adjoint operator> identity gives
$$
\|w\|^2=\langle Tv,w\rangle=\langle v,T^*w\rangle=0.
$$
Thus $Tv=0$ and $\ker T^2=\ker T$. Applying this implication successively to $T^{k-2}v,T^{k-3}v,\ldots,v$ proves
$$
\boxed{\ker T^k=\ker T\quad\text{for every integer }k\geq1}.
$$
In particular a <nilpotent> <normal operator> is zero. The argument uses no <spectral theorem>.
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