Solution
= Solution
Take $N=T-\lambda I$. It is a <polynomial> in the <normal operator> $T$, so part (b)(iii) makes $N$ a <normal operator>. Also $N^*=T^*-\bar\lambda I$. If $Tv=\lambda v$, part (b)(i), applied to $N$, gives
$$
\|(T^*-\bar\lambda I)v\|=\|N^*v\|=\|Nv\|=0.
$$
The <norm> vanishes only at the zero <vector>, so
$$
\boxed{T^*v=\bar\lambda v}.
$$
This is the <adjoint eigenvector identity for a normal matrix> expressed independently of <coordinates>.