= Solution
In fact the <adjoint operator>[adjoint] invariance follows from the <orthogonal decomposition> and the invariance under $T$, without using the <normal operator> property. For $x\in V_i$ and $y\in V_j$ with $j\ne i$,
$$
\langle T^*x,y\rangle=\langle x,Ty\rangle=0,
$$
since $Ty\in V_j$. Hence $T^*x$ is <orthogonal> to every other summand and therefore lies in $V_i$. Moreover, for $x,y\in V_i$, the <adjoint operator>[adjoint] identity restricted to that subspace shows
$$
\boxed{T_i^*=T^*|_{V_i}.}
$$
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