Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 / Algebra / 2 / c

In fact the 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 $j\ne i$,since $Ty\in V_j$. Hence $T^{\ast }x$ is orthogonal to every other summand and therefore lies in $V_i$. Moreover, for $x,y\in V_i$, the adjoint identity restricted to that subspace shows

Restricting $T^{\ast }T=T{T}^{\ast }$ to $V_i$, using (i), gives $T_i^{\ast }{T}_i=T_i{T}_i^{\ast }$, so the global normal operator property implies the normal operator property of every restriction. Conversely, if the restrictions are normal and the adjoint preserves each summand, then for $x={\sum }_i{x}_i$,Thus $\boxed{T\text{ is normal}⟺\text{every }{T}_i\text{ is normal}}$, with condition (i) automatic under the stated decomposition.