Show that $Tx = \sum_{n \in \Bbb N} \lambda_n \langle x, e_n \rangle e_n$ is normal for ONS

Question

Let $\mathcal H$ be a Hilbert space and $(e_n)_{n \in \Bbb N}$ be an orthonormal system and $(\lambda_n)_n$ a bounded sequence. Consider $Tx = \sum_{n \in \Bbb N} \lambda_n \langle x, e_n \rangle e_n$. My goal is to show that $T$ is normal. In order to do that, I first need to determine the adjoint operator.

I know that there have been answered similar questions before but they dealt with ONBs which is why here you can not use tricks such as $ \langle x, y \rangle = \sum_{n \in \Bbb N} \langle x, e_n \rangle \langle e_n, x \rangle$.

Answer 1

You can just extend $(e_n)$ to an orthonormal basis $(e_i)_{i\in I}$ (where $I$ is some index set that contains $\mathbb{N}$) and define $\lambda_i=0$ for $i\in I\setminus\mathbb{N}$. Then you are reduced to the case that you actually have an orthonormal basis.

Answer 2

The adjoint operator is simple to find as $$ ⟨T^*x,y⟩=⟨x,Ty⟩=\sum_{n∈\Bbb N}\bar λ_n\overline{⟨y,e_n⟩}⟨x,e_n⟩ =\Bigl\langle\sum_{n∈\Bbb N}\bar λ_n⟨x,e_n⟩e_n,y\Bigr\rangle \\~\\\implies T^* x=\sum_{n∈\Bbb N}\bar λ_n⟨x,e_n⟩e_n $$ Thus, now using $⟨e_m,e_n⟩=\delta_{mn}$, $$T^*Tx=\sum_{n∈\Bbb N}|λ_n|^2⟨x,e_n⟩e_n=TT^*x.$$