Why $S=\sum_{k=1}^n A_k$ is normal?

Question

Let $E$ be an infinite-dimensional complex Hilbert space.

Let $(A_1,\cdots,A_n)\in \mathcal{L}(E)^n$ be such that

• $A_iA_j=A_jA_i$ for all $i,j$.

• $A_k^*A_k=A_kA_k^*$ for all $k$.

Why $S=\sum_{k=1}^n A_k$ is normal?

Answer 1

Each $A_j$ is normal, and $A_i,\ A_j$ commute for all $i,j$, so by the Fuglede-Putnam-Rosenblum theorem, we have $A_i^*A_j=A_jA_i^*$ for all $i,j$.

Knowing this, we have $$S^*S=\left(\sum_{i=1}^nA_i^*\right)\left(\sum_{j=1}^nA_j\right)=\sum_{i=1}^n\sum_{j=1}^nA_i^*A_j=\sum_{i=1}^n\sum_{j=1}^nA_jA_i^*=SS^*$$

Answer 2

That $S$ is normal is a consequence of $A_iA_j=A_jA_i$ for $i,j$, the normality of all $A_k$ and Fuglede's theorem .

(https://en.wikipedia.org/wiki/Fuglede%27s_theorem)