If two normal operators commute, prove that their product is normal.
Solution
Let \(S\) and \(T\) be normal operators satisfying \[ ST=TS. \]
1. Commutation with a normal operator implies commutation with its adjoint.
By Exercise 13, there is a polynomial \(f\) such that \[ T^*=f(T). \] Since \(S\) commutes with \(T\), it commutes with every polynomial in \(T\). Therefore \[ ST^*=T^*S. \] Likewise, since \(S^*\) is a polynomial in \(S\) and \(T\) commutes with \(S\), we have \[ TS^*=S^*T. \] Taking adjoints also gives the remaining corresponding commutation relations. Thus the factors \(S,S^*,T,T^*\) commute whenever they come from different operators.
2. The product is normal.
Since \[ (ST)^*=T^*S^*, \] we obtain, using step 1 and normality of \(S\) and \(T\), \[ \begin{aligned} (ST)(ST)^* &=STT^*S^*\\ &=SS^*TT^*\\ &=S^*ST^*T\\ &=T^*S^*ST\\ &=(ST)^*(ST). \end{aligned} \] Hence \(ST\) commutes with its adjoint, so \[ \boxed{ST\text{ is normal}.} \]