If \(U\) and \(T\) are normal operators which commute, prove that \(U + T\) and \(UT\) are normal.
Solution
1. The four operators \(U,T,U^*,T^*\) commute pairwise.
Because \(T\) is normal on a finite-dimensional complex inner product space, its adjoint is a polynomial in \(T\): there is a polynomial \(f\) such that \[ T^*=f(T) \] [@HK71]. Since \(UT=TU\), the operator \(U\) commutes with every polynomial in \(T\), hence \[ UT^*=T^*U. \] Similarly, because \(U\) is normal, \(U^*\) is a polynomial in \(U\), so \[ U^*T=TU^*. \] Taking adjoints of either of these identities also gives \[ U^*T^*=T^*U^*. \] Normality itself supplies \[ UU^*=U^*U, \qquad TT^*=T^*T. \] Thus the four operators commute pairwise.
2. The sum \(U+T\) is normal.
Using step 1, \[ \begin{aligned} (U+T)(U+T)^* &=(U+T)(U^*+T^*)\\ &=UU^*+UT^*+TU^*+TT^*\\ &=U^*U+T^*U+U^*T+T^*T\\ &=(U^*+T^*)(U+T)\\ &=(U+T)^*(U+T). \end{aligned} \] Hence \(U+T\) is normal.
3. The product \(UT\) is normal.
Again using pairwise commutativity, \[ \begin{aligned} (UT)(UT)^* &=UTT^*U^*\\ &=UU^*TT^*\\ &=U^*UT^*T\\ &=T^*U^*UT\\ &=(UT)^*(UT). \end{aligned} \] Therefore \[ \boxed{U+T\text{ and }UT\text{ are normal}.} \]