Exercise 6.2.2

Solution True. Suppose two normal operators are $N_1=U_1{D}_1{U}_1^{\text{∗}},N_2=U_2{D}_2{U}_2^{\text{∗}}$. $U_1,U_2$ are unitary and $D_1,D_2$ are diagonal.

$N_1,N_2$ are normal, we need to prove $N_1^{\text{∗}}{N}_2+N_2^{\text{∗}}{N}_1=N_1{N}_2^{\text{∗}}+N_2{N}_1^{\text{∗}}$. In fact, $N_1^{\text{∗}}{N}_2=U_1{D}_1^{\text{∗}}{U}_1^{\text{∗}}{U}_2{D}_2{U}_2^{\text{∗}}$. $N_1{N}_2^{\text{∗}}=U_1{D}_1{U}_1^{\text{∗}}{U}_2{D}_2^{\text{∗}}{U}_2^{\text{∗}}$. As can be shown, $D_1^{\text{∗}}{U}_1^{\text{∗}}{U}_2{D}_2=D_1{U}_1^{\text{∗}}{U}_2{D}_2^{\text{∗}}$ because $D_1,D_2$ are diagonal matrices, $D_1^{\text{∗}}=\overline{D_1},D_2^{\text{∗}}=\overline{D_2}$. $D_1^{\text{∗}}{D}_2=D_1{D}_2^{\text{∗}}$ (for complex numbers $c_1,c_2,\overline{c_1}{c}_2=c_1\overline{c_2}$). By checking the entries of the product, one can conclude $D_1^{\text{∗}}{U}_1^{\text{∗}}{U}_2{D}_2=D_1{U}_1^{\text{∗}}{U}_2{D}_2^{\text{∗}}$ and $N_1^{\text{∗}}{N}_2=N_1{N}_2^{\text{∗}},N_2^{\text{∗}}{N}_1=N_2{N}_1^{\text{∗}}$. So the statement is true.