Prove \(T\) is normal if and only if \(T = T_1 + iT_2\), where \(T_1\) and \(T_2\) are self-adjoint operators which commute.
Solution
1. Every operator has canonical self-adjoint real and imaginary parts.
Define \[ T_1=\frac{T+T^*}{2}, \qquad T_2=\frac{T-T^*}{2i}. \] Then \[ T_1^*=T_1, \] and, since conjugating \(1/(2i)\) changes its sign, \[ T_2^*=\frac{T^*-T}{-2i}=T_2. \] Moreover \[ T_1+iT_2 =\frac{T+T^*}{2}+\frac{T-T^*}{2} =T. \]
2. For these canonical parts, commutation is equivalent to normality.
A direct calculation gives \[ \begin{aligned} T_1T_2-T_2T_1 &=\frac1{4i} \bigl((T+T^*)(T-T^*)-(T-T^*)(T+T^*)\bigr)\\ &=\frac1{2i}(T^*T-TT^*). \end{aligned} \] Therefore \[ T_1T_2=T_2T_1 \iff T^*T=TT^*. \] So if \(T\) is normal, the self-adjoint operators in step 1 commute.
3. Conversely, a sum of commuting self-adjoint real and imaginary parts is normal.
Suppose \[ T=T_1+iT_2, \] where \(T_1^*=T_1\), \(T_2^*=T_2\), and \(T_1T_2=T_2T_1\). Then \[ T^*=T_1-iT_2. \] Hence \[ TT^* =(T_1+iT_2)(T_1-iT_2) =T_1^2+T_2^2, \] because the cross terms cancel by commutativity. Similarly \[ T^*T=T_1^2+T_2^2. \] Thus \(TT^*=T^*T\), so \(T\) is normal.