Normal Operators as Commuting Self-Adjoint Parts

Separating the Real and Imaginary Parts

A complex operator can be separated into two self-adjoint operators, much as a complex number separates into real and imaginary parts. This is useful because self-adjoint operators have strong structure: for instance, their eigenvalues are real. The key question is how this decomposition relates to normality. The answer is exact: a complex operator is normal precisely when its self-adjoint parts commute.

Throughout, let \(V\) be a finite-dimensional complex inner product space, and let \(T:V\to V\) be linear. The adjoint \(T^*\) is the unique operator satisfying \(\langle T\mathbf{u},\mathbf{v}\rangle=\langle\mathbf{u},T^*\mathbf{v}\rangle\) for all \(\mathbf{u},\mathbf{v}\in V\). An operator \(S\) is self-adjoint if \(S^*=S\), and \(T\) is normal if \(T^*T=TT^*\). In an orthonormal basis, taking the adjoint of a matrix means taking its conjugate transpose, not just its transpose.

The word “real” here refers to the self-adjoint component, not to real-valued entries in a matrix. In particular, \(A\) and \(B\) can both have complex entries when represented in a complex basis. What matters is that each equals its own adjoint.

Why the Decomposition Is Valid and Unique

First check that \(A\) is self-adjoint. The adjoint conjugates scalar coefficients, so \((T+T^*)^*=T^*+T\); hence \(A^*=A\). For \(B\), conjugating the denominator \(2i\) changes its sign, and therefore \[ B^*=\frac{T^*-T}{-2i}=\frac{T-T^*}{2i}=B. \] Adding the two proposed parts gives \[ A+iB=\frac{T+T^*}{2}+\frac{T-T^*}{2}=T. \] This also gives uniqueness.

Proof. The calculations above show that the stated \(A\) and \(B\) are self-adjoint and satisfy \(T=A+iB\). To prove uniqueness, suppose also that \(T=C+iD\), where \(C^*=C\) and \(D^*=D\). Taking adjoints gives \(T^*=C-iD\). Adding the equations \(T=C+iD\) and \(T^*=C-iD\) yields \(T+T^*=2C\), so \(C=(T+T^*)/2=A\). Subtracting gives \(T-T^*=2iD\), so \(D=(T-T^*)/(2i)=B\). Thus no other self-adjoint pair can represent \(T\). \(\square\)

Normality Is Exactly Commutation

For any two operators \(A\) and \(B\), their commutator is the difference \(AB-BA\). They commute when this difference is the zero operator. The next result identifies this condition for the self-adjoint parts of \(T\); it is an equivalence, so both implications are essential.

Proof. Since \(A^*=A\), \(B^*=B\), and the adjoint conjugates \(i\) to \(-i\), we have \(T^*=A-iB\). Multiplying in the two orders, using associativity and distributivity, gives \[ T^*T=(A-iB)(A+iB)=A^2+B^2+i(AB-BA) \] and \[ TT^*=(A+iB)(A-iB)=A^2+B^2+i(BA-AB). \] Subtracting the second expression from the first yields \[ T^*T-TT^*=2i(AB-BA). \] If \(T\) is normal, the left side is zero. Since \(2i\ne0\), multiplying by its reciprocal gives \(AB-BA=0\), so \(AB=BA\). Conversely, if \(AB=BA\), then the right side of the difference equation is zero, so \(T^*T=TT^*\), which is precisely normality. This proves both directions. \(\square\)

The formula also describes the failure of normality: the difference between the two products is determined entirely by the failure of \(A\) and \(B\) to commute. The terms \(A^2+B^2\) appear in both orders and cancel in the difference. One must use the order of multiplication carefully; operators and matrices do not generally commute.

Reading the Criterion Without Confusing the Parts

The decomposition is determined by the operator, rather than by an arbitrary choice of two self-adjoint operators. Given \(T\), the formulas using \(T^*\) recover the only possible \(A\) and \(B\). Once they are found, the test has two equivalent forms: compare \(AB\) with \(BA\), or compare \(T^*T\) with \(TT^*\). The first can be simpler if the parts have a convenient structure; the second is a direct check from the definition of normality.

A normal operator need not itself be self-adjoint. In the diagonal example, \(\mathbf{T}\ne\mathbf{T}^*\), yet it is normal because its self-adjoint parts commute. Conversely, merely having two self-adjoint parts does not ensure normality: every complex operator has such parts, but those parts need not commute. The noncommuting example demonstrates why the condition in the theorem cannot be omitted.

Common Pitfall: Transpose Is Not Adjoint

For a complex matrix, the adjoint is the conjugate transpose. A tempting but wrong calculation for the matrix in the noncommuting example is to take only its transpose and conclude that \(\mathbf{T}\) is self-adjoint, since \(\mathbf{T}^{\top}=\mathbf{T}\). That calculation ignores the conjugation of \(i\). The correct adjoint is \(\mathbf{T}^*=\begin{bmatrix}1&-i\\-i&-1\end{bmatrix}\), which differs from \(\mathbf{T}\). The products also differ, as computed above, so the matrix is not normal either.

As a quick correction to this error, conjugate each entry first and then transpose: the entry \(i\) becomes \(-i\), while the real entries remain unchanged. For a real matrix, transpose and conjugate transpose agree; for a complex matrix, they need not. Any test involving self-adjointness or normality must use the conjugate transpose.

Why Commuting Self-Adjoint Parts Matter

This criterion converts a property of a complex operator into a relationship between two self-adjoint operators. It can make normality easier to analyze: instead of repeatedly expanding \(T^*T\) and \(TT^*\), find the two parts and check a single equality, \(AB=BA\). It also explains why normality is compatible with the spectral structure developed for complex inner product spaces. The test itself does not claim that arbitrary self-adjoint operators commute; they do not, as the last example shows. Commutation is an additional condition on the particular pair arising from \(T\).

When using the criterion, keep the logical direction straight. If \(AB=BA\), then \(T\) is normal. If \(T\) is normal, then its uniquely defined self-adjoint parts commute. Neither conclusion is being asserted for an arbitrary pair of self-adjoint operators without the relation \(T=A+iB\).