Norm Characterizations of Normality

Normality Through Norms

Normality is usually defined by an equation involving two operator products. A useful alternative asks what the operator and its adjoint do to the length of each vector. That turns an algebraic condition into a geometric comparison: for every input, do the two outputs have the same norm? The requirement that the equality hold for every vector is essential; agreement on a single vector does not certify normality.

Throughout, let \(V\) be a finite-dimensional inner product space over \(\mathbb{C}\). The inner product is linear in its first slot and conjugate-linear in its second, and the norm is \(\|\mathbf{x}\|=\sqrt{\langle\mathbf{x},\mathbf{x}\rangle}\). The adjoint \(T^*\) of a linear operator \(T:V\to V\) is defined by \(\langle T\mathbf{x},\mathbf{y}\rangle=\langle\mathbf{x},T^*\mathbf{y}\rangle\) for all \(\mathbf{x},\mathbf{y}\in V\).

When matrices are used below, they represent operators on \(\mathbb{C}^2\) with its standard inner product. In that setting, the adjoint matrix is the conjugate transpose. The identity operator is denoted by \(\mathbf{I}\). We will prove both directions of the characterization, including the step that turns equality of norms into equality of operators.

The Norm Characterization

Proof. Define \(D=T^*T-TT^*\). The adjoint laws \((AB)^*=B^*A^*\) and \((T^*)^*=T\) show that \(D^*=D\), so \(D\) is self-adjoint. By the defining property of the adjoint, for every \(\mathbf{x}\in V\),

If \(T\) is normal, then \(D=\mathbf{0}\), so these squared norms are equal for every \(\mathbf{x}\). Norms are nonnegative, and equal nonnegative squares have equal values. Therefore \(\|T\mathbf{x}\|=\|T^*\mathbf{x}\|\) for every \(\mathbf{x}\).

Conversely, suppose the norms are equal for every \(\mathbf{x}\). Their squares are equal, so \(\langle\mathbf{x},D\mathbf{x}\rangle=0\) for every \(\mathbf{x}\). Conjugating this equality gives \(q(\mathbf{x})=\langle D\mathbf{x},\mathbf{x}\rangle=0\) for every \(\mathbf{x}\). Fix arbitrary \(\mathbf{x},\mathbf{y}\in V\), and write \(z=\langle D\mathbf{x},\mathbf{y}\rangle\). Expanding \(q(\mathbf{x}+\mathbf{y})=0\), and using \(q(\mathbf{x})=q(\mathbf{y})=0\), gives \(\langle D\mathbf{x},\mathbf{y}\rangle+\langle D\mathbf{y},\mathbf{x}\rangle=0\). Because \(D^*=D\), \(\langle D\mathbf{y},\mathbf{x}\rangle=\langle\mathbf{y},D\mathbf{x}\rangle=\overline{z}\). Hence \(z+\overline{z}=0\), or \(\operatorname{Re}(z)=0\).

Now expand \(q(\mathbf{x}+i\mathbf{y})=0\). Linearity in the first slot and conjugate-linearity in the second give \(\langle D\mathbf{x},i\mathbf{y}\rangle=-iz\) and \(\langle iD\mathbf{y},\mathbf{x}\rangle=i\overline{z}\). The diagonal terms vanish, so \(i(\overline{z}-z)=0\), which gives \(\operatorname{Im}(z)=0\). Thus \(z=0\). Since \(\mathbf{x}\) and \(\mathbf{y}\) were arbitrary, \(\langle D\mathbf{x},\mathbf{y}\rangle=0\) for every pair. Taking \(\mathbf{y}=D\mathbf{x}\) yields \(\|D\mathbf{x}\|^2=0\), so \(D\mathbf{x}=\mathbf{0}\) for every \(\mathbf{x}\). Therefore \(D=\mathbf{0}\), meaning \(T^*T=TT^*\), and \(T\) is normal. \(\square\)

The key point in the converse is that a zero quadratic form for a self-adjoint operator forces the operator itself to be zero. The expansions with \(\mathbf{x}+\mathbf{y}\) and \(\mathbf{x}+i\mathbf{y}\) recover both the real and imaginary parts of the mixed inner product. Comparing norms only for a selection of vectors does not provide that information for all pairs.

Using the Identity to Test Matrices

What Normality Says About Eigenvectors

The norm characterization also compares the action of \(T\) and \(T^*\) on a candidate eigenvector. An eigenvalue \(\lambda\) of \(T\) has an eigenvector \(\mathbf{x}\ne\mathbf{0}\) satisfying \(T\mathbf{x}=\lambda\mathbf{x}\). The adjoint is conjugate-linear with respect to scalar multiplication in its scalar argument, which is why the corresponding eigenvalue in the result below is \(\overline{\lambda}\).

Proof. Set \(S=T-\lambda\mathbf{I}\). Its adjoint is \(S^*=T^*-\overline{\lambda}\mathbf{I}\). Expanding both products gives \(S^*S=T^*T-\lambda T^*-\overline{\lambda}T+|\lambda|^2\mathbf{I}\) and \(SS^*=TT^*-\overline{\lambda}T-\lambda T^*+|\lambda|^2\mathbf{I}\). Since \(T^*T=TT^*\), these products are equal; thus \(S\) is normal. The assumed eigenvector equation says \(S\mathbf{x}=\mathbf{0}\). Applying the norm characterization to \(S\) and \(\mathbf{x}\) gives \(\|S^*\mathbf{x}\|=\|S\mathbf{x}\|=0\), so \(S^*\mathbf{x}=\mathbf{0}\). Substituting \(S^*=T^*-\overline{\lambda}\mathbf{I}\) yields \(T^*\mathbf{x}=\overline{\lambda}\mathbf{x}\). \(\square\)

The theorem gives a way to compare proposed eigenvector data: if \(T\) is known to be normal and \(T\mathbf{x}=\lambda\mathbf{x}\), then \(T^*\mathbf{x}=\overline{\lambda}\mathbf{x}\) must also hold. For a vector that is not an eigenvector of \(T\), this conclusion does not apply. Nor does observing the relation for one eigenvector prove that the entire operator is normal; normality requires a condition on all vectors, or equivalently the operator equation \(T^*T=TT^*\).

Common Pitfall: Checking Just One Vector

A tempting but invalid argument is to compute equal norms for one vector and conclude that the operator is normal. For the non-normal matrix \(\mathbf{N}\) above, let \(\mathbf{u}=\begin{bmatrix}1\\1\end{bmatrix}\). Then

This equality is correct, but it does not establish normality: the earlier test vector \(\mathbf{e}_1\) gives norms \(1\) and \(\sqrt{2}\), so the identity fails. The correction is to keep the quantifier in view. Equal norms for one or several selected inputs provide evidence about those inputs only; to use the characterization as a proof of normality, establish equality for an arbitrary vector.

A second practical safeguard is to use the conjugate transpose when computing the adjoint of a complex matrix. For example, the adjoint of \(\begin{bmatrix}1+i&0\\0&2\end{bmatrix}\) has \(1-i\), not \(1+i\), in its first diagonal entry. Omitting conjugation can give an incorrect test for normality or an incorrect proposed adjoint eigenvalue.