# Chapter 67. Normal Operators # Chapter 67. Normal Operators Normal operators are the operators that commute with their adjoints. For a complex matrix \(A\), normality means $$ A^*A=AA^*. $$ This condition includes Hermitian matrices, unitary matrices, and skew-Hermitian matrices. It is the exact finite-dimensional condition for unitary diagonalization: a complex matrix is normal if and only if it has an orthonormal basis of eigenvectors. Equivalently, it can be written as \(A=UDU^*\), where \(U\) is unitary and \(D\) is diagonal. ## 67.1 Adjoint Operators Let \(V\) be a finite-dimensional complex inner product space. For a linear operator $$ T:V\to V, $$ the adjoint of \(T\), denoted \(T^*\), is the unique operator satisfying $$ \langle Tv,w\rangle=\langle v,T^*w\rangle $$ for all \(v,w\in V\). In an orthonormal basis, the matrix of \(T^*\) is the conjugate transpose of the matrix of \(T\). If $$ A= \begin{bmatrix} 1+i & 2 \\ 3i & 4-i \end{bmatrix}, $$ then $$ A^*= \begin{bmatrix} 1-i & -3i \\ 2 & 4+i \end{bmatrix}. $$ The adjoint is the complex analogue of transpose, but it also includes conjugation. ## 67.2 Definition of Normal Operator A linear operator \(T\) on a complex inner product space is normal if $$ T^*T=TT^*. $$ For a matrix \(A\), this becomes $$ A^*A=AA^*. $$ The condition says that \(A\) commutes with its adjoint. Normality is weaker than being Hermitian and weaker than being unitary, but it is strong enough to guarantee an orthonormal eigenbasis. ## 67.3 Main Examples Hermitian matrices are normal. If $$ A^*=A, $$ then $$ A^*A=A^2=AA^*. $$ Unitary matrices are normal. If $$ U^*U=I $$ and $$ UU^*=I, $$ then $$ U^*U=UU^*. $$ Skew-Hermitian matrices are normal. If $$ A^*=-A, $$ then $$ A^*A=(-A)A=-A^2 $$ and $$ AA^*=A(-A)=-A^2. $$ Thus $$ A^*A=AA^*. $$ These examples show that normality unifies several important matrix classes. | Matrix class | Defining condition | Eigenvalue behavior | |---|---|---| | Hermitian | \(A^*=A\) | Eigenvalues are real | | Unitary | \(A^*A=I\) | Eigenvalues have modulus \(1\) | | Skew-Hermitian | \(A^*=-A\) | Eigenvalues are purely imaginary | | Normal | \(A^*A=AA^*\) | Eigenvalues may be complex | ## 67.4 A Normal Matrix That Is Not Hermitian Consider $$ A= \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix}. $$ Then $$ A^*= \begin{bmatrix} 1 & 0 \\ 0 & -i \end{bmatrix}. $$ Compute $$ A^*A= \begin{bmatrix} 1 & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. $$ Also, $$ AA^*= \begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -i \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. $$ Thus \(A\) is normal. But $$ A^*\neq A, $$ so \(A\) is not Hermitian. Normal matrices may have complex eigenvalues. Hermitian matrices are the special normal matrices whose eigenvalues are real. ## 67.5 Normality and Norms A matrix \(A\) is normal if and only if $$ \|Ax\|=\|A^*x\| $$ for every vector \(x\). Indeed, $$ \|Ax\|^2=\langle Ax,Ax\rangle=\langle x,A^*Ax\rangle, $$ and $$ \|A^*x\|^2=\langle A^*x,A^*x\rangle=\langle x,AA^*x\rangle. $$ If $$ A^*A=AA^*, $$ then these quantities are equal for every \(x\). Conversely, if the equality holds for every \(x\), then $$ \langle x,(A^*A-AA^*)x\rangle=0 $$ for every \(x\). Since \(A^*A-AA^*\) is Hermitian, this implies $$ A^*A=AA^*. $$ Thus normality can be read as equality of the action of \(A\) and \(A^*\) on vector lengths. ## 67.6 Spectral Theorem for Normal Operators The spectral theorem for normal operators states: A complex matrix \(A\) is normal if and only if there exists a unitary matrix \(U\) and a diagonal matrix \(D\) such that $$ A=UDU^*. $$ The diagonal entries of \(D\) are the eigenvalues of \(A\). The columns of \(U\) are orthonormal eigenvectors of \(A\). This theorem is the central reason normal operators are important. It says that normal operators are exactly the operators that become diagonal in an orthonormal basis. The diagonal entries of \(D\) need not be real. They may be any complex numbers. ## 67.7 Comparison with Hermitian Operators Hermitian operators satisfy $$ A^*=A. $$ Normal operators satisfy $$ A^*A=AA^*. $$ Every Hermitian operator is normal, but not every normal operator is Hermitian. The difference appears in the eigenvalues. If \(A\) is Hermitian, then $$ A=U\Lambda U^* $$ where \(\Lambda\) is real diagonal. If \(A\) is normal, then $$ A=UDU^* $$ where \(D\) may have complex diagonal entries. Thus Hermitian operators are normal operators with real spectrum. ## 67.8 Comparison with Unitary Operators Unitary operators satisfy $$ U^*U=I. $$ Every unitary operator is normal, because $$ U^*U=I=UU^*. $$ If \(U\) is unitary and $$ Uv=\lambda v, $$ then $$ \|Uv\|=\|v\|. $$ But $$ \|Uv\|=\|\lambda v\|=|\lambda|\|v\|. $$ Since \(v\neq 0\), we get $$ |\lambda|=1. $$ Thus unitary matrices are normal matrices whose eigenvalues lie on the unit circle. ## 67.9 Orthogonality of Eigenvectors Eigenvectors of a normal operator corresponding to distinct eigenvalues are orthogonal. Let $$ Av=\lambda v $$ and $$ Aw=\mu w $$ with $$ \lambda\neq \mu. $$ Because \(A\) is normal, the spectral theorem gives an orthonormal eigenbasis. Distinct eigenspaces are orthogonal in that basis. There is also a direct argument. For a normal matrix, if $$ Av=\lambda v, $$ then $$ A^*v=\overline{\lambda}v. $$ Using this fact, $$ \lambda\langle v,w\rangle = \langle \lambda v,w\rangle = \langle Av,w\rangle. $$ By the adjoint identity, $$ \langle Av,w\rangle = \langle v,A^*w\rangle. $$ Since \(Aw=\mu w\), normality gives $$ A^*w=\overline{\mu}w. $$ Therefore, $$ \langle v,A^*w\rangle = \langle v,\overline{\mu}w\rangle = \mu\langle v,w\rangle $$ under the convention that the inner product is conjugate-linear in the second argument. Thus $$ \lambda\langle v,w\rangle=\mu\langle v,w\rangle. $$ Since $$ \lambda\neq\mu, $$ we obtain $$ \langle v,w\rangle=0. $$ ## 67.10 Why Normality Is the Right Condition Every complex matrix has a Schur decomposition: $$ A=UTU^*, $$ where \(U\) is unitary and \(T\) is upper triangular. If \(A\) is normal, then \(T\) is also normal. But a normal upper triangular matrix must be diagonal. Therefore $$ A=UDU^*. $$ This explains why normality is the exact condition for unitary diagonalization. General matrices can be made upper triangular by a unitary change of basis. Normal matrices can be made diagonal. ## 67.11 Example: A Rotation Matrix Consider the real matrix $$ R= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}. $$ This matrix represents rotation by \(90^\circ\) in the plane. Since \(R\) is real, $$ R^*=R^T= \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}. $$ Compute $$ R^*R= \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. $$ Also, $$ RR^*= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. $$ Thus \(R\) is normal. It is not symmetric, because $$ R^T\neq R. $$ Its eigenvalues are $$ i \qquad \text{and} \qquad -i. $$ Over \(\mathbb{R}\), it cannot be diagonalized. Over \(\mathbb{C}\), it is unitarily diagonalizable. ## 67.12 Example: A Non-Normal Matrix Consider $$ A= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}. $$ Since \(A\) is real, $$ A^*=A^T= \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}. $$ Compute $$ A^*A= \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}. $$ Compute $$ AA^*= \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}. $$ Since $$ A^*A\neq AA^*, $$ the matrix is not normal. It has only one eigenvalue, namely \(1\), and only one independent eigenvector. It cannot be diagonalized. This example shows that failure of normality often appears as nonorthogonal or insufficient eigenvector structure. ## 67.13 Normal Operators and Eigenspaces For a normal operator, eigenspaces corresponding to distinct eigenvalues are orthogonal. Moreover, the whole space decomposes as an orthogonal direct sum of eigenspaces: $$ V= E_{\lambda_1}\oplus E_{\lambda_2}\oplus\cdots\oplus E_{\lambda_k}. $$ The sum is orthogonal, so if $$ v_i\in E_{\lambda_i} $$ and $$ v_j\in E_{\lambda_j} $$ with $$ i\neq j, $$ then $$ \langle v_i,v_j\rangle=0. $$ On each eigenspace \(E_{\lambda_i}\), the operator acts as multiplication by \(\lambda_i\). This is the clean structural picture of a normal operator. ## 67.14 Spectral Decomposition If $$ A=UDU^* $$ and the columns of \(U\) are $$ u_1,u_2,\ldots,u_n, $$ then $$ A= \lambda_1u_1u_1^* + \lambda_2u_2u_2^* + \cdots + \lambda_nu_nu_n^*. $$ Each matrix $$ u_iu_i^* $$ is the orthogonal projection onto the line spanned by \(u_i\). If eigenvalues repeat, group the projections by eigenspace. If the distinct eigenvalues are $$ \alpha_1,\alpha_2,\ldots,\alpha_k, $$ then $$ A=\alpha_1P_1+\alpha_2P_2+\cdots+\alpha_kP_k, $$ where \(P_i\) is the orthogonal projection onto \(E_{\alpha_i}\). This is the spectral decomposition of a normal operator. It writes the operator as a weighted sum of orthogonal projections. ## 67.15 Functions of Normal Operators Normal operators allow functions to be applied directly to eigenvalues. If $$ A=UDU^* $$ and $$ D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n), $$ then define $$ f(A)=Uf(D)U^*, $$ where $$ f(D)=\operatorname{diag}(f(\lambda_1),\ldots,f(\lambda_n)). $$ For example, $$ A^k=UD^kU^*, $$ and $$ e^A=Ue^DU^*. $$ If \(A\) is invertible, then no eigenvalue is zero, and $$ A^{-1}=UD^{-1}U^*. $$ Normality makes matrix functions stable and transparent because the unitary matrices do not distort norms. ## 67.16 Hermitian and Skew-Hermitian Parts Every complex matrix can be decomposed into Hermitian and skew-Hermitian parts: $$ A=H+K, $$ where $$ H=\frac{1}{2}(A+A^*) $$ and $$ K=\frac{1}{2}(A-A^*). $$ Then $$ H^*=H $$ and $$ K^*=-K. $$ A matrix \(A\) is normal if and only if its Hermitian part and skew-Hermitian part commute: $$ HK=KH. $$ This gives another way to understand normality. The real-like part and imaginary-like part of the operator must be compatible. For scalars, every complex number decomposes as $$ z=\operatorname{Re}(z)+i\operatorname{Im}(z), $$ and these parts commute automatically. For matrices, commutation is an extra condition. ## 67.17 Normal Operators and Singular Values If \(A\) is normal with eigenvalues $$ \lambda_1,\ldots,\lambda_n, $$ then its singular values are $$ |\lambda_1|,\ldots,|\lambda_n|. $$ Indeed, if $$ A=UDU^*, $$ then $$ A^*A = UD^*U^*UDU^* = UD^*DU^*. $$ Since $$ D^*D= \operatorname{diag}(|\lambda_1|^2,\ldots,|\lambda_n|^2), $$ the eigenvalues of \(A^*A\) are $$ |\lambda_i|^2. $$ The singular values are their square roots. Thus, for normal matrices, eigenvalues and singular values are directly related by modulus. ## 67.18 Simultaneous Diagonalization If two normal matrices commute, then under suitable finite-dimensional complex hypotheses they can be simultaneously unitarily diagonalized. Suppose \(A\) and \(B\) are normal and $$ AB=BA. $$ Then there exists an orthonormal basis consisting of vectors that are eigenvectors for both \(A\) and \(B\). In that basis, both matrices are diagonal. This fact is important in quantum mechanics, representation theory, and multivariate spectral analysis. Commuting normal operators represent compatible observables or compatible measurements. ## 67.19 Real Normal Matrices For real matrices, the adjoint is the transpose. A real matrix is normal if $$ A^TA=AA^T. $$ Real symmetric matrices are normal. Real skew-symmetric matrices are normal. Real orthogonal matrices are normal. However, a real normal matrix may have complex eigenvalues. It may not diagonalize over \(\mathbb{R}\) into a real diagonal matrix. Instead, real normal matrices can often be represented using real block diagonal forms, with \(1\times 1\) blocks for real eigenvalues and \(2\times 2\) rotation-scaling blocks for complex conjugate pairs. The clean diagonal form belongs naturally to complex vector spaces. ## 67.20 Common Errors The first common error is to confuse normal with normalized. Normality concerns the equation $$ A^*A=AA^*. $$ It has nothing to do with a vector having length \(1\). The second common error is to assume normal means Hermitian. Hermitian matrices are normal, but normal matrices may have complex eigenvalues. The third common error is to assume diagonalizable implies normal. A matrix may be diagonalizable using a nonunitary matrix but fail to be normal. Normality requires an orthonormal eigenbasis. The fourth common error is to ignore the adjoint. Over complex spaces, the condition uses \(A^*\), not \(A^T\). The fifth common error is to forget the field. Complex normal matrices are unitarily diagonalizable over \(\mathbb{C}\). A real normal matrix may need complex eigenvectors for diagonalization. ## 67.21 Summary A normal operator satisfies $$ T^*T=TT^*. $$ A normal matrix satisfies $$ A^*A=AA^*. $$ Normal operators are exactly the finite-dimensional complex operators that can be diagonalized by an orthonormal basis. In matrix form, $$ A=UDU^*, $$ where \(U\) is unitary and \(D\) is diagonal. Hermitian, unitary, and skew-Hermitian matrices are all normal. Hermitian matrices have real eigenvalues. Unitary matrices have eigenvalues on the unit circle. General normal matrices may have arbitrary complex eigenvalues. Normality is the precise condition that allows spectral theory to work with orthonormal coordinates.