Normal Operators

A normal operator is a finite-dimensional operator that commutes with its adjoint. Normal operators are important because they are exactly the complex matrices that can be diagonalized by a unitary change of basis.

Do not confuse normal with normalized. A normalized vector has norm one. A normal operator satisfies an algebraic condition involving its adjoint.

Definition

Let $A$ be a linear operator on a finite-dimensional complex inner-product space. It is normal if

$$A^\dagger A = AA^\dagger.$$

Equivalently,

$$[A,A^\dagger]=0.$$

Normality depends on the inner product because the adjoint depends on the inner product.

Main Theorem

In finite-dimensional complex inner-product spaces, the following statements are equivalent:

• $A$ is normal.
• $A$ has an orthonormal eigenbasis.
• $A$ is unitarily diagonalizable.

Thus there exists a unitary matrix $U$ and a diagonal matrix $D$ such that

$$U^\dagger A U = D, \qquad A = UDU^\dagger.$$

This is stronger than ordinary diagonalization, where the diagonalizing matrix may be any invertible matrix. Normality says the eigenbasis can be chosen orthonormal.

The general algebraic diagonalization criterion is treated in Diagonalization.

Relation to Hermitian and Unitary Operators

Every Hermitian operator is normal. If $A^\dagger=A$, then

$$A^\dagger A = A^2 = AA^\dagger.$$

Every unitary operator is normal. If $U^\dagger U=UU^\dagger=I$, then

$$U^\dagger U = UU^\dagger.$$

Normal operators include more than these two classes. For example,

$$A = \begin{pmatrix} 2 & 0\\ 0 & i \end{pmatrix}$$

is normal because it is diagonal in an orthonormal basis. It is not Hermitian, because one eigenvalue is not real, and it is not unitary, because one eigenvalue has modulus $2$.

Real and Imaginary Parts

Every operator can be decomposed as

$$A = B+iC,$$

where

$$B = \frac{A+A^\dagger}{2}, \qquad C = \frac{A-A^\dagger}{2i}.$$

Both $B$ and $C$ are Hermitian. A direct calculation gives

$$A^\dagger A = B^2+C^2+i[B,C],$$

and

$$AA^\dagger = B^2+C^2-i[B,C].$$

Therefore

$$A\text{ is normal} \qquad\Longleftrightarrow\qquad [B,C]=0.$$

This is a useful physical interpretation: a normal operator is equivalent to a pair of commuting Hermitian operators, its real and imaginary parts. Its complex eigenvalues can be read as paired real eigenvalues for compatible observables.

Orthogonality of Eigenvectors

For a normal operator, eigenvectors belonging to distinct eigenvalues are orthogonal. This extends a familiar property of Hermitian operators to the full normal class.

The reason is that normal operators admit an orthonormal eigenbasis. In such a basis the matrix is diagonal, and different eigenspaces are mutually orthogonal. The same statement can also be proved directly from the normality condition.

For non-normal matrices, eigenvectors may be strongly nonorthogonal even when the matrix is diagonalizable. This matters in numerical work and in non-Hermitian effective models.

Non-Normal Example

The Jordan block

$$J = \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}$$

is not normal. Its adjoint is

$$J^\dagger = \begin{pmatrix} 1 & 0\\ 1 & 1 \end{pmatrix}.$$

Then

$$J^\dagger J = \begin{pmatrix} 1 & 1\\ 1 & 2 \end{pmatrix}, \qquad JJ^\dagger = \begin{pmatrix} 2 & 1\\ 1 & 1 \end{pmatrix}.$$

These are not equal. The same matrix also fails to be diagonalizable because it has only one independent eigenvector.

Non-normality can occur even when a matrix is diagonalizable. In that case the diagonalizing eigenbasis exists but is not orthonormal.

Worked Example

Consider

$$A = \begin{pmatrix} 1 & i\\ i & 1 \end{pmatrix}.$$

This matrix is not Hermitian, since its off-diagonal entries are not complex conjugates in transposed positions. It is normal because

$$A = I+i\sigma_x,$$

where $\sigma_x$ is Hermitian and commutes with $I$.

The normalized eigenvectors of $\sigma_x$ are

$$u_+ = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad u_- = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ -1 \end{pmatrix}.$$

They are also eigenvectors of $A$:

$$Au_+ = (1+i)u_+, \qquad Au_- = (1-i)u_-.$$

With

$$U = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix},$$

we get

$$U^\dagger A U = \begin{pmatrix} 1+i & 0\\ 0 & 1-i \end{pmatrix}.$$

The eigenvalues are complex, but the eigenvectors are orthonormal. That is the signature of normality.

Physical Interpretation

Hermitian operators represent observables in finite-dimensional quantum mechanics. Unitary operators represent ideal closed-system time evolution, symmetries, and basis changes. Both are normal, which explains why they admit orthonormal eigenbases and spectral decompositions.

Normal operators also appear when one packages two commuting Hermitian operators into a single complex operator. If $A=B+iC$ with $[B,C]=0$, the real and imaginary parts can be simultaneously diagonalized, and $A$ records their joint eigenvalues as complex numbers.

Normality is not a measurement postulate by itself. A normal operator with complex eigenvalues is not an ordinary observable. Its usefulness is mathematical: it is the largest finite-dimensional class for which unitary diagonalization works exactly.

Common Mistakes

• Confusing normal operators with normalized vectors.
• Thinking normal means Hermitian.
• Assuming every diagonalizable matrix is unitarily diagonalizable.
• Forgetting that normality uses the adjoint, so it depends on the inner product.
• Treating complex eigenvalues of a normal operator as measurement outcomes of a single ordinary observable.
• Ignoring non-normality in effective non-Hermitian models, where eigenvectors may be nonorthogonal.
• Extending finite-dimensional normal-operator facts to unbounded infinite-dimensional operators without domain care.

Cross-Links

• Diagonalization
• Hermitian Operators
• Unitary Operators
• Spectral Decomposition
• Matrix Functions and Exponentials
• Projectors
• Compatible Observables
• Spectral Decomposition
• Matrix Diagonalization

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• G. Strang, Linear Algebra and Its Applications, 4th ed., Brooks/Cole, 2006.
• P. R. Halmos, Finite-Dimensional Vector Spaces, 2nd ed., Springer, 1974.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Show directly that every Hermitian matrix is normal.

Solution

If $A$ is Hermitian, then $A^\dagger=A$. Therefore

$$A^\dagger A = AA = AA^\dagger.$$

Thus $A$ is normal.

2. Check whether

$$A = \begin{pmatrix} 0 & 2\\ 0 & 0 \end{pmatrix}$$

is normal.

Solution

Here

$$A^\dagger = \begin{pmatrix} 0 & 0\\ 2 & 0 \end{pmatrix}.$$

Then

$$A^\dagger A = \begin{pmatrix} 0 & 0\\ 0 & 4 \end{pmatrix}, \qquad AA^\dagger = \begin{pmatrix} 4 & 0\\ 0 & 0 \end{pmatrix}.$$

These are not equal, so $A$ is not normal.

3. Let $A=B+iC$, where $B$ and $C$ are Hermitian and commute. Show that $A$ is normal.

Solution

Since $B^\dagger=B$ and $C^\dagger=C$,

$$A^\dagger = B-iC.$$

Using $BC=CB$,

$$A^\dagger A = (B-iC)(B+iC) = B^2+C^2,$$

and

$$AA^\dagger = (B+iC)(B-iC) = B^2+C^2.$$

Thus $A^\dagger A=AA^\dagger$, so $A$ is normal.