Normal Forms, Spectra, and Projectors

Finite-dimensional operators admit several decompositions, but they answer different questions. A diagonalization describes eigenvectors when enough of them exist. The spectral theorem gives an orthonormal eigenbasis precisely for normal operators. Jordan form records the nilpotent data left when diagonalization fails. A singular-value decomposition exists for every linear map between inner-product spaces, but it uses separate bases in the domain and codomain and is not an eigenvalue decomposition. Keeping these statements separate prevents the spectrum from being asked to encode more than it actually does.

Required background. Vector Spaces, Duals, Linear Maps, and Bases supplies kernels, images, eigenvectors, and change-of-basis language.

Helpful background. Bilinear and Hermitian Forms, Adjoints, and Isometries supplies adjoints, orthogonality, and unitary maps for the normal-operator and singular-value results.

Unless stated otherwise, this page works on finite-dimensional complex vector spaces. Statements involving $A^\dagger$, orthogonal projectors, or unitary matrices assume a positive Hermitian inner product with bras conjugate-linear. A real operator can be complexified; genuinely real normal forms retain $2\times2$ blocks for nonreal conjugate eigenvalue pairs. Infinite-dimensional spectra and spectral measures belong to Functional and Spectral Analysis.

Eigenvalues and multiplicities

For an endomorphism $A:V\to V$, an eigenvalue $\lambda$ and eigenvector $v\neq0$ satisfy

$$Av=\lambda v.$$

In finite dimension, the spectrum is the finite set

$$\sigma(A) = \{\lambda\in\mathbb C: \det(\lambda\mathbb 1-A)=0\}.$$

The algebraic multiplicity of $\lambda$ is its multiplicity as a root of the characteristic polynomial. Its geometric multiplicity is

$$\dim E_\lambda, \qquad E_\lambda=\ker(A-\lambda\mathbb 1).$$

They obey

$$1 \leq \dim E_\lambda \leq \operatorname{mult}_{\mathrm{alg}}(\lambda).$$

The operator is diagonalizable exactly when the eigenspaces span $V$. Equivalently, the geometric and algebraic multiplicities agree for every eigenvalue, or the minimal polynomial splits into distinct linear factors. Having distinct eigenvalues is sufficient for diagonalizability, but it is not necessary.

If $A=SDS^{-1}$, then $D$ is diagonal in the eigenbasis formed by the columns of $S$. This is a similarity transformation: the same change of basis is used in the domain and codomain. It need not preserve an inner product.

The finite spectral theorem

An endomorphism of a complex inner-product space is normal when

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

The finite-dimensional spectral theorem states that the following are equivalent:

1. $A$ is normal.
2. $V$ has an orthonormal basis of eigenvectors of $A$.
3. There is a unitary $U$ and a diagonal $D$ such that $A=UDU^\dagger$.

This normal-operator formulation, together with the finite-dimensional functional calculus below, is developed in Axler 2024, Chapters 5 and 7 and Horn and Johnson 2013, Chapters 2 and 4.

Important special cases follow by restricting the diagonal entries:

• If $A^\dagger=A$, then every eigenvalue is real.
• If $A^\dagger=-A$, then every eigenvalue is purely imaginary.
• If $A^\dagger A=\mathbb 1$, then every eigenvalue has modulus one.
• If $A$ is positive semidefinite, so $\langle v|A v\rangle\geq0$ for every $v$, then every eigenvalue is nonnegative.

Conversely, a normal operator with real spectrum is Hermitian, and a normal operator with spectrum on the unit circle is unitary.

Normality is stronger than diagonalizability. A diagonalizable non-normal operator has an eigenbasis, but no orthonormal eigenbasis for the chosen inner product. It is also weaker than Hermiticity: unitary matrices and many other operators are normal without being Hermitian.

Spectral projectors and functions of an operator

For a normal operator, group an orthonormal eigenbasis by eigenvalue and let $P_\lambda$ be the orthogonal projector onto $E_\lambda$. Then

$$A = \sum_{\lambda\in\sigma(A)} \lambda P_\lambda,$$

with

$$P_\lambda^\dagger=P_\lambda, \qquad P_\lambda P_\mu = \delta_{\lambda\mu}P_\lambda, \qquad \sum_\lambda P_\lambda=\mathbb 1.$$

Individual eigenvectors within a degenerate eigenspace are not unique, but the projector onto the full eigenspace is. This is why degenerate physics is usually expressed with projectors rather than a preferred basis of states.

When the distinct eigenvalues are known, each projector is a polynomial in $A$:

$$P_\lambda = \prod_{\substack{\mu\in\sigma(A)\\\mu\neq\lambda}} \frac{A-\mu\mathbb 1}{\lambda-\mu}.$$

Acting on an eigenvector with eigenvalue $\nu$, every factor evaluates to zero if $\nu\neq\lambda$ and to one if $\nu=\lambda$. This formula works for any diagonalizable operator. For a non-normal diagonalizable operator the resulting projectors are generally oblique:

$$P_\lambda^2=P_\lambda, \qquad P_\lambda^\dagger\neq P_\lambda \quad\text{in general}.$$

For a function defined on the finite spectrum of a normal operator,

$$f(A) = \sum_\lambda f(\lambda)P_\lambda.$$

Thus

$$e^{tA} = \sum_\lambda e^{t\lambda}P_\lambda,$$

and, when $0\notin\sigma(A)$,

$$A^{-1} = \sum_\lambda\lambda^{-1}P_\lambda.$$

These finite sums are the matrix precursor of spectral functional calculus. They do not by themselves establish an infinite-dimensional spectral integral.

Jordan form records failed diagonalization

Over $\mathbb C$, every endomorphism is similar to a Jordan matrix:

$$A=SJS^{-1}, \qquad J=\bigoplus_{\alpha=1}^{m}J_{\lambda_\alpha,r_\alpha},$$

where a block of size $r$ is

$$J_{\lambda,r} = \lambda\mathbb 1_r+N_r, \qquad N_r^r=0.$$

The nilpotent matrix $N_r$ has ones on the superdiagonal in the standard Jordan convention. A block with $r>1$ signals a missing eigenvector. The corresponding Jordan chain $v_1,\ldots,v_r$ satisfies

$$(A-\lambda\mathbb 1)v_1=0, \qquad (A-\lambda\mathbb 1)v_{j+1}=v_j.$$

More generally, generalized eigenvectors satisfy

$$(A-\lambda\mathbb 1)^k v=0$$

for some $k>1$ and form chains that reconstruct the block.

The spectrum does not reveal the block sizes. For example,

$$A = \begin{pmatrix} \lambda&1\\ 0&\lambda \end{pmatrix}$$

has the same characteristic polynomial as $\lambda\mathbb 1_2$ but only one independent eigenvector. Since $A=\lambda\mathbb 1+N$ with $N^2=0$,

$$e^{tA} = e^{\lambda t} \left(\mathbb 1+tN\right) = e^{\lambda t} \begin{pmatrix} 1&t\\ 0&1 \end{pmatrix}.$$

The polynomial factor $t$ is dynamical information carried by the nilpotent part, not by the eigenvalue alone.

Jordan form is an exact classification, but it is often a poor numerical representation. Small perturbations can split a Jordan block, and the matrix of generalized eigenvectors can be badly conditioned. Schur decomposition and singular values are usually more stable computational tools.

Singular values use two orthonormal bases

Let $T:V\to W$ be any linear map between finite-dimensional positive inner-product spaces. The operator

$$T^\dagger T:V\longrightarrow V$$

is Hermitian and positive semidefinite because

$$\langle v|T^\dagger T v\rangle = \langle Tv|Tv\rangle \geq0.$$

Its eigenvalues are therefore nonnegative. Their nonnegative square roots

$$\sigma_a = \sqrt{\lambda_a(T^\dagger T)}$$

are the singular values of $T$. For every nonzero singular value, choose an orthonormal right singular vector $|v_a\rangle$ and define

$$|u_a\rangle = \frac{T|v_a\rangle}{\sigma_a}.$$

The resulting left singular vectors are orthonormal, and

$$T = \sum_{a=1}^{\operatorname{rank}T} \sigma_a|u_a\rangle\langle v_a|.$$

Extend the right singular vectors by an orthonormal basis of $\ker T$, and the left singular vectors by one of $(\operatorname{im}T)^\perp$; these completed bases define the unitary matrices $V$ and $U$.

In orthonormal coordinate bases this is

$$T=U\Sigma V^\dagger,$$

where $\Sigma$ is rectangular if $\dim V\neq\dim W$. The right vectors diagonalize $T^\dagger T$, while the left vectors diagonalize $TT^\dagger$.

Every finite-dimensional linear map has an SVD. This does not mean that every square matrix is diagonalizable: SVD uses one basis for $V$ and a different basis for $W$, whereas eigenvalue diagonalization uses a similarity transformation on one space. For a normal operator, the singular values are the absolute values of the eigenvalues. That statement need not hold for a non-normal operator.

The SVD also gives the polar decomposition

$$T=Q|T|, \qquad |T|=(T^\dagger T)^{1/2},$$

where $Q$ maps the support of $|T|$ isometrically onto $\operatorname{im}T$ and vanishes on $\ker T$. Thus $Q$ is a partial isometry. If $T:V\to W$ is a linear isomorphism, $Q$ is a unitary isomorphism $V\to W$; in particular, for an invertible endomorphism it is unitary.

Non-normal operators and basis sensitivity

For a diagonalizable operator $A=SDS^{-1}$, the eigenvectors can become nearly linearly dependent even when the entries of $A$ are moderate. Then $S^{-1}$ and the oblique spectral projectors can have large norms. Small perturbations may produce eigenvalue shifts much larger than one would expect from the normal case.

This is not a contradiction with exact diagonalizability. It is a warning that the eigenbasis can be ill-conditioned. Normal operators avoid this particular problem because their eigenvector matrix is unitary and hence condition number one in the induced Euclidean norm.

For example,

$$A = \begin{pmatrix} 1&\kappa\\ 0&2 \end{pmatrix}, \qquad P_1 = \begin{pmatrix} 1&-\kappa\\ 0&0 \end{pmatrix}, \qquad P_2 = \begin{pmatrix} 0&\kappa\\ 0&1 \end{pmatrix}.$$

The eigenvalues remain $1$ and $2$, and $A=P_1+2P_2$ is diagonalizable for every $\kappa$. Nevertheless, the oblique-projector norms grow with $|\kappa|$, exposing increasing eigenbasis sensitivity.

The distinction matters for effective evolution matrices, transfer operators, and open-system generators. A list of eigenvalues alone does not control transient growth or perturbation sensitivity for a non-normal operator; see Trefethen and Embree 2005, §§ 1–2 and 14–16.

QFT-facing examples

Hermitian mass mixing

Let $M^2=(M^2)^\dagger$. Then

$$M^2 = \sum_a \mu_aP_a, \qquad \mu_a\in\mathbb R,$$

with orthogonal projectors $P_a$. A unitary change of field basis separates the mass eigenspaces. If masses are degenerate, the basis inside a degenerate subspace remains arbitrary, while $P_a$ is invariant. Hermiticity alone does not imply stability. If $M^2\geq0$, write $\mu_a=m_a^2$.

For a kinetic matrix $K=K^\dagger>0$, solve the generalized eigenvalue problem

$$M^2v=\mu Kv.$$

The operator $K^{-1}M^2$ is self-adjoint with respect to $\langle v|w\rangle_K=v^\dagger Kw$, not necessarily with respect to the standard coordinate inner product. One may equivalently transform to a canonical kinetic basis before applying the ordinary spectral theorem. The same reduction underlies the standard numerical treatment of Hermitian definite generalized eigenproblems; see Anderson et al. 1999, “Generalized Symmetric Definite Eigenproblems,” Table 2.13.

Transverse polarization projector

For nonzero spatial momentum $\mathbf p$, define

$$\widehat p_j = \delta_{jk}\widehat p^{\,k}.$$

Here $i,j$ are Euclidean spatial indices; no spacetime-metric lowering is intended. The transverse projector is

$$(P_{\mathrm T})^i{}_j = \delta^i{}_j - \widehat p^{\,i}\widehat p_j, \qquad \widehat{\mathbf p} = \frac{\mathbf p}{|\mathbf p|}.$$

Then

$$P_{\mathrm T}^2=P_{\mathrm T}, \qquad P_{\mathrm T}^\dagger=P_{\mathrm T}.$$

This spatial projector is the finite-dimensional form of the transverse polarization completeness relation used in Tong 2006–2007, § 6.2, Eq. (6.523).

Its eigenspace with eigenvalue $1$ is the plane transverse to $\mathbf p$, and its eigenvalue-$0$ eigenspace is $\operatorname{span}\{\mathbf p\}$. In three spatial dimensions, $\operatorname{tr}P_{\mathrm T}=2$, so it selects two transverse directions. The field-theoretic distinction between massive and massless spin-one modes is developed on Massive and Massless Spin-One Polarizations.

Common pitfalls

Equating diagonalizable with normal. Diagonalizability requires some eigenbasis. Normality requires an orthonormal eigenbasis for the chosen positive inner product.

Reading degeneracy as a preferred set of eigenvectors. A degenerate eigenspace admits many orthonormal bases. The full spectral projector is the basis-independent object.

Treating SVD as similarity diagonalization. The two unitary matrices in $U\Sigma V^\dagger$ act on different sides and generally represent different spaces. Singular values are not eigenvalues.

Ignoring the nilpotent part. Two matrices can have identical eigenvalues and different Jordan blocks. Their exponentials and responses to forcing can therefore differ by polynomial factors.

Using the positive spectral theorem with an indefinite form. An operator that is self-adjoint relative to an indefinite metric need not have all the properties of a Hermitian matrix on a positive inner-product space. The form and its signature are hypotheses, not notation.

Exporting finite sums to infinite dimension. Continuous spectrum, residual spectrum, domains, unbounded operators, and projection-valued measures require the later functional-analytic framework.

Exercises

1. For

   $$A= \begin{pmatrix} 3&1\\ 0&3 \end{pmatrix},$$

   find the algebraic and geometric multiplicities of $3$ and compute $e^{tA}$.

   Solution

   The characteristic polynomial is $(\lambda-3)^2$, so the algebraic multiplicity is two. The kernel of $A-3\mathbb 1$ is spanned by $(1,0)^{\mathsf T}$, so the geometric multiplicity is one. Writing $A=3\mathbb 1+N$ with $N^2=0$ gives

   $$e^{tA} = e^{3t} \begin{pmatrix} 1&t\\ 0&1 \end{pmatrix}.$$

2. Let a diagonalizable operator have distinct eigenvalues $\lambda_1,\ldots,\lambda_r$. Verify directly that

   $$P_a = \prod_{b\neq a} \frac{A-\lambda_b\mathbb 1}{\lambda_a-\lambda_b}$$

   acts as the identity on $E_{\lambda_a}$ and vanishes on every other eigenspace.

   Solution

   On $v\in E_{\lambda_c}$, replace every occurrence of $A$ by $\lambda_c$. If $c=a$, each factor is one. If $c\neq a$, the factor with $b=c$ has zero numerator. Since the eigenspaces span $V$, this proves the projector properties.

3. Verify the projector identities for $P_{\mathrm T}=\mathbb 1-\widehat{\mathbf p} \widehat{\mathbf p}^{\mathsf T}$ and determine its trace in $d$ spatial dimensions.

   Solution

   Since $\widehat{\mathbf p}^{\mathsf T}\widehat{\mathbf p}=1$,

   $$P_{\mathrm T}^2 = \mathbb 1 -2\widehat{\mathbf p}\widehat{\mathbf p}^{\mathsf T} +\widehat{\mathbf p}\widehat{\mathbf p}^{\mathsf T} = P_{\mathrm T}.$$

   The matrix is real symmetric, so it is Hermitian. Its trace is $d-1$, matching the dimension of the subspace perpendicular to $\mathbf p$.

4. Let $T=|u\rangle\langle v|$ with nonzero $u$ and $v$. Find its only nonzero singular value.

   Solution

   One has

   $$T^\dagger T = \|u\|^2|v\rangle\langle v|.$$

   Its nonzero eigenvalue is $\|u\|^2\|v\|^2$, so the nonzero singular value is $\|u\|\,\|v\|$.

References

• E. Anderson et al., LAPACK Users’ Guide, 3rd ed., Society for Industrial and Applied Mathematics, 1999, “Generalized Symmetric Definite Eigenproblems”, for reducing $Av=\lambda Bv$ with Hermitian $A$ and positive-definite Hermitian $B$ to an ordinary Hermitian eigenproblem.
• Sheldon Axler, Linear Algebra Done Right, 4th ed., Springer, 2024, Chapters 5, 7, and 8, for eigenvalues, the finite spectral theorem, singular values, generalized eigenvectors, and Jordan form.
• Roger A. Horn and Charles R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2013, for normal matrices, canonical forms, projectors, and singular-value decomposition.
• David Tong, Lectures on Quantum Field Theory, Cambridge Part III lecture notes, University of Cambridge, 2006–2007, § 6.2, especially Eq. (6.523), for transverse polarization completeness.
• Lloyd N. Trefethen and Mark Embree, Spectra and Pseudospectra, Princeton University Press, 2005, §§ 1–2 for sensitivity and §§ 14–16 for transient growth of non-normal matrices.