Diagonalization and the Spectral Decomposition

Diagonalization answers a structural question underneath most of computational physics: is there a basis in which a linear operator does nothing but scale each axis independently? When a basis of eigenvectors exists, the matrix collapses to a diagonal list of eigenvalues, and every hard matrix operation — powers, exponentials, inverses, square roots — becomes the same scalar operation one eigenvalue at a time. A tangled coupled system falls apart into independent one-dimensional pieces the moment you view it in that basis. The same fact wears a second face, the spectral decomposition. Instead of $A=PDP^{-1}$ with an explicit change-of-basis matrix, we write $A=\sum_i\lambda_iP_i$ , a weighted sum of projectors onto the eigenspaces. The projectors $P_i$ are intrinsic to $A$ — idempotent, mutually orthogonal, resolving the identity. This form is the gateway to functional calculus and to the operator language of quantum mechanics, where observables are Hermitian operators and a measurement is a projection. Not every operator submits. The obstruction is defectiveness — a shortage of eigenvectors, measured exactly by the geometric multiplicity falling below the algebraic multiplicity, and organized by the Jordan canonical form. But the operators that dominate physics — Hermitian, anti-Hermitian, unitary, and generally normal operators satisfying $AA^\dagger=A^\dagger A$ — obey the spectral theorem, which guarantees diagonalizability by an orthonormal eigenbasis, i.e. by a unitary transformation. This article develops the algebraic statement, the projector statement, the sharp conditions for each, and the functional calculus that makes the machinery indispensable.

Key takeaways

An operator $A$ is diagonalizable when its eigenvectors form a basis; in that basis its matrix is diagonal with the eigenvalues on it.
Concretely $A=PDP^{-1}$ , the columns of $P$ being eigenvectors and $D=\mathrm{diag}(\lambda_1,\dots,\lambda_n)$ — a similarity into the eigenbasis.
Equivalently $A=\sum_i\lambda_iP_i$ , the spectral decomposition into eigenprojectors that are idempotent, sum to the identity, and (for normal $A$ ) mutually orthogonal.
Diagonalizability holds iff the minimal polynomial has no repeated root — i.e. geometric multiplicity equals algebraic multiplicity for every eigenvalue.
The spectral theorem makes every normal operator unitarily diagonalizable — real eigenvalues if Hermitian, unit-modulus if unitary.
Functional calculus is then immediate: $f(A)=\sum_if(\lambda_i)P_i$ , so $e^{A}$ , $A^{-1}$ , $\sqrt{A}$ all reduce to scalar functions of the eigenvalues.
Key idea: Diagonalizing an operator is choosing the eigenbasis as your coordinate system; the spectral decomposition $A=\sum_i\lambda_iP_i$ then exhibits $A$ as a superposition of independent scalings on orthogonal eigenspaces, turning any function of $A$ into the same function applied scalar-by-scalar to the eigenvalues.

Where this lives today

Diagonalization is the inner loop of an enormous amount of live computation. Quantum chemistry and condensed-matter physics spend much of their machine time diagonalizing Hamiltonians — the Lanczos and Davidson algorithms extract the low-lying eigenpairs of matrices with billions of rows for molecular energies and phonon spectra, and the normal modes of a vibrating molecule or crystal are literally the eigenvectors of its dynamical matrix. Quantum computing runs on the same idea: a variational quantum eigensolver estimates the smallest eigenvalue of a Hamiltonian, and every quantum gate is a unitary whose spectral decomposition says what it does. In data science, principal component analysis diagonalizes a covariance matrix so its eigenvectors become the axes of maximal variance, and Google's original PageRank is nothing but the dominant eigenvector of a stochastic web-link matrix. Spectral graph theory reads community structure off a Laplacian's eigenvalues; stability analysis asks one question of a linearized system — are the real parts of its eigenvalues negative? Wherever a linear operator must be understood rather than merely applied, the first move is to diagonalize it.

The physical picture

Picture $A$ as a deformation of space. A generic vector gets rotated and stretched in a way hard to read off the matrix; eigenvectors are the special directions that survive intact up to a scale factor — the natural axes of the deformation, the grain of the wood. Diagonalizing redraws the coordinate grid along these axes, after which $A$ does nothing but stretch each by its eigenvalue. For the symmetric matrix with rows (2, 1) and (1, 2) the axes lie at 45°: the diagonal is stretched by 3, the anti-diagonal left almost alone at 1, and the unit circle is carried to an ellipse aligned with those axes.

Figure

Eigenvectors are the invariant directions of A. The matrix with rows (2, 1) and (1, 2) carries the unit circle to an ellipse: the 45° axis is scaled by λ₊ = 3, the perpendicular axis by λ₋ = 1, so Av₊ = 3v₊ and Av₋ = 1v₋

The spectral form $A=\sum_i\lambda_iP_i$ turns that geometry into an algebraic recipe. To apply $A$ to any vector, resolve it into its component in each eigenspace — what the projector $P_i$ extracts — scale each by the matching $\lambda_i$ , and add the pieces back; the operator never mixes the eigenspaces. This is why functions of $A$ are trivial here: squaring $A$ squares each scale factor and exponentiating it exponentiates each one, the eigenspaces being inert containers on which a scalar acts. The second figure exhibits the decomposition as an actual sum of numeric projector matrices resolving the identity.

Figure

The spectral decomposition rebuilds A as 3 P₊ + 1 P₋ from two rank-one orthogonal projectors. The same two projectors resolve the identity, P₊ + P₋ = I, so every vector splits cleanly into eigenspace parts

For Hermitian operators the eigenspaces are mutually orthogonal, so the natural axes are perpendicular and the change of basis is a rigid unitary rotation, never a shear. That single fact is why measurement in quantum mechanics is clean: distinct outcomes live in orthogonal subspaces, read out by the projectors $P_i$ above.

The core equations

Everything begins with the eigenvalue equation, the statement that certain directions are merely scaled by the operator. It is the defining relation from which diagonalization, the spectral decomposition, and functional calculus all descend.

A\,v_i=\lambda_i\,v_i,\qquad i=1,\dots,n \qquad (1)

where: $A$ — linear operator, or an $n\times n$ matrix once a basis is fixed; $v_i$ — the $i$ -th eigenvector, a nonzero direction left invariant by $A$ ; $\lambda_i$ — its scale factor. The article is about when there are enough such axes to form a basis. Collect $n$ independent eigenvectors as the columns of a matrix $P$ . The whole set of eigenvalue equations then bundles into a single similarity, and $A$ is conjugate to a diagonal matrix.

A=P\,D\,P^{-1},\qquad D=\mathrm{diag}(\lambda_1,\dots,\lambda_n) \qquad (2)

where: $P$ — the eigenvector matrix, columns $v_1,\dots,v_n$ ; $D$ — the diagonal eigenvalue matrix; $P^{-1}$ — the inverse change-of-basis, existing precisely when the eigenvectors are independent. Read right to left: $P^{-1}$ rotates into the eigenbasis, $D$ scales, $P$ rotates back. The same operator repackaged geometrically is the spectral decomposition: a weighted sum of projectors, one per distinct eigenvalue, weighted by that eigenvalue.

A=\sum_i\lambda_i\,P_i \qquad (3)

where: $P_i$ — the orthogonal projector onto the eigenspace of $\lambda_i$ ; $\lambda_i$ — a distinct eigenvalue, the sum running over distinct values rather than a basis, with degeneracy absorbed into the rank of $P_i$ . The projectors obey a tight algebra encoding "the eigenspaces do not overlap and together fill the space." This algebra, not the explicit matrices, is what makes the decomposition compute.

P_iP_j=\delta_{ij}P_i,\qquad \sum_iP_i=I,\qquad P_i^\dagger=P_i\ (\text{normal }A) \qquad (4)

where: $\delta_{ij}$ — Kronecker delta, so the projectors are idempotent ( $P_i^2=P_i$ ) and mutually annihilating; $I$ — identity, the resolution of the identity that guarantees probabilities sum to one; $P_i^\dagger=P_i$ — self-adjointness, holding for the orthogonal projectors of a normal operator. Because powers of $A$ collapse under the projector algebra, any function defined on the spectrum is obtained by applying that scalar function to each eigenvalue — the single most useful consequence of the whole framework.

f(A)=\sum_if(\lambda_i)\,P_i \qquad (5)

where: $f$ — any function defined on the eigenvalues of $A$ ( $e^{x}$ , $1/x$ , $\sqrt{x}$ , $\ln x$ ). No power series is ever summed; one evaluates $f$ on a handful of numbers and reassembles — the functional calculus. For the operators that matter in physics the projectors are not just any idempotents but orthogonal ones, because the eigenbasis can be chosen orthonormal. The spectral theorem states exactly when.

A=U\,D\,U^\dagger,\qquad U^\dagger U=UU^\dagger=I\quad\Longleftrightarrow\quad AA^\dagger=A^\dagger A \qquad (6)

where: $U$ — unitary matrix of orthonormal eigenvectors; $U^\dagger$ — conjugate transpose, equal to $U^{-1}$ ; the two-sided condition on the right is normality. Hermitian ( $A=A^\dagger$ ) gives real $D$ , unitary ( $A^\dagger=A^{-1}$ ) gives $|D_{ii}|=1$ ; both are special cases of normal. Finally, the sharp criterion for the general (not necessarily normal) case is a count: diagonalizability is the statement that no eigenvalue is short of eigenvectors.

m_{\mathrm{geo}}(\lambda)=\dim\ker(A-\lambda I),\qquad A\ \text{diagonalizable}\ \Leftrightarrow\ m_{\mathrm{geo}}=m_{\mathrm{alg}}\ \ \forall\lambda \qquad (7)

where: $m_{\mathrm{geo}}$ — geometric multiplicity, the dimension of the eigenspace; $m_{\mathrm{alg}}$ — algebraic multiplicity, the order of $\lambda$ as a root of $\det(A-\lambda I)$ . Always $1\le m_{\mathrm{geo}}\le m_{\mathrm{alg}}$ ; equality for every $\lambda$ is the whole condition. The invariants $\det A=\prod_i\lambda_i$ and $\mathrm{tr}\,A=\sum_i\lambda_i$ (with multiplicity) give quick sanity checks on any claimed spectrum.

Derivation AWhy the eigenvector matrix diagonalizes the operator

The goal is to show that the abstract sentence " $A$ has a basis of eigenvectors" is literally the same statement as the matrix equation $A=PDP^{-1}$ , and to pin down where linear independence of the eigenvectors is the one hypothesis that makes the passage legal. We stack the $n$ eigenvalue equations into a single matrix identity, recognize the diagonal factor, and invert.

Step 1. Assemble the eigenvectors as columns, $P=[\,v_1\ \cdots\ v_n\,]$ , and act with $A$ . Matrix multiplication is column-wise, so $AP=[\,Av_1\ \cdots\ Av_n\,]$ . Why: multiplying $A$ into $P$ applies $A$ to every eigenvector at once, handling the $n$ eigenvalue equations in a single stroke.

Step 2. Replace each column using $Av_i=\lambda_iv_i$ , giving $AP=[\,\lambda_1v_1\ \cdots\ \lambda_nv_n\,]$ . Scaling the $i$ -th column by $\lambda_i$ is exactly right-multiplication by the diagonal matrix $D=\mathrm{diag}(\lambda_1,\dots,\lambda_n)$ , so $AP=PD$ . Why: right-multiplication by a diagonal rescales columns (left-multiplication would rescale rows), which is what pins the eigenvalues onto the diagonal in the order matching $P$ 's columns.

Step 3. The identity $AP=PD$ holds for any list of eigenvectors; the real content is invertibility of $P$ . If the $v_i$ are linearly independent, $P^{-1}$ exists and right-multiplying gives $A=PDP^{-1}$ ; conversely, reading $AP=PD$ backwards from any such factorization recovers $Av_i=\lambda_iv_i$ , so the columns of a diagonalizing $P$ must be eigenvectors. Why: independence of the eigenvectors is exactly the hypothesis that $P^{-1}$ exists, so diagonalizability and "the eigenvectors span the space" are one statement, not two.

Go deeper · Derivation B · 1 minBuilding the eigenprojectors of a normal operator

Derivation BBuilding the eigenprojectors of a normal operator

The aim is to construct the projectors $P_i$ explicitly from an orthonormal eigenbasis and to prove — not merely assert — the three properties the spectral calculus rests on: idempotence, mutual orthogonality, and completeness. We start from the unitary diagonalization the spectral theorem guarantees, expand it as a sum of rank-one outer products, and check the algebra of the blocks.

Step 1. Write the normal operator in its orthonormal eigenbasis $\{u_a\}$ as $A=\sum_a\lambda(a)\,u_au_a^\dagger$ , the column-by-column expansion of $UDU^\dagger$ . Each $u_au_a^\dagger$ is a rank-one projector onto a single eigen-direction. Why: the outer-product form is the bridge from the matrix $UDU^\dagger$ to the coordinate-free sum $\sum\lambda_iP_i$ , available only because the eigenvectors are orthonormal.

Step 2. Group the terms that share an eigenvalue: define $P_i=\sum_{a:\,\lambda(a)=\lambda_i}u_au_a^\dagger$ , so that $A=\sum_i\lambda_iP_i$ . The projector onto a degenerate eigenspace is the sum of the rank-one projectors of its orthonormal basis vectors. Why: collecting equal eigenvalues converts a basis-dependent sum over $n$ vectors into an intrinsic sum over the distinct eigenvalues, making $P_i$ independent of how the degenerate subspace was resolved.

Step 3. Idempotence and orthogonality follow from $u_a^\dagger u_b=\delta_{ab}$ : expanding $P_iP_j=\sum_{a,b}u_a(u_a^\dagger u_b)u_b^\dagger$ kills every cross term, leaving $P_i$ when $i=j$ and $0$ otherwise, so $P_iP_j=\delta_{ij}P_i$ . Completeness follows from the eigenbasis spanning the space: $\sum_iP_i=\sum_au_au_a^\dagger=UU^\dagger=I$ . Why: the entire projector algebra is a corollary of one orthonormality relation — it fails for non-normal operators with skew eigenvectors — and $\sum_iP_i=I$ is the geometric heart of "the observable has a complete set of states."

Go deeper · Derivation C · 1 minFrom powers to any function of an operator

Derivation CFrom powers to any function of an operator

The goal is to justify the workhorse formula $f(A)=\sum_if(\lambda_i)P_i$ — a function of an operator is that function applied to each eigenvalue — building from the projector algebra through powers and polynomials to any analytic function. This is what reduces $e^{A}$ , $A^{-1}$ , and $\sqrt{A}$ from infinite processes to a scalar evaluation on a handful of numbers.

Step 1. Square the spectral form: $A^2=\sum_{i,j}\lambda_i\lambda_jP_iP_j=\sum_i\lambda_i^2P_i$ , since $P_iP_j=\delta_{ij}P_i$ collapses the double sum to its diagonal, and inducting gives $A^{k}=\sum_i\lambda_i^{k}P_i$ for every power. By linearity every polynomial follows: $p(A)=\sum_ip(\lambda_i)P_i$ . Why: the cross terms vanishing is the whole mechanism — the eigenspaces never talk to each other, so each eigenvalue is simply raised to the power in place.

Step 2. Sum an analytic $f$ term by term: $f(A)=\sum_k c_kA^{k}=\sum_i\big(\sum_k c_k\lambda_i^{k}\big)P_i=\sum_if(\lambda_i)P_i$ , valid whenever every $\lambda_i$ lies in the disc of convergence. Read off the cases: $e^{A}=\sum_ie^{\lambda_i}P_i$ , $A^{-1}=\sum_i\lambda_i^{-1}P_i$ (iff all $\lambda_i\ne0$ ), $\sqrt{A}=\sum_i\sqrt{\lambda_i}\,P_i$ . Why: the eigenvalues are ordinary numbers, so an operator series poses no new analytic question — convergence at each $\lambda_i$ is all that is needed, and each function becomes a one-line substitution.

Limiting cases

Whether and how an operator diagonalizes is decided by its internal structure; the cases below run from the most generous to the outright forbidden.

Case	Result	Meaning
Distinct eigenvalues	$\lambda_i\ne\lambda_j\ (i\ne j)\ \Rightarrow\ A$ diagonalizable	$n$ distinct eigenvalues in dimension $n$ force $n$ independent eigenvectors — a sufficient, not necessary, guarantee
Degenerate but semisimple	$m_{\mathrm{geo}}=m_{\mathrm{alg}}$ with repeats	Repeated eigenvalues are fine as long as the eigenspace is full; the identity $I$ has one eigenvalue yet is already diagonal
Normal operator	$AA^\dagger=A^\dagger A\ \Rightarrow\ A=UDU^\dagger$	Spectral theorem: an orthonormal eigenbasis exists; covers Hermitian (real $D$ ) and unitary ( $\lvert D_{ii}\rvert=1$ )
Defective operator	$m_{\mathrm{geo}}(\lambda)<m_{\mathrm{alg}}(\lambda)$	Too few eigenvectors to span; not diagonalizable, and the Jordan form takes over
Minimal-polynomial test	$m_A(x)=\prod_i(x-\lambda_i)$ with distinct factors $\Leftrightarrow$ diagonalizable	The sharp iff criterion: the minimal polynomial must be square-free over the relevant field

Common trapDiagonalizable means distinct eigenvalues

This is the single most common error, and it is wrong in both directions. Distinct eigenvalues are sufficient but not necessary: the identity has every eigenvalue equal to 1 yet is perfectly diagonal, because its eigenspace is the whole space ( $m_{\mathrm{geo}}=m_{\mathrm{alg}}=n$ ). Conversely, a repeated eigenvalue does not doom you — it only raises the risk of a shortfall. The real criterion is whether the eigenvectors span the space, i.e. whether geometric multiplicity meets algebraic multiplicity for every eigenvalue. The clean counterexample is the shear with rows (1, 1) and (0, 1): its only eigenvalue is 1 with algebraic multiplicity 2, but $\ker(A-I)$ is the single line spanned by $(1,0)$ , so no $P$ diagonalizes it. Count eigenvectors, never eigenvalues.

Worked examples

1 · Full spectral decomposition and functional calculus

Take the symmetric matrix $A$ with rows $(2,1)$ and $(1,2)$ . Find its eigenvalues and eigenprojectors, verify the spectral decomposition, and use it to compute $e^{A}$ and $\sqrt{A}$ — quantities that would otherwise require summing an infinite matrix series.

Step 1. Characteristic polynomial: $\det(A-\lambda I)=(2-\lambda)^2-1=\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3)$ , so the eigenvalues are $\lambda_+=3$ and $\lambda_-=1$ . As a check, $\mathrm{tr}\,A=4=1+3$ and $\det A=3=1\cdot3$ . Eigenvalues 3 and 1.

Step 2. Eigenvectors. For $\lambda_+=3$ : $(A-3I)v=0$ gives $-x+y=0$ , so $v_+=\tfrac1{\sqrt2}(1,1)$ . For $\lambda_-=1$ : $x+y=0$ , so $v_-=\tfrac1{\sqrt2}(1,-1)$ . They are orthogonal, as the spectral theorem demands for a symmetric matrix. Orthonormal eigenbasis at ±45°.

Step 3. Eigenprojectors $P_i=v_iv_i^{\mathsf T}$ : $P_+=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}$ , $P_-=\tfrac12\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$ . They satisfy $P_++P_-=I$ , $P_+^2=P_+$ , $P_+P_-=0$ , and reassemble as $3P_++1P_-=\begin{pmatrix}2&1\\1&2\end{pmatrix}=A$ . Spectral decomposition verified.

Step 4. Matrix exponential by functional calculus: $e^{A}=e^{3}P_++e^{1}P_-$ . With $e^{3}=20.0855$ and $e^{1}=2.71828$ , the diagonal entry is $\tfrac12(e^3+e^1)=11.4019$ and the off-diagonal is $\tfrac12(e^3-e^1)=8.6836$ , so $e^{A}=\begin{pmatrix}11.402&8.684\\8.684&11.402\end{pmatrix}$ . Trace check: $22.804=e^3+e^1$ . e^A computed with no series summation.

Step 5. Principal square root the same way: $\sqrt{A}=\sqrt3\,P_++\sqrt1\,P_-$ , giving diagonal $\tfrac12(\sqrt3+1)=1.3660$ and off-diagonal $\tfrac12(\sqrt3-1)=0.3660$ , so $\sqrt A=\begin{pmatrix}1.366&0.366\\0.366&1.366\end{pmatrix}$ . Squaring it returns rows $(2,1)$ and $(1,2)$ exactly. Every function of A is one scalar evaluation per eigenvalue.

2 · Diagonalizable or defective? A multiplicity test

Two matrices share the same single eigenvalue 3 with algebraic multiplicity 2. Decide which is diagonalizable: $B=\begin{pmatrix}3&0\\0&3\end{pmatrix}$ and $C=\begin{pmatrix}3&1\\0&3\end{pmatrix}$ .

Step 1. Both have characteristic polynomial $(\lambda-3)^2$ , so $m_{\mathrm{alg}}(3)=2$ for each. Eigenvalue count alone cannot separate them — the decision lives entirely in the eigenvectors. Same spectrum, different fate.

Step 2. For $B$ : $B-3I=0$ , so $\ker(B-3I)$ is the whole plane and $m_{\mathrm{geo}}(3)=2=m_{\mathrm{alg}}(3)$ ; $B=3I$ is already diagonal, diagonalizable despite the repeat. B is semisimple.

Step 3. For $C$ : $C-3I=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has kernel only the line $(1,0)$ , so $m_{\mathrm{geo}}(3)=1<2=m_{\mathrm{alg}}(3)$ . The eigenvectors cannot span the plane; $C$ is defective and no similarity makes it diagonal. C fails the criterion.

Step 4. Confirm by the minimal polynomial: $m_B(x)=x-3$ is square-free, but $(C-3I)^2=0$ with $(C-3I)\ne0$ makes $m_C(x)=(x-3)^2$ — a repeated root, the sharp signal of defectiveness and the point where the Jordan form takes over. Minimal polynomial agrees: B diagonalizable, C not.

3 · Decoupling a linear system into normal modes

Solve $\dot{x}=Ax$ with $A$ the matrix of rows $(2,1)$ and $(1,2)$ and $x(0)=(1,0)$ by projecting onto the eigenbasis.

Step 1. In the eigenbasis the equations decouple, $\dot c_\pm=\lambda_\pm c_\pm$ , so with $c_\pm(0)=v_\pm^{\mathsf T}x(0)=\tfrac1{\sqrt2}$ the solution is $x(t)=e^{At}x(0)$ with $e^{At}=e^{3t}P_++e^{t}P_-$ , giving $x_1(t)=\tfrac12(e^{3t}+e^{t})$ and $x_2(t)=\tfrac12(e^{3t}-e^{t})$ . The eigenvectors are the normal modes — the $(1,1)$ mode grows like $e^{3t}$ , the $(1,-1)$ mode like $e^{t}$ . Eigenvectors are the normal modes; eigenvalues are their rates.

Physics of the portrait

The spectral decomposition is where nineteenth-century matrix algebra, integral equations, and quantum mechanics turn out to be one subject. The word that ties them together — spectrum — was coined not for atoms but for the eigenvalues of an operator, by the man who built the space they live in.

Origins · 1 minDavid Hilbert and how this was discovered

David Hilbert

1862–1943 · Königsberg and Göttingen · spectral theory and the naming of the spectrum

The finite-dimensional theorem came first and from elsewhere: Augustin-Louis Cauchy proved in 1829 that every real symmetric matrix has real eigenvalues and an orthogonal eigenbasis — the principal-axis theorem, born from the geometry of quadric surfaces and the secular equations of celestial mechanics. Karl Weierstrass (1868) and Camille Jordan (1870) then mapped the failures, classifying through elementary divisors and the Jordan canonical form when a matrix is defective. But it was David Hilbert who lifted the idea to infinite dimensions and gave it its language. In his 1904–1912 papers on integral equations he studied symmetric kernels as infinite matrices, proved a spectral theorem for them, and borrowed the word Spektrum for their eigenvalues — famously remarking that he had chosen the name with no suspicion it would later attach to the spectra of atoms; the space of square-summable sequences he worked in became "Hilbert space." The circle closed in 1929, when John von Neumann proved the spectral theorem for unbounded self-adjoint operators and made it the spine of quantum mechanics: an observable is its spectral decomposition, its spectrum the possible measurements, its projectors the machinery of collapse. The abstraction Hilbert built for integral equations became the exact language physics needed for the atom.

History

1829

— Cauchy proves the principal-axis theorem for real symmetric matrices

1868

— Weierstrass classifies elementary divisors, explaining defective forms

1870

— Jordan publishes the canonical form for non-diagonalizable operators

1904

— Hilbert builds the spectral theory of integral equations and names the spectrum

1929

— von Neumann extends the spectral theorem to unbounded operators for quantum mechanics

Going further

Texts. Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 5–6 (matrices, eigenvalue problems) · Hassani, Mathematical Physics, 2nd ed., Ch. 2–4 (operators and the spectral theorem) · Stone & Goldbart, Mathematics for Physics, Ch. on linear algebra and spectral theory · Nakahara, Geometry, Topology and Physics, §2 (linear algebra background) · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 3–4 · Horn & Johnson, Matrix Analysis, Ch. 1–2 (diagonalization, normal matrices, and the spectral theorem). The core results of this article — $A=PDP^{-1}$ , $A=\sum_i\lambda_iP_i$ , the projector algebra $P_iP_j=\delta_{ij}P_i$ with $\sum_iP_i=I$ , the functional calculus $f(A)=\sum_if(\lambda_i)P_i$ , and the spectral theorem $AA^\dagger=A^\dagger A\Leftrightarrow A=UDU^\dagger$ — are standard results of linear algebra. The anchor example throughout is the matrix with rows (2, 1) and (1, 2): eigenvalues 3 and 1, projectors $P_\pm=\tfrac12\begin{pmatrix}1&\pm1\\\pm1&1\end{pmatrix}$ , $e^{A}=\begin{pmatrix}11.402&8.684\\8.684&11.402\end{pmatrix}$ , $\sqrt A=\begin{pmatrix}1.366&0.366\\0.366&1.366\end{pmatrix}$ ; the defective shear with rows (1, 1) and (0, 1) has geometric multiplicity 1. No empirical constants enter.

Prerequisites. Eigenvalues, Eigenvectors, and the Characteristic Equation · Change of Basis and Similarity Transformations · Linear Independence, Basis, and Dimension · Linear Operators and Matrix Representations. Leads to. Adjoint Operators and Hermitian Operators · Unitary and Orthogonal Operators · Jordan Canonical Form and Defective Matrices · Orthogonality and the Gram–Schmidt Process.

Ask the tutor: Ask me to diagonalize any specific matrix, build its eigenprojectors, or compute a function like $e^{A}$ or $\sqrt{A}$ step by step, and I will walk the whole spectral decomposition with you.

What is the difference between diagonalization and the spectral decomposition?

They are two encodings of the same fact. Diagonalization is the matrix statement $A=PDP^{-1}$ , naming an explicit change-of-basis $P$ built from eigenvectors and a diagonal $D$ of eigenvalues. The spectral decomposition $A=\sum_i\lambda_iP_i$ is basis-free: it expresses $A$ as a weighted sum of projectors onto its eigenspaces, no coordinate system chosen. The projector form is more intrinsic — the $P_i$ do not depend on how you order or normalize eigenvectors within a degenerate subspace — and it generalizes cleanly to infinite dimensions and to functions of the operator. For a normal operator the two are related by $P_i=\sum_a u_au_a^\dagger$ over an orthonormal basis of the $i$ -th eigenspace.

Why do physicists care so much about normal operators specifically?

Because normality ( $AA^\dagger=A^\dagger A$ ) is exactly the condition for an operator to have an orthonormal eigenbasis, and orthonormality is what quantum mechanics needs. Hermitian operators (real eigenvalues, observables), unitary operators (symmetries, time evolution), and anti-Hermitian generators are all normal, so the spectral theorem covers every operator that appears as a physical quantity. The payoff is perpendicular eigenspaces: distinct measurement outcomes live in orthogonal subspaces, projector probabilities are non-negative and sum to one via $\sum_iP_i=I$ , and the change of basis is a rigid unitary rotation, not a distorting shear. A non-normal operator may still be diagonalizable, but its eigenvectors are skew and the clean probabilistic reading is lost.

Can a matrix with repeated eigenvalues still be diagonalized?

Yes — repeated eigenvalues are not the problem. The identity has a single eigenvalue repeated $n$ times and is already diagonal. What matters is whether each eigenspace is as large as its algebraic multiplicity: you need $m_{\mathrm{geo}}(\lambda)=m_{\mathrm{alg}}(\lambda)$ for every $\lambda$ . A degenerate eigenvalue just means several independent eigenvectors share a scale factor, which is fine and common (rotational symmetry produces it). Trouble arises only when an eigenvalue is "short" of eigenvectors, $m_{\mathrm{geo}}<m_{\mathrm{alg}}$ — the defective case handled by the Jordan form. Repeated eigenvalues raise the risk of non-diagonalizability but never cause it by themselves.

How does functional calculus let me compute e^A without an infinite series?

Once you have $A=\sum_i\lambda_iP_i$ , any function is applied eigenvalue-by-eigenvalue: $f(A)=\sum_if(\lambda_i)P_i$ . Powers collapse under the projector algebra — $A^k=\sum_i\lambda_i^kP_i$ because $P_iP_j=\delta_{ij}P_i$ — so the series $e^{A}=\sum_kA^k/k!$ just sums each scalar series $\sum_k\lambda_i^k/k!=e^{\lambda_i}$ in place. You diagonalize once, exponentiate a few numbers, and reassemble. For the worked matrix $e^{A}=e^3P_++e^1P_-$ gives entries $11.402$ and $8.684$ directly, and the same one-liner delivers $A^{-1}$ (use $1/\lambda_i$ ), $\sqrt A$ (use $\sqrt{\lambda_i}$ ), and $\ln A$ — provided $f$ is defined on every eigenvalue.

What exactly goes wrong for a defective matrix, and what replaces diagonalization?

A defective matrix has an eigenvalue whose eigenspace is too small — geometric multiplicity below algebraic multiplicity — so its eigenvectors fail to span the space and no eigenbasis exists. The shear with rows (1, 1) and (0, 1) is canonical: eigenvalue 1 twice over, but only the single direction $(1,0)$ , so no diagonal $A=PDP^{-1}$ is possible. The replacement is the Jordan canonical form $A=PJP^{-1}$ , with $J$ block-diagonal carrying 1's on the superdiagonal that couple generalized eigenvectors. Functional calculus still works, but each Jordan block contributes derivatives of $f$ , not just $f(\lambda)$ . Defectiveness never happens for normal operators, which is why physics rarely meets it.

Why are the eigenspaces of a Hermitian operator automatically orthogonal?

It follows in two lines from $A=A^\dagger$ and the reality of Hermitian eigenvalues. Take eigenpairs $Au=\lambda u$ , $Av=\mu v$ with $\lambda\ne\mu$ . Comparing $\langle Au|v\rangle=\lambda\langle u|v\rangle$ with $\langle u|Av\rangle=\mu\langle u|v\rangle$ — equal because $A=A^\dagger$ — gives $(\lambda-\mu)\langle u|v\rangle=0$ , and since $\lambda\ne\mu$ we conclude $\langle u|v\rangle=0$ . Eigenvectors for distinct eigenvalues are orthogonal; within a degenerate eigenspace you orthonormalize by Gram–Schmidt. The result is a complete orthonormal eigenbasis — the concrete content of the spectral theorem and the reason the diagonalizing $U$ is unitary.