The spectral theorem

Theorem

The spectral theorem

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Statement

Let \( \mathbb{F} \in \{\mathbb{R}, \mathbb{C}\} \), let \( V \) be a finite-dimensional inner product space over \( \mathbb{F} \) with \( \dim V = n \ge 1 \) (inner product \( \langle \cdot, \cdot \rangle \) linear in the first argument, conjugate-linear in the second, positive definite), and let \( T : V \to V \) be a linear operator that is self-adjoint: \( \langle Tu, v \rangle = \langle u, Tv \rangle \) for all \( u, v \in V \). Then there exist an orthonormal basis \( e_1, \dots, e_n \) of \( V \) and real numbers \( \lambda_1, \dots, \lambda_n \) such that \( T e_i = \lambda_i e_i \) for each \( i \). Equivalently, \( T = \sum_{i=1}^{n} \lambda_i \, \langle \cdot , e_i \rangle \, e_i \); in matrix form, every Hermitian matrix \( A \in \mathbb{C}^{n \times n} \) (respectively every symmetric matrix \( A \in \mathbb{R}^{n \times n} \)) factors as \( A = Q \Lambda Q^{*} \) with \( Q \) unitary (respectively \( A = Q \Lambda Q^{\top} \) with \( Q \) orthogonal) and \( \Lambda \) real diagonal.

Why it matters

The spectral theorem is the structural endpoint of a first linear algebra sequence: it says that the operators compatible with the inner product geometry — the self-adjoint ones — are exactly as simple as operators can be. In a well-chosen orthonormal frame they act by independent real rescalings along perpendicular axes. Every quadratic form can be brought to a sum of signed squares by a rotation (principal axes), every real symmetric matrix can be diagonalised without leaving the world of orthogonal coordinate changes, and functions of the operator (\( T^k \), \( e^{T} \), \( \sqrt{T} \)) can be defined and computed one eigenvalue at a time.

Beyond the course, it is the finite-dimensional prototype of the spectral theory that runs through analysis and physics: observables in quantum mechanics are self-adjoint precisely so that their spectra — the possible measurement outcomes — are real and their eigenstates orthogonal. Principal component analysis, the singular value decomposition, Sturm–Liouville theory and the theory of compact self-adjoint operators on Hilbert space are all descendants of this statement.

Hypotheses

Hypothesis (finite dimension).Drop it and even bounded self-adjoint operators can have no eigenvectors at all: the multiplication operator \( (Mf)(x) = x f(x) \) on \( L^2[0,1] \) is bounded and self-adjoint, but \( x f(x) = \lambda f(x) \) forces \( f = 0 \) almost everywhere, so \( M \) has empty point spectrum and certainly no eigenbasis. The infinite-dimensional theorem must be restated with spectral measures.

Hypothesis (self-adjointness).Drop it and both conclusions fail. The nilpotent matrix \( \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \) has no basis of eigenvectors at all; the rotation \( \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \) on \( \mathbb{R}^2 \) has no real eigenvalue whatsoever; and \( \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \) is diagonalisable with real eigenvalues but its eigenlines \( \operatorname{span}(1,0) \) and \( \operatorname{span}(1,1) \) are not perpendicular, so no orthonormal eigenbasis exists.

Hypothesis (genuine inner product, i.e. positive definiteness).Over an indefinite form the argument "\( \|v\|^2 \gt 0 \), hence \( \lambda = \bar{\lambda} \)" dies. On \( \mathbb{R}^2 \) with the Minkowski form \( \langle u, v \rangle = u_1 v_1 - u_2 v_2 \), the operator \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \) is self-adjoint for that form yet has no real eigenvalues: self-adjoint operators on indefinite spaces need not be diagonalisable.

Hypothesis (field \( \mathbb{R} \) or \( \mathbb{C} \)).The existence step leans on the fundamental theorem of algebra and on the order structure of \( \mathbb{R} \). Over \( \mathbb{Q} \) the symmetric matrix \( \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} \) has characteristic polynomial \( z^2 - z - 1 \) with irrational roots \( \frac{1 \pm \sqrt{5}}{2} \), so it has no eigenvectors in \( \mathbb{Q}^2 \) at all.

Convention (Hermitian, not merely complex-symmetric).Over \( \mathbb{C} \) the correct hypothesis is \( A = A^{*} = \overline{A}^{\top} \), not \( A = A^{\top} \). The complex-symmetric matrix \( \begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix} \) has trace \( 0 \) and determinant \( -1 - i^2 = 0 \), so both eigenvalues vanish; a diagonalisable matrix with only the eigenvalue \( 0 \) would be zero, so it is not diagonalisable at all.

Proof

Throughout, \( \langle \cdot, \cdot \rangle \) is linear in the first slot and conjugate-linear in the second; over \( \mathbb{R} \) the conjugations are vacuous. We prove three lemmas (reality of eigenvalues, existence of an eigenvalue, invariance of orthogonal complements) and then induct on \( \dim V \).

\[ \textbf{Lemma A (eigenvalues are real).}\quad Tv = \lambda v,\; v \ne 0 \;\Longrightarrow\; \lambda \in \mathbb{R}. \]

Compute \( \lambda \|v\|^2 = \langle \lambda v, v \rangle = \langle Tv, v \rangle = \langle v, Tv \rangle = \langle v, \lambda v \rangle = \bar{\lambda} \|v\|^2 \), using linearity in the first slot, self-adjointness, and conjugate-linearity in the second slot. Positive definiteness gives \( \|v\|^2 \gt 0 \), so \( \lambda = \bar{\lambda} \), i.e. \( \lambda \in \mathbb{R} \). B

\[ \textbf{Matrix model.}\quad \text{Fix any orthonormal basis of } V \text{ and let } A \text{ be the matrix of } T. \text{ Then } A = A^{*}. \]

An orthonormal basis exists by Gram–Schmidt applied to any basis. In an orthonormal basis the matrix of the adjoint \( T^{*} \) is the conjugate transpose of the matrix of \( T \) (entries are \( A_{ij} = \langle T e_j, e_i \rangle \), and \( \langle T^{*} e_j, e_i \rangle = \overline{\langle T e_i, e_j \rangle} \)). Since \( T = T^{*} \), the matrix satisfies \( A = A^{*} \): Hermitian if \( \mathbb{F} = \mathbb{C} \), real symmetric if \( \mathbb{F} = \mathbb{R} \). A

\[ \textbf{Lemma B, complex case.}\quad \mathbb{F} = \mathbb{C} \;\Longrightarrow\; T \text{ has an eigenvalue}. \]

The characteristic polynomial \( \chi_A(z) = \det(zI - A) \) is a monic polynomial of degree \( n \ge 1 \) over \( \mathbb{C} \); by the fundamental theorem of algebra it has a root \( \lambda \in \mathbb{C} \). Then \( \det(\lambda I - A) = 0 \), so \( \lambda I - A \) is singular and its kernel contains a nonzero vector, an eigenvector for \( \lambda \). A

\[ \textbf{Lemma B, real case.}\quad \mathbb{F} = \mathbb{R} \;\Longrightarrow\; T \text{ has a (real) eigenvalue and a real eigenvector}. \]

This is the step a rotation of \( \mathbb{R}^2 \) fails, so symmetry must be used. Regard the real symmetric matrix \( A \) from Step 2 as an operator on \( \mathbb{C}^n \) with the standard Hermitian inner product; since \( A = A^{\top} = A^{*} \) it is self-adjoint there. By Step 3 it has an eigenvalue \( \lambda \in \mathbb{C} \), and by Lemma A (applied on \( \mathbb{C}^n \)) \( \lambda \in \mathbb{R} \). Now \( \chi_A \in \mathbb{R}[z] \) and \( \lambda \in \mathbb{R} \), so \( A - \lambda I \) is a singular matrix with real entries; Gaussian elimination over \( \mathbb{R} \) produces a nonzero \( v \in \mathbb{R}^n \) with \( Av = \lambda v \). Transporting back along the orthonormal basis gives an eigenvector of \( T \) in \( V \). C

\[ \textbf{Lemma C (invariant complements).}\quad W \subseteq V \text{ a } T\text{-invariant subspace} \;\Longrightarrow\; T(W^{\perp}) \subseteq W^{\perp}. \]

Let \( u \in W^{\perp} \) and \( w \in W \). Then \( \langle Tu, w \rangle = \langle u, Tw \rangle = 0 \), because \( Tw \in W \) by invariance and \( u \perp W \). As \( w \in W \) was arbitrary, \( Tu \in W^{\perp} \). This is the decisive structural use of self-adjointness: for a general operator only \( T^{*} \)-invariance of \( W^{\perp} \) follows. B

\[ \textbf{Restriction stays self-adjoint.}\quad S := T|_{W^{\perp}} : W^{\perp} \to W^{\perp} \text{ satisfies } \langle S u, u' \rangle = \langle u, S u' \rangle \; \forall u, u' \in W^{\perp}. \]

By Step 5, \( S \) genuinely maps \( W^{\perp} \) into itself, and the displayed identity is the identity for \( T \) read on the subspace, with the inner product restricted. So \( S \) is a self-adjoint operator on the inner product space \( W^{\perp} \). A

\[ \textbf{Induction on } n = \dim V. \quad \text{Base } n = 1. \]

Pick any unit vector \( e_1 \in V \). Since \( \dim V = 1 \), \( T e_1 \in \operatorname{span}(e_1) \), so \( T e_1 = \lambda_1 e_1 \) for some scalar \( \lambda_1 \), which is real by Lemma A. Then \( \{e_1\} \) is the required orthonormal eigenbasis. A

\[ \textbf{Inductive step.}\quad n \ge 2, \text{ assuming the theorem for all self-adjoint operators on spaces of dimension } n - 1. \]

By Lemma B, \( T \) has an eigenvalue \( \lambda_1 \) (real by Lemma A) with eigenvector \( v \ne 0 \); set \( e_1 = v / \|v\| \), so \( T e_1 = \lambda_1 e_1 \) and \( \|e_1\| = 1 \). The line \( W = \operatorname{span}(e_1) \) is \( T \)-invariant, so \( W^{\perp} \) is \( T \)-invariant by Lemma C. For a finite-dimensional subspace \( V = W \oplus W^{\perp} \) (projection theorem, proved via Gram–Schmidt), hence \( \dim W^{\perp} = n - 1 \). By Step 6 the restriction \( S = T|_{W^{\perp}} \) is self-adjoint, so the inductive hypothesis yields an orthonormal basis \( e_2, \dots, e_n \) of \( W^{\perp} \) with \( S e_i = \lambda_i e_i \), \( \lambda_i \in \mathbb{R} \). B

\[ \textbf{Assembly.}\quad \{e_1, e_2, \dots, e_n\} \text{ is an orthonormal basis of } V \text{ of eigenvectors of } T. \]

Each \( e_i \) is a unit vector; \( e_2, \dots, e_n \) are pairwise orthogonal by the inductive hypothesis and each is orthogonal to \( e_1 \) because they lie in \( W^{\perp} \). An orthonormal list of \( n \) vectors in an \( n \)-dimensional space is a basis (orthonormal lists are linearly independent). Finally \( T e_i = S e_i = \lambda_i e_i \) for \( i \ge 2 \) and \( T e_1 = \lambda_1 e_1 \), with all \( \lambda_i \) real. \( \blacksquare \) A

\[ \textbf{Matrix form.}\quad A = A^{*} \in \mathbb{C}^{n \times n} \;\Longrightarrow\; A = Q \Lambda Q^{*}, \; Q^{*}Q = I, \; \Lambda = \mathrm{diag}(\lambda_1, \dots, \lambda_n) \in \mathbb{R}^{n \times n}. \]

Apply the theorem to \( T x = A x \) on \( \mathbb{F}^n \) with the standard inner product and let \( Q \) be the matrix whose columns are the orthonormal eigenvectors \( e_1, \dots, e_n \). Column orthonormality says exactly \( Q^{*} Q = I \), and \( A Q = Q \Lambda \) restates \( A e_i = \lambda_i e_i \) column by column; hence \( A = Q \Lambda Q^{*} \). Over \( \mathbb{R} \), \( Q^{*} = Q^{\top} \) and \( Q \) is orthogonal. Equivalently \( A = \sum_i \lambda_i e_i e_i^{*} \), the spectral decomposition into rank-one orthogonal projections. C

Result

\[ T = T^{*} \text{ on finite-dimensional } V \;\Longrightarrow\; \exists \text{ orthonormal basis } e_1, \dots, e_n \text{ and } \lambda_1, \dots, \lambda_n \in \mathbb{R} : \; T e_i = \lambda_i e_i . \]

Reading. An operator that respects the inner product symmetry (you can move it across the pairing) is, in the right rotated frame, nothing but a list of real stretch factors along mutually perpendicular axes. All the apparent complexity of a symmetric matrix is an artefact of looking at it in the wrong orthonormal basis.

Scope. Exactly: finite-dimensional real or complex inner product spaces, operator self-adjoint with respect to that (positive-definite) inner product. Over \( \mathbb{C} \) the conclusion extends verbatim to normal operators (\( TT^{*} = T^{*}T \)), except that eigenvalues may then be complex. In infinite dimensions the statement survives for compact self-adjoint operators (Hilbert–Schmidt theorem) and, with eigenbases replaced by spectral measures, for bounded and for unbounded self-adjoint operators on Hilbert space.

Corollaries & converses

Orthogonal diagonalisation. Every real symmetric matrix satisfies \( A = Q \Lambda Q^{\top} \) with \( Q \) orthogonal, \( \Lambda \) real diagonal; every Hermitian matrix satisfies \( A = Q \Lambda Q^{*} \) with \( Q \) unitary.
All eigenvalues real; eigenspaces orthogonal. \( V = \bigoplus_{\lambda} \ker(T - \lambda I) \) is an orthogonal direct sum: eigenvectors for distinct eigenvalues are automatically perpendicular (Problem 1).
Spectral resolution and functional calculus. \( T = \sum_{\lambda} \lambda P_{\lambda} \) with \( P_{\lambda} \) the orthogonal projections onto the eigenspaces, \( P_{\lambda} P_{\mu} = 0 \) for \( \lambda \ne \mu \), \( \sum_{\lambda} P_{\lambda} = I \). For any \( f : \mathbb{R} \to \mathbb{R} \) one may define \( f(T) = \sum_{\lambda} f(\lambda) P_{\lambda} \); this recovers powers, \( e^{T} \), and square roots of positive semidefinite operators (Problem 4).
Variational description. \( \lambda_{\max}(T) = \max_{\|v\| = 1} \langle Tv, v \rangle \) and \( \lambda_{\min}(T) = \min_{\|v\| = 1} \langle Tv, v \rangle \) (Problem 3); this seeds the Courant–Fischer min-max theorem.
Converse over \( \mathbb{R} \): true. If \( T \) on a real inner product space has an orthonormal eigenbasis, its matrix in that basis is real diagonal, hence symmetric, so \( T = T^{*} \). Orthonormal eigenbasis \( \Longleftrightarrow \) self-adjoint, over \( \mathbb{R} \).
Converse over \( \mathbb{C} \): false as stated. \( T = iI \) has an orthonormal eigenbasis (any orthonormal basis) but \( T^{*} = -iI \ne T \). The correct complex characterisation: \( T \) has an orthonormal eigenbasis \( \Longleftrightarrow \) \( T \) is normal; it is self-adjoint iff additionally all eigenvalues are real (Problem 5).

Fails without

Without self-adjointness (no eigenbasis at all): \( N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \) has the single eigenvalue \( 0 \) with one-dimensional eigenspace \( \operatorname{span}(1,0) \); two independent eigenvectors do not exist, so no basis of eigenvectors, orthonormal or otherwise. Indeed \( N^{*}N = \mathrm{diag}(0,1) \ne \mathrm{diag}(1,0) = NN^{*} \), so \( N \) is not even normal.
Without self-adjointness (eigenbasis exists but cannot be made orthonormal): \( B = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \) is diagonalisable with real eigenvalues \( 1, 2 \), but the eigenspaces are \( \operatorname{span}(1,0) \) and \( \operatorname{span}(1,1) \), meeting at \( 45^{\circ} \). Diagonalisable \( \ne \) orthogonally diagonalisable.
Without finite-dimensionality: \( (Mf)(x) = x f(x) \) on \( L^2[0,1] \) is bounded, self-adjoint (\( \int_0^1 x f \bar{g} = \int_0^1 f \overline{x g} \)), yet \( Mf = \lambda f \) forces \( (x - \lambda) f(x) = 0 \), so \( f = 0 \) a.e.: no eigenvectors, no eigenbasis. The spectrum \( [0,1] \) is purely continuous.
Over \( \mathbb{R} \) without symmetry: the rotation \( R_{\pi/2} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \) has characteristic polynomial \( z^2 + 1 \) with no real roots: Lemma B genuinely uses \( A = A^{\top} \), not just realness of the entries.
With an indefinite "inner product": on \( \mathbb{R}^2 \) with \( \langle u, v \rangle = u_1 v_1 - u_2 v_2 \), the operator \( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \) satisfies \( \langle Tu, v \rangle = \langle u, Tv \rangle \) for this form but has no real eigenvalues; positive definiteness is what turns symmetry into diagonalisability.

Common errors

Conflating "diagonalisable" with "orthogonally diagonalisable". \( \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \) is diagonalisable with real eigenvalues but not symmetric; writing \( A = Q \Lambda Q^{\top} \) for it with an invertible-but-not-orthogonal \( Q \) is false the moment \( Q^{\top} \) is used in place of \( Q^{-1} \).
Skipping Gram–Schmidt inside a repeated eigenspace. The theorem guarantees eigenvectors for distinct eigenvalues are orthogonal; a computed basis of a two-dimensional eigenspace (e.g. the \( \lambda = -1 \) space in Problem 2) is usually not orthogonal until you orthonormalise it by hand.
Forgetting to normalise. \( Q^{\top} A Q = \Lambda \) requires unit columns; orthogonal-but-unnormalised eigenvectors give \( Q^{\top} Q = D \ne I \) and a silently wrong factorisation.
Using \( A = A^{\top} \) over \( \mathbb{C} \). The complex hypothesis is Hermitian, \( A = \overline{A}^{\top} \). The complex-symmetric matrix \( \begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix} \) is not diagonalisable at all (both eigenvalues \( 0 \), matrix nonzero).
Believing every normal operator has real eigenvalues. Reality of the spectrum is a theorem about self-adjoint operators specifically; unitary operators are normal with spectrum on the unit circle.
Deducing symmetry from real eigenvalues. "All eigenvalues real" plus "diagonalisable" does not imply self-adjoint; orthonormality of the eigenbasis is the missing (and essential) ingredient.

Discussion

Historically the theorem predates linear algebra as we teach it. Cauchy (1829) proved that the eigenvalues of a real symmetric matrix are real while studying the principal axes of quadric surfaces — the observation that any ellipsoid \( \sum a_{ij} x_i x_j = 1 \) has mutually perpendicular symmetry axes is exactly orthogonal diagonalisation of \( (a_{ij}) \). The word spectrum is Hilbert's, coined around 1904 in his work on integral equations, a generation before physicists discovered — to Hilbert's own surprise — that the spectra of his operators would model the literal spectra of atoms.

The structural heart of the proof is Lemma C: self-adjointness converts one invariant subspace into a pair of complementary ones, and this is what lets induction bite. For a general operator, \( W \) invariant gives only \( W^{\perp} \) invariant under \( T^{*} \), and the induction collapses — as it must, since Jordan blocks exist. The second load-bearing wall is Lemma B for the real case: existence of one eigenvalue is analytic input (the fundamental theorem of algebra smuggled in through the complexification), and rotations of \( \mathbb{R}^2 \) show it cannot be had for free. An alternative route to Lemma B, older and more geometric, maximises the Rayleigh quotient \( \langle Tv, v \rangle / \langle v, v \rangle \) on the unit sphere and shows a maximiser is an eigenvector — that variational proof is the one that survives the passage to infinite dimensions.

In quantum mechanics the theorem is a postulate-shaped fact: observables are self-adjoint operators precisely because self-adjointness delivers real measurement values (Lemma A) and an orthonormal system of eigenstates into which any state decomposes, with \( |\langle \psi, e_i \rangle|^2 \) the outcome probabilities. In statistics, diagonalising the (symmetric, positive semidefinite) covariance matrix is principal component analysis. And the singular value decomposition of an arbitrary \( m \times n \) matrix \( A \) is nothing but the spectral theorem applied to the self-adjoint operators \( A^{*}A \) and \( AA^{*} \).

Over \( \mathbb{C} \) the sharp statement is: \( T \) admits an orthonormal eigenbasis iff \( T \) is normal (\( TT^{*} = T^{*}T \)); self-adjoint, skew-adjoint and unitary operators are the normal operators with spectrum in \( \mathbb{R} \), in \( i\mathbb{R} \), and on the unit circle respectively. Over \( \mathbb{R} \), normality buys only a block-diagonal form with \( 2 \times 2 \) rotation-scaling blocks. In infinite dimensions the eigenbasis formulation fails (multiplication operators have no eigenvectors), and the theorem is reborn twice: for compact self-adjoint operators the finite-dimensional statement survives verbatim with a sequence \( \lambda_k \to 0 \) (Hilbert–Schmidt), while for general bounded self-adjoint operators one gets a projection-valued measure \( E \) with \( T = \int_{\sigma(T)} \lambda \, dE(\lambda) \), and von Neumann's extension to unbounded operators underwrites the mathematical consistency of quantum mechanics.

Common misconceptions. The theorem does not say the orthonormal eigenbasis is unique — any eigenvalue of multiplicity \( m \ge 2 \) admits infinitely many orthonormal bases of its eigenspace (what is unique is the list of eigenvalues with multiplicities and the eigenspace decomposition itself). It does not say symmetric matrices have distinct eigenvalues (\( I \) is symmetric). And it does not say every diagonalisable operator is self-adjoint — orthonormality of the eigenbasis, together with reality of the eigenvalues, is exactly what separates the two.

Worked examples

Example 1 (real symmetric: spectral decomposition and matrix powers). Let \( A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \). Find an orthogonal diagonalisation and a closed formula for \( A^k \).

\[ A^{\top} = A, \text{ so the spectral theorem applies: an orthonormal eigenbasis of } \mathbb{R}^2 \text{ exists.} \]

Check the hypothesis before anything else; symmetry is visible by inspection. A

\[ \chi_A(\lambda) = \det(\lambda I - A) = (\lambda - 2)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda - 1)(\lambda - 3). \]

Direct expansion of the \( 2 \times 2 \) determinant; eigenvalues \( \lambda = 1, 3 \), both real as the theorem guarantees. A

\[ \lambda = 1 : (A - I)v = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} v = 0 \Rightarrow v \parallel (1, -1); \qquad \lambda = 3 : (A - 3I)v = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix} v = 0 \Rightarrow v \parallel (1, 1). \]

Kernel computations by row reduction. Note \( (1,-1) \cdot (1,1) = 0 \): orthogonality across distinct eigenvalues arrives automatically, as the theorem predicts. A

\[ e_1 = \tfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix}, \quad e_2 = \tfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad Q = \tfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, \quad Q^{\top} A Q = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}. \]

Normalisation (divide by \( \sqrt{2} \)) is required for \( Q \) to be orthogonal; verify \( Q^{\top} Q = I \). A

\[ A = 1 \cdot P_1 + 3 \cdot P_3, \quad P_1 = e_1 e_1^{\top} = \tfrac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}, \quad P_3 = e_2 e_2^{\top} = \tfrac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}. \]

Spectral resolution: \( P_1, P_3 \) are the orthogonal projections onto the eigenlines; check \( P_1 + P_3 = I \) and \( P_1 P_3 = 0 \). B

\[ A^k = 1^k P_1 + 3^k P_3 = \tfrac{1}{2} \begin{pmatrix} 3^k + 1 & 3^k - 1 \\ 3^k - 1 & 3^k + 1 \end{pmatrix} \quad (k \in \mathbb{N}). \]

Because \( P_1, P_3 \) are idempotent with product zero, powers pass to the eigenvalues: \( A^k = \sum \lambda^k P_{\lambda} \) — the functional calculus in action. Sanity check \( k = 1 \) recovers \( A \). B

\[ A = Q \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix} Q^{\top}, \quad Q = \tfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, \qquad A^k = \tfrac{1}{2} \begin{pmatrix} 3^k + 1 & 3^k - 1 \\ 3^k - 1 & 3^k + 1 \end{pmatrix}. \]

Reading. In the rotated frame at \( 45^{\circ} \), \( A \) just stretches one axis by \( 3 \) and leaves the other alone; every power of \( A \) is read off instantly.

Scope. The same recipe computes \( f(A) \) for any function defined on \( \{1, 3\} \).

Example 2 (Hermitian: real spectrum and an operator square root). Let \( A = \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix} \) acting on \( \mathbb{C}^2 \) with the standard inner product \( \langle u, v \rangle = u_1 \overline{v_1} + u_2 \overline{v_2} \). Diagonalise \( A \) unitarily and find a positive semidefinite \( S \) with \( S^2 = A \).

\[ A^{*} = \overline{A}^{\top} = \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix} = A, \]

so \( A \) is Hermitian (note it is not complex-symmetric: \( A^{\top} \ne A \)); the spectral theorem applies and promises real eigenvalues and an orthonormal eigenbasis. A

\[ \chi_A(\lambda) = \det(\lambda I - A) = (\lambda - 1)^2 - (-i)(i) = (\lambda - 1)^2 - 1 = \lambda(\lambda - 2). \]

\( \det \begin{pmatrix} \lambda - 1 & -i \\ i & \lambda - 1 \end{pmatrix} = (\lambda - 1)^2 - (-i)(i) = (\lambda-1)^2 - 1 \). Eigenvalues \( 0 \) and \( 2 \): real, exactly as guaranteed, despite the complex entries. A

\[ \lambda = 0 : \; A v = 0 \Rightarrow v_1 + i v_2 = 0 \Rightarrow v \parallel \begin{pmatrix} -i \\ 1 \end{pmatrix}; \qquad \lambda = 2 : \; (A - 2I) v = 0 \Rightarrow -v_1 + i v_2 = 0 \Rightarrow v \parallel \begin{pmatrix} i \\ 1 \end{pmatrix}. \]

Kernel computations; verify: \( A(-i, 1)^{\top} = (-i + i, \; (-i)(-i) + 1)^{\top} = (0, 0)^{\top} \) and \( A(i, 1)^{\top} = (2i, 2)^{\top} = 2 (i, 1)^{\top} \). A

\[ \left\langle \begin{pmatrix} -i \\ 1 \end{pmatrix}, \begin{pmatrix} i \\ 1 \end{pmatrix} \right\rangle = (-i)\overline{(i)} + 1 \cdot 1 = (-i)(-i) + 1 = -1 + 1 = 0; \quad e_1 = \tfrac{1}{\sqrt{2}} \begin{pmatrix} -i \\ 1 \end{pmatrix}, \; e_2 = \tfrac{1}{\sqrt{2}} \begin{pmatrix} i \\ 1 \end{pmatrix}. \]

Orthogonality must be checked with the Hermitian inner product (conjugate the second vector); each vector has norm \( \sqrt{|{\pm i}|^2 + 1} = \sqrt{2} \), hence the normalisation. B

\[ U = \tfrac{1}{\sqrt{2}} \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix}, \qquad U^{*} U = I, \qquad U^{*} A U = \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix}, \qquad A = 0 \cdot P_0 + 2 P_2, \quad P_2 = e_2 e_2^{*} = \tfrac{1}{2} \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix}. \]

Unitary diagonalisation assembled from the orthonormal eigenbasis; note the pleasant accident \( P_2 = \tfrac{1}{2} A \), consistent with \( A = 2 P_2 \). B

\[ S := \sqrt{0}\, P_0 + \sqrt{2}\, P_2 = \tfrac{1}{\sqrt{2}} \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix}, \qquad S = S^{*}, \quad \langle S v, v \rangle \ge 0 \; \forall v, \quad S^2 = 2 P_2 = A. \]

Functional calculus with \( f(t) = \sqrt{t} \), legitimate because both eigenvalues are \( \ge 0 \); \( S^2 = 0 \cdot P_0 + 2 P_2 = A \) since \( P_2^2 = P_2 \) and \( P_0 P_2 = 0 \). Positive semidefiniteness of \( S \) holds because its eigenvalues \( 0, \sqrt{2} \) are nonnegative. (This \( S \) is the unique psd square root — Problem 4.) C

\[ A = U \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix} U^{*}, \quad U = \tfrac{1}{\sqrt{2}} \begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix}, \qquad \sqrt{A} = \tfrac{1}{\sqrt{2}} A. \]

Reading. A Hermitian matrix with visibly complex entries still has a purely real spectrum \( \{0, 2\} \) and perpendicular eigenvectors; its square root is computed by taking square roots of eigenvalues only.

Scope. The square-root construction works for every positive semidefinite operator and is the engine behind the polar decomposition and the SVD.

Problems

(Routine.) Let \( T \) be self-adjoint on an inner product space and suppose \( Tu = \lambda u \), \( Tv = \mu v \) with \( \lambda \ne \mu \). Prove \( \langle u, v \rangle = 0 \).
Solution

Both \( \lambda, \mu \) are real (Lemma A of the proof). Compute, using self-adjointness and conjugate-linearity in the second slot:
\[ \lambda \langle u, v \rangle = \langle Tu, v \rangle = \langle u, Tv \rangle = \langle u, \mu v \rangle = \bar{\mu} \langle u, v \rangle = \mu \langle u, v \rangle . \]
Hence \( (\lambda - \mu) \langle u, v \rangle = 0 \), and since \( \lambda \ne \mu \) we conclude \( \langle u, v \rangle = 0 \). \( \blacksquare \)
(Computational.) Orthogonally diagonalise \( A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \): find an orthogonal \( Q \) and diagonal \( \Lambda \) with \( Q^{\top} A Q = \Lambda \).
Solution

Write \( A = J - I \) where \( J \) is the all-ones matrix. \( J \) has rank \( 1 \), so \( 0 \) is an eigenvalue of \( J \) of multiplicity \( 2 \); also \( J (1,1,1)^{\top} = 3 (1,1,1)^{\top} \). Since \( A = J - I \), the eigenvalues of \( A \) are \( 3 - 1 = 2 \) (simple) and \( 0 - 1 = -1 \) (double).

Eigenvector for \( \lambda = 2 \): \( u_1 = \tfrac{1}{\sqrt{3}} (1, 1, 1)^{\top} \).

The eigenspace for \( \lambda = -1 \) is \( \ker(A + I) = \ker J = \{ x : x_1 + x_2 + x_3 = 0 \} \), the plane orthogonal to \( (1,1,1) \). A basis is \( (1, -1, 0)^{\top}, (1, 1, -2)^{\top} \) — chosen orthogonal; in general one must apply Gram–Schmidt inside a repeated eigenspace. Normalising:
\[ u_2 = \tfrac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \qquad u_3 = \tfrac{1}{\sqrt{6}} \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}. \]
Then with
\[ Q = \begin{pmatrix} \tfrac{1}{\sqrt{3}} & \tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{6}} \\ \tfrac{1}{\sqrt{3}} & -\tfrac{1}{\sqrt{2}} & \tfrac{1}{\sqrt{6}} \\ \tfrac{1}{\sqrt{3}} & 0 & -\tfrac{2}{\sqrt{6}} \end{pmatrix}, \qquad Q^{\top} A Q = \begin{pmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}. \]
Verification: \( Q^{\top} Q = I \) (columns orthonormal by construction) and \( A u_i = \lambda_i u_i \) is checked directly, e.g. \( A(1,-1,0)^{\top} = (-1, 1, 0)^{\top} \). \( \blacksquare \)
(Medium: variational characterisation.) Let \( T \) be self-adjoint on finite-dimensional \( V \) with eigenvalues \( \lambda_1 \le \dots \le \lambda_n \). Prove \[ \lambda_n = \max_{\|v\| = 1} \langle Tv, v \rangle \qquad \text{and} \qquad \lambda_1 = \min_{\|v\| = 1} \langle Tv, v \rangle . \]
Solution

By the spectral theorem choose an orthonormal eigenbasis \( e_1, \dots, e_n \) with \( T e_i = \lambda_i e_i \). Any unit vector expands as \( v = \sum_{i=1}^n c_i e_i \) with \( \sum_i |c_i|^2 = \|v\|^2 = 1 \) (Parseval for orthonormal bases). Then, using orthonormality,
\[ \langle Tv, v \rangle = \left\langle \sum_i c_i \lambda_i e_i , \sum_j c_j e_j \right\rangle = \sum_{i=1}^{n} \lambda_i |c_i|^2 . \]
This is a weighted average of the \( \lambda_i \) with nonnegative weights summing to \( 1 \), hence
\[ \lambda_1 = \lambda_1 \sum_i |c_i|^2 \le \sum_i \lambda_i |c_i|^2 \le \lambda_n \sum_i |c_i|^2 = \lambda_n . \]
Both bounds are attained: \( v = e_n \) gives \( \langle T e_n, e_n \rangle = \lambda_n \) and \( v = e_1 \) gives \( \lambda_1 \). Since the extrema are attained, \( \max \) and \( \min \) (not merely \( \sup, \inf \)) are justified. \( \blacksquare \)

Remark. Note \( \langle Tv, v \rangle \) is real for every \( v \) when \( T = T^{*} \), since \( \langle Tv, v \rangle = \langle v, Tv \rangle = \overline{\langle Tv, v \rangle} \); the statement would not typecheck otherwise.
(Harder: positive square roots.) Call a self-adjoint \( T \) positive semidefinite (psd) if \( \langle Tv, v \rangle \ge 0 \) for all \( v \). Prove that every psd operator \( T \) on finite-dimensional \( V \) has exactly one psd square root: a psd operator \( S \) with \( S^2 = T \).
Solution

Eigenvalues of a psd operator are nonnegative. If \( Tv = \lambda v \) with \( \|v\| = 1 \), then \( \lambda = \langle Tv, v \rangle \ge 0 \).

Existence. By the spectral theorem write \( T = \sum_{\lambda \in \sigma(T)} \lambda P_{\lambda} \) with \( P_{\lambda} \) the orthogonal projections onto the eigenspaces, \( P_{\lambda} P_{\mu} = 0 \) for \( \lambda \ne \mu \), \( \sum_{\lambda} P_{\lambda} = I \). Define
\[ S := \sum_{\lambda \in \sigma(T)} \sqrt{\lambda} \, P_{\lambda} , \]
legitimate since every \( \lambda \ge 0 \). Then \( S^2 = \sum_{\lambda, \mu} \sqrt{\lambda}\sqrt{\mu}\, P_{\lambda} P_{\mu} = \sum_{\lambda} \lambda P_{\lambda} = T \) by orthogonality and idempotence of the projections. \( S \) is self-adjoint (real combination of self-adjoint projections) and psd because its eigenvalues \( \sqrt{\lambda} \) are nonnegative: for \( v = \sum_\lambda P_\lambda v \), \( \langle Sv, v \rangle = \sum_{\lambda} \sqrt{\lambda}\, \|P_{\lambda} v\|^2 \ge 0 \).

Uniqueness. Suppose \( R \) is psd with \( R^2 = T \). It suffices to show \( R v = \sqrt{\lambda}\, v \) for every eigenvector \( v \) of \( T \) with \( T v = \lambda v \), since such eigenvectors span \( V \) (spectral theorem) and this forces \( R = S \).

Case \( \lambda \gt 0 \): factor
\[ 0 = (T - \lambda I) v = (R^2 - \lambda I) v = \left(R + \sqrt{\lambda}\, I\right) \left(R - \sqrt{\lambda}\, I\right) v . \]
The operator \( R + \sqrt{\lambda} I \) is invertible: for any \( w \ne 0 \), \( \langle (R + \sqrt{\lambda} I) w, w \rangle = \langle Rw, w \rangle + \sqrt{\lambda} \|w\|^2 \ge \sqrt{\lambda} \|w\|^2 \gt 0 \), so its kernel is trivial. Applying its inverse gives \( (R - \sqrt{\lambda} I) v = 0 \), i.e. \( Rv = \sqrt{\lambda}\, v \).

Case \( \lambda = 0 \): \( R^2 v = 0 \) gives \( \|Rv\|^2 = \langle R^2 v, v \rangle = 0 \) (self-adjointness of \( R \)), so \( Rv = 0 = \sqrt{0}\, v \).

Hence \( R \) and \( S \) agree on a spanning set, so \( R = S \). \( \blacksquare \)
(Conceptual: the exact converse.) (a) Let \( V \) be a finite-dimensional real inner product space and suppose \( T : V \to V \) has an orthonormal basis of eigenvectors. Prove \( T \) is self-adjoint. (b) Show by example that the analogous statement over \( \mathbb{C} \) is false, and state (with a one-line proof of the easy direction) the property of \( T \) that an orthonormal eigenbasis over \( \mathbb{C} \) actually characterises.
Solution

(a) Let \( e_1, \dots, e_n \) be an orthonormal eigenbasis, \( T e_i = \lambda_i e_i \) with \( \lambda_i \in \mathbb{R} \) (all scalars are real here). For arbitrary \( u = \sum_i a_i e_i \) and \( v = \sum_j b_j e_j \), orthonormality gives
\[ \langle Tu, v \rangle = \sum_{i} \lambda_i a_i b_i = \langle u, Tv \rangle , \]
since both sides reduce to the same sum \( \sum_i \lambda_i a_i b_i \) by \( \langle e_i, e_j \rangle = \delta_{ij} \). Hence \( T = T^{*} \). (Matrix version: the matrix of \( T \) in this basis is real diagonal, hence symmetric, and symmetry of the matrix in one orthonormal basis is equivalent to self-adjointness of the operator.)

(b) Over \( \mathbb{C} \), take \( T = iI \) on \( \mathbb{C}^n \). Every orthonormal basis is an eigenbasis (eigenvalue \( i \)), but \( T^{*} = -iI \ne T \), so \( T \) is not self-adjoint. The reality of the eigenvalues in (a) was essential.

The correct characterisation: \( T \) admits an orthonormal eigenbasis over \( \mathbb{C} \) if and only if \( T \) is normal, \( T T^{*} = T^{*} T \). Easy direction: if \( U^{*} T U = D \) is diagonal with \( U \) unitary, then
\[ T T^{*} = U D D^{*} U^{*} = U D^{*} D U^{*} = T^{*} T , \]
because diagonal matrices commute with each other (in particular \( D D^{*} = D^{*} D \), both equal to \( \mathrm{diag}(|d_1|^2, \dots, |d_n|^2) \)). The converse — every normal operator is unitarily diagonalisable — is the complex spectral theorem for normal operators; one proof runs exactly parallel to the induction above once one shows \( \ker(T - \lambda I) = \ker(T^{*} - \bar{\lambda} I) \) for normal \( T \). Within the normal operators, the self-adjoint ones are precisely those whose eigenvalues are all real (one direction is Lemma A; the other is part (a)'s computation transported to \( \mathbb{C} \) with \( \lambda_i \in \mathbb{R} \)). \( \blacksquare \)