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Coercive non-symmetric forms need not have an orthonormal eigenbasis

Statement refuted

Every bounded coercive sesquilinear form on a finite-dimensional Hilbert space has an orthonormal basis of eigenvectors.

Facts & Assumptions

Given: the field $K \in {R, C}$ , the matrix $M = (\begin{array}{cc} 2 & 2 \\ 0 & 1 \end{array})$ acting on $K^{2}$ , and the form $a (u, v) : = (M u, v)$ .

[F1]

Sesquilinear forms, boundedness and coercivity: a form is bounded when $∣ a (u, v) ∣ \leq C ∥ u ∥ ∥ v ∥$ and coercive with constant $α > 0$ when $Re a (u, u) \geq α ∥ u ∥^{2}$ ; the adjoint form is $a^{*} (u, v) = \overline{a (v, u)}$ (Bounded, coercive and symmetric sesquilinear forms, Self-adjoint, positive, unitary and normal operators).

[F2]

Finite-dimensional Hilbert-space data: $K^{2}$ carries the standard inner product, linear in the first argument and conjugate-linear in the second, and $(M u, v)$ is computed by matrix multiplication (Real and complex inner-product spaces and their induced length, Hilbert space, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

[F3]

Eigenvalue data for endomorphisms of finite-dimensional spaces: eigenvalues form $σ_{K} (T) = {λ : χ_{T} (λ) = 0}$ for the characteristic polynomial $χ_{T}$ of the matrix of $T$ , and an eigenvalue of algebraic multiplicity one spans a one-dimensional eigenspace (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism, For $A \in M_{n} (F)$ , the characteristic polynomial is $χ_{A} (x) = \det (x I_{n} - A)$ when $n \geq 1$ , with $χ_{A} (x) = 1$ for the unique $0 \times 0$ matrix, For every finite-dimensional space, $σ_{F} (T)$ is exactly the set of roots in $F$ of $χ_{T}$ , An eigenvalue of algebraic multiplicity one has a one-dimensional eigenspace).

[F4]

Cauchy--Schwarz: $∣ (u, v) ∣ \leq ∥ u ∥ ∥ v ∥$ in an inner product space (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

Counterexample

1.1F1F2givenalgebra

The Hermitian part of $M$ is $H : = \frac{1}{2} (M + M^{*}) = (\begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array})$ and $Re a (u, u) = (H u, u)$ for $u = (u_{1}, u_{2})$ . Writing $x = ∣ u_{1} ∣$ and $y = ∣ u_{2} ∣$ , the elementary bound $2 Re (u_{2} \overline{u_{1}}) \geq - 2 x y$ gives $Re a (u, u) \geq 2 x^{2} - 2 x y + y^{2}$ . With $φ : = \frac{1 + \sqrt{5}}{2}$ one has $α : = \frac{3 - \sqrt{5}}{2} = 2 - φ > 0$ and $φ^{- 1} = φ - 1 = 1 - α$ , hence $2 x^{2} - 2 x y + y^{2} - α (x^{2} + y^{2}) = φ (x - y / φ)^{2} \geq 0$ . Therefore $Re a (u, u) \geq α ∥ u ∥^{2}$ , so $a$ is coercive with constant $α$ .

1.2F2F4algebra

Boundedness: for all $u, v$ one has $∣ a (u, v) ∣ = ∣ (M u, v) ∣ \leq ∥ M u ∥ ∥ v ∥$ by [F4], and the coordinate estimate $∥ M u ∥^{2} = ∣ 2 u_{1} + 2 u_{2} ∣^{2} + ∣ u_{2} ∣^{2} \leq (4 + 4 + 1) (∣ u_{1} ∣^{2} + ∣ u_{2} ∣^{2}) = 9 ∥ u ∥^{2}$ obtained from Cauchy--Schwarz in the two-dimensional index gives $∥ M u ∥ \leq 3 ∥ u ∥$ ; hence $∣ a (u, v) ∣ \leq 3 ∥ u ∥ ∥ v ∥$ and $a$ is bounded.

1.3F2F3givenalgebra

Eigenvalues and eigenvectors: the characteristic polynomial of $M$ is $χ_{M} (λ) = (2 - λ) (1 - λ)$ , whose roots are $1$ and $2$ ; by [F3] the spectrum is ${1, 2}$ , both roots are simple, and each eigenspace is one-dimensional. Solving $(M - 2 I) u = 0$ gives $u_{2} = 0$ , so $E_{2} = K (1, 0)$ , and solving $(M - I) u = 0$ gives $u_{1} + 2 u_{2} = 0$ , so $E_{1} = K (2, - 1)$ . The two exhibited eigenvectors satisfy $((2, - 1), (1, 0)) = 2 \neq 0$ .

2.1F1F3step 1.1step 1.2step 1.3∎

No orthonormal eigenbasis exists. Suppose $u, v$ were an orthogonal pair of nonzero eigenvectors; since $E_{1}$ and $E_{2}$ are one-dimensional and distinct, after relabelling $u \in E_{2}$ and $v \in E_{1}$ , so $u = c (1, 0)$ and $v = d (2, - 1)$ with $c, d \neq 0$ , and step 1.3 gives $(u, v) = 2 c \overline{d} \neq 0$ , a contradiction. Thus no orthogonal pair of eigenvectors exists, although $a$ is bounded and coercive by steps 1.1 and 1.2 and all its eigenvalues $1, 2$ are real. The displayed form is therefore a counterexample to the refuted statement: the symmetry hypothesis of the symmetric elliptic spectral theorem is not redundant.

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Sources

Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, 2020, complete 158-page graduate notes) (standard reference, not scraped)
Richard S. Laugesen, Spectral Theory of Partial Differential Equations (University of Illinois lecture notes, arXiv:1203.2344, complete 120 pages) (standard reference, not scraped)