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A coercive non-symmetric form can have non-real Galerkin eigenvalues

Example

On C2 let β>0, M=(1β−β1) and a(u,v):=(Mu,v). Then a is bounded and coercive with constant 1, since Re⁡a(u,u)=∣u1∣2+∣u2∣2; but the eigenvalues of M are 1±iβ, which are non-real. Consequently for every real λ the equation a(u,v)=λ(u,v) for all v∈C2 has no nonzero solution: this non-symmetric coercive form has no weak eigenpair with a real eigenvalue, and its 2×2 Galerkin matrix has a conjugate pair of non-real eigenvalues. This shows that the reality of the eigenvalues in the discrete spectral theorem is a consequence of symmetry and not of coercivity.

Facts & Assumptions

Given: a real β>0, the matrix M=(1β−β1), and the sesquilinear form a(u,v)=(Mu,v) on C2.

[F1]

Boundedness and coercivity of a sesquilinear form on a Hilbert space are defined by ∣a(u,v)∣≤C∥u∥∥v∥ and Re⁡a(u,u)≥α∥u∥2, with the form linear in the first argument and conjugate-linear in the second (Bounded, coercive and symmetric sesquilinear forms).

[F2]

C2 is a complex Hilbert space with the standard inner product, and (Mu,v) is computed by matrix multiplication and conjugation accordingly (Real and complex inner-product spaces and their induced length, Hilbert space, Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes, Real and imaginary parts, complex conjugation, and modulus).

[F3]

For an endomorphism of a finite-dimensional complex vector space the spectrum is the root set of its characteristic polynomial, the characteristic polynomial of a matrix A is χA(λ)=det⁡(λI−A), and a weak eigenpair identity a(u,v)=λ(u,v) for all v is equivalent to Mu=λu when a(u,v)=(Mu,v) (Eigenvalues, eigenvectors, eigenspaces Eλ(T)=ker⁡(T−λI), and the spectrum σF(T) of an endomorphism, For A∈Mn(F), the characteristic polynomial is χA(x)=det⁡(xIn−A) when n≥1, with χA(x)=1 for the unique 0×0 matrix, For every finite-dimensional space, σF(T) is exactly the set of roots in F of χT).

Verification

technique · direct
1.1F1F2givenalgebra

Coercivity. For u=(u1,u2) one computes Mu=(u1+βu2, −βu1+u2) and Re⁡a(u,u)=Re⁡(∣u1∣2+βu2u1‾−βu1u2‾+∣u2∣2)=∣u1∣2+∣u2∣2=∥u∥2, because u2u1‾−u2u1‾‾ is purely imaginary and β is real. Hence Re⁡a(u,u)=∥u∥2≥1⋅∥u∥2 and a is coercive with constant 1.

1.2F1F2F4givenalgebra

Boundedness. Applying Cauchy--Schwarz in the index, ∣u1+βu2∣2≤(1+β2)(∣u1∣2+∣u2∣2) and ∣−βu1+u2∣2≤(1+β2)(∣u1∣2+∣u2∣2), so ∥Mu∥2=∣u1+βu2∣2+∣−βu1+u2∣2≤2(1+β2)∥u∥2. Thus ∥Mu∥≤2+2β2 ∥u∥, and [F4] gives ∣a(u,v)∣=∣(Mu,v)∣≤2+2β2 ∥u∥∥v∥, so a is bounded.

1.3F2F3givenalgebra

Spectrum. The characteristic polynomial is χM(λ)=det⁡(λI−M)=(1−λ)2+β2, and because β>0 its two roots are λ=1±iβ, which are not real. By [F3] the spectrum of the endomorphism u↦Mu is exactly {1+iβ,1−iβ}, a conjugate pair of non-real eigenvalues of the Galerkin matrix M.

2.1F1F3step 1.1step 1.2step 1.3∎

No real-eigenvalue weak eigenpair. Let λ∈R be real and suppose a nonzero u∈C2 satisfies a(u,v)=λ(u,v) for every v. Subtracting, (Mu−λu,v)=0 for every v, and testing with v=Mu−λu gives ∥Mu−λu∥2=0, so Mu=λu; by [F3], λ would be a real eigenvalue of M, contradicting step 1.3. Hence there is no real λ with a nonzero weak eigenpair. Since a is nevertheless bounded and coercive by steps 1.1 and 1.2, this two-dimensional model shows that coercivity alone does not force real eigenvalues. Its complex eigenpairs at 1±iβ do exist; the exclusion just proved concerns real eigenvalues.

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