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A damped oscillator and its first companion pencil

Example

Consider the scalar damped oscillator with mass $M = 1$ , damping coefficient $C = 1 / 2$ , and stiffness $K = 2$ . Its quadratic polynomial is $P (λ) = λ^{2} + (1 / 2) λ + 2$ . The first companion pencil is $L (λ) = [\begin{array}{cc} λ + 1 / 2 & 2 \\ - 1 & λ \end{array}] .$ This example checks its finite roots and the companion eigenvector lift.

Facts & Assumptions

Given: The scalar grade-two polynomial $P (λ) = λ^{2} + (1 / 2) λ + 2$ and its first companion pencil.

[F1]

Regular matrix polynomials and homogeneous eigenpairs: a finite eigenpair satisfies $P (λ) x = 0$ with $x \neq 0$ .

[F2]

First Frobenius companion pencil of a matrix polynomial: for grade two the lift is $z = (λ x, x)^{T}$ and the pencil has top row $(λ A_{2} + A_{1}, A_{0})$ and lower row $(- 1, λ)$ .

[F3]

The first Frobenius companion pencil preserves the finite and infinite polynomial spectrum: finite polynomial eigenvectors lift to companion-pencil eigenvectors, and $\det L (λ) = \det P (λ)$ .

Verification

technique · direct calculation

1.1givenF1algebra

The discriminant is $(1 / 2)^{2} - 4 (1) (2) = 1 / 4 - 8 = - 31 / 4$ . The quadratic formula therefore gives the two roots $λ_{\pm} = - 1 / 4 \pm i \sqrt{31} / 4$ . Neither is infinite because the leading coefficient $A_{2} = 1$ is invertible.

2.1givenF2F3step 1.1algebra

Direct expansion gives $\det L (λ) = λ (λ + 1 / 2) + 2 = λ^{2} + (1 / 2) λ + 2 = P (λ)$ , so these two values are precisely the finite companion-pencil eigenvalues.

3.1givenF1F2step 1.1algebra∎

For either root, $P (λ_{\pm}) \cdot 1 = 0$ , and its companion lift is $z = (λ_{\pm}, 1)^{T}$ . Multiplication gives $L (λ_{\pm}) z = ((λ_{\pm} + 1 / 2) λ_{\pm} + 2, - λ_{\pm} + λ_{\pm})^{T} = (P (λ_{\pm}), 0)^{T} = (0, 0)^{T}$ .

Depends on

Used by

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Sources

Françoise Tisseur and Karl Meerbergen, The Quadratic Eigenvalue Problem, §§1 and 3.4 (standard reference, not scraped)