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Polynomial eigenproblems and linearization — Examples

1 · Prerequisites

Polynomial eigenproblems and linearization

2 · Summary

The examples compute both roots and companion eigenvectors for a damped scalar oscillator, distinguish algebraic multiplicity at infinity from the nullity of a singular leading matrix, and show a rational transfer matrix whose pole becomes a false polynomial root after clearing its denominator. Each calculation uses the domain and reversal conventions from the A page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

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 $L (λ_{\pm}) z = ((λ_{\pm} + 1 / 2) λ_{\pm} + 2, - λ_{\pm} + λ_{\pm})^{T} = (P (λ_{\pm}), 0)^{T} = (0, 0)^{T}$ .

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A singular leading coefficient gives an infinite polynomial eigenvalue

Example

Take the grade-two matrix polynomial $P (λ) = diag (1, λ^{2})$ , so $n = m = 2$ , $\det P (λ) = λ^{2}$ , and $A_{2} = diag (0, 1)$ is singular. The finite root $λ = 0$ and the infinite root are different projective eigenvalues. Their algebraic and geometric multiplicities are computed directly below.

Facts & Assumptions

Given: $A_{0} = diag (1, 0)$ , $A_{1} = 0$ , and $A_{2} = diag (0, 1)$ for a regular grade-two polynomial.

[F1]

Regular matrix polynomials and homogeneous eigenpairs: $P_{h} (1, 0) = A_{2}$ , and the algebraic multiplicity at infinity is the zero order at $0$ of $\det P_{2}^{r e v}$ .

[F2]

First Frobenius companion pencil of a matrix polynomial: at infinity the homogeneous companion pencil is $- B_{c} = diag (A_{2}, I_{2})$ , with block eigenvectors $(x, 0)^{T}$ for $x \in \ker A_{2}$ .

[F3]

The first Frobenius companion pencil preserves the finite and infinite polynomial spectrum: the finite and infinite eigenspaces correspond to those of the polynomial, and the infinite algebraic multiplicity is $m n - \deg \det P$ .

Verification

technique · direct calculation

1.1givenF1F3algebra

The determinant is $\det P (λ) = λ^{2} \equiv̸ 0$ , so $P$ is regular; its degree is $2$ , while $m n = 4$ . Thus the theorem's degree-defect formula gives infinite algebraic multiplicity $4 - 2 = 2$ . Also $P_{2}^{r e v} (μ) = diag (μ^{2}, 1)$ , whose determinant has a zero of order two at $μ = 0$ .

2.1givenF1F2step 1.1algebra

The leading matrix $A_{2}$ has rank one and kernel $span {e_{1}}$ , so the infinite eigenspace has geometric multiplicity one and is represented in the companion pencil by $(e_{1}, 0)^{T}$ . This differs from its algebraic multiplicity two.

3.1givenF1F2F3step 1.1algebra∎

At the finite root $λ = 0$ , $P (0) = diag (1, 0)$ has kernel $span {e_{2}}$ . The finite lift is $(0, e_{2})^{T}$ , which the companion theorem maps to a pencil eigenvector. Thus the finite root zero is not the infinite eigenvalue represented by $(e_{1}, 0)^{T}$ .

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A transfer-function pole becomes a spurious root after denominator clearing

Example

The rational transfer matrix $R (λ) = [\begin{array}{cc} (λ - 1)^{- 1} & 0 \\ 0 & 1 \end{array}]$ has a pole at $λ = 1$ . Here a finite-dimensional state-space transfer matrix means a rational matrix of the form $C (λ I - A)^{- 1} B + D$ ; the displayed $R$ has such a realization with $A = [1]$ , $B = [1 0]$ , $C = [1 0]^{T}$ , and $D = diag (0, 1)$ . It has no rational eigenpair on its domain, but clearing the common denominator yields a polynomial matrix singular at the excluded pole.

Facts & Assumptions

Given: $R (λ) = diag ((λ - 1)^{- 1}, 1)$ with domain $Ω = C ∖ {1}$ .

[F1]

Rational matrix eigenproblem on its pole-free domain: rational eigenpairs require a nonzero vector and a parameter in the pole-free domain; a reduced common denominator $q$ gives $N = q R$ .

[F2]

Clearing a common denominator preserves rational eigenpairs only away from poles: $R (λ) x = 0$ is equivalent to $N (λ) x = 0$ only for $λ \in Ω$ , and poles are excluded even if $N (λ)$ is singular.

Verification

technique · direct calculation

1.1givenF1algebra

A state-space realization is $A = [1]$ , $B = [1 0]$ , $C = [1 0]^{T}$ , and $D = diag (0, 1)$ , since $C (λ I - A)^{- 1} B + D = diag ((λ - 1)^{- 1}, 1)$ . Its reduced common denominator is $q (λ) = λ - 1$ , so $N (λ) = q (λ) R (λ) = diag (1, λ - 1)$ .

2.1givenF1F2step 1.1algebra∎

For every $λ \neq 1$ , $\det R (λ) = 1 / (λ - 1) \neq 0$ , so no nonzero vector lies in its kernel. At $λ = 1$ , $N (1) = diag (1, 0)$ has kernel $span {e_{2}}$ , but $R (1)$ is undefined because its first entry has a pole. By [F2], the cleared root at $1$ is spurious and must be filtered out.

Polynomial eigenproblems and linearization — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A damped oscillator and its first companion pencil

Example

Facts & Assumptions

Verification

A singular leading coefficient gives an infinite polynomial eigenvalue

Example

Facts & Assumptions

Verification

A transfer-function pole becomes a spurious root after denominator clearing

Example

Facts & Assumptions

Verification

Sources