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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.

Depends on

Used by

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Dependency tree · two levels

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Sources

Timo Betcke, Nicholas J. Higham, Volker Mehrmann, Christian Schröder, and Françoise Tisseur, NLEVP: A Collection of Nonlinear Eigenvalue Problems, §§1 and 2.1 (standard reference, not scraped)